Solutions to Tutorial 1

Transcription

1 Solutions to Tutorial. Picture the Problem: The free body diagram for this problem is shown on the right. Strategy: Use the free body diagram to determine the net force on the rock, then apply Newton s Second Law to find the acceleration of the rock. Let upward be the positive direction. Solution:. Find the net force:. Now apply Newton s second law to find a r : r F ( 40.0 N) yˆ + ( 46. N) yˆ ( 6. N) yˆ r r F ( 6. N) ˆ m y a 5.00 kg (. m/s ) yˆ F r W r Insight: If the astronaut were to eert less than 40.0 N of upward force on the rock, it would accelerate downward.. Picture the Problem: The skier is accelerated along a straight line. Strategy: Use a constant acceleration equation of motion to determine the acceleration of the skier, then use Newton s Second Law to find the force eerted on the skier. Solution:. Find the acceleration of the skier from his final velocity and the distance covered:. Find the net force on the skier using: v a v 0 v v m/s 0 0 F ma m 9 kg 60 N 5 m 0 ( ) Insight: This moderate force accelerates the skier at a rate of.9 m/s or 0.9g. 3. Picture the Problem: The parachutist is moving straight downward, slowing down and coming to rest over a distance of m. Strategy: Use a constant acceleration equation of motion to determine the acceleration of the parachutist, and then use Newton s Second Law to find the net force on her. Solution:. (a) Find the acceleration of the parachutist: v v + a y y 0 y vy v0 y 0 ( 3.85 m/s) a r ˆ ˆ 9.88 m/s ˆ y y ( m) y y r r F a yˆ yˆ. Find the net force: m ( 4.0 kg)( 9.88 m/s ) ( 45 N) 3. (b) If the parachutist comes to rest in a shorter distance, the acceleration will be greater and the force will therefore be greater. Insight: There are two other forces on the parachutist, the upward force of her parachute and the downward force of gravity. Those two forces must cancel if she is moving downward at constant speed (no acceleration), so in this case the net force also equals the force the ground eerts on her. of 89

2 4. Picture the Problem: The 747 is accelerated horizontally in the direction opposite its motion in order to slow it down from its initial speed of 7.0 m/s. Strategy: Find the acceleration from the known force and mass, then find the speed and distance traveled. Solution:. (a) Find a r r 5 r F ( ) ˆ N : a 5 (.3 m/s ) ˆ m kg. Find the final speed: v v at 3. (b) Find the distance travelled by the 747 as it slows down: m/s +.3 m/s 7.50 s 7.8 m/s v 0 + v t m/s 7.50 s 68 m Insight: The landing speed of a Boeing is 7.9 m/s (6 mi/h) and it has a specified landing roll distance of, m, requiring an average landing acceleration of. m/s. 5. Picture the Problem: The free body diagram of the brick is shown at right. Strategy: Use the vectors depicted in the free body diagram to answer the questions. Solution:. (a) There are two forces acting on the brick.. (b) The forces acting on the brick are due to gravity and your hand. 3. (c) Yes, these forces are equal and opposite, because the brick remains at rest: r r r r r r F F + W ma 0 so that F W 4. (d) No, these forces are not an action-reaction pair, because they are acting on the same object. Insight: Action reaction pairs always act on different objects, and they can therefore never cancel each other. 6. Picture the Problem: The free body diagram for the car is shown at right. For the trailer there is only one force acting on it in the forward direction, the force eerted by the car on the trailer. Strategy: In order to determine the forces acting on an object, you must consider only the forces acting on that object and the motion of that object alone. For the trailer there is only one force F r eerted on it by the car, and it has the same acceleration (.90 m/s ) as the car. For the car there are two forces acting on it, the engine F r and the trailer F r. Apply Newton s second and third laws as appropriate to find the requested forces. Solution:. (a) Write Newton s Second Law for the trailer:. (b) Newton s Third Law states that the force the trailer eerts on the car is equal and opposite to the force the car eerts on the trailer: 3. (c) Write Newton s Second Law for the car: r r r F F ( ma 540 kg.90 m/s ) ˆ (.0 kn) ˆ r F.0 kn ˆ F r F r F r r r F M a ( 300 kg)(.90 m/s ) ˆ (.5 kn) ˆ Insight: The engine force F r must be ( 3.5 kn) ˆ because it must both balance the.0-kn force from the trailer and accelerate the car in the forwards, requiring an additional.5 kn of force. F r W r of 89

3 7. Picture the Problem: The force pushes on bo in the manner indicated by the figure at right. Strategy: The boes must each have the same acceleration, but because they have different masses the net force on each must be different. These observations allow you to use Newton s Second Law for each individual bo to determine the magnitudes of the contact forces. First find the acceleration of all the boes and then apply equation 5- to find the contact forces. Solution:. (a) The 7.50 N force accelerates all the boes together:. Write Newton s Second Law for the first bo: 3. (b) Write Newton s Second Law for the third bo: Insight: Another way to solve the problem is to note that F F ( m + m + m3 ) a a m + m + m N a m/s.30 kg kg kg F r F Fc m a Fc F m a 7.50 N.30 kg m/s 6.46 N F r Fc3 m3 a 4.90 kg m/s 3.9 N Fc accelerates boes and 3: F c kg m/s 6.46 N, the same result as in step. 8. Picture the Problem: The light bo of mass m sits adjacent to the heavy bo of mass M as depicted in the figure at right. Strategy: The boes must each have the same acceleration, but because they have different masses the net force on each must be different. These observations allow you to use Newton s Second Law for each individual bo to determine the magnitudes of the contact forces. First find the acceleration of both boes and then find the contact forces. Solution:. (a) The 7.50 N force accelerates all the boes together:. Find the contact force by writing Newton s Second Law for the heavy bo only: F 5.0 N F m + M a a m + M kg 0.40 m/s F r Fc M a 7.4 kg 0.40 m/s 3.0 N 3. (b) When the force is applied to the heavier bo the contact force between them will be less than it was before, because the lighter bo requires less force for the same acceleration, and the contact force is the only force on the lighter bo. 4. (c) Find the contact force by F r Fc m a ( 5. kg)( 0.40 m/s ). N writing Newton s Second Law for the lighter bo only: Insight: Another way to view the answer to (b) is to say the inertia of the heavier bo shields the lighter bo from eperiencing some of the pushing force. In case (a) the lighter bo provides less shielding and the contact force is greater. 3 of 89

4 9. Picture the Problem: The trailer is pulled up the incline at constant speed. The free body diagram of the trailer is depicted in the diagram on the right: y F r Strategy: Because the trailer is moving at constant speed, the net force on the trailer must be zero, and the force eerted by the tractor on the trailer must equal the component of the trailer s weight that is pointing down the incline and parallel to it. Solution: Write Newton s Second Law in the direction for the trailer: F + F mg sinθ ma 0 F mg sinθ 3900 kg 9.8 m/s sin6, 000 N kn Insight: The steeper the hill, the larger the force the tractor must eert. If the tractor were pulling straight upward (θ 90 ) it would have to support the entire 38-kN weight. On level ground with no friction, the required force is zero. 0. Picture the Problem: The trolley is pushed partly into the incline and partly up the incline by the pushing force F r, as shown in the figure at right. θ mg r mg sinθ Strategy: Write Newton s Second Law for the direction, where ˆ points up the incline and parallel to it. Solve the resulting equation for the magnitude of F r. Solution: The component of the force pushing up the incline is F cosθ and the component of the weight pushing down the incline is mg sinθ : F F cosθ mg sinθ ma ma + mg sinθ 7.5 kg m/s sin3 F cosθ cos3 F 8 N Insight: They d better offer double coupons at this store, because a 3 incline is a 3% slope; danger territory for over-the-road truckers and a lot of etra work for the average grocery shopper!. Picture the Problem: The two men pull on the barge in the directions indicated by the figure at right. Strategy: Place the -ais along the forward direction of the boat. Use the vector sum of the forces to find the force F such that the net force in the y-direction is zero. Solution: Set the sum of the forces in the y direction equal to zero: Fy 30 N sin 34 + F sin 45 0 sin 34 F ( 30 N) 00 N 0.0 kn sin 45 Insight: The second crewman doesn t have to pull as hard as the first because a larger component of his force is pulling in the y direction. However, his force in the forward direction (73 N) is not as large as the first crewman (0 N). 4 of 89

5 . Picture the Problem: The two teenagers pull on the sled in the directions indicated by the figure at right. Strategy: Write Newton s Second Law in the direction (parallel to a r ) in order to find the acceleration of the sled. Solution : Write Newton s Second Law in the direction : F F cos35 57 N msled + mchild a F cos35 57 N a msled + mchild ( 55 N) cos35 57 N.5 m/s a kg Insight: Some of the force eerted by the teenagers is eerted in the y direction and cancels out; only the components of the forces move the sled. 3. Picture the Problem: The skier glides down the incline without friction Strategy: Let the y direction point perpendicular to the snow trail and the direction point parallel to the slope and downhill. Since the skier is not accelerating in the y direction, the two forces in that direction, N and W, cancel each other. The net force on the skier is then the only unbalanced force, W. Fnd the net force on the skier. Solution:. (a) The net force on the skier is W : F W mg sinθ 65 kg 9.8 m/s sin F 0.4 kn The force points downhill, parallel to the slope.. (b) As the slope becomes steeper, θ increases and the net force W increases. Insight: If θ were to increase to 90, the skier would be falling straight downward and the force in him would be maimum. If θ were to decrease to 0, there would be no net force on the skier and no acceleration. 4. Picture the Problem: The force vectors acting on the Moon are depicted on the right: Strategy: Find the vector sum of the forces by using the component method, and use the components to find the magnitude and direction of the net force. Solution:. (a) Add the r r F FSM + FEM 0 0 force vectors: ( N) ˆ + (.98 0 N). Find the direction of the net force: 3. (b) Find the magnitude of the net force: r r ( F) θ r N F y N tan y tan 4.5 above the + ais 0 yˆ F r N.98 0 N N 5 of 89

6 4. (c) Find the acceleration from Newton s Second Law: r 0 r F N a m/s m kg Insight: It was a calculation of the acceleration of the Moon that led Isaac Newton to conclude the same force that causes an apple to fall is the force that keeps the Moon in its orbit. 5. Picture the Problem: The astronaut is accelerated straight upward by the force of the rocket engine. Strategy: There are two forces acting on the astronaut: the applied force F r of the seat acting upward and the force of gravity W r acting downward. The force F represents the apparent weight of the astronaut since that is both the force the seat eerts on him and the force he eerts on the seat. Use Newton s Second Law together with the known acceleration to determine F. Solution: Write out Newton s Second Law in the vertical direction and solve for F : F F mg ma y a a ( 88 kg)( m/s ) W F ma + mg m a + g + W 3500 N 3.5 kn Insight: The astronaut s weight on Earth is 0.86 kn, so this astronaut is eperiencing 3.5/ g s of acceleration. 6. Picture the Problem: The elevator accelerates up and down, changing your apparent weight W a. A free body diagram of the situation is depicted at right. Strategy: There are two forces acting on you: the applied force r r r F Wa of the scale acting upward and the force of gravity W acting downward. The force W a represents your apparent weight because it is both the force the scale eerts on you and the force you eert on the scale. Use Newton s Second Law together with the known force W a acceleration to determine the acceleration a. Solution:. (a) The direction of acceleration is upward. An upward acceleration results in an apparent weight greater than the actual weight.. (b) Use Newton s Second Law together with the known forces to determine the acceleration a. Fy Wa W ma Wa W Wa W N a ( 9.8 m/s ).9 m/s m W g 60 N 3. (c) The only thing we can say about the velocity is that it is changing in the upward direction. That means the elevator is either speeding up if it is travelling upward, or slowing down if it is travelling downward. Insight: You feel the effects of apparent weight twice for each ride in an elevator, once as it accelerates from rest and again when it slows down and comes to rest. 7. Picture the Problem: The bowling ball is lifted straight upward by the applied force. Strategy: There are two forces acting on the bowling ball: the applied force F r acting upward and the force of gravity W r acting downward. Write Newton s Second Law for each case in order to obtain two equations with two unknowns, and then use algebra to find W and a. 6 of 89

7 Solution:. (a) Write Newton s Second Law for each case to obtain two equations and two unknowns: F F mg ma y, F F mg ma y,. Divide the second equation by the first and solve for m: F mg ma F mg F mg F mg ma m F F 8 N 9 N g 9.8 m/s 3. Fnd the weight W: W mg 4. (b) Use the first Newton s Second Law equation to find a: 7.3 kg 9.8 m/s 7 N F mg 8 7 N.4 m/s a m 7.3 kg 7.3 kg Insight: In the first case, 7 N of force supported the weight of the ball and 0 N accelerated it at a rate of.4 m/s. In the second case, 0 N of net force accelerated the ball at a rate of.8 m/s. 8. Picture the Problem: The free body diagram of the suitcase is shown at right: Strategy: Write Newton s Second Law in the vertical direction to determine the magnitude of the applied force F. Solution: Use Newton s Second Law in the vertical direction to find F: F N W + F sinθ 0 y W N 3 kg 9.8 m/s 80 N F sinθ sin 5 F 0 N 0. kn Insight: If the upward force applied by the handle were zero, the normal force N would equal the weight W of 30 N. 9. Picture the Problem: The child sits on a chair and the chair sits on the floor. The free body diagrams of the child and the chair are shown at right. Strategy: There are two forces acting on the child: the normal force N r of the chair acting upward and the force of gravity W r acting downward. There are three forces child acting on the chair: the normal force N r of the floor acting upward, the weight of the baby acting downward, and the force of gravity W r chair acting downward. Write Newton s Second Law in the vertical direction for each case and then use the equations to find N. Solution:. (a) Write out Newton s Second Law in the vertical direction and solve for N: F N mg 0 y N mg 9 N ( 9.3 kg)( 9.8 m/s ) (a) child (b) chair 7 of 89

8 . (b) Write out Newton s Second Law in the vertical direction for the chair and solve for N: F N m g m g 0 y baby chair ( baby chair ) ( kg)( 9.8 m/s ) N m + m g + N 30 N 0.3 kn Insight: The normal force is larger in case (b) because the floor must support the weight of the child plus the weight of the chair, whereas the chair must only support the weight of the child. 0. Picture the Problem: The free body diagram of the potatoes is shown on the right: Strategy: There are two forces acting on the potatoes: the normal force N r of the shopping cart acting upward and the force of gravity W r acting downward. Solution:. (a) The free body diagram for the potatoes is shown at right.. (b) The free-body diagram does not change. A constant velocity implies zero acceleration and therefore zero net force on the potatoes. Insight: Note that as far as Newton s Second Law is concerned, zero velocity is no different than constant, nonzero velocity. This is the essence of Newton s First Law.. Picture the Problem: The ant walks along the surface of the bowling ball until the normal force between it and the ball becomes too small. Strategy: Write Newton s Second Law for the ant along the radial direction, remembering that the ant s weight always acts straight downward. The ant walks slowly enough that there is no acceleration in any direction. Solution:. Write Newton s Second Law for the radial direction:. Set N equal to half the weight and F N mg cosθ 0 solve for θ: θ ( ) r mg mg cosθ 0 cos 60 Insight: The normal force itself will not be zero until θ 90, at which point nothing will stop the ant s fall unless it has sticky feet!. Picture the Problem: The child slides down the incline that is tilted 3.0 above the horizontal. A free-body diagram of the situation is depicted at right: Strategy: Choose the -ais along the direction of motion. Write Newton s Second Law in the y direction to find the normal force, and then write Newton s Second Law in the direction and solve for µ k. Solution:. Write Newton s Second Law in the y direction:. Now write Newton s Second Law in the direction: F N mg cosθ 0 y N mg cosθ F mg sinθ µ N ma y mg sinθ ma µ N k k k ( cos ) mg sinθ ma µ mg θ 8 of 89

9 3. Solve the equation for µ k : g sinθ a 9.8 m/s sin m/s µ k g cosθ 9.8 m/s cos3.0 Insight: The coefficient of friction between the child and the slide depends upon many factors, including the clothing the child wears, the material from which the slide is constructed, and whether the slide is wet or not. 3. Picture the Problem: The book slides in a straight line across the top of the tabletop. Strategy: The minimum force required to get the book moving is related to the maimum coefficient of static friction, and the force required to keep the book sliding at constant speed is equal to the magnitude of the kinetic friction force, from which µ k can be determined. Solution:. When the book begins sliding, the applied force equals the maimum static friction force:. When the book is sliding at constant speed, the applied force equals the kinetic friction force: F f app s µ smg Fapp.5 N µ s 0.7 mg (.80 kg)( 9.8 m/s ) Fapp fk µ kmg Fapp.50 N µ k mg (.80 kg)( 9.8 m/s ) Insight: The coefficient of kinetic friction is usually smaller than the coefficient of static friction. This is the basic idea behind antilock brakes, which seek to keep the tire of a car rolling so that the friction between the tire and the road remains in the static regime, where there is a greater force to stop the car and improved handling during the stop. 4. Picture the Problem: The free-body diagram of this situation is depicted at right. Strategy: Write Newton s Second Law in the vertical direction to determine the normal force of the floor on the crate. Then write Newton s Second Law in the horizontal direction to determine the minimum force necessary to start the crate moving. Solution:. Write Newton s Second Law in the vertical direction:. Write Newton s Second Law in the horizontal direction: 3. Now substitute for N and solve for the maimum force F that would produce no acceleration: F F N mg F sinθ ma 0 y N mg + F sinθ F F cosθ µ N ma 0 ( cos sin ) s s F cosθ µ N s s s y ( sin ) F cosθ µ mg + F θ θ µ θ µ mg F µ mg s ( θ µ θ ) ( 0.57)( 3 kg)( 9.8 m/s ) cos s sin cos 0.57 sin F 50 N 0.5 kn Insight: The required pushing force is 50N and represents about 78% of the crate s weight. If the person simply pushed horizontally they would need only 80 N of force. f r s W r N r θ F r 9 of 89

10 5. Picture the Problem: The coffee cup is accelerated in a straight line due to the static friction between it and the roof of the car. Strategy: The static friction between the coffee cup and the roof of the car provides the forward force needed to accelerate the coffee cup. First write Newton s Second Law to find the maimum acceleration of the car, and then find the smallest amount of time in which the car can accelerate to 5 m/s. Solution:. (a) Write Newton s Second Law for the coffee cup:. Solve the equation for a: F f µ mg ma s s a µ s g m/s.35 m/s.4 m/s 3. (b) Find the minimum time: v v0 5 m/s 0 t 6.4 s a.35 m/s Insight: If the person owned a Ferrari 575M Maranello capable of going from zero to 60 mi/h in 4. seconds (6.4 m/s ), she would need a coefficient of friction of 0.65 to prevent the cup from slipping. Not likely, given the smooth aerodynamic finish of the sports car! 6. Picture the Problem: Your car travels in a straight line, slowing down from its initial speed v and skidding to a stop due to the kinetic friction between the tires and the road. Strategy: Use Newton s Second Law to find the acceleration of the car, and then use equation - to find the distance the car travels during the time it is slowing down. Solution:. (a) Solve Newton s Second Law for the acceleration of the car:. Find the travel F µ N µ mg ma a µ g vf v 0 v v d a µ g µ g distance of the car: ( ) 3. (b) The stopping distance depends upon the square of the initial speed, so if your speed is doubled the stopping distance quadruples. 4. (c) If the truck has the same initial speed and coefficient of kinetic friction, its stopping distance will be the same as your car s stopping distance. Insight: The stopping distance of trucks is greater than cars, but not because the coefficient of friction for truck tires is any different. It is mainly due to the braking system of the truck, as the brakes must safely dissipate a much larger amount of kinetic energy than car brakes. If the car and truck each skid, the stopping distance is the same. 7. Picture the Problem: The free-body diagram for the person in the hammock is depicted at right. Strategy: There are three forces acting on the person, the two tensions and gravity. The horizontal components of the tensions cancel out, and the vertical components of the tensions must balance the downward weight. Write Newton s Second Law in the vertical direction to find the tension. Solution: Write Newton s Second Law in the vertical direction and solve for the tension T: F T sinθ mg 0 y ( 50.0 kg)( 9.8 m/s ) mg T sinθ sin N Insight: Notice that the tension in each rope is almost double the person s 49 N weight. That s because only a small portion (sin %) of the tension is supporting the person s weight. T r θ T cosθ W r θ T r T sinθ 0 of 89

11 8. Picture the Problem: The spring stretches while the backpack remains at rest. Strategy: There are two forces acting on the backpack, the spring force to the left and the friction force to the right. The magnitudes of these two forces must be equal for the backpack to remain at rest. Use Newton s Second Law to relate the magnitudes of the forces and Hooke s Law (equation 6-4) to find the spring force and hence the friction force. Solution:. (a) Write Newton s Second Law in the horizontal direction:. Use Hooke s Law to find the spring force and solve for the friction force. Note that the spring stretches in the negative direction. F f F ma 0 s fs F k 50 N/m m 3.0 N 3. (b) No, the answer to (a) doesn t depend on the mass of the backpack, it only depends upon the spring force and the fact that the backpack remains at rest. Insight: The static friction force changes with the applied force, always matching its magnitude until the maimum friction force is reached. If the applied force is increased even more, the backpack will slide. 9. Picture the Problem: The spring is capable of being either stretched or compressed by the eternal force. Strategy: Use Hooke s law to find the force required to stretch the spring to a length that is twice its equilibrium length. In that case the stretch distance is L L L. Then use Hooke s law again to find the force required to compress the spring an amount equal to L L L. Solution:. (a) Use Hooke s Law to find the magnitude of the force required to stretch the spring by the amount L :. (b) Use Hooke s Law to find the magnitude of the force required to compress the spring by the amount L : F k k L 50 N/m 0.8 m 45 N F k k L 50 N/m 0.8 m 3 N 3. Because the force depends upon the stretch distance, the magnitude of the force required to compress the spring to half its equilibrium length is less than the force found in part (a). Insight: The magnitudes of the two forces would be the same if in part (a) the spring were stretched to a length of.5l. 30. Picture the Problem: The free-body diagram for the contact point between the two strings is depicted at right. Strategy: The horizontal components of the string tensions must be equal because the picture is not accelerating. The same is true of the vertical components of the forces. Use Newton s Second Law in the horizontal direction to find the tension in string, and in the vertical direction to find the weight of the picture. Solution:. (a) The tension in string is less than the tension in string, because it provides mostly a sideways component of force that is balanced by the horizontal component of string. That means string must support most of the weight of the of 89

12 picture plus balance string s horizontal component, giving it a larger tension than string.. (b) Write Newton s Second Law in the horizontal direction in order to find the F T cosθ + T cosθ 0 cosθ cos 65.7 N 0.85 N cosθ cos 3 tension in string : T T 3. (c) Write Newton s Second Law Fy T sinθ + T sinθ W 0 in the vertical direction in order to W find the picture s weight:.7 N sin N sin 3.0 N Insight: As the angle of string approaches 90 and the angle of string approaches 0, the tension in string drops to zero and the entire.0 N weight of the picture is supported by string. 3. Picture the Problem: The free-body diagram for each pulley is depicted at right. Strategy: The tension in the rope is the same everywhere and is equal to F as long as the pulleys rotate without friction. Use the free-body diagrams of the pulleys to write Newton s Second Law in the vertical direction for each pulley, and use the resulting equations to find the tensions. The acceleration is zero everywhere in the system. Solution:. (a) Write Newton s Second Law for the bo:. Now write Newton s Second Law for the lower pulley and solve for F: 3. (b) Write Newton s Second Law for the upper pulley and solve for C upper 4. (c) Use the result of step to find C lower : C C F C mg 0 y lower lower mg F F + F C 0 y lower F C mg upper lower ( 5 kg)( 9.8 m/s ) 55 N 0.6 kn downwards F C F F 0 y upper F 0.55 kn 0.5 kn Clower mg 5 kg 9.8 m/s 50 N 0.5 kn Insight: Note that there are two rope tensions pulling upward on the lower pulley. Therefore each one supports half the weight of the crate and the tension in the rope is half that of the chains. 3. Picture the Problem: The physical situation is depicted in the figure at right. The tension in the string between the two blocks is T and the tension in the string that is tied to the wall is T. A free-body diagram for the upper block is also shown at right.. Strategy: Set the -ais parallel to and pointing up the incline, and the y-ais perpendicular to the incline. Write Newton s Second Law in the direction for each block and use the resulting equations to find T and T. The acceleration is zero everywhere in the system. F Lower Pulley C lower F F C upper Upper Pulley F of 89

13 Solution:. (a) Write Newton s Second Law for block and block F m g sinθ + T 0 solve for T :. (b) Write Newton s Second Law for block and solve for T : block T m g sinθ.0 kg 9.8 m/s sin 3 T 5. N F T m g sinθ + T 0 T T + m g sinθ 5. N +.0 kg 9.8 m/s sin 3 5 N Insight: As epected the tension T is larger than T because it must support the weight of both blocks. If θ were to increase so would the tension, until at θ 90 the blocks hang straight down and T ( m + m ) g 9 N. 33. Picture the Problem: The forces acting on the small pulley are depicted at right. Strategy: The rope tension is equal to the weight of the.50 kg mass and will everywhere be the same because the pulleys are assumed frictionless. Write Newton s Second Law in the direction and solve for the magnitude of F r. Solution:. Find the rope tension:. Write Newton s Second Law in the direction and solve for F: T T mg F F + T cos T cos F mg cos kg 9.8 m/s cos N 9.54 lb Insight: The traction device is arranged to produce a force that is parallel to the leg bone so that it can heal straight. However, the force of gravity on the leg has been ignored here and in real life there would have to be upward component of the force eerted on the leg. 34. Picture the Problem: The free-body diagram for the wood block is depicted at right. Strategy: Write Newton s Second Law in the vertical direction for the block to find the magnitude of the normal force. Then write Newton s Second Law in the horizontal direction for the block to find the magnitude of the applied force. In both cases the acceleration of the block is zero. Solution:. (a) Write Newton s Second Law in the vertical direction:. Write Newton s Second Law in the horizontal direction: 3. Solve for the applied force F: F f mg µ N mg 0 y s mg N µ s mg F F N F 0 µ s s (.6 kg)( 9.8 m/s ) s mg F 0 N 0.00 kn 4.5 lb µ ( 0.79) Free-body diagram for the upper block 4. (b) The force of static friction would remain the same if you push with a greater force because it must eactly balance the weight of the wood board. Insight: Pushing with a force > 0 N, the ma static friction force increases. T r mg r θ T r mg sinθ 3 of 89

14 35. Picture the Problem: The free-body diagrams for each mass are shown at right. Strategy: The positive ais is along the line of the string and points up the incline for the 5.7 kg mass and in the downward direction for the hanging mass. Let M represent the mass on the incline and m the hanging mass. Write Newton s Second Law for each mass and combine the equations to find the acceleration of the hanging mass (which is the same as the acceleration of M because they are connected by a string). Solution:. (a) Write Newton s Second Law for M:. Write Newton s Second Law for m and substitute the epression for T found in step : F Mg sinθ + T M a ( sinθ ) T M a g ( sinθ ) F T + mg ma M a + g + mg ma 3. Solve for a: Mg sinθ + mg ( m + M ) a m M sinθ 3. kg 5.7 kg sin 35 a g m + M kg a m/s ( 9.8 m/s ) 4. The negative value means that the acceleration of the hanging mass is in the upward direction. 5. (b) The magnitude of the acceleration is m/s. Insight: Verify for yourself that the angle of the incline that balances the two masses is 34 when m 3. kg. 36. Picture the Problem: Refer to the figure at right: Strategy: Write Newton s Second Law for each block and add the equations to eliminate the unknown tension T. Solve the resulting equation for the acceleration a, and use the acceleration to find the tension. Let be positive in the direction of each mass s motion, m be the mass on the table, and m be the hanging mass. Solution:. (a) The tension in the string is less than the weight of the hanging mass. If it were equal to the weight, the hanging mass would not accelerate.. (b) Write Newton s Second Law for each block and block block F T m a add the equations: ( + ) F T + m g m a m g m m a 3. Solve the resulting equation for a: a m m + m.80 kg ( 9.8 m/s ) 4.36 m/s 6.30 kg g 4 of 89

15 4. (c) Use the first equation to find T: T ma 3.50 kg 4.36 m/s 5.3 N Insight: Note that the blocks move as if they were a single block of mass 6.30 kg under the influence of a force equal to mg 7.5 N. The tension in the string would be zero if m fell freely, 7.5 N if m (and the entire system) were at rest. 37. Picture the Problem: Two buckets are attached to either end of a rope that is hung over a pulley, as shown in the figure at right. Strategy: Use Newton s Second Law to determine the tension in the rope for all three cases. For the case when the masses accelerate, write Newton s Second Law for each bucket and combine the equations to find the tension. Let m represent the lighter bucket and m the heavier bucket (on the right in the figure). The positive ais points upward for m and downward for m. Solution:. (a) Write Newton s Second Law for m with a 0 :. (b) Write Newton s Second Law for m : 3. Write Newton s Second Law for m and substitute for the acceleration from step : 4. Substitute W m g and m m W W : 5. (c) Write F 0 for m with a 0 : T F T + m g 0 T m g 0 N F T m g m a a T m g F T + m g m a m T m g m g T + m m ( 0 N) W + W W + 0 N 63 N 80 N F T m g 0 T m g 63 N Insight: Another way to solve this problem is to add the two versions of Newton s Second Law in steps and 3 to eliminate T and solve for a, then use a to find T. Verify for yourself that a.7 m/s in part (b). 38. Picture the Problem: Your car travels along a circular path at constant speed. Strategy: Static friction between your tires and the road provides the centripetal force required to make the car travel along a circular path. Set the static friction force equal to the centripetal force and calculate its value. Solution: Set the static friction force equal to the centripetal force: Insight: The maimum static friction force is ( 00 kg)( 5 m/s) mv fs Fcp macp 4.7 kn r 57 m µ smg kg 9.8 m/s 0.4 kn which corresponds to a maimum cornering speed (without skidding) of m/s. 5 of 89

16 39. Picture the Problem: The test tube travels along a circular path at constant speed. Strategy: Solve for the speed required to attain the desired acceleration. Solution: Solve for the speed: v racp r ( 5, 000g ) ( m)( 5, 000)( 9.8 m/s ) 00 m/s 0.0 km/s Insight: This speed corresponds to 5,000 revolutions per minute for the centrifuge, or 45 revolutions per second. 40. Picture the Problem: The car follows a circular path at constant speed as it passes over the bump. Strategy: The centripetal acceleration is downward, toward the center of the circle, as the car passes over the bump. Write Newton s Second Law in the vertical direction and solve for the normal force N, which is also the apparent weight of the passenger. Solution:. Write Newton s Second Law for the passenger and solve for N: Fy N mg macp mv r N m g v r. Insert numerical values: m/s N ( 67 kg) 9.8 m/s 380 N 0.38 kn 35 m 4. Picture the Problem: The car follows a circular path at constant speed as it passes over the bump. Strategy: The centripetal acceleration is downward, towards the center of the circle, as the car passes over the bump. Write Newton s Second Law in the vertical direction and set the normal force N, which is also the apparent weight of the passenger, equal to zero. Then solve for the speed of the car. Solution:. Write Newton s Second Law for the passenger, and set N to zero:. Now solve for v: Fy N mg macp mv r 0 mg mv r v rg 35 m 9.8 m/s 9 m/s Insight: The car has zero normal force in it as well, meaning it is now airborne! 6 of 89

17 Solutions to Tutorial. Picture the Problem: The farmer pushes the hay horizontally. Strategy: Multiply the force by the distance because in this case the two point along the same direction. Solution: Apply W Fd directly: W Fd 86 N 3.4 m 90 J 0.9 kj Insight: The 3-kg mass is unneeded information unless we needed to know the amount of friction or the acceleration of the bale. 86 N 3.4 m. Picture the Problem: The suitcase is pushed horizontally. Strategy: Determine the applied force and solve for d. Solution: Solve for d: W W 640 J d 3.6 m f µ mg k k ( 0.6)( 70.0 kg)( 9.8 m/s ) Insight: The applied force equals the friction force as long as the suitcase does not accelerate. 3. Picture the Problem: The paint can is lifted vertically. Strategy: Multiply the force by the distance because the two vectors point in the same direction in part (a). In part (b) the distance travelled is zero, and in part (c) the force and distance are antiparallel. Solution:. (a) Apply W Fd:. (b) Now the force and distance are perpendicular: W Fd mgd W 3.4 kg 9.8 m/s.8 m 60 J W 0 3. (c) Now the force and distance are W mgd 60 J kj antiparallel: Insight: The applied force equals the weight as long as the paint can does not accelerate. The can gains potential energy as it is lifted and loses potential energy as it is lowered. F d mg.8 m 4. Picture the Problem: The water skier is pulled horizontally. Strategy: Multiply the force by the distance because in this case the two point along the same direction. Solution:. (a) The work is positive because the force is along the direction of motion (θ 0 ).. (b) Apply W Fd directly: W Fd 0 N 65 m 7800 J 7.8 kj Insight: While the work done by the rope is positive, the work done by friction is negative, so as long as the skier moves at constant speed she doesn t gain or lose kinetic energy. 5. Picture the Problem: The wagon rolls horizontally but the force pulls upward at an angle. Strategy: Keep in mind the angle between the force and the direction of motion. θ F d 7 of 89

18 Solution: Use W Fdcosθ: o W Fd cosθ 6 N 0.0 m cos 5 50 J 0.5 kj Insight: Only the component of the force along the direction of the motion does any work. The vertical component of the force reduces the normal force a little. 6. Picture the Problem: The mop head is being pushed downward into the floor. Strategy: Keep in mind the angle between the force and the direction of motion. θ F d Solution:. (a) Use W Fdcosθ: o W Fd cosθ 50.0 N 0.50 m cos 55 4 J. (b) If the angle is increased to 65, a smaller component of the force will be along the direction of motion and therefore the work done by the janitor will decrease. Insight: Only the component of the force along the direction of the motion does any work. The vertical component of the force increases the normal force. 7. Picture the Problem: The plane and glider must be at different altitudes. Since the altitudes are constant, both are moving horizontally. Strategy: Solving for the angle between the force and the direction of motion. Solution:. Solve W Fdcosθ for the angle: W W Fd cos θ or cosθ Fd glider F θ d airplane. Calculate the angle: W.00 0 J θ cos cos 57.4 Fd ( 560 N)( 45 m) 5 o Insight: Only the component of the force along the direction of the motion does any work. The vertical component of the force helps to lift the glider a little. 8. Picture the Problem: The force and distance vectors for the woman are depicted on the right. Strategy: Multiply the distance by the component of the force that is parallel to the distance. N skateboard 7 N F r bicycle r r d vt Solution:. (a) First find the distance travelled:. Now multiply only the - components: 3. (b) The force on the bicycle is opposite that on the skateboard by Newton s third law. 4. (c) Multiply only the - components: r r d vt ( 4. m/s)( 5 s) ˆ 0.5 m ˆ W F d 7 N 0.5 m 700 J r Fbike ( 7 N) ˆ + ( N) yˆ W Fbike, d 7 N 0.5 m 700 J Insight: The bicyclist must do work while pedalling or she will lose kinetic energy and come to a stop. The force of friction on the skateboard must do 700 J of work because her velocity (and therefore kinetic energy) is constant. 8 of 89

19 9. Picture the Problem: The boat and skier are both moving toward the left but the rope is pulling at an angle. Strategy: Solve for the angle between the force and the direction of motion. Solution:. Solve for the angle: W W Fd cos θ cosθ Fd. Calculate the angle: W 3500 J θ cos cos Fd ( 75 N)( 50 m) o Insight: Only the component of the force along the direction of the motion does any work. The work the boat does on the skier is balanced by the negative work friction does on the skier, so that the kinetic energy of the skier is constant. 0. Picture the Problem: The fragment moves at high speed in a straight line. Strategy: Calculate the kinetic energy is using K ½ mv. Solution: Apply K½mv directly: K mv 770 kg 0 m/s.7 0 J.7 MJ 7 Insight: The energy came from the work the rocket motor did in order to place Skylab into orbit.. Picture the Problem: The pine cone falls straight down under the influence of gravity. Strategy: The work done by gravity equals the change in kinetic energy. The work done by gravity is always W mgh. Solution:. (a) The work done by gravity on the pine cone equals the increase in its kinetic energy. Set the energies equal and solve for v: W K mgh mv v gh 9.8 m/s 6 m 8 m/s. (b) Air resistance did negative work because the speed and therefore the kinetic energy of the pine cone when it landed was reduced. Air resistance removed energy from the pine cone. Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion.. Picture the Problem: The object falls straight down under the influence of gravity. Strategy: Use the dependence of kinetic energy upon mass and speed to answer parts (a) and (b). The work done by gravity can be found from the change in the kinetic energy. Solution:. (a) Apply K½mv directly: K mv 0.40 kg 6.0 m/s 7. J. (b) Solve for speed: K ( 5 J) v m/s m 0.40 kg 3. (c) Calculate W K : W K Kf Ki 5 J 7. J 8 J Insight: As an object falls, the work done by gravity increases the kinetic energy of the object. 3. Picture the Problem: The car slows down as it rolls horizontally a distance of 30.0 m through the sand. Strategy: The kinetic energy of the car is reduced by the amount of work done by friction. The work done is the force times the distance, so once we know the work done and the distance, we can find the force. 9 of 89

20 Solution:. (a) The net work done on the car must have been negative since the kinetic energy decreased.. (b) The work done by friction equals the average force of friction times the distance the car travelled. Apply equations, including a minus sign to indicate the force and distance are in opposite directions: W Fd K mv mv so that F f i ( 300 kg)( 5 8 m /s ) ( 30.0 m) ( f vi ) m v F 00 N. kn Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion. The actual force eerted on the car would be 00 N if the distance travelled is taken to be m. 4. Picture the Problem: The bicycle rolls horizontally on level ground, slows down, and comes to rest. Strategy: The work done by the brakes equals the change in the kinetic energy of the bicycle. Find the distance travelled, and the magnitude of the braking force. Solution:. (a) Calculate W K : W K mv mv 0 ( kg)( 4 m/s) f i 700 J 7. kj. (b) Find the distance: ( v v ) t 3. (c) The force can be found from W Fd: m/s 4.0 s 8 m 0 W W 700 J F 60 N 58 lb d 8 m Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion. The average velocity is half the initial velocity as long as the acceleration is constant. d 5. Picture the Problem: The spring is stretched to m by pulling to the right or compressed to m by pushing to the left. Strategy: The work to stretch or compress a spring a distance is k. Solution :. (a) Apply equation 7-8:. (b) Apply equation 7-8 again: W k W ( N/m)( m) 4 44 J 4 k N/m m 44 J Insight: The work done on the spring is the same in either case because the spring force is opposite the applied force in each case. The applied force and distance vectors are parallel in each case, so the works are positive. 0 of 89

21 6. Picture the Problem: The block slides toward the right and into the spring. After compressing the spring the block comes to rest. Strategy: The work to stretch or compress a spring a distance is k. The work done on the block by the spring equals the kinetic energy lost by the block. The work done on the block is negative because the force on the block is toward the left while the motion is toward the right. Solution:. Apply equations 7-7 and 7-8: W K 0 mv k on block block i. Now solve for k: v. m/s k m i ( 0.3 m).8 kg 9 N/m Insight: The kinetic energy of the block is transformed into the energy stored in the spring as it is compressed. 7. Picture the Problem: The work done by the force is the area under the force versus position graph. Strategy: The total work done by the force is the total area under the graph from zero to 0.75 m. The work done from 0.5 m to 0.60 m is the area shaded in grey in the figure at right. Add the works done in each of the three segments to find the total work. Solution:. (a) The total area under the graph:. (b) The area of the graph shaded in grey: Wtotal A 0.5 m N 0.45 J ( m)( 0.6 N) ( 0.5 m)( 0.4 N) ( m)( 0.8 N) W + + W 0.4 J Insight: The work is positive as long as the object moves from left to right (from small to large ). Therefore the object gains energy as it moves from left to right. 8. Picture the Problem: The spring is compressed horizontally. Strategy: The work and stretch distance can be used to find the spring constant by applying W ½ k. Solution:. (a) Solve for k : W k ( 60 J) ( 0.4 m) N/m 6 kn/m. (b) It would take more than 60 J of work because W is proportional to : W k k N/m 0.8 m 0.4 m 480 J of 89

22 Insight: The etra work required to stretch the spring an additional 0.4 m can be pictured as the difference in area of two triangles, one with base 0.8 m and one with base 0.4 m, both bounded by the line given by k. 9. Picture the Problem: The work done by the force is the area under the force versus position graph. Strategy: Determine the geometric area under the graph for the various given starting and ending positions. The area under the graph equals the work done on the block. Solution:. (a) The work done on the object is the area under the graph between 0 and 0.30 m:. The magnitude of the force at 0.0 m can be found from the given formula: 3. (b) The work done is the area under the graph between 0 and 0.40 m minus the area of the triangle between 0 and 0.0 m: W + 0. m 400 N m 400 N 0.8 kj 4 F k.0 0 N/m 0.0 m 000 N ( 0. m)( 400 N) ( m)( 400 N) ( 0.0 m)( 000 N). kj W + Insight: When the force varies as a function of it is often useful to break the area under the graph into simple geometric shapes to aid in the calculation of work. 0. Picture the Problem: The fly does work against gravity as it elevates its centre of mass. Strategy: The power required is the force times the velocity, where the force is just the weight of the fly. Solution: Apply equation P Fv: ( kg)( 9.8 m/s )( 0.05 m/s) P Fv mg v P W 0.3 mw Insight: The energy and power required of the fly is higher than this because it isn t 00% efficient at converting food energy into mechanical energy.. Picture the Problem: The bucket is lifted vertically upward. Strategy: The power required is the force times the velocity, where the force is just the weight of the bucket. P P 08 W Solution: Solve P Fv for v: v.0 m/s F mg 5.00 kg 9.8 m/s Insight: Lifting faster than this would require more power. If the rope s mass were not ignored it would require additional power since its centre of mass is also lifted. If the force eceeded the weight, the bucket would accelerate. of 89

23 . Picture the Problem: The kayaker paddles horizontally in a straight line at constant speed. Strategy: The kayaker does positive work on the kayak as she paddles, and friction does negative work at the same time. The two works are equal because the kayak does not change its kinetic energy. Therefore the force she eerts must be equivalent to the force of friction. Solution:. (a) Solve P Fv for F: P 50.0 W F 33.3 N 7.5 lb v.50 m/s. (b) Since the speed of the kayak is proportional to P / F, doubling the power would double the speed for the same F. Insight: Newton s Second Law F ma states that the net force on the kayak must be zero since it is not accelerating. That s another way of figuring that the magnitude of the paddling force equals the magnitude of the friction force. 3. Picture the Problem: The weight slowly descends straight down. Strategy: The power delivered is the force (the weight) times the speed m 3.5 d s/d Solution:. (a) Apply P Fv: P Fv mgv ( 4.5 kg)( 9.8 m/s ) P W 0.04 mw. (b) To increase the power delivered you must either increase the force or the velocity. In this case, the time it takes for the mass to descend should be decreased so the velocity will increase and so will the delivered power. Insight: The weight delivers energy to the clock by doing work. The downward force it eerts on the clock is parallel to its displacement, so it is doing positive work on the clock. 4. Picture the Problem: The car accelerates horizontally in a straight line. Strategy: The power required is the work required to change the kinetic energy divided by the time. Use ratios to easily find the desired quantities. Solution:. (a) Combine equations:. Now divide top and bottom by m and substitute for the velocities: W K mv mv t P P mv T f i i v v v v t 3T v T v T f i i 3. (b) Again combine equations: W K mvf Pt ( T ) 4. Now solve for v f : mv mvf mv so v f v and thus vf v Insight: The assumption that the power remains constant is not realistic because car engines only generate their rated horsepower at high engine rpm, so less power is generated when the car first begins to accelerate. T 3 of 89

24 5. Picture the Problem: The three paths of the sliding bo are depicted at right. Strategy: The work done by friction is W µ kmgd, where d is the distance the bo is pushed irregardless of direction, because the friction force always acts in a direction opposite the motion. Sum the work done by friction for each segment of each path.. Solution:. Calculate the work for path : k [ ] [ m] W µ kmg d + d + d3 + d4 + d5 µ mg [ ] W kg 9.8 m/s.0 m 73 J. Calculate W for path : W µ mg [ d + d + d ] k kg 9.8 m/s.0 m +.0 m +.0 m 33 J 3. Calculate the work for path 3: [ ] W µ mg d + d + d 3 k kg 9.8 m/s.0 m m m 46 J Insight: The amount of work done depends upon the path because friction is a nonconservative force. 6. Picture the Problem: The two paths of the object are shown at right. Strategy: Find the work done by gravity W mgy when the object is moved downward, W mgy when it is moved upward, and zero when it is moved horizontally. Sum the work done by gravity for each segment of each path. Solution:. (a) Calculate the work for path : ( 0 ) W mg + y 5. kg 9.8 m/s.0 m 5 J. Calculate the work for path : W mg ( y + 0).6 kg 9.8 m/s.0 m 5 J 3. (b) If you increase the mass of the object the work done by gravity will increase because it depends linearly on m. Insight: The work is path-independent because gravity is a conservative force. 7. Picture the Problem: The physical situation is depicted at right. Strategy: Use Wsp k ( i f ) for the work done by the spring. That way the work will always be negative if you start out at i 0 because the spring force will always be in the opposite direction from the stretch or compression. The work done by kinetic friction is Wfr µ kmgd, where d is the distance the bo is pushed irregardless of direction, because the friction force always acts in a direction opposite the motion. 4 of 89

25 Solution:. (a) Sum the work done by the spring for each segment of path :. Sum the work done by friction for each segment of path : 3. (b) Sum the work done by the spring for the direct path from A to B: 4. Sum the work done by friction for the direct path from A to B: W k + W sp sp { } ( 480 N/m) 0 ( 0.00 m) ( 0.00 m) ( 0.00 m ) J + 0 J J µ ( ) W mg d + d fr k kg 9.8 m/s m 0.5 J W k sp A B 480 N/m m J Wfr µ kmgd kg 9.8 m/s 0.00 m J Insight: The work done by friction is always negative, and increases in magnitude with the distance travelled. 8. Picture the Problem: The climber stands at the top of Mt. Everest. Strategy: Find the gravitational potential energy by using U mgy. 6 Solution: Calculate U mgy : U mgy ( 83 kg)( 9.8 m/s )( 8848 m) 7. 0 J 7. MJ Insight: You are free to declare that the climber s potential energy is zero at the top of Mt. Everest and 7. MJ at sea level! 9. Picture the Problem: The mass is suspended from a vertical spring. As the spring is stretched it stores potential energy. Strategy: Doubling the mass doubles the force eerted on the spring, and therefore doubles the stretch distance due to Hooke s Law F k. Use a ratio to find the increase in spring potential energy when the stretch distance is doubled. Solution:. (a) If the mass attached to the spring is doubled the stored potential energy in the spring will increase by a factor of four because the stretch distance will double. A doubling of the mass will double the etension of the spring. Doubling the etension of the spring will increase the potential energy by a factor of 4.. (b) Find an epression for the stretch distance as a function of U and m: 3. Doubling the mass doubles the force and therefore doubles the stretch distance: 4. Calculate the ratio U U : ( 0.96 J) U k mg mg U m mg mg k mg k U U U ( 3.0 kg)( 9.8 m/s ) ( 0.3 m) ( m) k 4 k 4U J 3.85 J Insight: Note that the change in gravitational potential energy also quadruples as the new mass is hung on the spring. It doubles because there is twice as much mass and it doubles again because the spring stretches twice as far. 5 of 89

26 30. Picture the Problem: The spring in the soap dispenser is compressed by the applied force. Strategy: Find the spring constant using the given energy and compression distance data. Solve the same equation for in order to answer part (b). Solution:. (a) Solve for k:. (b) Solve for : ( J) ( m) U k 00 N/m 0.0 kn/m U J 0.9 cm k 00 N/m Insight: To compress the spring of this dispenser 0.50 cm requires.0 N of force. 3. Picture the Problem: A graph of the potential energy vs. stretch distance is depicted at right. Strategy: The work that you must do to stretch a spring is equal to minus the work done by the spring because the force you eert is in the opposite direction from the force the spring eerts. Find the spring constant and the required work to stretch the spring the specified distance. Solution:. (a) Because the stored potential energy in a spring is proportional to the stretch distance squared, the work required to stretch the spring from 5.00 cm to 6.00 cm will be greater than the work required to stretch it from 4.00 cm to 5.00 cm.. (b) Find k: 3. Use k to find the new W : req W W U U U req spring 5 4 k k k ( m) ( m) W 30.5 J k N/m req Wreq k N/m m m 37.3 J Insight: Using the same procedure we discover that it would take 44. J to stretch the spring from 6.00 cm to 7.00 cm. 3. Picture the Problem: The pendulum bob swings from point A to point B and loses altitude and thus gravitational potential energy. See the figure at right. Strategy: Use the geometry of the problem to find the change in altitude y of the pendulum bob, and then find its change in gravitational potential energy. Solution:. Find the height change y of the pendulum bob: ( θ ) y L cosθ L L cos. Use y to find U : U mg y mgl ( cosθ ) 0.33 kg 9.8 m/s. m cos 35 U 0.70 J Insight: Note that y is negative because the pendulum swings from A to B. Likewise, positive and the pendulum gains potential energy if it swings from B to A. y is 6 of 89

27 Solutions to Tutorial 3. Picture the Problem: The owner walks slowly toward the northeast while the cat runs eastward and the dog runs northward. Strategy: Sum the momenta of the dog and cat using the component method. Use the known components of the total momentum to find its magnitude and direction. Let north be in the y direction, east in the direction. Solution:. Use the component method of vector addition to find the owner s momentum:. Divide the owner s momentum by his mass to get the components of the owner s velocity: 3. Use the known components to find the direction and magnitude of the owner s velocity: r r r r r p p + p m v + m v total d c d d c c total r r r p m v p r p v total ( 0.0 kg)(.50 m/s yˆ ) ( 5.00 kg)( 3.00 m/s ˆ ) ( 5.0 kg m/s) ˆ ( 50.0 kg m/s) ˆ + r p + y total 0 m0 ( 5.0 kg m/s) ˆ + ( 50.0 kg m/s) 70.0 kg ( 0.4 m/s) ˆ ( 0.74 m/s) θ tan v 0.43 m/s m/s m/s Insight: We bent the rules of significant figures a bit in step 3 in order to avoid rounding error. The owner is moving much slower than either the cat or the dog because of his larger mass.. Picture the Problem: The two carts approach each other on a frictionless track at different speeds. Strategy: Add the momenta of the two carts and set it equal to zero. Solve the resulting epression for v. Then find the total kinetic energy of the two-cart system. Let cart travel in the positive direction. Solution:. (a) Set p r r r r 0 and p m v + mv 0 solve for v : mv ( 0.35 kg)(. m/s) v 0.69 m/s m 0.6 kg. (b) No, kinetic energy is always greater than or equal to zero. 3. (c) Sum the kinetic energies of the two carts: K m v + m v yˆ ( 0.35 kg)(. m/s) ( 0.6 kg)( 0.69 m/s) J Insight: If cart is travelling in the positive ˆ direction, then its momentum is ( and the momentum of cart is ( 0.4 kg m/s) ˆ. yˆ 0.4 kg m/s 3. Picture the Problem: The baseball drops straight down, gaining momentum due to the acceleration of gravity. Strategy: Determine the speed of the baseball before it hits the ground, then find the height from which it was dropped. Solution:. Use p mv to find the speed of the ball when it lands: p kg m/s v 5.0 m/s m 0.50 kg ) ˆ 7 of 89

28 . (b) Solve for y 0. Let y 0 and v 0 0 : ( 5.0 m/s) v v g y y 0 0 v y0.38 m g 9.8 m/s Insight: Another way to find the initial height is to use conservation of energy, setting mgy mv and solving for y Picture the Problem: The ball falls vertically downward, landing with a speed of.5 m/s and rebounding upward with a speed of.0 m/s. Strategy: Find the change in momentum of the ball when it rebounds. r r r r r p p p m v v Solution:. (a) Find p r : f i ( f i ) ( 0.0 kg) (.0 m/s) (.5 m/s) ( 0.99 kg m/s). (b) Subtract the magnitudes of the momenta: r p 0.99 kg m/s ( 0.0 kg)(.0 m/s.5 m/s) p p m v v f i f i p p 0. kg m/s f i yˆ yˆ yˆ 3. (c) The quantity in part (a) is more directly related to the net force acting on the ball during its r r collision with the floor, first of all because F p t and as we can see from above that p r p p. Secondly, we epect the floor to eert an upward force on the ball but we f i calculated a downward (negative) value in part (b). Insight: If the ball were to rebound at.5 m/s upward we would find p r mv. kg m/s and p f pi 0. Such a collision with the floor would be called elastic. 5. Picture the Problem: The individual momenta and final momentum vectors are depicted at right. Strategy: The momenta of the two objects are perpendicular. Because of this we can say that the momentum of object is equal to the -component of the total momentum and the momentum of object is equal to the y-component of the total momentum. Find the momenta of objects and in this manner and divide by their speeds to determine the masses. Solution:. Find divide by v : p total, and. Find p total, y and divide by v : total, total v.80 m/s p p p cosθ 7.6 kg m/s cos kg m/s m m p p 7.0 kg m/s p.5 kg ( 7.6 kg m/s)( sin 66.5 ) sinθ total v v 3.0 m/s 5. kg Insight: Note that object has the larger momentum because the total momentum points mostly in the ŷ direction. The two objects have similar speeds, so object must have the larger mass in order to have the larger momentum. 6. Picture the Problem: The two skaters push apart and move in opposite directions without friction. Strategy: By applying the conservation of momentum we conclude that the total momentum of the two skaters after the push is zero, just as it was before the push. Set the total momentum of the system to zero and solve for m. Let the velocity v r point in the negative direction, v r in the positive direction. y p r total p r p r 8 of 89

29 Solution: Set p total 0 and solve for m mv ( 45 kg)( 0.6 m/s ) : p + p 0 m v + m v m v 0.89 m/s Insight: An alternative way to find the mass is to use the equations of kinematics. 3 kg 7. Picture the Problem: The two pieces fly in opposite directions at different speeds.. Strategy: As long as there is no friction the total momentum of the two pieces must remain zero, as it was before the eplosion. Combine the conservation of momentum with the given kinetic energy ratio to determine the ratio of the masses. Let m represent the piece with the smaller kinetic energy. Solution:. Set p r 0 and solve p + p 0 mv + mv for m m : m v m v m v m v. Set K K : 3. Combine the epressions from steps and : mv m v K v m K v m v m m m v m m m 4. The piece with the smaller kinetic energy has the larger mass. Insight: The smaller mass carries the larger kinetic energy because kinetic energy increases with the square of the velocity but is linear with mass. Its higher speed more than compensates for its smaller mass. 8. Picture the Problem: The astronaut and the satellite move in opposite directions after the astronaut pushes off. The astronaut travels at constant speed a distance d before coming in contact with the space shuttle. Strategy: As long as there is no friction the total momentum of the astronaut and the satellite must remain zero, as it was before the astronaut pushed off. Use the conservation of momentum to determine the speed of the astronaut, and then multiply the speed by the time to find the distance. Assume the satellite s motion is in the negative -direction. Solution:. Find the speed of the astronaut using conservation of momentum:. Find the distance to the space shuttle: pa + ps 0 mava + msvs msvs va m s s d vat t ma a ( 00 kg)( 0.4 m/s) ( 9 kg) m v 7.5 s 4 m Insight: One of the tricky things about spacewalking is that whenever you push on a satellite or anything else, you yourself get pushed! (Newton s Third Law). Conservation of momentum makes it easy to predict your speed. 9. Picture the Problem: The lumberjack moves to the right while the log moves to the left. Strategy: As long as there is no friction the total momentum of the lumberjack and the log remains zero, as it was before the lumberjack started trotting. Combine vector addition for relative motion with the epression from the conservation of momentum to find v L, s speed of lumberjack relative to the shore. Let v L, log speed of lumberjack relative to the log, and v log, s speed of the log relative to the shore. 9 of 89

30 Solution:. (a) Write out the equation for relative motion. Let the log travel in the negative direction:. Write out the conservation of momentum with respect to the shore: 3. Substitute the epression from step into step and solve for v L, s r r r v v + v L, s L, log log, s v v v L, s L, log log, s v v v log, s L, log L, s p r 0 mlvl, s mlogvlog, s ( ml + mlog ) m v m v m v v L L, s log log, s log L, log L, s v m + m m v L, s L log log L, log v L, s ( 380 kg)(.7 m/s) ( kg) mlogvl, log. m/s 4. (b) If the mass of the log had been greater, the lumberjack s speed relative to the shore would have been greater than that found in part (a), because the log would have moved slower in the negative direction. 5. (c) Use the epression from step 3 to find the new speed of the lumberjack: v L, s ( ml + mlog ) ( 450 kg)(.7 m/s) ( kg) mlogvl, log.3 m/s Insight: Taking the argument in (b) to its etreme, if the mass of the log equalled the mass of the Earth the lumberjack s speed would be eactly.7 m/s relative to the Earth (and the log). If the mass of the log were the same as the mass of the lumberjack, the speed of each relative to the Earth would be half the lumberjack s walking speed. 0. Picture the Problem: The vector diagram at right indicates the momenta of the three pieces. y Strategy: Since the plate falls straight down its momentum in the y plane is zero. That means the momenta of all three pieces must sum to zero. Choose the motion of the two pieces at right angles to one another to be in the ˆ and ŷ directions. Set the total momentum equal to zero and solve for v r 3. Solution:. Set p r r r 0 and p mv ˆ + mv yˆ + mv solve for v r : r ( mv) ˆ + ( mv) yˆ 3 m. Find the speed 3 : v ( v) ˆ ( v) v + yˆ 3 + v v v v 5 p r p r p r 3 3. Find the direction of 3 : v v θ tan tan v v v 3, y 3, Insight: As long as the first two pieces have equal masses the direction of v r 3 will always be the same. For instance, if the third piece has four times the mass of either piece or, its speed would be v 8 but θ would remain of 89

31 Solutions to Tutorial 4. Picture the Problem: The tire rotates about its ais through a certain angle. Strategy: Use equation s rθ to find the angular displacement. Solution: Solve 0- for θ: θ s.75 m 5.3 rad r 0.33 m Insight: This angular distance corresponds to 304 or 84% of a complete revolution.. Picture the Problem: The Earth travels in a nearly circular path around the Sun, completing one revolution per year. Strategy: Convert the known angular speed of rev/yr into units of rev/min. Solution: Convert the units: rev yr day h ω yr 365 days 4 h 60 min rev/min Insight: This angular speed corresponds to about deg/day or rad/s. A good rule of thumb in astronomy is that the Sun appears to move /day against the background of the fied stars. 3. Picture the Problem: The Earth rotates once on its ais every 4 hours. Strategy: Solution: Convert the units: Convert the known angular speed of rev/day into units of radians per second. rev π rad day h ω day rev 4 h 3600s rad/s Insight: This angular speed corresponds to about 5 / hour. A rule of thumb in astronomy is that the fied stars will move across the sky at this rate ( every 4 minutes, or 5 arcsec/s) due to Earth s rotation. 4. Picture the Problem: The pulsar rotates about its ais, completing revolution in 0.33 s. Strategy: Divide one revolution or π radians by the period in seconds to find the angular speed. Solution: Calculate ω using equation 0-3: ω rad rad θ π π 90 rad/s t T 0.033s Insight: The rotation rate of the pulsar can also be described as 800 rev/min. 5. Picture the Problem: The floppy disk rotates about its ais at a constant angular speed. Strategy: Use equation ω π/t to relate the period of rotation to the angular speed. Then use v rω to find the linear speed of a point on the disk s rim. Solution:. (a) Solve for ω :. (b) Apply v rω directly: π π ω 3.4 rad/s T 0.00s vt rω 3.5 in 3.4 rad/s 55 in/s m 39.4 in.4 m/s 3. (c) A point near the centre will have the same angular speed as a point on the rim because the rotation periods are the same. Insight: While the angular speed is the same everywhere on the disk, the linear speed is greatest at 3 of 89

32 the rim. The read/write circuitry must compensate for the different speeds at which the bits of data will move past the head. 6. Picture the Problem: The propeller rotates about its ais with constant angular acceleration. Strategy: Use the kinematic equations for rotating objects and the given formula to find the average angular speed and angular acceleration during the specified time intervals. By comparison of the formula given in the problem, ( 5 rad/s) t ( 4.5 rad/s ) can identify ω 0 5 rad/s and θ + t,with α. 4.5 rad/s θ θ θ ω0t + αt θ 0 0 Solution:. (a) ωav Use equations t t t to find ω av : ( 5 rad/s)( 0.00 s) + ( 4.5 rad/s )( 0.00 s) s ω 5 rad/s.3 0 rad/s av θ θ ω α 0 + 0t + t, we. (b) Use equations to find ω : av 3. (c) Use equations to find ω : av θ ω0t αt 5 rad/s.00 s 4.5 rad/s.00 s rad θ0 ω0t0 αt 0 ( 5 rad/s)(.000 s) ( 4.5 rad/s )(.000 s) rad θ θ θ rad ω 0 rad/s. 0 rad/s 0 av t t t s θ ω0t αt 5 rad/s.00 s 4.5 rad/s.00 s rad θ0 ω0t0 αt 0 ( 5 rad/s)(.000 s) ( 4.5 rad/s )(.000 s) rad θ θ θ rad ω 95 rad/s rad/s 0 av t t t s 4. (d) The angular acceleration is positive because the angular speed is positive and increasing with time. ω ω0 0 5 rad/s 5. (e) Find α av : αav 85 rad/s t s ω ω rad/s 6. Find α av : αav 85 rad/s t s Insight: We violated the rules of significant figures in order to report answers with two significant figures. Such problems arise whenever you try to subtract two large but similar numbers to get a small difference. The answers are only known to one significant figure, but we reported two in order to show clearly that the angular acceleration is constant. Of course we could also have figured from the equation given in the problem that since α 4.5 rad/s, it must be true that α 85.0 rad/s. 7. Picture the Problem: The propeller rotates about its ais, increasing its angular velocity at a constant rate. Strategy: Solution: Solve for α : Use the kinematic equations for rotation to find the angular acceleration. ( π ) ω ω 6 rad/s rad/s 0 α 7 rad/s θ.5 rev rad rev Insight: A speed of 6 rad/s is equivalent to 50 rev/min, indicating the motor is turning pretty slowly. A typical outboard motor is designed to operate at 5000 rev/min at full throttle. 3 of 89

33 8. Picture the Problem: The propeller rotates about its ais, increasing its angular velocity at a constant rate. Strategy: Use the kinematic equations for rotation to find the angle through which the propeller rotated. Solution: Solve for θ : θ ( ω + ω ) t ( + 6 rad/s)(.5 s) 48 rad 7.5 rev 0 Insight: A speed of 6 rad/s is equivalent to 50 rev/min, indicating the motor is turning pretty slowly. A typical outboard motor is designed to operate at 5000 rev/min at full throttle. 9. Picture the Problem: The bicycle wheel rotates about its ais, slowing down with constant angular acceleration before coming to rest. Strategy: Use the kinematic equations for rotation to find the angular acceleration and the time elapsed. Solution:. (a) Solve for α : 0 0 ( 6.35 rad/s) ω ω0 α 0.6 rad/s θ θ 4. rev π rad rev ω ω rad/s. (b) Solve for t: t α 0.6 rad/s 8. s Insight: The greater the friction in the ale, the larger the magnitude of the angular acceleration and the sooner the wheel will come to rest. 0. Picture the Problem: The ceiling fan rotates about its ais, slowing down with constant angular acceleration before coming to rest. Strategy: Use the kinematic equations for rotation to find the number of revolutions through which the fan rotates during the specified intervals. Because the fan slows down at a constant rate of acceleration, it takes eactly half the time for it to slow from 0.90 rev/s to 0.45 rev/s as it does to come to a complete stop. Solution:. (a) Find θ: θ ( ω ω ) t rev/s. min 60 s min 59 rev. (b) Find θ: θ ( ω ω ) t rev/s. min 60 s min 45 rev Insight: An alternative way to solve the problem is to find α rev/s and use α to find θ for each of the specified intervals. Note that you can stick with units of rev/s to find θ in units of revolutions instead of converting to radians and back again.. Picture the Problem: The discus thrower rotates about a vertical ais through her centre of mass, increasing her angular velocity at a constant rate. Strategy: Use the kinematic equations for rotation to find the number of revolutions through which the athlete rotates and the time elapsed during the specified interval. Solution:. (a) Solve for θ : ω ω 6.3 rad/s θ θ θ 9.0 rad rev π rad α. rad/s.4 rev ω ω rad/s. (b) Solve for t: t.9 s α. rad/s Insight: Notice the athlete turns nearly one and a half times around. Therefore, she should begin her spin with her back turned toward the range if she plans to throw the discus after reaching 6.3 rad/s. If she does let go at that point, the linear speed of the discus will be about 6.3 m/s (for a.0 m long arm) and will travel about 4.0 m if launched at 45 above level ground. Not that great compared with a championship throw of over 40 m for a college woman. 33 of 89

34 . Picture the Problem: The centrifuge rotates about its ais, slowing down with constant angular acceleration and coming to rest. Strategy: Use the kinematic equations for rotation to find the angular acceleration and the number of revolutions through which the centrifuge rotates before coming to rest. Solution:. (a) Solve for α :. (b) Solve for θ ω ω rev/min min 60 s 0 α 6.9 rev/s t 0. s : θ ( ω ω ) t rev/min min 60 s 0. s 37 rev Insight: Another way of epressing the angular acceleration is to say that it slows down at a rate of 377 rev/min/s. 3. Picture the Problem: The compact disk rotates about its ais, increasing its angular speed at a constant rate. Strategy: Use the kinematic equations for rotation to find the average angular speed during the time interval and then the angle through which the disk spins during this interval. Solution:. (a) Find the average angular speed over the time interval and find θ.. (b) Find θ: θ ( ω ω ) t rev/min 3.0 s min 60 s 7.8 rev 0 Insight: An alternative way to solve the problem is to find α.7 rev/s and use α to find θ 7.8 rev for the specified interval. Note that you can stick with units of rev/s to find θ in units of revolutions instead of converting to radians and back again. 4. Picture the Problem: The drill bit rotates about its ais, increasing its angular speed at a constant rate. Strategy: Use the kinematic equations for rotation to find the angular acceleration, average angular speed during the time interval, and the angle through which the drill bit spins during this interval. Solution:. (a) Solve for α :. (b) Find θ ( ) 3 ω ω 350, 000 rev/min min 60 s 0 0 α.8 0 rev/s t.s : θ ( ω + ω ) t ( 350, rev/min )(. s min 60 s) rev Insight: The angular acceleration could also be epressed as rad/s. Note that the bit spins thousands of times during the. seconds it is coming up to speed. At full speed it spins over,000 times in. seconds! 5. Picture the Problem: The hour hand rotates about its ais at a constant rate. Strategy: Convert the angular speed of the tip of the hour hand into a linear speed by multiplying by its radius. Solution: Apply v rω: v r t rev π rad h ω 8. cm 0.00 cm/s µ m/s h rev 3600 s Insight: The tip of a minute hand travels much faster, not only because its angular speed is times faster than the hour hand, but also because the minute hand is longer than the hour hand. 34 of 89

35 6. Picture the Problem: The Frisbee rotates at a constant rate about its central ais. Strategy: Find the angular speed from the knowledge of the linear speed and the radius. Solution: Solve v rω for ω : v r 3.4 m/s t ω ( 0.9 m ) 3 rad/s Insight: The rotation of a Frisbee produces its unique, stable flight characteristics. 7. Picture the Problem: The two horses are located at different places on the same carousel, which is rotating about its ais at a constant rate. Strategy: Find the angular speed of the horses by dividing π radians (for a complete circle) by the time it takes to complete a revolution. Then v rω together with the angular speed to find the linear speed. π rad Solution:. (a) Find ω and ω : ω ω 45 s 0.4 rad/s. Apply v rω directly: v rω t.75 m 0.4 rad/s 0.38 m/s 3. (b) Apply v rω directly: v r ω t.75 m.4 rad/s 0.4 m/s Insight: The outer horse eperiences a greater linear speed and greater centripetal acceleration because it is at a larger radius. 8. Picture the Problem: The compact disk rotates about its central ais at a constant angular speed. Strategy: Use v rω to find the linear speed of a point on the outer rim of the CD, and then use a cp rω to find the centripetal acceleration. Use ratios to determine the linear speed and centripetal acceleration for a point that is half the distance to the rotation ais. Solution:. (a) Apply v rω directly: v t rω 0.0 m 5.5 rad/s 0.35 m/s. (b)apply a cp rω directly: 3. (c) Use a ratio to find the new linear speed: 4. (d) Use a ratio to find the new a cp : acp rω 0.0 m 5.5 rad/s.65 m/s v r ω r ( 0.35 m/s) 0.58 m/s v r r v v ω a a r ω r.65 m/s 0.87 m/s a a r ω r Insight: The angular velocity is the same for all points on the CD regardless of the distance to the rotation ais. 9. Picture the Problem: The Ferris wheel rotates at a constant rate, with the centripetal acceleration of the passengers always pointing toward the ais of rotation. The acceleration of the passenger is thus upward when they are at the bottom of the wheel and downward when they are at the top of the wheel. Strategy: Use a cp rω to find the centripetal acceleration. The centripetal acceleration remains constant (as long as the angular speed remains the same) and points toward the ais of rotation. Solution:. (a) Apply a cp rω directly: a rω cp π rad 9.5 m 0.37 m/s 3 s. When the passenger is at the top of the Ferris wheel, the centripetal acceleration points downward toward the ais of rotation. 3. (b) The centripetal acceleration remains 0.37 m/s for a passenger at the bottom of the wheel because the radius and angular speed remain the same, but here the acceleration points upward 35 of 89

36 toward the ais of rotation. Insight: In order to double the centripetal acceleration you need to increase the angular speed by a factor of or decrease the period by a factor of ; in this case a period of 3 seconds will double the centripetal acceleration. 0. Picture the Problem: The Ferris wheel rotates anticlockwise but is slowing down at a constant rate. The Ferris wheel has a radius of 9.5 m and rotates once every 3 s. Strategy: Find the tangential acceleration of the passenger at the top of the Ferris wheel and combine it with the centripetal acceleration to find the total acceleration. a r Top of Ferris Wheel v r a r t φ a r cp r Solution:. Use a cp rω to find a cp :. Use equation a t rα to find a : t 3. (c) Combine the components to find a : a cp π rad rω 9.5 m 0.37 m/s 3 s at rα 9.5 m 0. rad/s. m/s (downward) (to the left) a acp + at 0.37 m/s +. m/s. m/s 4. (d) Find the angleφ and relate it to the direction of motion (which is to the left) a cp 0.37 m/s φ tan tan 0 or 70 below the at. m/s direction of motion. Insight: In this case the tangential acceleration is 5.7 times greater than the centripetal acceleration. The passengers will notice the slowing down more than they noticed the centripetal acceleration when it was rotating at a constant rate.. Picture the Problem: The ball moves in a circle of constant radius at constant speed. Strategy: The motion is approimately horizontal so we can neglect the fact that the rope would be inclined a little bit below horizontal in order to support the weight of the ball. Set the rope tension equal to the centripetal force required to keep the ball moving in a circle and solve for the angular speed. Solution:. (a) Set the string force F ma cp and solve for ω : F ma mrω cp F N ω. rad/s mr ( 0.5 kg)( 4.5 m). (b) Since ω is inversely proportional to r, the maimum angular velocity will increases if the rope is shortened. Insight: This is a fairly weak rope. Still, the problem illustrates well how the centripetal force increases linearly with the distance from the rotation ais. Decreasing r decreases the force, or allows a higher ω for the same amount of force.. Picture the Problem: The sanding disk rotates about its ais at a constant rate. Strategy: Convert the angular speed of the disk into the linear speed of its rim by multiplying by its radius v rω. Use the same equation together with equation 0-5 to determine the period of rotation for the given rim speed. 36 of 89

37 4 Solution:. (a) Apply v rω directly: v rω. (b) Substitute ω π T and solve t m.5 0 rad/s 68.8 m/s v rω π r T for T: π r π ( m) t t s 73. s T v 75 m/s Insight: An angular speed of rad/s is equivalent to 05,000 rev/min, or 340 rev/s! Such high speeds are necessary to get the linear speed of the rim of such a small tool up to a value where it polishes well. 3. Picture the Problem: The wheel rotates about its ais, increasing its angular velocity at a constant rate. Strategy: Set the tangential and centripetal accelerations equal to each other for a single point on the rim. Find an epression for the angular speed as a function of time ( ω 0 0 since the wheel starts from rest), and substitute the epression into the result of the first step. Then solve the equation for t. Solution:. Set at acp : a rα rω a t α ω cp µ. (b) Substitute ω 0 + αt and solve for t: t t α α α t t α α Insight: The greater the angular acceleration, the shorter the elapsed time before the angular and centripetal accelerations equal each other. After that the centripetal acceleration dominates because it is proportional to α. 4. Picture the Problem: The force is applied in a direction perpendicular to the handle of the wrench and at the end of the handle. Strategy: Find the force from a knowledge of the torque and the length of the wrench. Solution: Solve for F: ( sinθ ) τ r F τ 5 N m F 60 N r sinθ 0.5 m sin 90 Insight: A longer wrench can eert a larger torque for the same amount of force. 5. Picture the Problem: The weed is pulled by eerting a downward force on the end of the tool handle. Strategy: Set the torque on the tool equal to the force eerted by the weed times the moment arm and solve for the force. Solution: Solve for F: F τ F weed weed weed τ.3 N m 3 N r m weed r Insight: The torque must be the same everywhere on the tool. Therefore, the hand must eert a.3 N m 0. m 5.6 N force to produce a 3-N force at the weed. The force is multiplied by a factor of of 89

38 6. Picture the Problem: The arm etends out either horizontally or at some angle below horizontal, and the weight of the trophy is eerted straight downward on the hand. Strategy: The torque equals the moment arm times the force. In this case the moment arm is the horizontal distance between the shoulder and the hand, and the force is the downward weight of the trophy. Find the horizontal distance in each case and multiply it by the weight of the trophy to find the torque. In part (b) the horizontal distance is r r cosθ ( m) cos m. Solution:. (a) Multiply the moment arm by the weight:. (b) Multiply the moment arm by the weight: τ r mg m.5 kg 9.8 m/s 8.96 N m τ r mg m.5 kg 9.8 m/s 8.8 N m Insight: The torque on the arm is reduced as the arm is lowered. The torque is eactly zero when the arm is vertical. 7. Picture the Problem: The arm etends out either horizontally and the weight of the crab trap is eerted straight downward on the hand. Strategy: The torque equals the moment arm times the force. In this case the moment arm is the horizontal distance between the shoulder and the hand, and the force is the downward weight of the crab trap. Solution: Multiply the moment arm by the weight: τ r mg 0.70 m 3.6 kg 9.8 m/s 5 N m Insight: If the man bent his elbow and brought his hand up net to his shoulder, the torque on the shoulder would be zero but the force on his hand would remain 35 N. 8. Picture the Problem: The biceps muscle, the weight of the arm, and the weight of the ball all eert torques on the forearm as depicted at right. Strategy: Determine the torques produced by the biceps muscle, the weight of the forearm, and the weight of the ball. Sum the torques together to find the net torque. According to the sign convention, torques in the anticlockwise direction are positive, and those in the clockwise direction are negative. Solution:. (a) Compute the individual torques and sum them: τbiceps r F m.6 N N m forearm r mg ( 0.70 m)(.0 kg)( 9.8 m/s ).00 N m τball r W ball ( m)(.4 N) N m τ τbiceps + τforearm + τball N m.4 N m τ. (b) Negative net torque means the clockwise direction; the forearm and hand will rotate 38 of 89

39 downward. 3. (c) Attaching the biceps farther from the elbow would increase the moment arm and increase the net torque. Insight: The biceps would need to eert a force of at least 90.3 N in order to prevent the arm from rotating downward. 9. Picture the Problem: The adult pushes downward on the left side of the teeter-totter and the child sits on the right side as depicted in the figure: Strategy: Calculate the torques eerted by the weight of the child and the force of the parent s hands and sum them. The sign of the net torque indicates the direction in which the teeter-totter will rotate. Solution:. (a) Find the torque the child eerts on the teetertotter.. Find the torque eerted by the parent and sum the torques to find the direction of travel: 3. (b) Repeat step with the new r for the adult: 4. (c) Repeat step with the new r for the adult: τ τ child child child child r m g (.5 m)( 6 kg)( 9.8 m/s ) 35 N m τ adult r adult Fadult 3.0 m 95 N 85 N m. Here τ adult + τ child > 0 so the teeter-totter will rotate anticlockwise and the child will move up. τ adult r adult Fadult.5 m 95 N 38 N m. Here τ adult + τ child > 0 so the teeter-totter will rotate anticlockwise and the child will move up. τ adult r adult Fadult.0 m 95 N 90 N m. Here τ adult + τ child < 0 so the teeter-totter will rotate clockwise and the child will move down. Insight: The parent would have to eert the 95-N force eactly.48 m from the pivot point in order to balance the teeter-totter. We bent the rules for significant figures slightly to more easily compare the magnitudes of the torques. 30. Picture the Problem: The ceiling fan rotates about its ais, decreasing its angular speed at a constant rate. Strategy: Determine the angular acceleration and then find the moment of inertia of the fan. Solution: Solve for I: Insight: Friction converts the fan s initial kinetic energy of ( 0.0 N m)( 9.5 s) ( 0.65 rad/s) τ τ τ t I kg m α ω t ω Iω J into heat. 3. Picture the Problem: The ladder rotates about its centre of mass, increasing its angular speed at a constant rate. Strategy: Find the moment of inertia of a uniform rod of mass M and length L that is rotated about its centre of mass: I M L. Then find the required torque to produce the acceleration. Solution:. Find I F adult r adult M L : I M L 8. kg 3.5 m 6.80 kg m. Apply τ Iα directly: τ Iα r child mchild g 6.80 kg m 0.30 rad/s.05 N m 39 of 89

40 3. Picture the Problem: The wheel rotates about its ais, decreasing its angular speed at a constant rate, and comes to rest. Strategy: Find the moment of inertia of a uniform disk and calculate I. Then find the angular acceleration from the initial angular speed and the angle through which the wheel rotated. Use I and α together to find the torque eerted on the wheel. Solution:. (a) Find. Solve for α : I MR : I MR 6.4 kg 0.7 m.6 kg m ( π ) ω ω 0. rad/s 0 α 0.58 rad/s θ 0.75 rev rad rev 3. Apply τ Iα directly: τ Iα.6 kg m 0.58 rad/s 0.5 N m 4. (b) If the mass of the wheel is doubled and its radius is halved, the moment of inertial will be cut in half (doubled because of the mass, cut to a fourth because of the radius). Therefore the magnitude of the angular acceleration will increase if the frictional torque remains the same, and the angle through which the wheel rotates before coming to rest will decrease. Insight: If the moment of inertia is cut in half, the angular acceleration will double to 0.3 rad/s and the angle through which the wheel rotates will be cut in half to 0.38 rev. This is because the wheel has less rotational inertia but the frictional torque remains the same. We bent the rules for significant figures in step to avoid rounding error in step Picture the Problem: The object consists of three masses that can be rotated about any of the, y, or z aes, as shown in the figure at right. Strategy: Calculate the moments of inertia about the, y, and z aes, and then find the required torque to give the object an angular acceleration of.0 rad/s about the various aes. Let m 9.0 kg, m. kg, and m 3.5 kg. Solution:. (a) Calculate I :. Find I m r + m r + m r I kg.0 m kg m τ : τ I α 3. (b) Calculate I y : 4. Find y 9.0 kg m.0 rad/s N m I kg.0 m 0 kg m y τ : τ y I yα 5. (c) Calculate I z : 6. Find z 0 kg m.0 rad/s N m I 9.0 kg.0 m kg.0 m 9 kg m z τ : τ z I zα 9 kg m.0 rad/s 3 N m Insight: When the ais of rotation passes through a particular mass, that mass does not contribute to the moment of inertia because r 0. The most torque is required to rotate the masses about the z ais because that ais passes through the least amount of mass (only the.-kg mass). 34. Picture the Problem: The fish eerts a torque on the fishing reel and it rotates with constant angular acceleration. Strategy: Determine the moment of inertia of the fishing reel assuming it is a uniform cylinder ( MR ). Find the torque the fish eerts on the reel. Then apply Newton s Second Law for rotation to find the angular acceleration and find the amount of line pulled from the reel. 40 of 89

41 Solution:. (a) Find I: I MR 0.84 kg m kg m. Apply τ rf directly to find τ : r F τ m. N 0. N m 3. Solve τ Ia for α : τ 0. N m α I kg m 9 rad/s 4. (b) Solve for s: ( ) s rθ r αt 0.6 m m 9 rad/s 0.5 s Insight: This must be a small fish because it is not pulling very hard. Or maybe the fish is tired? 35. Picture the Problem: The fish eerts a torque on the fishing reel and it rotates with constant angular acceleration. Strategy: Determine the moment of inertia of the fishing reel assuming it is a uniform cylinder ( MR ). Find the net torque on the reel by subtracting the torque from the friction clutch from the torque due to the force the fish eerts. Then apply Newton s Second Law for rotation to find the angular acceleration and find the amount of line pulled from the reel. Solution:. (a) Find I: I MR 0.84 kg m kg m. Apply τ rf directly to find τ : r F C τ m. N N m N m 3. Solve τ Iα for α : τ N m α I kg m 54 rad/s 4. (b) Solve for s: ( ) s rθ r αt m 54 rad/s 0.5 s m 9.3 cm Insight: Less line is pulled because the friction clutch reduces the net torque and angular acceleration of the reel. We bent the rules for significant figures in steps and to avoid rounding errors in subsequent steps. For instance, if we follow the rules of subtraction in step, τ N m 0.07 N m, just one significant figure. 4 of 89

42 Solutions to Tutorial 5. Picture the Problem: The position of the mass oscillating on a spring is given by the equation of motion. Strategy: The oscillation period can be obtained directly from the argument of the cosine function. The mass is at one etreme of its motion at t 0, when the cosine is a maimum. It then moves toward the center as the cosine approaches zero. The first zero crossing will occur when the cosine function first equals zero, that is, after one-quarter period. Solution:. (a) Identify T with the time 0.58 s:. (b) Multiply the period by one-quarter to find the first zero crossing: Since π π cos t cos t, therefore T 0.58 s. T 0.58 s t 0.58 s 0.5 s. 4 Insight: A cosine function is zero at ¼ and ¾ of a period. It has its greatest magnitude at 0 and ½ of a period.. Picture the Problem: As the mass oscillates on the spring we can find its position at any given time from the equation of motion. Strategy: The oscillation period can be obtained directly from the argument of the cosine function. The frequency is the inverse of the period. The mass is at one etreme of its motion at t 0, when the cosine is a maimum. The mass is at the point of interest when the cosine function is equal to. This occurs one-half a period later. Solution:. (a) Observe that in the argument of cosine the period is in the denominator: π π Since cos t cos t, therefore T T 0.68 s 0.68 s. Invert the period to obtain the frequency: f 0.68 s.5 Hz. 3. (b) The time the mass is at 7.8 cm is half a period: t T (0.68 s) 0.34 s. Insight: This problem could also be solved by setting the motion equation equal to 7.8 cm t 0.68 s π cos 7.8 cm 7.8 cm 0.34 s. and solving for the time: 3. Picture the Problem: When two or more atoms are bound in a molecule they are separated by an equilibrium distance. If the atoms get too close to each other the binding force is repulsive. When the atoms are too far apart the binding force is attractive. The nature of the binding force therefore is to cause the atoms to oscillate about the equilibrium distance. Strategy: Since the mass starts at A at time t 0, this is a cosine function given Acos ωt. From the data given we need to identify the constants A and ω. A cosine by function is at its maimum at t 0, but a sine function equals zero at t 0. Solution:. (a) Identify the amplitude as A:. Calculate the angular frequency from the frequency: 3. Substitute the amplitude and angular frequency into the cosine equation: 4. (b) It will be a sine function, Asin ( ωt) 0. A 3.50 nm ω π f 4.00π 0 s 4 4 ( 3.50 nm) cos ( 4.00π 0 s ), since sine satisfies the initial condition of 0 at t Insight: A cos function has a maimum amplitude at t 0. A sin function has 0 amplitude at t 0. t 4 of 89

43 4. Picture the Problem: One period of oscillation is shown in the figure. Strategy: Since the mass is at 0 at t 0, this will be a sine function. Substitute the amplitude and period into the sine equation to determine the general equation of motion. Finally substitute in the specific times to determine the position at each time. Solution:. Write the equation sine equation in terms of the given amplitude and period:. (a) Substitute t T/8 into the sine equation and evaluate: 3. (b) Substitute t T/4 into the sine equation and evaluate: 4. (c) Substitute t T/ into the sine equation and evaluate: 5. (d) Substitute t 3T/4 into the sine equation and evaluate: π ( 0.48 cm) cos t T π T π ( 0.48 cm) cos ( 0.48 cm) cos 0.34 cm T 8 4 π T π ( 0.48 cm) cos ( 0.48 cm) cos 0.48 cm T 4 π T ( 0.48 cm) cos ( 0.48 cm) cos( π ) 0 T π 3T ( 0.48 cm) cos T 4 3π ( 0.48 cm) cos 0.48 cm 6. (e) Sketch a plot with the four data points: Insight: The sine curve has been included in the sketch of part (e) to show that the four positions are consistent with a sine function. 5. Picture the Problem: A mass is attached to a spring. The mass is displaced from equilibrium and released from rest. The spring force causes the mass to oscillate about the equilibrium position in harmonic motion. Strategy: Since the mass starts from rest at t 0, the harmonic equation will be a sine function. We will use the amplitude and period to determine the general equation, which can be evaluated for any specific time. During the first half of each period the mass will be moving in the negative -direction toward the minimum and during the second half the mass will move in the positive direction, back toward the maimum. Therefore we can determine the direction of motion by finding in which half of a period the time is located. Solution:. (a) Insert the period and amplitude to create the general harmonic equation:. Insert t 6.37 s into the general equation and evaluate the position: 3. (b) Divide the time by one period: π t ( m) cos. 3.5 s π 6.37 s ( m) cos m 3.5 s 6.37 s s 4. Since this is slightly greater than two full periods, the mass is headed toward the origin from its maimum displacement and is moving in the negative direction. 43 of 89

44 Insight: This problem could also be solved by inserting a time slightly later than t 6.37 s (such as t 6.38 s) and evaluating the position ( m). Since this result is smaller than m, the mass is moving in the negative direction. 6. Picture the Problem: One velocity over one period is shown in the figure. The region for which the speed ( v ) is greater than v ma / is shaded gray. Strategy: Since speed is the magnitude of the velocity it will be greater than v ma / twice during each period. The two regions are symmetric; the total time is double the time interval over which the velocity is positive. The end points of the regions are found when the sine function is equal to one-half. Solution:. Set v -Aωsin(ωt) equal to v ma / and divide out v ma : v ma π t sin v T π t sin T ma. Take the arcsine of each side and solve for t: π 5π 3. Evaluate using sin and : 6 6 π t sin T T t sin π T π 5 5 or T π t T or T π 6 π 6 4. Subtract the first time from the second time and multiply by two: 5T T 4T T. 3 The mass s speed is greater than v ma / for two-thirds of a cycle. Insight: If the problem had asked for the time that the velocity was greater than +v ma /, then only the crest of the cycle would have been included and the time would have been one-third of a cycle. 7. Picture the Problem: When an object is oscillating in simple harmonic motion it eperiences a maimum acceleration when it is displaced at its maimum amplitude. As the object moves toward the equilibrium position, the acceleration decreases and the velocity of the object increases. The object eperiences its maimum velocity as it passes through the equilibrium position. Strategy: The maimum velocity and acceleration can both be written in terms of the amplitude and angular speed, vma Aω, ama Aω. We can rearrange these equations to solve for the amplitude and angular speed. Then we can use the angular speed to determine the period. Solution:. (a) Divide the square of the velocity by the acceleration to find the amplitude:. (b) Divide the acceleration by the velocity to determine the angular speed: Aω v A Aω Aω a ω Aω v ma ma a ma ma 3. Divide π by the angular speed to calculate the period: π π π v T ω a a ma vma Insight: When two or more quantities are functions of the same variables, it is often possible to ma ma 44 of 89

45 rearrange the equations to isolate one or more of the variables. This can be a useful mathematical procedure. 8. Picture the Problem: When an object is oscillating in simple harmonic motion it eperiences a maimum acceleration when it is displaced at its maimum amplitude. As the object moves toward the equilibrium position the acceleration decreases and the velocity of the object increases. The object eperiences its maimum velocity as it passes through the equilibrium position. Strategy: The maimum velocity and acceleration can both be written in terms of the amplitude and angular speed, vma Aω, ama Aω. Rearrange these equations to solve for the amplitude and angular speed. Then use the angular speed to determine the period. Solution:. (a) Divide the square of the velocity by the acceleration to find the amplitude:. (b) Divide the acceleration by the velocity to determine the angular speed: 3. Divide π by the angular speed to calculate T: ( Aω ) v ( 4.3 m/s) A Aω Aω a ω Aω v T ma ama ma ma π π ω a ma vma ( 0.65 m/s ) π v 8 m ma a ma π 4.3 m/s ( 0.65 m/s ) Insight: When two or more quantities are functions of the same variables, it is often possible to rearrange the equations to uniquely determine those variables. 9. Picture the Problem: As the child rocks back and forth on the swing, her speed increases as she approaches the equilibrium of the swing and then decreases back to zero at the end of the swing. The maimum speed occurs when the swing is vertical. Strategy: The maimum velocity equals the amplitude times the angular speed, which in turn depends upon the period. Solution:. Write the maimum velocity in terms of amplitude and period:. Insert the amplitude and period into the equation for maimum speed: 4 s π π vma Aω A ( 0.04 m) m s T.80 s π π 0.04 m vma A T.80 s m s Insight: The girl s motion has an amplitude of 0.34cm and a maimum speed of.64 km/h. Not a very eciting swing. 0. Picture the Problem: A structural beam is a metal rod that is necessary to maintain the shape and integrity of the spacecraft. Large forces, which could be caused by small, but rapid oscillations, could damage the support beam, jeopardizing the integrity of the spacecraft. Strategy: The maimum acceleration can be written in terms of the amplitude and angular speed, a Aω, where angular speed is π times the frequency. ma Solution:. Multiply the amplitude by the square of π times the frequency to get a ma :. Factor out g 9.8 m/s to obtain the acceleration as a multiple of g: ma ω ( π ) ( π ) a A A f a 0.5 mm 0 Hz 9 m/s g 9 m/s g 9.8 m/s ma Insight: Since the acceleration is proportional to the square of the frequency, a large frequency will result in a very large acceleration. This is true even for small amplitude oscillations. 45 of 89

46 . Picture the Problem: The figure shows a turntable rotating with a peg on its outer rim. The table is illuminated on one side. The shadow of the peg moves with simple harmonic motion along the wall. Strategy: The shadow of the peg moves along the wall with simple harmonic motion. The period is the time for the peg (and as such the shadow) to complete a full revolution. The amplitude is the same as the radius of the circle, and is also the maimum distance from the center. Knowing the period and amplitude we can calculate the maimum velocity, remembering that the angular frequency is inversely related to the period. Finally, use the amplitude and period to calculate the maimum acceleration. Solution:. (a) Divide the C π r π ( 0.5 m) circumference of the turntable by T the peg s tangential velocity to get v v 0.67 m/s the period of rotation:. (b) Set the amplitude equal to the radius of the turntable: 3. (c) Insert the period and amplitude into the maimum velocity acceleration: A r 0.5 m.3 s π A Av vma Aω v 0.67 m s T r 4. (d) Insert the period and π v v ( 0.67 m/s) ama Aω A A.8 m/s amplitude into the maimum T r r 0.5 m acceleration equation: Insight: The maimum speed of the shadow is equal to the tangential speed of the peg. The shadow and the peg travel at the same speed when the peg travels perpendicular to the light source. The shadow travels slower than the peg when a component of the peg s velocity is parallel to the light.. Picture the Problem: In an engine the moving pistons compress the fuel in the chamber and epand after the fuel has been ignited. This motion provides the power to the car. The frequency of the piston motion is measured by the number of revolutions of the crankshaft per minute (rev/min). Strategy: To solve for the maimum acceleration and speed we must first convert the angular speed from rev/min to rad/s. Then we can use the amplitude and angular speed in the equations for maimum acceleration and maimum speed. Solution:. Convert the angular speed to rad/s:. (a) Use the amplitude and angular speed to solve for maimum acceleration: 3. (c) Use the amplitude and angular speed to solve for maimum speed: rev π rad min s min rev 60 s ama Aω 3.5 cm 78 s. km/s vma Aω 3.5 cm 78 s 6. m/s Insight: The maimum acceleration of the pistons is over 00 times the acceleration due to gravity. Therefore the gravitational force has negligible effect on the motion of a working piston. 3. Picture the Problem: An air cart is attached to the end of a spring and pulled slightly away from equilibrium and released. It oscillates about the equilibrium position at a frequency determined by the mass of the cart and the stiffness of the spring. Strategy: The maimum kinetic energy can be obtained by inserting the maimum velocity into the kinetic energy equation. The maimum force can be obtained by substituting the maimum acceleration into Newton s Second Law. From the position equation we see that the amplitude is A 0.0 cm and the angular speed is ω.00 rad/s. We will use these values to calculate the - 46 of 89

47 maimum velocity and acceleration. Solution:. (a) Write the kinetic energy in terms of mass, amplitude, and angular speed:. Insert the values of mass, K amplitude and angular speed to calculate K : ma 3. (b) Write the force equation in terms of mass, amplitude, and angular speed: 4. Insert the values of mass, amplitude and angular speed to calculate the maimum force: Kma mvma m Aω ma ma ( 0.64 kg )( 0.00 m ) (.00 s m Aω ) 0.3 J ma F ma m Aω Fma m Aω 0.64 kg 0.00 m.00s 0.6 N Insight: The phase shift of +π in the displacement equation has no effect on either K ma or F ma, but it reverses the sign of the displacement. Therefore, at t 0 the cart starts 0 cm instead of at 0 cm. 4. Picture the Problem: A mass attached to a spring is pulled slightly away from equilibrium and released. The mass then oscillates about the equilibrium position at a frequency determined by the stiffness of the spring. Strategy: We can determine the spring constant by solving the equation for the period of a mass on a spring for the spring constant, and substituting in the given period and mass. Solution:. Solve the period equation for the spring constant: m π T π k m k T. Insert the numeric values for T π π k m ( 0.4 kg) 9 N m and m: T 0.75 s Insight: Measuring the period of oscillation is in many cases the most accurate way of measuring a spring constant. 5. Picture the Problem: The picture shows the unstretched spring and the spring with a 0.50-kg mass attached to it. Strategy: We can use the displacement of the spring to calculate the spring constant. The spring constant and period can then be inserted into the period equation to solve for the necessary mass. Solution:. Use the spring F mg force equation F ky k y y to solve for the spring constant:. Insert numeric values to obtain k: 3. Solve the period equation for the mass: 0.50 kg 9.8 m/s k 3.7 N/m 5 0 m m T T π m k k π 0.75 s 4. Insert numeric values to m ( 3.7 N/m) 0.47 kg obtain the mass: π Insight: Since the period is proportional to the square root of the mass, increasing the mass will increase the period. 47 of 89

48 6. Picture the Problem: When the two people enter the car they compress the springs. The distance that the springs are compressed is regulated by their mass and the stiffness of the spring. When the car hits a bump in the road the car begins to oscillate up and down at a frequency determined by the total mass of the car and riders and the stiffness of the springs. Strategy: Using the mass of the two people and the amount the springs compressed, we can calculate the spring constant. The total load can be obtained by solving the period of oscillation equation for the mass. The mass of the car is found by subtracting the mass of the two people from the total mass. Solution:. Solve the force equation for the spring constant:. Enter numeric values for the spring constant: 3. (a) Solve the period equation for the total load (M+m): F mg F ky k y y 5 kg 9.8 m/s k m N/m M + m T T π M + m k k π 4. Enter numeric values for the total mass: 4 M m 5. (b) Subtract the mass of the two people to obtain M:.65 s N/m 060 kg π M ( M + m) m 057 kg 5 kg 93 kg Insight: The period of oscillation is an ecellent method of determining the mass of an object. This is especially useful in orbit, where conventional scales do not work. 7. Picture the Problem: A mass attached to a vertical spring, pulled slightly down from the equilibrium position and released will oscillate in simple harmonic motion. The acceleration of the mass will be a maimum when the spring is at maimum displacement. As the mass moves back to the equilibrium position the speed increases and the acceleration decreases. The maimum speed is at equilibrium position. As the mass moves away from equilibrium the velocity decreases as the deceleration increases until the mass stops at the opposite amplitude. Strategy: The period can be found from the spring constant and mass. From the maimum speed and the period we can calculate the amplitude. We can calculate the maimum acceleration from the period and the amplitude. Solution:. (a) Insert the mass and spring constant into the period equation:. (b) Solve the maimum velocity equation for the amplitude: m 0.85 kg T π π 0.47 s k 50 N/m v s 0.35 m/s ma Tvma A ω π π.6 cm 3. (c) Insert the amplitude and period into the maimum acceleration equation: π π ama Aω A m 4.6 m s T s Insight: Another way to solve this problem is to calculate the angular speed ω the period. Then A vma ω and ama vmaω. k m instead of 8. Picture the Problem: If the motorcycle is pushed down slightly on its springs it will oscillate up and down in harmonic motion. A rider sitting on the motorcycle effectively increases the mass of the motorcycle and oscillates also. Strategy: We can use the equation for the period of a mass on a spring. Writing this equation for 48 of 89

49 the motorcycle without rider and again for the motorcycle with rider we can calculate the percent difference in the periods. Solution:. (a) The period increases, because the person s mass is added to the system and T m.. (b) Write the equation for the period of the motorcycle without the rider: T π m k 3. Write the equation for the period of the motorcycle with the rider: 4. Calculate the percent difference between the two periods: 5. Simplify by factoring out π m k from the numerator and denominator: T m + M π k m + M m π π T T k k T m π k T T m + M T m 5+ kg % 5 kg Insight: The percent change in the period does not depend on the spring constant. It only depends on the fractional increase in mass. 9. Picture the Problem: The mass on the left is hung from one spring. The mass on the right is hung from two identical springs. Strategy: When a mass is attached to a single spring it stretches by a distance. When two identical springs are connected end to end and the same mass is attached, each spring will stretch by the same amount and the total stretch will be. Since the force has not changed, but the total stretch is twice as much, the effective spring constant will be half the single spring constant. Use this information to find a relationship between the two periods. Solution:(a) The period is more than the period of a single spring because the effective spring constant is smaller and T k.. (b) Write the combined spring constant as onehalf of the single spring constant: 3. Replace k with k/ in the period equation: k k m T π π k m ( k / ) 4. Rearrange the equation: m m T π π k k 5. Replace the term in parentheses with the original period: m T π T k Insight: If the spring had been replaced by 3 identical springs the resulting spring constant would be k k / 3, giving T 3T. In general the period will be proportional to the square-root of the length of the spring. 49 of 89

50 0. Picture the Problem: A mass is attached to the end of a.5-meter-long string, displaced slightly from the vertical and released. The mass then swings back and forth through the vertical with a period determined by the length of the string. Strategy: Use the period of the pendulum and its length to calculate the acceleration of gravity. Solution:. Solve the period equation for gravity: L π T π g L g T π. Insert numeric values: g (.5 m) 9.6 m/s 5 ( 6 s) Insight: The small variations in gravity around the surface of the Earth are measured using the period of a pendulum.. Picture the Problem: A simple pendulum is a mass attached to a string. The mass is displaced so the string is slightly away from the vertical and released. The mass then oscillates about the vertical with a period determined by the length of the string and gravity. Strategy: Calculate the length of the pendulum from its period. Solution:. Solve the period equation for length: L T T π L g g π. Insert numeric values: L.00 s 9.8 m/s 4.8 cm π Insight: This is the length of the pendulum in many older clocks. Larger clocks, such as a grandfather clock, have pendulums about a meter long with a period of seconds.. Picture the Problem: The pendulum on the Moon is the same length string and mass, with the mass displaced from the vertical and released. The period is determined by the length of the string and the acceleration due to gravity. Strategy: Since the period of a pendulum is inversely proportional to the square-root of gravity, the smaller gravitational pull on the Moon would increase the period of the pendulum. The period of the pendulum on the Moon can be calculated by replacing the acceleration of gravity on the Earth with the acceleration of gravity on the Moon, ( g g ). Solution:. Write the period on the Moon in terms of the period on Earth: T Moon Moon 6 Earth 6 Earth L L π 6 π 6 T g g. Calculate the period on the Moon: T Moon 6 (.00 s).45 s Insight: A grandfather clock taken to the Moon would run.45 times slower than one on the Earth. To run properly, the pendulum in the clock would need to be shortened to one-sith of its original length. 3. Picture the Problem: A pendulum is made by attaching a mass to the end of a string. The opposite end of the string is attached to the ceiling of an elevator. The mass is displaced slightly from vertical and released. The pendulum oscillated back and forth with a period determined by the length of the string and the effective gravity eperienced in the elevator. Strategy: The acceleration of gravity is that felt by the pendulum. As an object accelerates upward it eperiences an effective gravity g + a. When the elevator accelerates downward the effective gravity is or g a. To solve for the accelerated period we can substitute the effective gravity into the period equation. Earth Earth Solution:. (a) Replace g by g + a in the equation for the period of a pendulum: T π L g + a 50 of 89

51 . (b) ) Replace g by g a in the equation for the period of a pendulum: T π L g a Insight: Consider the effect on the pendulum if the elevator were to be in free fall. According to our answer to part (b), as the downward acceleration approaches g, the period increases. In the limit that a g, T. If the elevator were in free fall the tension in the pendulum string would be zero and the pendulum would not oscillate. 5 of 89

52 Solutions to Tutorial 6. Picture the Problem: The image shows a wave with the given wave dimensions. Strategy: Set the wavelength equal to the horizontal crest-to-crest distance, or double the horizontal crest-totrough distance. Set the amplitude equal to the vertical crest-to-midline distance, or half the vertical crest-totrough distance. Solution:. (a) Double the horizontal crest-totrough distance:. (b) Halve the vertical crest-to-trough distance: λ ( 6 cm) 5 cm A cm 5.5 cm Insight: Note the difference in wavelength and amplitude. The wavelength is the entire distance from crest to crest, but amplitude is only from the equilibrium point to the crest.. Picture the Problem: A surfer measures the frequency and length of the waves that pass her. From this information we wish to calculate the wave speed. Strategy: Write the wave speed as the product of the wavelength and frequency. Solution: Multiply wavelength by frequency: min v λ f ( 34 m)( 4 /min) 7.9 m/s 60 sec Insight: The wave speed can increase by either an increase in wavelength or an increase in frequency. 3. Picture the Problem: The image shows water waves passing to a shallow region where the speed decreases. We need to calculate the wavelength in the shallow area. Strategy: Calculate the frequency in the deep water. Then use the constant frequency and the speed in the shallow water to calculate the new wavelength. Solution:. Calculate the frequency: f v λ.0 m/s.5 m.333 Hz v.6 m/s. Calculate the new wavelength: λ. m f.333 Hz Insight: Note that decreasing the speed, with constant frequency, will decrease the wavelength. 4. Picture the Problem: The speed and wavelength of a tsunami are given and we wish to calculate the frequency. Strategy: Solve for the frequency. Solution: Calculate the frequency: ( 750 km/h) h 4 v f Hz λ 30 km 3600 s Insight: Although the tsunami has a very high speed, the long wavelength gives the tsunami a low frequency. 5 of 89

53 5. Picture the Problem: A wave of known amplitude, frequency, and wavelength travels along a string. We wish to calculate the distance travelled horizontally by the wave in 0.5 s and the distance travelled by a point on the string in the same time period. Strategy: Multiply the time by the wave speed to calculate the horizontal distance travelled by the wave. A point on the string travels up and down a distance four times the amplitude during each period. Calculate the fraction of a period by dividing the time by the time of a full period. Set the period equal to the inverse of the period and multiply by four times the amplitude to calculate the distance traveled by a point on the string. Solution:. (a) Calculate the horizontal distance:. (b) Calculate the vertical distance: ( λ ) dw vt f t 7 0 m 4.5 Hz 0.50 s 0.6 m t dk ( 4A) 4Aft 4( 0 m)( 4.5 Hz)( 0.50 s). m T 3. (c) The distance travelled by a wave peak is independent of the amplitude, so the answer in part (a) is unchanged. The distance travelled by the knot varies directly with the amplitude, so the answer in part (b) is halved. Insight: A point on the string travels four times the wave amplitude in the same time that the crest travels one wavelength. 6. Picture the Problem: Using the equation for the speed of deep water waves given in the problem we want to calculate the speed and frequency of the waves. Strategy: Insert the given data into the equation v gλ π to solve for the speed of the waves. Then calculate the wave frequency. Solution:. (a) Insert the frequency into the deep water velocity equation:. (b) Solve for the frequency: v gλ / π 9.8 m/s 4.5 m / π.65 m s v.65 m/s f λ 4.5 m 0.59 Hz Insight: Since the velocity is proportional to the square-root of the wavelength, the frequency is inversely proportional to the square-root of the wavelength. Increasing the wavelength by a factor of four will double the wave speed and cut the frequency in half. 7. Picture the Problem: The speed of shallow water waves is proportional to the square-root of the water depth. We wish to calculate the speed and frequency of some shallow water waves. Strategy: Use the speed equation v gd given in the problem, where d is the water depth, to calculate the wave speed. Then calculate the wave frequency. Solution:. (a) Calculate the wave speed:. (b) Calculate the wave frequency: v gd 9.8 m/s 0.06 m 0.5 m s v m/s f λ m 67 Hz Insight: As the wave approaches shallower water, with constant frequency, its wavelength decreases. In this problem, if the depth drops to.3 cm, the wavelength will decrease to 0.59 cm. 8. Picture the Problem: The string tension is changed until the wave speed doubles. Strategy: The speed of a wave on a string is given by v (F/µ). Solve the equation for the tension in the string. Then use a ratio to find the factor by which the tension increases. Solution:. Solve for the tension: v F µ F v µ. Divide the tension at higher velocity by the initial tension: F v µ v 3 m/s 4 F v µ v 6 m/s The tension increases by a factor of 4. Insight: The tension increases by a factor equal to the square of the fractional increase in velocity. 53 of 89

54 9. Picture the Problem: The image shows two people talking on tin can telephone. The cans are connected by a 9.5-meter-long string weighing 3 grams. We wish to calculate the time it takes for a message to travel across the string. Strategy: Set the time equal to the distance divided by the velocity. The linear mass density is the total mass divided by the length. Solution:. Set the time equal to the distance divided by velocity:. Substitute µ m d and insert numerical values: d t d v µ F ( 0.03 kg)( 9.5 m) m / d md t d 0.9 s F F 8.6 N Insight: The message travels the same distance in the air in 0.08 seconds, about 7 times faster. 0. Picture the Problem: The image shows two people talking on tin can telephone. The cans are connected by a 9.5-meter-long string weighing 3 grams. We wish to determine how the tension in the string affects the time for the message to travel across the string. Strategy: In problem 9, we found that the travel time across the string is given by t md / F. Use this equation to calculate the time for the different tensions. Solution:. (a) Since the time is inversely related to the tension, increasing the tension will result in less time.. (b) Set the tension equal to 9.0 N: t ( 0.03 kg)( 9.5 m ) / ( 9.0 N) 0.8 s 3. (c) Set the tension equal to 0.0 N: t ( 0.03 kg)( 9.5 m ) / ( 0.0 N) 0.7 s Insight: As predicted, increasing the tension decreases the time for the message to travel the string.. Picture the Problem: Sound takes 0.94 seconds to travel across a wire of known length and density. We want to calculate the tension in the wire. Strategy: Solve for the tension in the wire. The velocity is given by the length of the wire divided by the time for the sound to travel across it. The linear mass density is the mass divided by the length. Solution:. (a) Solve for the tension: F v µ F v µ m L ml L t t 54 of 89

55 . Insert the given mass, length and time: ( kg)( 7.3 m) ( 0.94 s) F 0.70 N 3. (b) The mass is proportional to the tension (if L and t remain constant). So increased mass means increased tension. 4. (c) Solve with a mass of kg. ( kg)( 7.3 m) ( 0.94 s) F 0.78 N Insight: A heavier string requires greater tension for a wave to travel across it in the same time.. Picture the Problem: Waves travel down two strings, made of the same material and having the same length, but having different diameters and tensions. We wish to calculate the ratio of the wave speeds on these two strings. Strategy: Calculate the ratio of the velocities. Set the linear mass densities equal to the density of steel times the cross-sectional area of the wires. v Solution:. (a) Write the ratio of the velocities: A FA / µ A FA µ B v F / µ F µ B B B B A. Write the linear mass density in terms of density and area: 3. Write the area in terms of the diameter: va FA AB FA A ρ v F ρ A F A B B A B ( db ) ( d ) v F π F d v F F d A A A B B π A B B A B A va 40 N.0 mm 4. Insert the given tensions and diameters:.4 v 80 N 0.50 mm Insight: The ratio of the velocities is proportional to the square root of the tensions and inversely proportional to the diameters. 3. Picture the Problem: The speed of a wave on a string depends on the tension, radius and density of the string. We wish to use dimensional analysis to create an equation relating the speed to these parameters. Strategy: Set the dimensions of speed, [L]/[T] equal to the dimensions of tension, [M][L]/[T] ; radius, [L]; and density [M]/[L] 3 each raised to the powers α, β, and γ respectively and solve for the powers. B Solution:. Write the dimensions of speed in terms of the product of powers of the dimensions of force, diameter, and density:. Use the dimensions of time to calculate α: 3. Use the dimensions of mass to determine γ: [ ] [ ] [ ][ ] [ ] [ L] [ ] [ ] L M L M β 3 T T L [ L] [ T ] [ M ] [ L] [ T ] α + γ α + β 3γ α [ ] [ ] T T α α α 0 [ ] [ ] M M α γ α + γ 0 + γ α α γ 55 of 89

56 4. Use the dimensions of length to determine β: [ ] [ ] 3 L L α + β γ α + β 3γ β α + 3γ + 3( ) 5. Use the dimensions to write a dimensionally correct equation for velocity: v T R ρ T R ρ Insight: The eact velocity equation cannot be derived from dimensional analysis because of nondimensional constants. However, from this analysis we can determine that doubling the radius would cut the velocity in half. Doubling the tension, increases the velocity by a factor of. 4. Picture the Problem: The picture depicts a person shouting toward a distant cliff and hearing her echo. We want to calculate the distance to the cliff based on the time to hear the echo. Strategy: Between the shout and hearing the echo the sound has travelled to the cliff and back, or twice the distance to the cliff. Multiply the speed of sound by the time lapse to calculate the distance the sound has travelled. The cliff will be one-half of this distance away. Solution: Calculate the distance to the cliff: d vt ( 343 m/s )(.85 s ) 37 m Insight: On a cold morning, when the speed of sound is only 30 m/s, it would take the echo.98 seconds to be heard. 5. Picture the Problem: The dolphin sends a signal to the ocean floor and hears its echo. Strategy: We want to calculate the time the elapses before the dolphin hears the echo and the wavelength of the sound in the ocean. The wave must travel to the ocean floor and back before it is heard. So the distance travelled is twice the distance to the floor. Divide this distance by the speed of sound in water to calculate the time. Calculate the wavelength. Solution:. (a) Divide the distance by the speed of sound in water:. (b) Solve for the wavelength: d 75 m t v 530 m/s s v 530 m/s λ f 55 khz m 8 mm Insight: In air the wavelength would be 6. mm. The wavelength is longer in the water because the wave travels faster in water, while the frequency is the same. 6. Picture the Problem: We need to calculate the wavelength of sound in air from its frequency. Strategy: Solve for the wavelength, using 343 m/s for the speed of sound in air. Solution:. (a) Solve for the wavelength: λ v 343 m/s m f 45 Hz 56 of 89

57 . (b) Eamine the relationship between wavelength and frequency: Wavelength is inversely related to frequency so, if the frequency increases the wavelength decreases. 3. (c) Calculate the wavelength at 450 Hz: 343 m/s λ 475 Hz 0.7 m Insight: As predicted, an increase in frequency corresponds to a decrease in wavelength. 7. Picture the Problem: The figure represents you dropping a rock down a well and listening for the splash. From the time lapse between dropping the rock and hearing the splash we want to calculate the depth of the well. Strategy: The time to hear the splash, t.5 s, is the sum of the time for the rock to fall to the water, t, and the time for the sound of the splash to reach you, t. Solve the free-fall equation for the time to fall and displacement at constant velocity to calculate the time for the sound to return. Set the sum of these times equal to the time to hear the splash and solve for the distance. Solution:. (a) Solve for the falling time: t. Solve for the time for the sound to travel up the well: d t v d d 3. Sum the two times to equal the total time: t t + t g + v 4. Rewrite as a quadratic equation in terms of the variable d : 5. Solve for d using the quadratic formula and square the result: d g s 0 d + d t v g s s 0 ( d ) + d.s 343 m/s 9.8 m/s d m d m m 6. (b) The time to hear the sound would be less then 3.0 seconds because, although the sound travel time would double, the fall time would less than double. Insight: The time to hear the sound for a -meter-deep well is. s, which is indeed less than 3.0 s. 8. Picture the Problem: The figure shows a person throwing a rock down an 8.80-m deep well. The sound of the splash reaches the person s ear.0 seconds after the rock is thrown. We want to calculate the speed of the rock. Strategy: Solve for the initial velocity of the rock, where the fall time is equal to the total time minus the sound travel time. The sound travel time is the depth of the well divided by the speed of sound. Solution:. Calculate t s : d 8.80 m ts s v 343m/s. Subtract t s from the total time: t f. s s.743 s 57 of 89

58 3. Solve for v 0 : y y + v t + at y y v at 0 0 f f 0 0 f tf 8.8 m ( 9.8 m/s )(.743 s).7 m s.743 s The initial velocity of the rock is.7 m/s downward Insight: Even though the speed of sound is much larger than the speed of the rock, the time for the sound to travel up the well is significant. If the sound travel time was not included, the initial velocity of the rock would incorrectly be calculated as.4 m/s, which is 5% off of the actual velocity. 58 of 89

59 Solutions to Tutorial 7. Picture the Problem: A gold ring has a density equal to its mass divided by its volume. Strategy: If the ring is pure gold, its density will be equal to the density of gold. Since the mass and volume of the ring are known, use ρ M/V to calculate the density. Compare the result with the density of gold given in tables. Solution:. Divide the volume by the mass:. Compare with the density of gold from the table: ρ m 0.04 g 6.4 g cm 3 V 0.00 cm 3 ρ gold 9.3 g cm. Therefore, the ring is not solid gold. Insight: If the ring were pure gold of the same volume given in the problem, its mass would be 4.5 g.. Picture the Problem: A cube has a mass of kg and sides of 3. cm each. Strategy: Calculate the density of the cube. Compare the resulting density with the densities given determine the likely composition. Solution:. Calculate the density of m kg ρ.05 0 kg m 3 00 cm the cube: V m 3. cm. Compare with the densities: The cube has the density of silver Insight: Cubes made of different materials could have considerably different masses. For eample, a cube of gold (with the same volume as the silver cube) would have a mass kg, while a cube of aluminium would have a mass of kg. 3. Picture the Problem: A pressure of dyne per square centimetre needs to be converted to the units of pascals and atmospheres. Strategy: The pressure is given as a force divided by area. Convert the units of force and area to the standard SI units of Newton and square meter to write the pressure in pascals. Then convert to atmospheres. Solution:. (a) Convert the pressure to pascals: 5 dyne 0 N 00 cm cm dyne m 0 Pa 6. (b) Convert to atmospheres: atm 0 Pa 0 atm Pa Insight: A dyne/cm is the standard unit of pressure in the cgs system of units, which is commonly used in chemistry. The pascal is the standard unit of pressure in the mks system. 4. Picture the Problem: When a person sits in a four-legged chair the weight of the person and chair is distributed over each leg of the chair, increasing the pressure each leg eerts on the ground. Strategy: Calculate the pressure each leg eerts on the floor. Set the force equal to the sum of the weights of the person and chair and the area equal to four times the cross-sectional area of each leg. Solution:. Set the pressure equal to the weight divided by area: F mg mg P A d π d 4 π 59 of 89

60 (7 kg kg) ( 9.8 m/s ). Insert given data: P π ( 0.03 m) Pa Insight: Leaning back in the chair, so that is rests on only two legs, doubles the pressure those legs eert on the floor. 5. Picture the Problem: When walking with crutches, a person supports a large portion of her weight on the crutch. If the end of the crutch did not have a rubber tip, the entire weight would be supported over the small area of the crutch. The rubber tip increases the area over which the weight is distributed, thus decreasing the pressure. Strategy: Let P wo represent the pressure without the rubber tip and P w represent the pressure with the rubber tip. Calculate the ratio of P w to P wo. The force on the crutches is the same and the cross-sectional area is the area of a circle. Solution:. Find the ratio of the pressure with the tip to the pressure without:. Invert the ratio to find the factor by which the pressure decreases: (. cm) (.5 cm) F w Aw wo π wo F wo Awo w π w P A r 0.3 P A r Insight: Since area is proportional to the square of the radius, the pressure is decreased by the square of the fractional increase in radius. 6. Picture the Problem: When you ride a bicycle, your weight and the weight of the bicycle are supported by the air pressure in both tires spread out over the area of contact between the tires and the road. Strategy: To calculate your weight, first solve for the supporting force of the air pressure on the tires. Set this force equal to the sum of your weight and the weight of the bicycle. Subtract the weight of the bicycle to determine your weight. Solution:. Multiply the tire pressure by the contact area to calculate the supporting force on the bicycle:. Set the supporting force equal to the sum of your weight and the weight of the bicycle: F PA Pa m 70.5 lb/in 7.3 cm N 4.7 lb/in 00 cm F Wyou + Wbicycle Wyou + mbicycle g 3. Solve for your weight: W F m g N you bicycle kg 9.8m/s 65 N Insight: When popping a wheelie on the bicycle, such that only one wheel is touching the ground, that wheel must support the entire weight of the bicycle and rider. Therefore, since the tire pressure has not changed, the area of contact for the single tire would double. In this problem the area would increase to 4.6 cm. 7. Picture the Problem: The weight of the car is supported by the air pressure in all four tires spread out over the area of contact between the tires and the road. Strategy: Find the necessary contact area to support the weight of the car for the given tire pressure. Divide the area by four to calculate the area of contact for each tire. Since the weight of the car does not change significantly as the tire pressure is increased, the tire pressure and contact area are inversely proportional to each other. Finally, solve for the air pressure. Solution:. (a) Solve equation 5- for the total contact area: F mg A P P 60 of 89

61 . Solve for the contact area on one tire: 3. Insert given values: mg mg A 4 Atire Atire P 4P A tire 30 kg 9.8 m/s lb/in Pa ( 4.7 lb/in ) m 4. (b) Since the area and pressure are inversely proportional, as the pressure increases the area of contact decreases. 5. (c) Solve for the pressure: mg P 4A tire ( 00 cm ) 30 kg 9.8 m/s 4.7 lb/in 6. Insert the given values: P (.79 0 Pa) 40.6 lb/in 5 m 4 6cm.0 0 Pa 5 Insight: The pressure in part (c) was greater than the pressure given in part (a). As predicted, the increase in pressure resulted in a decrease in contact area. 8. Picture the Problem: The soft drink can has zero pressure inside and atmospheric pressure pushing inward from the outside. Strategy: Calculate the net force, with the pressure being atmospheric pressure and the area the area of a cylinder, A Dπ h. Solution:. Calculate vertical area: ( m) π ( 0. m) m A Dπ h 5. Solve for inward force: F P A at.0 0 N/m m.5 kn Insight: This force is equal to over 500 lbs, which will easily crush the can. 9. Picture the Problem: Atmospheric pressure will cause a column of mercury to rise 760 mm into a vacuum. Changes in air pressure are measured by the changes in height of the barometer. Atmospheric pressure will cause a column of water to rise much higher because of is lower density. Strategy: Convert the height of the mercury column to pascals by using the relation for 5 atmospheric pressure atm.0 0 Pa 760 mmhg. To calculate the height of the water column, use equation 5-7 with P equal to zero and P equal to the answer to part (a). Solution:. (a) Convert the height of the mercury column to pascals: Pa P (736 mmhg) 97.8kPa 760 mmhg. (b) Solve for the height of the water: P P + ρ gh h P P ρ g Pa 0 3 ( 000 kg/m )( 9.8 m/s ) 9.97 m Insight: The water column rises over 3 times further than the mercury column because its density is over 3 times smaller than the density of mercury. The height of the water column makes it impractical to use as a barometer. 6 of 89

62 0. Picture the Problem: Two pistons are supported by a fluid, as shown in the figure. The pressure in the fluid at the bottom of the left piston is equal to the pressure in the right piston at the same vertical level, which is a distance h below the right piston. Strategy: Set the pressures in the two columns equal at the depth of the left piston. Calculate the pressure due to the pistons and calculate the increase in pressure due to the fluid in the righthand column. Solution:. Set the pressures PL PR equal: ml g mr g + ρ gh A A L R. Solve for the height h: ml m R ml m R 4 ml m R h π π ρ AL AR ρ D 4 L D 4 R πρ DL DR 3. Insert the given values: 4.7 kg 3. kg h.0 m 3 π ( 750kg/m ) ( m) ( 0. m) Insight: The height difference does not depend on the height of the fluid in the left column.. Picture the Problem: As water is poured into the tube shown in the figure, the pressure inside the barrel increases. When the upward force on the barrel lid eceeds 643 N, the barrel will burst. Strategy: The force on the barrel top is the pressure at the surface times the area of the top. Calculate the height of the water column when the barrel will burst. Calculate the weight of the water column from the height, cross-sectional area, and density of water. Solution:. Calculate the bursting pressure of the lid:. Solve for the height of the water column: 3. Solve for the weight: F 643 N P 460 Pa A π P ρ gh ( 0.75 m ) P 455 Pa h 0.48 m 3 ρ g 000 kg/m 9.8 m/s d W mg ρ π hg m 000 kg/m π ( 0.48 m)( 9.8 m/s ) 0. N Insight: A very short column of water is able to increase the pressure sufficiently to burst the barrel. This is one reason why rain barrels always have a hole at the top to allow ecess water to flow out. 6 of 89

63 . Picture the Problem: A cylinder is filled with a fluid, as shown in the diagram. The pressure at the bottom of the fluid is greater than the atmospheric pressure at the top. We wish to find the depth of the fluid that will result in a pressure at the bottom of 6 kpa. Adding additional fluid to the container will increase the pressure at the bottom. We wish to calculate the increase in pressure when m 3 are added. Strategy: Calculate the depth of the fluid. To calculate the pressure when additional fluid has been added, divide the volume of the fluid by the cross-sectional area to find the additional height of the fluid. Then insert the total height into P P at + ρgh for the total pressure. Solution:. (a) Solve for h: P P + ρ gh at P P h ρ g at. Insert given values: 3. (b) Divide the volume by A: Pa.0 0 Pa h 3 ( 806 kg/m )( 9.8 m/s ) m m m h.90 m 4. Add the heights: htotal h + h.897 m m. m 5. Insert data: P P + ρ gh at Pa 806 kg/m 9.8 m/s (. m) 8 kpa Insight: Part (b) could also have been solved by adding the additional depth to the bottom of the 5 3 cylinder, such as: P.6 0 Pa kg/m 9.8 m/s 0.34 m 8 kpa. 3. Picture the Problem: As a submarine dives, the pressure difference between the interior and eterior increases. To be safe, this pressure difference cannot eceed 0.0 N/mm. We need to solve for the maimum depth that the submarine can dive. Strategy: Solve for the depth to which the submarine can descend. Use the density of sea water from the table. Eamine the resulting equation to determine how the density affects the maimum depth. Solution:. (a) Solve for h: P Patm + ρw gh P Patm h ρ g. Insert the given values: w 3 0 mm 0.0 N/mm ( m ) 3 ( 05 kg/m )( 9.8 m/s ) h 995 m 3. (b) Fresh water is less dense than sea water, so the maimum safe depth in fresh water is greater than in salt water. Insight: The maimum depth in fresh water is 00 m. 63 of 89

64 4. Picture the Problem: A water tower is filled with water. The pressure in the tank increases as the water depth increases. We wish to calculate the pressure at specific depths. Strategy: Solve for the pressure at the given depths. Use the density of water given in the table 5-. Solution:. (a) Apply P P at + ρgh directly:. (b) Repeat for a depth of 5.5 m: P P + ρ gh at Pa 000 kg/m 9.8 m/s 4.5 m.45 0 Pa P Pa 000 kg/m 9.8 m/s 5.5 m.54 0 Pa 3. (c) The bands are closer together near to bottom because pressure increases with depth. A greater confining force is needed near the bottom than near the surface of the water. Insight: The pressure at the bottom of the tank reaches a maimum of Pa. 5. Picture the Problem: Water is held in a glass on an elevator that is accelerating upward. A free body diagram for the water is shown at right. We want to calculate the additional pressure at the bottom of the glass due to the acceleration of the elevator. Strategy: Calculate the acceleration of the water from its change in speed and time. Use Newton s Second Law to write the additional force necessary to accelerate the water. Divide this force by the cross-sectional area of the glass to find the added pressure in the water. Solution:. (a) The force that the glass eerts upward on the water is greater than the weight of the water in order to provide upward acceleration. By Newton s Third Law, the water eerts an equal force downward on the glass bottom, so the pressure is greater than it was before the elevator began to move.. (b) Calculate the v. m/s 0 a 0.70 m/s acceleration of the t 3. s elevator: 3. Write the change in pressure as the added force divided by the area of the glass: 4. Write the mass as density times volume: F ma P A A ( ρ Ah) 3 P a ρha 000 kg/m m 0.70 m/s 46 Pa A Insight: If the elevator were accelerating downward at 0.70 m/s the pressure in the glass would decrease by 46 Pa. 6. Picture the Problem: A -cm-tall column of water lies under a 7.-cm-tall column of olive oil, as shown in the figure. We wish to calculate the pressure at the bottom of the water. Strategy: Establish the relationship between pressure and depth within a fluid of known density. The pressure under the oil, P, can be calculated with the density of the oil and height of the oil. To find the pressure at the bottom of the water, insert the pressure P and add the pressure change from the density of water and height of the water. Solution:. Calculate the pressure at the bottom of the oil: P P + ρ gh atm Pa 90kg/m 9.8m/s 0.07m Pa 64 of 89

65 . Calculate the pressure at the bottom of the water: P P + ρ gh water water Pa 000 kg/m 9.8 m/s 0. m.03 0 Pa Insight: Since the density of oil is less than the density of water, the pressure at the bottom of the water is slightly less than the pressure would be if the entire column were water. 7. Picture the Problem: A straw sits in a glass of water. When you suck on the straw, the water rises in the water. We want to know, theoretically, what is the highest the water can rise in the straw. Strategy: The minimum pressure that you could cause inside the straw would be a pure vacuum. The pressure outside the straw is atmospheric pressure. Solve for the height of the water in the straw with no pressure above the water and atmospheric pressure at the bottom. Solution:. (a) The atmospheric pressure that is eerted on the surface of the water creates an upward force on the water column in the straw that overcomes the force of gravity.. (b) Set the pressure above the fluid equal to zero: 3. Solve for the height of the water column: P P + ρ gh ρ gh h at P ρ g 5 at.0 0 Pa 3 ( 000 kg/m )( 9.8 m/s ) 0.3 m Insight: No amount of suction can cause the water to rise higher than 0.3 meters in the straw. 8. Picture the Problem: As shown in the figure, an IV solution is elevated above the injection point on a patient. The pressure in the bag is atmospheric pressure, while the pressure at the injection point is 09 kpa. We need to calculate the height of the bag. Strategy: Solve for the height of the solution above the injection point. Solution:. (a) Solve for h: P P P Pat + ρ gh h ρ g at. Insert the given 09 kpa 0.3 kpa h data: ( 00 kg/m 3 )( 9.8 m/s ) m 3. (b) From the equation in step, we see that the height is inversely proportional to the density of the fluid. Therefore if a less dense fluid is used, the height must be increased. Insight: If the density of the fluid were reduced to 90 kg/m 3, the bag would need to be suspended at a height of m, which is, as predicted, higher than the meters. 65 of 89

66 9. Picture the Problem: A cylinder is filled with mercury up to a depth d, and then filled the rest of the way with water, as shown in the figure. The pressure at the bottom of the cylinder is two atmospheres. Strategy: Set the pressure at the bottom of the cylinder equal to the pressure at the top (atmospheric) plus the pressure increases due to the water and the mercury. Calculate the pressure increases. The height of the water is one meter minus the height of the mercury, h.0 m d. Solution:. Write the pressure at the bottom of the cylinder:. Solve for d: 3. Insert the given values: P P + ρ gh + ρ gh at w w Hg Hg P P + ρ g(.0 m d) + ρ gd at at w Hg Pat ρw (.0 m) g d ρ ρ Hg w Pa w kg/m.0 m d 9.8 m/s kg/m 000 kg/m 0.74 m Insight: Atmospheric pressure is 760 mmhg. Since the density of mercury is much greater than the density of water, the height of the mercury is almost the same as if it were a vacuum above the mercury column. 66 of 89

67 Solutions to Tutorial 8. Picture the Problem: When the SR-7 Blackbird is in flight, its surface heats up significantly. This increase in temperature causes the plane to epand in length. Strategy: We want to calculate the temperature of the plane using the known epansion amount. Establish the relationship between the epansion of the plane and its change in temperature. Solve for the final temperature of the plane. Solution: Solve for T: L α L0 T L 0.50 ft T T0 + 3 C + 0 C 6 5 α L (4 0 K )(07 ft) 0 Insight: Note that in this problem it was not necessary to convert the lengths to metric units.. Picture the Problem: The Akashi Kaikyo Bridge is Japan is made of steel. When steel is heated it epands and when it is cooled it contracts. Strategy: In this problem we wish to find the change in length of the bridge between a cold winter day and a warm summer day. Determine the change in length. The coefficient of linear epansion for steel is given in the table. Solution: Insert the given L α L T 0 5 values: [ ]. 0 (C ) 390 m 30.0 C ( 5.00 C).6 m Insight: This change in length is about the height of a person. If there were no epansion joints in the bridge this increase in length would be sufficient to buckle the bridge. 3. Picture the Problem: An aluminium plate has a hole cut in its centre. The plate epands as it is heated. Strategy: We want to find the size of the hole after the temperature has increased to 99.0 C. The hole will epand at the same rate as the aluminium. Since the diameter of the hole is a unit of length, calculate the diameter as a function of the increase in temperature. The coefficient of linear epansion is given in the able 6-. Solution:. (a) Solve for the final diameter:. Insert the given data: d d d αd T ( α ) d d + αd T d + T 6 d.78 cm K 99.0 C 3.00 C.83 cm 3. (b) Solve for the change in temperature: 4. Solve for the final temperature: d d d αd T d d T T T0 αd d d T T0 + αd 3.00 C +.76 cm.78 cm 48 C 6 ( 4 0 K )(.78 cm) Insight: Since the final diameter (.76 cm) is smaller than the diameter at 3 C we would epect that the final temperature would be below 3 C. The calculations show that this is the case. 67 of 89

68 4. Picture the Problem: A steel bar has a diameter that is cm larger than the inner diameter of an aluminium ring that you would like to slip over the bar. Strategy: Calculate the temperature at which the ring s inner diameter will equal the diameter of the bar. The coefficient of thermal epansion is given in the table. Solution:. (a) The ring should be heated. Imagine that the ring is cut and unrolled. It would be a rectangle. If the rectangle is heated, it will epand along its length and width. Its length is the circumference of the ring. Since the length of the rectangle increases, the circumference of the circle increases, and therefore, so does its diameter.. (b) Solve for the change in temperature: 3. Solve for the final temperature: d αd T d T T T0 αd d T T0 + αd C + T 430 C cm cm 5 (.4 0 ( C ) )( cm) Insight: When the aluminium is heated to 430 C it will slip over the steel rod. As it cools back down it will shrink to form a tight bond with the steel. 5. Picture the Problem A brass sleeve has an inner diameter slightly smaller than the diameter of a steel bar. To shrink-fit the sleeve over the bar, you must either heat the sleeve or cool the bar. Strategy: Solve for the temperature at which the change in diameter is equal to the difference in diameters of the brass sleeve and the steel rod. For the case of heating the brass sleeve use the coefficient of thermal epansion of brass, the initial inner diameter of the brass and a positive change in diameter. For the case of cooling the steel rod, use the coefficient of thermal epansion of steel, the diameter of the steel rod, and a negative change in diameter. The coefficients of thermal epansion of brass and steel aluminium are given in the table. Solution:. Solve for the final temperature:. (a) Insert the data for heating the L α L T α L T T L T T0 + α L 5 brass sleeve: (.9 0 K )(.965 cm) 3. (b) Insert the data for cooling the cm.965 cm T.5 C + 76 C.965 cm.9893 cm T.5 C + 89 C 6 steel: ( 0 K )(.9893 cm) Insight: Since the coefficient of thermal epansion for brass is greater than the coefficient of 68 of 89

69 thermal epansion for steel, the brass does not have to be heated through as large of a temperature difference as the steel has to be cooled to achieve the same change in diameter. 6. Picture the Problem: A steel gasoline tank is completely filled with gasoline, such that the gasoline and the tank have the same initial volumes. When the gas and tank are heated, the gas epands more than the tank, causing some of the gas to spill out of the tank. Strategy: Since the initial volumes of the gas and tank are equal, the amount that will spill out is the difference in the increase in volume of the gas and tank, namely: The volume of spilled gasoline V V V Calculate the changes in volume for the gas and tank. The coefficient of spill gas tank. volume epansion for steel is 3 times the coefficient of linear epansion, which is given in the table. The coefficient of volume epansion for gas is given in the table. Solution:. Write the volume difference:. Insert the given data: ( β 3α ) ( β 3α ) V V V V V T V T spill gas tank gas 0 tank 0 gas tank V spill C 5 L C 0.93 L Insight: 0.93 L is about a quarter of a gallon. Most commercial gas pumps shut off before your car s tank is completely filled to prevent spillover due to the epansion of gas. 7. Picture the Problem: When at room temperature, a stainless steel pot has the same diameter as the pot s copper bottom. When the pot is heated, the copper epands faster than the steel, causing a difference in diameters. Strategy: Calculate the difference in diameters of the steel and copper when the temperature is 60 C. The coefficients of epansion are given in the problem. Solution:. Write the equation for the increases in diameters:. Subtract the two differences: d α d T Cu Cu 0 d α d T st st 0 ( α α ) 5 5 d d d T st Cu st Cu C 8.0 in. 60 C in. Insight: Since the coefficients of epansion between stainless steel and copper are similar (less than % difference) the difference in epansion is small. If instead of stainless steel, normal steel were used, the difference in diameters would be 0.3 in; more than sufficient to the break the pan apart. 8. Picture the Problem: Two cubes are constructed of aluminium and copper wire. The cubes initially enclose equal volumes. When the two cubes are heated they epand at different rates, resulting in different enclosed volumes. Strategy: Calculate the increase in volume of each wire cube. The coefficients of volume epansion are three times the coefficients of linear epansion given in the table. Solution:. (a) Aluminium has the larger coefficient of volumetric epansion, therefore, the aluminium cube will enclose a greater volume.. (b) Write the changes in volume for each cube: V β V T 3α V T Al Al 0 Al 0 V β V T 3α V T Cu Cu 0 Cu 0 69 of 89

70 3. Subtract the changes in volume to calculate the difference in volumes of the two cubes: ( α α ) 5 ( 3 ) V V 3 V T Al Cu Al Cu C 0.06m 97 C.6 0 m 5 3 Insight: This difference in volume is 6 ml, about two tablespoons, and is about 0.6% of the initial volume of 4. gallons. 9. Picture the Problem: A copper ball epands as it is heated, and in proportion to its increase in temperature. Strategy: The copper ball epands equally in all directions when heated. Relate the increase in diameter to the change in temperature. Solve the equation for the final temperature. The coefficient of linear epansion for copper is given in the table. Solution:. Solve for the L L α L0 T T T0 + final temperature: α L. Insert the given data: m T C C ( C ) (.6 0 m ) Insight: This problem could equivalently be solved by relating the change in volume of the sphere to the temperature Picture the Problem: The volume of an aluminium saucepan epands as the temperature of the pan increases. Water, which initially fills the saucepan to the brim, also increases in temperature and epands. If the water epands more than the saucepan, the water will spill over the top. If the saucepan epands more that the water, the water level will drop from the brim of the pan. Strategy: Calculate the change in volumes of the saucepan and of the water. Subtract the change in volume of the saucepan from the change in volume of the water to determine the volume of water that overflows the saucepan. The coefficient of volume epansion for water is given in Table 6-. The coefficient of volume of epansion of aluminium is three times its coefficient of linear epansion, which is also given in the table. Solution:. (a) Because water has a larger coefficient of volumetric epansion, its volume will increase more than the volume of the aluminium sauce pan. Therefore, water will overflow from the pan.. (b) Calculate the initial volumes of the saucepan and water: 3. Write the changes in volume: 4. Subtract the change on volume of the pan from the water to calculate the volume of water spilled: 3 cm V0 π r h π V β V T w w cm 493c m 3 V 3α V T Al Al 0 V V V ( β 3 α ) V T V spill w Al w Al 0 spill 5 ( 3 ) C 493 cm 88 9 C 4 cm Insight: This is the same principle that enables a mercury thermometer to work. The mercury epands faster than the surrounding glass, causing the mercury column to rise of 89

71 . Picture the Problem: Heat is added to a glass ball, resulting in an increase in temperature. Strategy: Solve for the heat necessary to increase the temperature. The specific heat of glass is given in the table. Solution: Solve for the heat: Q mc T kg 837 J/ kg K 5C 0.69 kj Insight: The change in temperature is proportional to the heat added. Doubling the heat added would result in a temperature change of 5 C.. Picture the Problem: A lead bullet travelling at 50 m/s has kinetic energy. As the bullet encounters a fence post it slows to a stop, converting the kinetic energy to heat. Half of the energy heats the bullet resulting in an increase in bullet temperature. Strategy: Solve for the change in temperature. Set the heat equal to one half of the initial kinetic energy of the bullet. The specific heat of lead is given in the table. Solution:. Solve equation for T :. Set the heat equal to half the initial kinetic energy: Q T mc ( 50 m/s) ( ) ( mv ) Q K v T mc mc mc 4c 0 K 4 8 J/ kg K Insight: The relatively small specific heat of lead leads to this large increase in temperature. A silver bullet travelling at the same speed would only heat up by 68 K. 3. Picture the Problem: As the hot silver pellets are dropped into the cool water, heat transfers from the pellets to the water. This results in a decrease in the temperature of the pellets and an increase in the temperature of the water. Strategy: Use conservation of energy, setting the sum of the heat lost by the silver and the heat gained by the water equal to zero. Solve the resulting equation for the mass of the silver which gives a final temperature of 5 C. Divide the resulting mass by the mass of each silver pellet to calculate the number of pellets needed. For the copper pellets, repeat the same calculation, but with the specific heat of copper. The specific heats of water, silver, and copper are found in the table. Solution:. (a) Sum the heat transfers to zero:. Solve for the mass of silver: 3. Divide by the mass of one pellet: Q Ag + Q 0 w m c T T + m c T T 0 m Ag Ag Ag w w w Ag Ag ( ) mwcw Tw T c T T ( Ag ) ( ) ( ) 0.0 kg 486 J/ kg K 4 5 C 0.7 kg 34 J/ kg K 5 85 C mag 0.7 kg 7. 0 n pellets m 0.00 kg 4. (b) Copper has a higher specific heat, so the required number of pellets decreases. 5. (c) Solve the conservation of energy equation for the mass of copper: m Cu pellet Cu ( ) ( Cu ) 387 J/ ( kg K) ( 5 85) C mwcw Tw T c T T (0.0 kg) 486 J/ kg K 4 5 C kg 7 of 89

72 6. Divide by the mass of one pellet: n m Cu kg pellets m pellet 0.00 kg Insight: The amount of heat needed to increase the water s temperature does not depend on whether silver or copper pellets provide the heat. Since the copper has a higher specific heat, each pellet is able to transfer more heat to the water, so fewer copper pellets are needed. 4. Picture the Problem: Heat transfers from the hot lead ball to the cool water, causing the lead to cool and the water to heat up. Eventually the water and lead will come to the same equilibrium temperature. Strategy: Calculate the equilibrium temperature. The specific heats of water and lead are given in the table. Solution: Insert given data: m c T T m c Pb Pb Pb w w w Pb Pb w w ( ) ( ) + ( ) 0.35 kg 8 J/ kg K 84. C kg 486 J/ kg K.5 C 0.35 kg 8 J/ kg K 0.77 kg 486 J/ kg K T 3.9 C + m c T + m c Insight: Since the specific heat of water is greater than the specific heat of lead, the final temperature is much closer to the initial temperature of the water. 5. Picture the Problem: As heat is added to an object its temperature rises. The ratio of the heat to the change in temperature is the heat capacity. The specific heat is the heat capacity per unit mass. Strategy: Calculate the heat capacity by dividing the heat by the change in temperature. Calculate the specific heat. Solution:. (a) Divide the heat by the change in temperature: Q 00 J C 0.8 kj C T C. (b) Divide the heat capacity by the mass: Q C c m T m 0.83 kj/ C 0.90 kg 0.96 kj kg C 0.96 kj kg K Insight: When the heat capacity is known, you can calculate the amount of heat necessary to produce a specific temperature change. For eample, to increase the temperature of the object by 30 C, 5.4 kj of heat should be added. 6. Picture the Problem: A lead ball is dropped from the specified height. During the fall the gravitational potential energy is converted to kinetic energy and finally to heat. The resulting heat increases the temperature of the ball. Strategy: Use conservation of energy to calculate the kinetic energy of the ball just before it hits the ground. Set the heat entering the lead ball equal to the kinetic energy. Solve for the change in temperature Solution:. Set the initial and final energies equal and solve for the kinetic energy:. Solve for the change in temperature: U + K U + K 0 0 mgh K K mgh Q Q mgh gh T mc mc c 0.46 K 0.46 C ( 9.8 m/s )( 5.43 m) 8 J/ ( kg K) 7 of 89

73 Insight: The mass of the ball did not affect the change in temperature. A ball with a larger mass would have a greater amount of heat available, but would require the additional heat to increase the temperature of the additional mass. 7. Picture the Problem: A hot object is immersed in water in a calorimeter cup. Heat transfers from the hot object to the cold water and cup, causing the temperature of the object to decrease and the temperature of the water and cup to increase. Strategy: Since the heat only transfers between the water, cup, and object, we can use conservation of energy to calculate the heat given off by the object by summing the heats absorbed by the water and cup. Use the heat given off by the object and its change in temperature to calculate its specific heat. Solution:. Let Q 0 and solve for Q : Ob 0 Q + Q + Q Ob w Al Q Q + Q m c + m c T Ob w Al w w Al Al w. Solve for the specific heat: c c c Ob Ob Ob Q ob mob ( T TOb ) ( mwcw + malca )( Tw T ) mob ( T TOb ) { } kg (.0 00) C 0.03 kg 486 J/ kg K kg 900 J/ kg K 0 C 385 J (kg C ) 385 J (kg K) 3. Look up the specific heat in the table: The object is made of copper. Insight: It is important to include the effect of the aluminium cup in this calculation. If the contribution of the cup were ecluded, the specific heat of the object would have been calculated as 9 J/(kg K). 8. Picture the Problem: Heat transfers from a hot horseshoe to the cold water. This decreases the temperature of the horseshoe and increases the temperature of the water until the water and horseshoe are at the same equilibrium temperature. Strategy: Since only two objects are transferring heat, calculate the equilibrium temperature. To determine which object would cause a larger final temperature, you should compare the heat capacities heats of the two objects. The object with the higher heat capacity will have more heat to transfer to the water, causing the final temperature to be greater. Solution:. (a) Insert given data:. (b) Write the heat capacities of the lead and iron: 3. Compare the heat capacities: m c T T m c h h h w w w h h w w ( 0.50 kg) 448 J/ ( kg K) + ( 5 kg) 486 J/ ( kg K) 0.50 kg 448 J/ kg K 450 C + 5 kg 486 J/ kg K 3 C T 4 C C C Pb Fe + m c T + m c kg 8 J/ kg K 8 J/K 0.50 kg 448 J/ kg K 4 J/K Since the heat capacity of the lead is less than the heat capacity of the iron, the final temperature will be less than 4 C. Insight: Even though the lead had twice the mass of the iron, the specific heat of lead is small enough that the heat capacity of the iron was larger than the heat capacity of the lead. 73 of 89

74 9 Picture the Problem: As coffee and cream are poured and mied in a ceramic cup, heat echanges between the three objects until they come to the same equilibrium temperature. Strategy: Set the sum of the heat echanges between the coffee, cream, and cup equal to zero since no heat leaves the system. Then using the specific heat equation, solve for the equilibrium temperature. The specific heat of ceramic is given in the problem. Use the specific heat of water for the specific heat of the coffee and cream. Solution:. Set the sum of the heats equal to zero:. Solve for the equilibrium temperature: 0 Q + Q + Q cup cof crm m c T T + m c T T + m c T T cup cup cup cof w cof crm w crm 0 T m c + m + m c m c T + m T + m T c T cup cup cof crm w cup cup cup cof cof crm crm w mcupccupt cup + ( mcof Tcof + mcrmtcrm ) cw mcupccup + ( mcof + mcrm ) cw ( 0.6 kg) 090 J/ ( kg K) ( 4.0 C) + ( 0.5 kg)( 80.3 C) + ( 0.0 kg)( 5.00 C) 486 J/ ( kg K) ( 0.6 kg) 090 J/ ( kg K) + ( 0.5 kg kg) 486 J/ ( kg K) 70.5 C Insight: The comparatively large heat capacity of the coffee, compared with the heat capacities of the cream and cup, causes the equilibrium temperature to be much closer to the initial temperature of the coffee than to the initial temperature of the cream or cup. 0. Picture the Problem: Heat conducts through a lead brick from the warm end to the cooler end. Strategy: Calculate the heat flow through the brick. The thermal conductivity of lead is found in the table. Solution: Apply the equation directly: T C Q ka t 34.3 W/ ( m K)(. 0 m ) (.0 s).3 J L 0.5 m Insight: An identical copper brick would transfer 7 J because of its higher thermal conductivity.. Picture the Problem: Two metal bars, of equal lengths and diameters, are connected in parallel. Heat transfers across both bars. We want to know what length is necessary for the rate of heat transfer to be 7.5 J/s. Strategy: Set the total rate of heat transfer equal to the sum of the heat transfers through each bar. Solve for the length of the bars. Solution:. (a) Sum the heat transfer rates for each bar: Q Q Q T T T t t t L L L total Al st + kal A + kst A A kal + kst. Solve for the A T ( kal + kst ) L length of the bars: ( Qtotal t) 4 π ( m) ( 8 0) C ( ) W/ ( m K) ( 7.5 J/s) m 74 of 89

75 3. (b) Part (a) shows that the rate of heat transfer is inversely proportional to the length of the bar. Therefore, doubling the length of the bars causes the rate to decrease by a factor of. Insight: The thermal conductivity of aluminium is much greater than the thermal conductivity of stainless steel, so most of the heat transfers through the aluminium; Q t 5.6 J s and Q t.9 J s. st. Picture the Problem: Two metal bars, of equal lengths, are connected in parallel. Heat transfers across both bars. The diameter of the lead rod is known. We want to know what diameter of the copper rod is necessary for the rate of heat transfer through both rods to be 33. J/s. Strategy: Set the total rate of heat transfer equal to the sum of the heat transfers through each rod. Solve for the diameter of the copper rod. Solution:. Sum the heat transfer rates of the two rods: Qtotal QCu QPb T T + kcu ACu + kpb APb t t t L L π π T π T kcu dcu + kpb dpb kcudcu + kpbd Pb 4 4 L 4L Al. Solve for the diameter of the copper rod: d Cu Q 4L t π T k total Cu k d Pb Pb 4( m) ( ) ( ) ( ) 33. J/s 34.3 W/ m K m π C 395 W/ m K.64 cm Insight: The diameters of both rods are about the same. However, since the thermal conductivity of the copper is over 0 times the thermal conductivity of lead, over 90% of the heat passes through the copper. 3. Picture the Problem: Two metal rods are connected in series. Heat flows from the high temperature source through the copper rod and then through the lead rod before reaching the cold temperature end. Strategy: Use the heat flow through the rods to calculate the temperature at the junction of the rods. Solution:. (a) The temperature of the junction is greater than 54 C. Since lead has a smaller thermal conductivity than copper, it must have a greater temperature difference across it to have the same heat flow.. (b) Solve for the junction T T L Q k A t temperature: QL (.4 J)( 0.55 m) T T k At 395W/ ( m K) ( 0.05 m) ( s ec) 06 C 98 C Insight: This problem can also be solved by finding the temperature difference across the lead: QL (.4 J)( 0.55 m) T T + C + 98 C k At 34.3W/ ( m K) ( 0.05 m) ( sec) 75 of 89

76 4. Picture the Problem: Two metal rods are connected end to end as shown in the diagram. Heat flows from the hot temperature end through the aluminium and then through the lead to reach the cold side. Strategy: To calculate the length of the aluminum rod, set the heat flow through the aluminum rod equal to the heat flow through the lead rod and solve for the length. Solution:. (a) Since the heat must flow through one rod before it passes through the other, the heat flow rate through both rods must be the same.. (b) Set the rate of heat flow through the rods equal: 3. Solve for the length of the aluminum rod: T Al kal A kpb A LAl L T L Pb Pb ( ) k T 7 W/ m K 80.0 C 50.0 C L ( 4 cm ) 89 cm Al Al Al Pb kpb TPb 34.3 W/ m K 50.0 C 0.0 C Insight: Since heat flows more easily through the aluminium rod, and both rods eperience the same change in temperature, the aluminium rod must be longer than the lead rod. 5. Picture the Problem: A copper poker is kept hot at one end and cool at the other as shown in the figure. Strategy: We wish to find the temperature 3 cm from the cold end. The rate of heat flow through the copper rod is constant throughout the rod. Set the heat flow through the entire rod equal to the heat flow through the final 3 cm and solve for the temperature 3 cm from the end. Solution:. Set the heat flow through the rod equal to the heat flow through the last 3 cm: 05 C C T3 C. Substitute for the heat flows: ka ka 85 cm 3 cm 3. Now solve for T 3 : T Q t 3 Q 3 t 3 cm C + 84 C C + 3 C 44 C 85 cm Insight: Note that the thermal conductivity of copper cancelled out of this problem. The temperature would be 44 C at 3 cm from the end for any type of thermally conductive poker. 6. Picture the Problem: Two identical objects at different temperatures radiate heat into a room, which is at a lower temperature than the object. Strategy: We wish to find the ratio of power radiated by the hotter object to power radiated by the colder object. Divide the net power radiated by the hotter object by the power radiated by the cooler object. Solution: Write the ratio: 4 4 ( s ) P eσ A T T K K 4 4 P eσ A T Ts K K Insight: Since the cooler object s temperature is close to the room temperature, its net power 6 76 of 89

77 radiated is much smaller than the power radiated by the warmer object. If the room temperature were increased by only degrees (to 3 C), the ratio would increase to Picture the Problem: A rectangular cube has sides L, L, and 3L, as shown in the figure. When the L L faces are held at fied temperature, the cube is a conductor with cross-sectional area A L and length 3L. Changing the sides that are held at fied temperature changes the cross-sectional area and length of the conductor. Strategy: Write P in terms of the thermal conductivity, length L, and change in temperature. For each rotation of the cube, use the appropriate the cross-sectional area and length to write the rate of heat flow in terms of P. Solution:. Set P equal to the heat flow through the length 3L and A L L :. (a) Set P a equal to the heat flow through the length L and A 3L L : 3. (b) Set P b equal to the heat flow through the length L and A 3L L : T T P ka k ( L ) k L T L 3L 3 T 3 9 Pa k ( 3L ) k L T P L 4 T Pb k ( 6L ) 6k L T 9P L Insight: Note that the heat flow is greatest when the length is the shortest and the cross-sectional area is the greatest. The heat flow is least when the length is the greatest. 3 L L L 77 of 89

78 Solutions to Tutorial 9. Picture the Problem: A proton moves in a magnetic field that is directed at right angles to its velocity. Strategy: Combine Newton's Second Law with the magnetic force (equation -) to find the acceleration of the particle. Solution: Set the magnetic force equal to the mass multiplied by the F ma evb sin 90 acceleration and solve for a: evb ( C)( 9.5 m/s)(.6 T) a.5 0 m/s 7 m kg Insight: If the magnetic field were parallel to the velocity, the angle θ 0 and the force and acceleration would be zero.. Picture the Problem: An electron moves at right angles to a magnetic field and eperiences a magnetic force. Strategy: Solve for the speed of the electron that would produce the specified force. Solution: Solve for v: 5 F N v e B sin C 0. T m/s Insight: Doubling the magnetic field would cut the required speed in half in order for the electron to eperience the same magnetic force. 3. Picture the Problem: A negatively charged ion moves due north while immersed in Earth s magnetic field at the equator. Strategy: The magnetic field of Earth points due north at the equator. Therefore, the velocity of the particle is parallel to the magnetic field, which means that the angle θ 0 and the force on the particle is zero. Solution: Solve directly: F qvb sin 0 0 Insight: If the ion had the same charge as an electron, and moved due east instead of north, and the magnetic field of Earth pointed north and had a magnitude of T at that location, the magnetic force on the ion would be. 0 7 N in the downwards direction (remember the ion is negatively charged!). 4. Picture the Problem: A proton moves straight downwards from a location above the equator, moving at right angles to the magnetic field that is horizontal and points due north. Strategy: Combine Newton's Second Law with the magnetic force to find the acceleration of the proton. Solution: Set the magnetic force equal to the mass multiplied by the acceleration and solve for a: evb sin 90 a m.38 0 m/s 6 ( C)( 355 m/s)( T) kg Insight: The proton would eperience the very same acceleration if it were travelling due east or due west, or any other direction that is perpendicular to the horizontal magnetic field that points due north. 5. Picture the Problem: A charged particle moves in a region in which a magnetic field eists. Strategy: Solve the magnetic force equation for the angle θ that would produce the specified force. F Solution:. (a) Solve for θ : θ sin qvb 78 of 89

79 . Insert the numerical values for F N θ sin µn: ( C)( 6 m/s)( 0.95 T) 3. (b) Insert the numerical values for F θ N sin µn: ( C)( 6 m/s)( 0.95 T) 4. (c) Insert the numerical values for F µn: ( C)( 6 m/s)( 0.95 T) N θ sin. Insight: The magnetic force is the largest when the velocity and the magnetic field are perpendicular, and it is zero when v r and B r are parallel. 6. Picture the Problem: A charged particle moves in a region in which a magnetic field eists. Strategy: Use a ratio to determine the force the particle eperiences after changing its speed and the angle its velocity makes with the magnetic field. Solution:. Make a ratio:. Now solve for new old old old old old Fnew qvnew Bsinθ 6.3 m/s sin 5 new vnew sinθ new F qv B sinθ v sinθ 7 m/s sin 90 F : Fnew Fold N. 0 N Insight: The speed decreased by a factor of 4.3 and the angle changed to 5, decreasing the force by another factor of.4. Together these two effects decreased the force by a factor of Picture the Problem: An ion moves with constant speed in a magnetic field. Strategy: The ion eperiences no magnetic force when it is moving in the ŷ direction, so we conclude that the magnetic field also points in the ŷ direction. When the ion travels in the y plane and along the line y, it moves at an angle of 45 with respect to B r. When it moves in the ˆ direction, it eperiences the maimum magnetic force. Solution: Letting Fma q vb : F F 6 6 ma sin 45 ( 6. 0 N) sin N Insight: The ion won t travel along the line y for very long, because the magnetic force will cause the charged particle to travel in a circle. 8. Picture the Problem: An electron moves in a region in which a magnetic field eists. Strategy: The ion eperiences no magnetic force when it is moving in the ˆ direction, so we conclude that the magnetic field also points along either the + ˆ or ˆ direction. We can use either the Left-Hand Rule for negative charges, or use the Right-Hand Rule and remember to reverse the direction of the force because the charge on the electron is negative. Try pointing your left thumb downwards ( z ˆ direction) and the fingers of your left hand in the ŷ direction to find that the magnetic field must point in the it moves in the Solution: Find : ˆ direction. Then use the force eperienced by the electron when +y ˆ direction to find the magnitude of B r. B r 3 F sinθ (.0 0 N) sin 90 B 9 5 ev (.6 0 C)( 9. 0 m/s) Insight: If instead the electron were to travel in the r 3 F.0 0 N yˆ. r.4 T B.4 T ˆ +z ˆ direction, it would eperience a force 79 of 89

80 9. Picture the Problem: Two charged particles travel in a magnetic field along the same direction and eperience the same magnetic force, but they travel at different speeds. Strategy: Calculate the ratio of the speeds of the particles. Solution:. (a) Since the magnetic force is directly proportional to both the charge and the speed of the particles, and since the particles eperience the same force, particle must have a greater speed because particle has the greater charge. v F qb sinθ q q. (b) Find the ratio: v F q Bsinθ q 4q 4 Insight: Suppose both charges were allowed to travel in circles as described in section -3. r mv qb vq v q Assuming both charges have the same mass, we find that ; r mv qb vq 4v 4q 6 that is, the slower particle with the larger charge has a much smaller radius of motion in the magnetic field. 0. Picture the Problem: The magnetic force on an electron travelling at constant speed causes it to move in a circle. Strategy: Find the radius of the electron s circular orbit. Solution: Apply equation directly: 3 5 ( 9. 0 kg)( m/s) mv r eb 9 (.6 0 C)( 0.45 T) µ m 7.9 m Insight: A proton has the same magnitude charge but a much larger mass, which causes it to orbit in a circle of much larger radius than an electron if the two particles have the same speed.. Picture the Problem: The magnetic force on a proton travelling at constant speed causes it to move in a circle. Strategy: Find the radius of the proton s circular orbit. Solution: Apply equation directly: 7 5 ( kg)( m/s) mv r eb 9 (.6 0 C)( 0.45 T).5 0 m.5 cm Insight: An electron has the same magnitude charge but a much smaller mass, which causes it to orbit in a circle of much smaller radius than a proton if the two particles have the same speed. Picture the Problem: The magnetic force on an electron travelling at constant speed causes it to move in a circle. Strategy: The magnetic force provides the centripetal force required to keep the electron moving in a circle. The radius of the circle in terms of m, v, q, and B is given. We must first find the speed v of the electron by using conservation of energy, then solve for B. Solution :. Find v:. Solve for B: e V v m 7. 0 m/s 9 ( ).60 0 C 40 V 3 ( 9. 0 kg) 3 7 ( 9. 0 kg)(. 0 m/s) mv B er 9 (.60 0 C)( 0.7 m) T 0.40 mt Insight: The electron s speed is only 4.0% of the speed of light, so that we can safely neglect relativistic effects. The magnetic field of 0.40 mt is about eight times larger than Earth s magnetic field (0.50 G mt). 80 of 89

81 3. Picture the Problem: The magnetic force on a charged particle travelling at constant speed causes it to move in a circle. Strategy: The magnetic force provides the centripetal force required to keep the particle moving in a circle. The radius of the circle in terms of m, v, q, and B is given. We must first solve for the speed v of the particle, then find the time it takes the particle to complete one orbit by dividing the circumference by the speed. Solution:. (a) Solve for v:. (b) Divide the circumference by the speed: qrb v m. m/s ( C)( 6.8 m)(.0 T) kg π (. m/s) π r π m 6.8 m t 3.9 s v qb Insight: A fairly large field (.0 T is 0,00 times stronger than Earth s field) is required to keep this particle travelling in a circle of large radius (6.8 m) because the charge to mass ratio q / m is relatively small. 4. Picture the Problem: The magnetic force on a charged particle travelling at constant speed causes it to move in a circle. Strategy: Apply the Right-Hand Rule to the diagram at the right in order to determine whether the particle is positively or negatively charged. Then find the radius of the circle in terms of m, v, q, and B, in order to find the mass m of the particle. Solution:. (a) According to the RHR, a positively charged particle would eperience a force to the left. Since the particle is eperiencing a force to the right, it must be negatively charged.. (b) Solve for m: 9 ( ) erb.60 0 C 0.50 m 0.80 T.0 u m.5 u 7 v.67 0 kg 6 ( m/s) Insight: Another way to answer part (a) is to point the fingers of your hand along v r and curl them into the page in the direction of B r. For which hand does the thumb (and therefore F r ) point to the right? Since it is the left hand that works, the particle must be negatively charged, since negative particles follow the Left-Hand Rule. 5. Picture the Problem: The magnetic force on a proton travelling at constant speed causes it to move in a circle. Strategy: The magnetic force provides the centripetal force required to keep the electron moving in a circle. The radius of the circle in terms of m, v, q, and B. We must first find the speed v of the proton by using the definition of kinetic energy, then find the radius r. Solution:. Solve for v:. Find r: K m K mv v 7 6 ( )( ) mv m K mk kg J r 3. cm eb eb m eb 9 (.6 0 C)( 0.6 T) Insight: The magnetic field must be increased in order to decrease the radius of the proton s path. 8 of 89

82 6. Picture the Problem: The magnetic force on an alpha particle travelling at constant speed causes it to move in a circle. Strategy: The magnetic force provides the centripetal force required to keep the particle moving in a circle. The radius of the circle in terms of m, v, q, and B is given. We must first find the time it takes the particle to travel halfway through a complete circle by dividing half the circumference by the speed, then substitute for the speed v of the particle in the epression. Solution:. (a) Find half of the orbit period, then substitute for v:. Insert numerical values: T π r π r π mv π m t v v v qb qb 7 ( kg) π t 9 (.60 0 C)( 0.55 T) s 4 ns 3. (b) The time does not depend upon the speed of the particle, so the answer to part (a) will stay the same. 4. (c) The time stays the same: t s 4 ns Insight: Since the charge-to-mass ratio is a constant for alpha particles, the only way to change the period of its orbit (or T f, which is sometimes called the cyclotron frequency) is to change the magnitude of B. 7. Picture the Problem: The magnetic force causes both an electron and a proton to move in circles at constant speed. Strategy: Form a ratio r electron r proton for each of two cases: (a) the particles have equal momenta, and (b) the particles have equal kinetic energies. Solution:. (a) Calculate r e r when e e e e e p p : rp mpv p p qp B p e B e p p. (b) Calculate r e r p when K K : e p r m v q B p e B.00 r m v q B m K m e B m K m r m v q B m K m e B m K m e e e e e e e e e e p p p p p p p p p p kg kg Insight: If the kinetic energies are the same, the proton has a much larger radius. If the momenta are the same, the two radii are equal. 8. Picture the Problem: A current-carrying wire eperiences a force due to the presence of a magnetic field. Strategy: Find the magnetic force that is eerted on the wire. Solution: Find F: F I L B sin 90 ( A)(.5 m)( 0.70 T).08 N Insight: This force amounts to 0.50 N or.8 ounces of force per meter of wire. If the current were increased to 5 A, the force would increase to N per meter of wire. 8 of 89

83 9. Picture the Problem: A current-carrying wire eperiences a force due to the presence of a magnetic field. Strategy: Find the magnetic force that is eerted on the wire. Solution: Find F: F I L B θ sin.8 A.5 m 0.88 T sin N Insight: The maimum force occurs at θ 90, at which angle the force amounts to.5 N of force per meter of wire. 0. Picture the Problem: A current-carrying wire eperiences a force due to the presence of a magnetic field. Strategy: Solve for the angle θ that would produce the given magnetic force on the wire. Solution: Solve for θ: F sinθ I L B θ F I L B.6 N sin sin 63 ( 3.0 A)(. m)( 0.50 T) Insight: If the angle between the wire and the magnetic field were θ 90, the force on the wire would be.8 N.. Picture the Problem: A magnetic field eerts forces on the four sides of a square loop of current-carrying wire. Strategy: Find the force on each of the four sides of the square current loop. Solution:. The top and bottom wires run parallel to the field:. The left and right wires are perpendicular to the field: 0 or Ftop Fbottom I L B sin 0 80 Fleft Fright I L Bsin A 0.46 m 0.34 T.5 N Insight: The force on the left wire points out of the page and the force on the right wire points into the page.. Picture the Problem: A wire carries a current in the positive direction while immersed in a magnetic field that eerts an upward force on the wire. Strategy: Set the magnitude of the magnetic force equal to the weight of the wire to find the magnitude of B r. Then use the Right-Hand Rule to determine the direction that B r must point in order for the magnetic force to be eerted in the upward direction on the wire. Solution:. Set the magnetic force equal to the weight:. Solve for B: F mg I L Bsin 90 I L B ( 0.7 kg)( 9.8 m/s ) mg B 0.34 T I L ( A)( 0.45 m) 3. Let upward in the figure be the positive y-direction, and the positive -direction be to the right. The RHR stipulates that B r must point into the page, or in the z ˆ direction. Therefore, 83 of 89

84 r B 0.34 T. zˆ Insight: This physics setup is related to that used for magnetic levitation trains. A pretty substantial field (0.34 T is 6800 times stronger than Earth s magnetic field) is required to lift a pretty small wire (0.7 kg) in this arrangement. 3. Picture the Problem: A current-carrying wire eperiences a force due to the presence of a magnetic field. Strategy: Solve for B to answer the question of part (a), and solve the same equation for θ to answer the question of part (b). Solution:. (a) Write equation in terms of force per unit length:. Solve for B: F I Bsinθ L F L N/m B 0.04 T 4 mt I sin θ 6. A sin 7.5. (b) Solve for θ : F L 0.05 N/m θ sin sin 3.4 I B ( 6. A)( 0.04 T) Insight: At small angles the sinθ function is roughly linear with θ, so that cutting the force per unit length approimately in half requires cutting the angle in half. 4. Picture the Problem: A wire carries a current in a region where the magnetic field eerts an upward force on the wire. Strategy: Set the magnitude of the magnetic force equal to the weight of the wire to find the current required to levitate the wire. The force is maimum when the current is perpendicular to the field. Therefore, the minimum required current occurs in the perpendicular configuration. Solution:. Set the magnetic force equal to the weight:. Solve for I: F mg I L Bsin 90 I L B ( 0.75 kg)( 9.8 m/s ) mg I 3. A LB (.7 m)( 0.84 T) Insight: This physics setup is related to that used for magnetic levitation trains. A pretty substantial field (0.84 T is 6,800 times stronger than Earth s magnetic field) is required to lift a pretty small wire (0.75 kg) in this arrangement. 5 Picture the Problem: Two current-carrying wires eperience a force due. to Earth s magnetic field. Strategy: Use the Right-Hand Rule to determine the direction and magnitude of the magnetic force on the currents. Let ˆ point east, ŷ point north, and ẑ point upward. Converting the magnetic field from gauss to 4 5 tesla, we find that 0.59 G.0 0 T/G T. Solution:. (a) According to the RHR, the magnetic force points toward north, 8 above the horizontal. 84 of 89

85 . Calculate F: F I L B sinθ 5 0 A 50 m T sin 90.6 N 3. (b) According to the RHR, the magnetic force points toward the east Calculate F: F I L B θ sin 0 A 50 m T sin 7.5 N Insight: These are the forces if the wires carry direct current. In reality, the electrical current in highvoltage wires is alternating current, so that the net force on the wire due to the magnetic field of Earth is zero. 6. Picture the Problem: A metal bar is suspended from two conducting wires and immersed in a magnetic field that points straight downward. Strategy: Looking down the bar from the left, so that the current I r points into the page, the magnetic force I L B points to the left, mg points downward, and the tension in the wires points up and to the right. Use Newton's Second Law in the vertical and horizontal directions to find the angle θ at which the rod is in equilibrium. Solution:. Set the net forces equal to zero: F 0 T cosθ mg y F 0 T sinθ I LB mg I L B. Solve each epression from step for T: T cosθ sinθ 3. Rearrange to solve for θ : tanθ θ Insight: The angle θ can be increased by increasing I or B and decreased by increasing m. I LB mg tan I LB mg 85 of 89

86 Solutions to Tutorial 0. Picture the Problem: The image show a ring of radius 3. cm oriented at an angle of θ 6º from a B T magnetic field. Strategy: Solve for the flu. Solution: Calculate the flu: Φ BAcosθ T π 0.03 m cos6 Φ Wb Insight: The maimum flu through this coil, Wb, occurs when the angle θ is zero.. Picture the Problem: The image shows a bo immersed in a vertical magnetic field. Strategy: Calculate the flu through each side. Solution:. The sides of the bo are parallel to the field, so the magnetic flu through the sides is zero.. Calculate the flu through the bottom: 4 Φ BA cosθ T 0.35 m 0.0 m cos Wb. Insight: The height of the bo is not important in this problem. 3. Picture the Problem: The image shows a rectangular loop oriented 4 degrees from a magnetic field. Strategy: Solve for the magnetic field. Solution: Calculate the magnetic field: 5 Φ Tm B 7 mt A cosθ m m cos 4 Insight: The minimum magnetic field that would produce this flu would occur when the rectangle is parallel to the magnetic field. 4. Picture the Problem: A house has a floor of dimensions m by 8 m. The local magnetic field due to Earth has a horizontal component T and a downward vertical component T. Strategy: The horizontal component of the magnetic field is parallel to the floor, so it does not contribute to the flu. Calculate the flu using the vertical component. Solution: Calculate the Φ BA θ B A magnetic flu: 5 cos 4. 0 T m 8 m.7 0 Wb Insight: The flu through the vertical walls of the house is determined by the horizontal component of the magnetic field instead of the vertical component. 5. Picture the Problem: A solenoid of diameter. m produces a magnetic field of.7 T. Strategy: Multiply the magnetic field by the cross-sectional area of the solenoid to calculate the magnetic flu. Solution: Calculate the magnetic flu:. m Φ BA cos θ.7 T π cos 0.9 Wb Insight: Note that the length of the solenoid does not affect the flu through the solenoid. 86 of 89

87 6. Picture the Problem: A magnetic field of magnitude T is directed 7 below the horizontal and passes through a horizontal region 30 cm by 8 cm. Strategy: Calculate the flu, where the angle from the vertical is θ Solution: Calculate the flu: Φ BA 5 5 cosθ T.30 m 0.8 m cos Wb Insight: Increasing the angle from the horizontal increases the flu through the desk top. For eample, if the angle were increased to 80 from the horizontal the total flu would increase to Wb. 7. Picture the Problem: A coil of radius 5 cm and 53 turns is oriented perpendicular to a magnetic field. The magnetic field changes from 0.5 T to zero in 0. s. Strategy: Calculate the induced emf, with the flu. Solution: Calculate the Φ BA T π 0.5 m ε N N V emf: t t 0. s Insight: Note that the emf is inversely proportional to the time it takes for the magnetic field to change. In this case, if it dropped to zero in only seconds (half the time), the average emf would be twice as large, or 5.6 V. 8. Picture the Problem: The figure shows the flu through a single loop coil as a function of time. Strategy: Calculate the emf at the times t 0.05 s, 0.5 s, and 0.50 s. Use the graph to find the change in flu. Solution:. (a) Calculate the emf at t 0.05 s:. (b Calculate the emf at t 0.5 s: 3. (c) Calculate the emf at t 0.50 s: ε ε ε Φ 0 Wb 0 N 0. kv t 0. s 0 5Wb 0 Wb 0.6 s 0. s 0.04 kv Insight: When the slope of the flu is constant, the emf is constant. The emf is 0.04 kv from t 0. s to t 0.6 s. 9. Picture the Problem: The image shows the emf through a single loop as a function of time. Strategy: Calculate the emf at the times t 0.5 s and t 0.55 s Solution:. (a) The flu at t 0.5 s is about 8 Wb. This is greater than the flu at t 0.55 s, which is about 3 Wb.. (b) The two emf values are the same, because at those two times the flu is changing at the same rate. 3. (c) Calculate the slope of the flu between 0. s and 0.6 s: Φ 5 Wb 0 Wb 37.5 Wb/s t 0.6 s 0. s 87 of 89

88 Φ 37.5 Wb/s 0.04 kv t 4. Calculate the induced emf : ε N Insight: Note that the emf is zero for times 0. s < t < 0. s and t > 0.6 s. The voltage is not determined by the magnitude of the flu but by the slope of the flu vs. time graph. For these two time periods the slope is zero. 0. Picture the Problem: The image shows a single loop of area m and resistance 0 Ω. The loop is perpendicular to a magnetic field. Strategy: Solve Ohm s Law for the necessary emf. Then insert the emf to calculate the rate of change in the magnetic field. Solution:. Calculate the emf : ε IR 0.3 A 0 Ω 35. V. Solve for the change in magnetic field: Φ A B ε N N t t B ε 35. V t ΝΑ m T/s Insight: The magnitude of the magnetic field (0.8 T) is not important in this problem, only the change in the field.. Picture the Problem: A coil with 0 loops is oriented perpendicular to a changing magnetic field. Strategy: Solve for the emf, where the magnetic flu is given. Solution: Calculate the emf : ε Φ BA BA BA N N N t t t 0.0 T m 0 7. V 0.34 s Insight: This emf is double the value the coil would eperience if the magnetic field simply dropped to zero in the same time period. Picture the Problem: The image shows a square loop of wire with. circumference. m that is in a 0.05 T magnetic field. The square is changed into a circle with the same perimeter in 4.5 s. Strategy: Calculate the flu through the circle and the rectangle. Then insert the flues to calculate the induced emf. Solution:. Calculate the radius of the circle: C π r C r m π. Calculat e the flu through the circle: Φ Wb π circle BA Bπ r 0.05 T m 88 of 89