Study Questions/Problems Week 4

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1 Study Questions/Problems Week 4 Chapter 6 treats many topics. I have selected on average less than three problems from each topic. I suggest you do them all. Likewise for the Conceptual Questions and exercises, which you should do first. Chapter 6: Conceptual Questions 1, 2, 3, 5, 8, 9, 12, 15, 22 Conceptual Exercises 1, 2, 3, 4, 7, 10, 12 Problems 2, 3, 4, 9, 16, 19, 21, 22, 23, 27, 32, 34, 40, 42, 43, 44, 46 Answers/solutions for even numbered CQs and CEs, and for all the problems listed above are on the following pages don t peek until you have done your best to solve a problem.

2 Chapter 6: Applications of Newton s Laws Answers to Even-Numbered Conceptual Questions 2. Spinning the wheels is likely to decrease the force exerted by the Jeep. The reason is that the force exerted by the spinning wheels is kinetic friction, and the coefficient of kinetic friction is generally less than the coefficient of static friction. The spinning wheels look more exciting in the movie, however. 4. The maximum acceleration is determined by the normal force exerted on the drive wheels. If the engine of the car is in the front, and the drive wheels are in the rear, the normal force is less than it would be with front-wheel drive. During braking, however, all four wheels participate including the wheels that sit under the engine. 6. Friction is beneficial whenever you want to start, stop, or turn a car. It is also beneficial when you tune a guitar, tie your shoes, or sprint from rest to catch a bus. 8. This is possible because if you spin the bucket rapidly enough, the force needed to produce circular motion is greater than the force of gravity. In this case, a force in addition to gravity must act at the top of the circle to keep the water moving in its circular path. This force is provided by the bottom of the bucket. Therefore, the bottom of the bucket pushes against the water, and the water pushes back against the bucket this keeps the water from falling out of the bucket. 10. Yes. Equilibrium simply means that the net force acting on an object is zero. Therefore, an object moving with constant velocity can be considered to be in equilibrium. In a frame of reference moving with the same velocity, the object would be at rest and would have zero net force acting on it which is the way we usually think of equilibrium. 12. Astronauts feel weightless because they are in constant free fall as they orbit, just as you would feel weightless inside an elevator that drops downward in free fall. 14. A passenger on this ride feels pushed against the wall, which means that the wall, in turn, exerts a normal force on the passenger. If the corresponding force of static friction is greater than the passenger s weight, the passenger stays put as the floor is lowered away. (Why do all passengers stay put at the same rotational speed, regardless of their weight?) 16. If the parking brake is applied while the car is in motion, the rear wheels begin to skid across the pavement. This means that the friction acting on the rear wheels is kinetic friction, which is smaller in magnitude than the static friction experienced by the front wheels. As a result, the rear wheels will overtake the front wheels, causing the car to spin around and begin moving rear wheels first. This is standard procedure for stunt drivers wishing to spin a car around in a chase scene. 18. As the basket within a washing machine rotates, the clothes collect on the rim of the basket. Here the basket exerts an inward force on the clothes, causing them to follow a circular path. The water contained in the clothes, however, is able to pass through the holes of the basket where it can be drained from the machine. 20. People on the outer rim of a rotating space station must experience a force directed toward the center of the station in order to follow a circular path. This force is applied by the floor of the station, which is really its outermost wall. Because people feel an upward force acting

3 on them from the floor, just as they would on Earth, the sensation is like an artificial gravity. 22. The physics of this scene is somewhere between bad and ugly. When the rope burns through, Robin is moving horizontally. This horizontal motion should continue as Robin falls, leading to a parabolic trajectory rather than the straight downward drop shown in the movie. Answers to Even-Numbered Conceptual Exercises 2. At the equator, you are moving in a circular path. Therefore, part of the force of gravity acting on you is providing your centripetal acceleration; the rest shows up as a reduced weight on the scale. At the poles, the scale reads your full weight. (In addition, the Earth bulges at the equator, due to the same effects just discussed. Therefore, you are farther from the center of the Earth when you are at the equator, and this too diminishes your weight, as we shall see in Chapter 12.) 4. The acceleration of the masses is proportional to the difference in the masses divided by the sum of the masses. Adding the same amount to both masses leaves the difference in masses unchanged, but increases the sum of the masses. Therefore, adding the same amount to each mass results in a decrease in the acceleration. The only exception to this is the special case where the two masses are equal. In this case, the difference in mass is zero, resulting in zero acceleration. Adding the same amount to each mass still results in zero acceleration. Note that if each mass is increased by the same multiplicative factor, however, the acceleration is always unchanged. 6. In principle, the reading on the scale would be greater than 700 N if the Earth were to stop rotating. See the discussion for Conceptual Exercise 2 for further details. 8. The force of kinetic friction is independent of the area of contact, but proportional to the normal force. The two cases discussed here have the same normal force, and hence the force of kinetic friction in case 1 is equal to the force of kinetic friction in case The answer is C < B < A. To see this, note that the likelihood a car might skid is directly related to the centripetal force the road must exert on the car to keep it on the track. The speed of the car is constant, and therefore the force the track must exert on the car is greatest where the curvature of the track is greatest. The track is relatively straight at point C, more curved at point B, and most sharply curved at point A. This leads to the indicated ranking. 12. (a) If the spring is cut in half, it stretches half as much for a given applied force as it stretched before being cut. Therefore, the force constant of the half spring is twice what it was for the full spring. (b) When two springs are connected end-to-end, they stretch twice as much for a given applied force. It follows that the force constant in this case is half what it was for a single spring. 14. The path that describes the subsequent motion of the puck is B. Once the string breaks, the puck has zero net force acting on it. Therefore, it will move with constant velocity from that point on. At the moment the string broke, the puck was moving tangentially to the circular path, in the direction of path B it will continue to move in this direction with constant speed after the string breaks. See following pages for Problem solutions

4 Solutions to Ch. 6 Problems 2. Picture the Problem: The child slides down the incline that is tilted 31.0 above the horizontal. A free-body diagram of the situation is depicted at right: Strategy: Choose the x-axis 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 x direction and solve for k. Solution: 1. Write Newton s Second Law in the y direction: 2. Now write Newton s Second Law in the x direction: 3. Solve the equation for k : N mg cos 0 μ k N mg cos mg sin μ k N ma mg sin ma μ k N mg sin ma μ k mg cos g sin a g cos ( 9.81 m/s 2 )sin m/s 2 ( 9.81 m/s 2 )cos 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 Porsche accelerates in a straight line due to the static friction between the tires and the road. Strategy: The static friction between the tires and the road provides the forward force on the Porsche needed to accelerate it at 12 m/s 2. Use Newton s Second Law to find the minimum coefficient of static friction. Solution: 1. Write Newton s Second Law for the Porsche: 2. Solve the equation for s : a g f s mg ma 12 m/s m/s Insight: There is no physical reason why the coefficient of friction cannot be greater than one. In fact, tires designed for Formula One racing cars have coefficients of friction of 1.6 or better to obtain superior cornering and acceleration. 4. 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 maximum 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 k can be determined.

5 Solution: 1. When the book begins sliding, the applied force equals the maximum static friction force: F app f s mg F app mg 2.25 N 1.80 kg m/s 2 2. When the book is sliding at constant speed, the applied force equals the kinetic friction force: F app f k μ k mg μ k F app mg 1.50 N 1.80 kg m/s 2 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. 9. Picture the Problem: The crate slides down the incline when the angle is 23. A free-body diagram of the situation is depicted at right. Strategy: Choose the x-axis to be parallel to the ramp, pointing uphill, and the y axis to point perpendicular to the ramp. Write Newton s Second Law in the y direction to find the normal force N. Then write Newton s Second Law in the x direction with a x 0 (the box just begins to slide) to find the coefficient of static friction. Solution: 1. (a) Write Newton s Second Law in the y direction: 2. Write Newton s Second Law in the x direction, setting a x 0 (the box just begins to slide) and solving for : N mg cos 0 N mg cos N mg sin ma x 0 ( mg cos ) mg sin 0 tan tan (b) Since only depends upon, changing the mass will not change and the angle remains 23. Insight: Increasing the mass does increase the normal force and hence the friction force, but the component of the crate s weight that pulls it down the ramp is also increased. The two effects cancel and depends upon only. 16. Picture the Problem: The weight of the mass exerts a downward force on the spring, compressing it. Strategy: The weight of the 9.29-kg mass is the force responsible for compressing the spring. Use equation 5-5 to find the weight and equation 6-4 to find the spring constant. Let the x axis point straight upward, with x 0 at the top of the uncompressed spring m/s 2 Solution: Solve equation 6-4 for the spring constant and use the k F mg 9.29 kg weight of the mass as the x x 2220 N/m 2.22 kn/m ( m) compressing force: Insight: The negative sign in Hooke s law indicates the force exerted by the spring is in the direction opposite the stretch or compression. In this case, the compression is towards negative x (downward) but the force exerted by the spring on the mass is positive (upward).

6 19. 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: 1. (a) Write Newton s Second Law in the horizontal direction: 2. Use Hooke s Law to find the spring force and solve for the friction force. Note that the spring stretches in the negative x direction. f s F ma x 0 f s F kx ( 150 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 maximum friction force is reached. If the applied force is increased even more, the backpack will slide (see the next problem). 21. Picture the Problem: The spring is capable of being either stretched or compressed by the external 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 x 2L L L. Then use Hooke s law again to find the force required to compress the spring an amount equal to x 1 L L 1 L. 2 2 Solution: 1. (a) Use Hooke s Law to find the magnitude of the force required to stretch the spring by the amount x L : 2. (b) Use Hooke s Law to find the magnitude of the force required to compress the spring by the amount x 1 L : 2 F k x kl ( 250 N/m) ( 0.18 m) 45 N F k x k ( 1 L 2 ) ( 250 N/m) 1 ( 0.18 m ) 23 N 2 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 1.5L. 22. Picture the Problem: Illinois Jones is lifted straight upward by the tension in a rope. Strategy: There are two forces acting on Illinois Jones, the rope tension T acting straight upward and the force of gravity W mg acting straight downward. Write Newton s Second Law for Mr. Jones (with upward being the positive direction) and solve for the acceleration a that corresponds to the maximum tension. Use equation 2-11 and the known acceleration to find the minimum time that is required to pull him from the pit. Solution: 1. (a) Write Newton s Second Law for Mr. Jones and solve for a: T mg ma a T m g

7 2. Use equation 2-11, with v 0 0, to find the minimum time required to pull him out of the pit: y v 0 t at at2 t 2 y a 2 y T m g ( 755 N) 70.0 kg m 9.81 m/s s 3. (b) A shorter time would require a larger acceleration, which in turn would require a larger tension in the rope because T ma + mg ma+ ( g). Thus the tension would become larger than 755 N and the rope would break. Insight: A tension of 755 N is only 170 lb, suggesting the rope is not a very strong one. Being rescued from a snake pit with a flimsy rope only adds to the adventure! 23. 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 spring force and thus the compression distance x. In both cases the acceleration of the block is zero. Solution: 1. (a) Write Newton s Second Law in the vertical direction: 2. Write Newton s Second Law in the horizontal direction: f s mg N mg 0 N mg F N kx mg 0 3. Solve for the compression distance x: mg 0.27 kg x k m/s 2 ( 120 N/m) m 4.8 cm 4. (b) Yes; the spring compression would double in order to create the larger friction force needed to balance m g. Insight: The equation in step 3 indicates that the compression distance will double if you double the mass of the block but would be cut in half if you double the spring constant k. 27. 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: 1. (a) Write Newton s Second Law for the box: C lower mg 0 C lower mg Lower Pulley F F C lower C upper F F Upper Pulley

8 2. 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 F + F C lower 0 F 1 C 2 lower 1 mg ( 52 kg) 9.81 m/s N 0.26 kn downwards C upper F F 0 C upper 2F 2( kn) 0.51 kn 4. (c) Use the result of step 1 to find C lower : C lower mg ( 52 kg) ( 9.81 m/s 2 ) 510 N 0.51 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. 32. Picture the Problem: The forces exerted on the left and right blocks are depicted at right. Strategy: Because the pulley is ideal, the tension in the string is equal to the weight of the hanging block. This can be verified by Newton s Second Law in the vertical direction for the hanging block: T mg 0. Write Newton s Second Law along the direction parallel to the incline for the 6.7 kg block and substitute T mg into the resulting equation to find m. Solution: 1. Write Newton s Second Law along the direction parallel to the incline: 2. Substitute T mg into the resulting equation to find m. T Mg sin 0 T mg Mg sin m M sin ( 6.7 kg)sin kg Insight: A larger m is required if the angle is increased. If it is increased all the way to 90, the large mass will be hanging straight down and the mass m required to maintain equilibrium would be 6.7 kg. 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: 1. (a) Write Newton s Second Law in the vertical direction: f s mg N mg 0 N mg

9 2. Write Newton s Second Law in the horizontal direction: F N F mg 0 3. Solve for the applied force F: F mg 0.79 ( 1.6 kg) 9.81 m/s 2 20 N kn 4.5 lb 4. (b) The force of static friction would remain the same if you push with a greater force because it must exactly balance the weight of the wood board. Insight: If you push with a force greater than 20 N, the maximum possible static friction force increases, so you could support a larger block of wood without slipping. 40. 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 x be positive in the direction of each mass s motion, m 1 be the mass on the table, and m 2 be the hanging mass. Solution: 1. (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. 2. (b) Write Newton s Second Law for each block and add the equations: 3. Solve the resulting equation for a: T m 1 a F x block 1 block 2 T + m 2 g m 2 a m 2 g ( m 1 + m 2 )a m a kg m 1 + m g 2 ( 9.81 m/s2 ) 4.36 m/s kg 4. (c) Use the first equation to find T: T m 1 a ( 3.50 kg) ( 4.36 m/s 2 ) 15.3 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 m 2 g 27.5 N. The tension in the string would be zero if m 2 fell freely, 27.5 N if m 2 (and the entire system) were at rest.

10 42. 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 1 represent the lighter bucket and m 2 the heavier bucket (on the right in the figure). The positive x axis points upward for m 1 and downward for m 2. Solution: 1. (a) Write Newton s Second Law for m 2 with a 0 : 2. (b) Write Newton s Second Law for m 1 : 3. Write Newton s Second Law for m 2 and substitute for the acceleration from step 2: 4. Substitute W 2 m 2 g and m 2 m 1 W 2 W 1 : 5. (c) Write 0 for m 1 with a 0 : T + m 2 g 0 T m 2 g 110 N F x T m 1 g m 1 a a T m 1 g T + m 2 g m 2 a m 2 T m 1 g T 2W 2 1+ W 2 W 1 2m 2 g T 1+ m 2 m 1 2( 110 N) ( 63 N) 80 N N T m 1 g 0 T m 1 g 63 N Insight: Another way to solve this problem is to add the two versions of Newton s Second Law in steps 2 and 3 to eliminate T and solve for a, then use a to find T. Verify for yourself that a 2.7 m/s 2 in part (b). 43. 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: f s F cp ma cp mv2 r ( 1200 kg) 15 m/s 57 m 2 Insight: The maximum static friction force is mg ( 0.88) ( 1200 kg) ( 9.81 m/s 2 ) 10.4 kn which corresponds to a maximum cornering speed (without skidding) of 22 m/s. 4.7 kn 44. Picture the Problem: The test tube travels along a circular path at constant speed. Strategy: Solve equation 6-15 for the speed required to attain the desired acceleration. Solution: Solve equation 6-15 for the speed: v ra cp r ( 52,000g) ( m) ( 52, 000) 9.81 m/s m/s 0.20 km/s Insight: This speed corresponds to 25,000 revolutions per minute for the centrifuge, or 415 revolutions per second.

11 46. Picture the Problem: The forces acting on the car are depicted at right. Strategy: Write Newton s Second Law in the horizontal and vertical directions and combine them to obtain the radius of the curve. The procedure is similar to Example 6-9 in the text. Solution: 1. Write Newton s Second Law in the x direction and solve for r: 2. Write Newton s Second Law in the y direction and solve for N: 3. Substitute for N in the equation from step 1: r N sin ma cp mv2 r mv2 N sin N cos mg 0 N mg cos r mv2 cos sin mg v 2 g tan ( 22.7 m/s) m ( 9.81 m/s 2 )tan 31.5 Insight: Note that the car follows the curved path without any help from friction at all. This is because the inward component of the normal force is sufficient to provide the necessary centripetal force.

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