Pearson Physics Level 20 Unit II Dynamics: Unit II Review Solutions

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1 Pearson Physics Level 0 Unit II Dynamics: Unit II Review Solutions Student Book pages Vocabulary 1. action-at-a-distance force: a force that acts on objects whether or not the objects are touching action force: a force initiated by object A on object B apparent weight: the negative of the normal force acting on an object coefficient of friction: proportionality constant relating the magnitude of the force of friction to the magnitude of the normal force field: a three-dimensional region of influence surrounding an object free-body diagram: a vector diagram of an object in isolation showing all the forces acting on it free fall: a situation in which the only force acting on an object that has mass is the gravitational force gravitational field strength: gravitational force per unit mass at a specific location gravitational force: attractive force between any two objects due to their masses gravitational mass: mass measurement based on comparing the known weight of one object to the unknown weight of another inertia: property of an object that resists acceleration inertial mass: mass measurement based on the ratio of a known net force on an object to the acceleration of the object kinetic friction: force exerted on an object in motion that opposes its motion as it slides on another object net force: vector sum of two or more forces acting simultaneously on an object Newton s first law: an object will continue either being at rest or moving at constant velocity unless acted upon by an external non-zero net force Newton s law of gravitation: any two objects, A and B, in the universe exert gravitational forces of equal magnitude but opposite direction on each other; the forces are directed along the line joining the centres of both objects. Newton s second law: when an external non-zero net force acts on an object, the object accelerates in the direction of the net force; the magnitude of the acceleration is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object Newton s third law: if object A exerts a force on object B, then B exerts a force on A that is equal in magnitude and opposite in direction normal force: a force on an object that is perpendicular to a common contact surface reaction force: force exerted by object B on object A static friction: force exerted on an object at rest that prevents the object from sliding on another object tension: magnitude of a force T exerted by a rope or string on an object at the point where the rope or string is attached to the object 1

2 true weight: gravitational force acting on an object that has mass Knowledge Chapter 3. Given 1 = 60 N [.0 ] = 36 N [110 ] 3 = 83 N [300 ] net force on object ( net ) Draw a free-body diagram for the object. Resolve all forces into x and y components.

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4 Vector x component y component 1 (60 N)(cos.0 ) (60 N)(sin.0 ) (36 N)(cos 70.0 ) (36 N)(sin 70.0 ) 3 (83 N)(cos 60.0 ) (83 N)(sin 60.0 ) 4

5 Add the x and y components of all force vectors in the vector addition diagram. x direction net x y direction net y = 1x + x + 3 x F net x = F 1x + F x + F 3 x = (60 N)(cos.0 ) + { (36 N)(cos 70.0 )} + (83 N)(cos 60.0 ) = 84.8 N = 1y + y + 3 y F net y = F 1y + F y + F 3 y = (60 N)(sin.0 ) + (36 N)(sin 70.0 ) + { (83 N)(sin 60.0 )} = 15.6 N Use the Pythagorean theorem to find the magnitude of net. F net = ( F ) + ( F ) net x net y = (84.8 N) + ( 15.6 N) = 86 N 5

6 Use the tangent function to find the direction of net. tan θ = opposite adjacent = 15.6 N 84.8 N = θ = tan 1 (0.1836) = 10.4 From the vector addition diagram, this angle is between net and the positive x-axis. So the direction of net measured counterclockwise from the positive x-axis is = 350. net = 86 N [350 ] The net force on the object is 86 N [350 ]. 3. If an object experiences zero net force, it may be either stationary or moving at constant velocity. 4. A person with a plaster cast on an arm or leg experiences extra fatigue because the cast adds mass to the arm or leg. Every time the person moves the limb with the cast, the limb accelerates. Since the limb with the cast has greater mass, it requires a greater net force to cause the same acceleration than without the cast. This additional net force is supplied by the muscles which become fatigued. 5. During the spin cycle, the drum of a washing machine exerts a net force inward on the wet clothes to change their direction of motion. The result is the clothes move in a circle at high speed while excess water continues to move in a straight line through the small holes in the drum. The motion of the water out of the drum is tangent to the drum. Since the extracted water is drained while the drum is spinning, the water cannot come in contact with the clothes again when the machine stops spinning. 6. Given m c = 1.5 kg app = 6.0 N [left] a = 3.0 m/s [left] mass of load (m l ) The load and cart are a system because they move together as a unit. Find the total mass of the system. m T = m l + m c = m l kg 6

7 Draw a free-body diagram for the system. The system is not accelerating up or down. So in the vertical direction, F net = 0 N. v Write equations to find the net force on the system in both the horizontal and vertical directions. horizontal direction vertical direction net h = app F net h = app Apply Newton s second law. m T a = F app m T = F app a m l kg = F app a netv F netv = N F = 0 + g Calculations in the vertical direction are not required in this problem. 7

8 m l = 6.0 N 3.0 m/s 1.5 kg = m 6.0 kgi s 1.5 kg m 3.0 s = 0.50 kg The load has a mass of 0.50 kg. 7. (a) If the mass is constant and the net force quadruples, the magnitude of the acceleration will quadruple. a = F net m = 4F net m = 4 F net m (b) If the mass is constant and the net force is divided by 4, the acceleration will be 1 4 of its original magnitude. a = F net m 1 = 4 F net m = 1 4 F net m (c) If the mass is constant and the net force becomes zero, the acceleration will be zero. 8. Given a = F net m = 0 m = 0 A [along rope] B = 5.0 N [along rope] θ 1 = 50 θ = 345 = 55.4 N [6 ] net 8

9 magnitude of person A s applied force (F A ) Draw a free-body diagram for the wagon. Resolve all forces into x and y components. Vector x component y component A F A (cos 50 ) F A (sin 50 ) B (5.0 N)(cos 15 ) (5.0 N)(sin 15 ) net (55.4 N)(cos 6 ) (55.4 N)(sin 6 ) Add the x components of all force vectors in the vector addition diagram. 9

10 x direction net x = Ax + B x F net x = F Ax + F B x (55.4 N)(cos 6 ) = F A (cos 50 ) + (5.0 N)(cos 15 ) F A = 1 cos 50 {(55.4 N)(cos 6 ) (5.0 N)(cos 15 )} = 39.9 N Person A is applying a force of magnitude 39.9 N on the wagon. 9. This example illustrates Newton s third law which states that if the table exerts a force on the book, the book exerts a force on the table of equal magnitude and opposite direction. The action-reaction forces act on two different objects. 10. The reaction force is the force exerted by the book on the pencil, which is 15 N [up]. 11. The coefficients of static and kinetic friction are numerals without units because they are ratios of two physical quantities that have the same units. Both quantities are forces with units of newtons. These units cancel out, leaving just a numeral. μ s = F f N F N static N and μ k = F f N F N kinetic N 10

11 1. (a) From the free-body diagram below, the block is stationary because fstatic have the same magnitude but opposite directions. and g (b) A free-body diagram helps you visualize the situation of the problem. All the forces acting on the block are included, so it is easy to see why no motion occurs in this situation. 13. Since μ s for wet concrete is greater than that for wet asphalt, the car will be able to slow down more quickly on wet concrete, assuming the car is not skidding. So the stopping distance and stopping time will be shorter. If the car is skidding, the car will slide with equal ease on both surfaces because μ k is the same for wet concrete and wet asphalt. Chapter Since gravitational field strength is equivalent to the acceleration due to gravity, both values will be the same at the top of a tall skyscraper. 15. The inertia of an object is directly proportional to its mass. The greater the mass, the greater the inertia. Since both gravitational force and gravitational field strength vary with mass, mass is the fundamental quantity that affects the inertia of an object. 16. Since the magnitude of the gravitational field strength is slightly less at the top of a mountain, an athlete would weigh slightly less at the top than at the base. A ski jumper may be able to jump higher at top of the mountain than at the base. Also, the density of air decreases as you go up in altitude. So the air resistance acting on an athlete would be slightly less at the top of a mountain than at the base, resulting in slightly faster bobsled times. 11

12 1 17. From Newton s law of gravitation, F g m A m B and F g. To double the r gravitational force, you could double the mass of one of the bags. The figure below represents this situation. F g (m)(m) m Another way to double the gravitational force is to reduce the separation distance. The figure below represents this situation. 1 F g 1 r 1 ( ) 1 r r 1

13 18. Substitute the definition 1 N = 1 kg m s into the student s answer. Then simplify N s kgi m s m = 57.3 s m = 57.3 kg The student was solving for mass. 19. If you express the statement as an equation, you get F g 1 r. If the probe is twice as far away from Mars, the gravitational force would be 1 = 1 4 of its original magnitude. If the probe is 1 3 of its original distance from Mars, the gravitational force 1 would be 1 = 9 times its original magnitude An object is weightless only if there is no gravitational force acting on it. An object in free fall is experiencing the gravitational force of a celestial body. So an object in free fall is not weightless, unless it is in deep space where the gravitational force is not measurable. 1. Use Newton s law of gravitation to express the gravitational force on each satellite. The gravitational force on satellite m is (F g ) m = GmM r. The gravitational force on satellite m is (F g ) m = G(m)M r or GmM r. So the gravitational force on satellite m is twice that on satellite m, for the same separation distance, because it has twice the mass.. From Figure 4.34 on page 1, the magnitude of Earth s gravitational field strength is greater at the North Pole than at the equator. So an object in free fall will experience a slightly greater acceleration due to gravity at the North Pole. 3. (a) The mass of an object has no effect on the acceleration due to gravity, provided that air resistance is negligible. For an object in free fall, the net force on the object is given by net = g + air F net = F g + F air ma = mg + F air If F air = 0, a = g, so the acceleration is independent of mass. 13

14 (b) Gravitational field strength is the force per unit mass at a specific location. Gravitational field strength can be calculated using g = GM source r, which is independent of the mass of the object. Applications 4. Given 1 = 150 N [40.0 ] = 0 N [330 ] net force on the soccer player ( net ) Draw a free-body diagram for the soccer player. 14

15 Resolve all forces into x and y components. Vector x component y component 1 (150 N)(cos 40.0 ) (150 N)(sin 40.0 ) (0 N)(cos 30.0 ) (0 N)(sin 30.0 ) Add the x and y components of all force vectors in the vector addition diagram. x direction net x y direction net y = 1x + x F net x = F 1x + F x = (150 N)(cos 40.0 ) + (0 N)(cos 30.0 ) = N = 1y + y F net y = F 1y + F y = (150 N)(sin 40.0 ) + { (0 N)(sin 30.0 )} = (150 N)(sin 40.0 ) (0 N)(sin 30.0 ) = N Use the Pythagorean theorem to find the magnitude of net. F net = ( F ) + ( F ) net x net y = (305.4 N) + ( N) = 306 N Use the tangent function to find the direction of net. 15

16 tan θ = opposite adjacent = N N = θ = tan 1 (0.0445) =.5 From the vector addition diagram, this angle is between net and the positive x- axis. So the direction of net measured counterclockwise from the positive x-axis is = 357. net = 306 N [357 ] The net force on the soccer player is 306 N [357 ]. 5. Given m = 1478 kg net = 3100 N [W] acceleration of car ( a ) The car is not accelerating up or down. So in the vertical direction, F = 0 N. netv In the horizontal direction, the acceleration of the car is in the direction of the net force. So use the scalar form of Newton s second law. F net h = ma F net a = h m = 3100 N 1478 kg m 3100 kg i = s 1478 kg =.097 m/s a =.097 m/s [W] The car will have an acceleration of.097 m/s [W]. 16

17 6. Car stopped Car speeding up from stoplight 17

18 Car cruising at city speed limit Car going on highway ramp 18

19 Car cruising at highway speed limit 7. Given magnitude of net = 8.0 N m = 4.0 kg v i = 10 m/s [right] = 18 m/s [right] time interval during which net force acts (Δt) The acceleration of the object is in the direction of the net force. So use the scalar form of Newton s second law. F net = ma v f 19

20 a = F net m = 8.0 N 4.0 kg m 8.0 kg i = s 4.0 kg =.00 m/s a =.00 m/s [right] Calculate the time interval. a = Δv Δt Δt = Δv a = v f v i a 18 m/s 10 m/s =.00 m/s = 4.0 s The net force is applied for 4.0 s. 8. (a) 0

21 (b) Given T 1 = 65.0 N [along rope] T θ 1 = θ = 30.0 magnitude of f = 65.0 N [along rope] = 104 N net force on boat ( net ) Resolve all forces into x and y components. Vector x component y component T 1 (65.0 N)(cos 30.0 ) (65.0 N)(sin 30.0 ) T (65.0 N)(cos 30.0 ) (65.0 N)(sin 30.0 ) From the chart, F T1 y = F T y. So net y = T 1y + T y F net y = F T 1y + F T y = 0 Add the x components of all force vectors in the vector addition diagram. 1

22 x direction net x = T 1x + T x + f F net x = F T 1x + F T x + F f = (65.0 N)(cos 30.0 ) + (65.0 N)(cos 30.0 ) + ( 104 N) = (65.0 N)(cos 30.0 ) + (65.0 N)(cos 30.0 ) 104 N = 8.58 N From the vector addition diagram, net is along the positive x-axis. net = 8.58 N [0 ] The net force on the boat is 8.58 N [0 ]. 9. Given magnitude of a A = 0.40 m/s m c A = kg m = kg c B m l A = kg m = kg l B acceleration of train B ( a B ) The locomotive and cars of each train are a system because they move together as a unit. Find the total mass of each train. m T A = m l A + m c m A T = m B l B + m c B = kg kg = kg kg = kg = kg Train A is not accelerating up or down. So in the vertical direction, F = 0 N. netv In the horizontal direction, the net force on train A is in the direction of its acceleration. So use the scalar form of Newton s second law. F net h = mt A a A = ( kg)(0.40 m/s ) = N Train B is not accelerating up or down. So in the vertical direction, F = 0 N. netv In the horizontal direction, the acceleration of train B is in the direction of the net force. So use the scalar form of Newton s second law. F net h = mt B a B

23 a B = F net h m TB = N kg 5 m kg i = s kg = 0.56 m/s a B = 0.56 m/s [in same direction as train A] Train B will have an acceleration of 0.56 m/s [in same direction as train A]. 30. Given m = 8. t or kg g = 9.81 m/s v = 10 cm/s [down] force exerted by water and cable on chamber ( T Draw a free-body diagram for the chamber. + water ) Since the chamber is not accelerating, net = 0 N in both the horizontal and vertical directions. For the vertical direction, write an equation to find the net force on the chamber. net v = T + water + g 0 = F T + F water + F g 3

24 F T + F water = F g = mg = { ( kg)(9.81 m/s )} = N T + water = N [up] The water and cable exert a force of N [up] on the chamber. 31. Given m b = 40 kg m r = 70 kg air = 180 N [backward] f = 1950 N [forward] static acceleration of system ( a ) The motorcycle and rider are a system because they move together as a unit. Find the total mass of the system. m T = m b + m r = 40 kg + 70 kg = 310 kg Draw a free-body diagram for the system. 4

25 The system is not accelerating up or down. So in the vertical direction, F = 0 N. netv Write equations to find the net force on the system in both the horizontal and vertical directions. horizontal direction vertical direction net h = f static + air netv = N F net h = F f static + F air F net = 0 v = 1950 N + ( 180 N) Calculations in the vertical direction are not required = 1950 N 180 N in this problem. = 670 N Apply Newton s second law to the horizontal direction. F net h = m T a Fnet a = h m T = 670 N 310 kg + g kg im 670 = s 310 kg =. m/s a =. m/s [forward] The system will have an acceleration of. m/s [forward]. 3. Given m = 0.5 kg v i = 15 m/s [up] force exerted by escaping gas ( net ) Calculate the acceleration of the rocket. a = Δv Δt = v f v i Δt 40 m/s 15 m/s = 0.60 s = 41.7 m/s Δt = 0.60 s v f = 40 m/s [up] 5

26 a = 41.7 m/s [up] The acceleration of the rocket is in the direction of the net force. So use the scalar form of Newton s second law. F net = ma = (0.5 kg)(41.7 m/s ) = 10 N net = 10 N [up] The escaping gas exerts a force of 10 N [up] on the rocket. 33. (a) Given m A = 60 kg m B = 90 kg app = 800 N [right] acceleration of both boxes ( a ) Both boxes are a system because they move together as a unit. Find the total mass of the system. m T = m A + m B = 60 kg + 90 kg = 150 kg Draw a free-body diagram for the system. The system is not accelerating up or down. So in the vertical direction, F = 0 N. netv Write equations to find the net force on the system in both the horizontal and vertical directions. 6

27 horizontal direction net h = app F net h = F app netv = 800 N Apply Newton s second law to the horizontal direction. F net h = m T a Fnet a = h m T = 800 N 150 kg vertical direction net = v N + g F = 0 Calculations in the vertical direction are not required in this problem. kg im 800 = s 150 kg = 5.3 m/s a = 5.3 m/s [right] The acceleration of both boxes is 5.3 m/s [right]. (b) Given m A = 60 kg m B = 90 kg app = 800 N [right] a = 5.33 m/s [right] from part (a) magnitude of action-reaction forces between the boxes (F A on B ) Draw a free-body diagram for box B. Box B is not accelerating up or down. 7

28 So in the vertical direction, F net = 0 N. v Write equations to find the net force on box B in both the horizontal and vertical directions. horizontal direction vertical direction net h = A on B net = v N + g F net h = F A on B F net = 0 v Apply Newton s second law. Calculations in the vertical direction are not required m B a = F A on B in this problem. F A on B = (90 kg)(5.33 m/s ) = N The magnitude of the action-reaction forces between the boxes is N. 34. Given app = 1.5 N [right] g Moon = 1.6 m/s m =.0 kg coefficient of kinetic friction (μ k ) Draw a free-body diagram for the glass block. Since the block is not accelerating, net = 0 N in both the horizontal and vertical directions. Write equations to find the net force on the block in both directions. horizontal direction vertical direction net h = app F net h = app + fkinetic F + fkinetic F v net = v N + g F net = F N + F g 8

29 0 = 1.5 N + ( μ k F N ) 0 = F N + ( mg Moon ) = 1.5 N μ k F N = F N mg Moon μ k F N = 1.5 N F N = mg Moon Substitute F N = mg Moon into the last equation for the horizontal direction. μ k mg Moon = 1.5 N μ s = 1.5 N mg Moon = 1.5 N (.0 kg) 1.6 m s m 1.5 kg i s = m (.0 kg ) 1.6 s = 0.46 The coefficient of kinetic friction for the glass block on the surface is (a) Given m A = 4.0 kg m B = 6.0 kg m C = 3.0 kg g = 9.81 m/s a = 1.4 m/s [forward] μ s = 0.5 from Table 3.4 (dry oak on dry oak) applied force on blocks ( app ) The three blocks are a system because they move together as a unit. Find the total mass of the system. m T = m A + m B + m C = 4.0 kg kg kg = 13.0 kg Draw a free-body diagram for the system. 9

30 Since the system is not accelerating in the vertical direction, F net = 0 N. h Write equations to find the net force on the system in both the horizontal and vertical directions. horizontal direction vertical direction net h = app + f static net = v N + g F + F f F static net = F v N + F g m T a = F app + F f 0 = F static N + ( m T g) F net h = app = F app + ( μ s F N ) = F N m T g = F app μ s F N F N = m T g F app = m T a + μ s F N Substitute F N = m T g into the last equation for the horizontal direction. F app = m T a + μ s m T g = m T (a + μ s g) = (13.0 kg){1.4 m/s + (0.5)(9.81 m/s )} = N 30

31 app = N [forward] An applied force of N [forward] will cause the blocks to accelerate at 1.4 m/s [forward]. (b) Given m A = 4.0 kg m B = 6.0 kg m C = 3.0 kg g = 9.81 m/s a = 1.4 m/s [forward] μ s = 0.5 from Table 3.4 (dry oak on dry oak) force exerted by block B on block C ( B on C ) Draw a free-body diagram for block C. Block C is not accelerating up or down. So in the vertical direction, F = 0 N. netv Write equations to find the net force on block C in both the horizontal and vertical directions. horizontal direction vertical direction net h = B on C + f on C netv = N + g F net h = F B on C + F f on C F net = F v N + F g m C a = F B on C + ( μ s F N ) 0 = F N + ( m C g) = F B on C μ s F N = F N m C g F B on C = m C a + μ s F N F N = m C g Substitute F N = m C g into the last equation for the horizontal direction. F B on C = m C a + μ s m C g = m C (a + μ s g) = (3.0 kg){1.4 m/s + (0.5)(9.81 m/s )} = 10 1 N B on C = 10 1 N [forward] 31

32 Block B exerts a force of 10 1 N [forward] on block C. (c) Given m A = 4.0 kg m B = 6.0 kg m C = 3.0 kg g = 9.81 m/s a = 1.4 m/s [forward] μ s = 0.5 from Table 3.4 (dry oak on dry oak) app = N [forward] from part (a) force exerted by block B on block A ( B on A ) Draw a free-body diagram for block A. Block A is not accelerating up or down. So in the vertical direction, F = 0 N. netv Write equations to find the net force on block A in both the horizontal and vertical directions. horizontal direction vertical direction net h = app + B on A + f on A net = v N + g F net h = F app + F B on A + F f on A F net = F v N + F g m A a = F app + F B on A + ( μ s F N ) 0 = F N + ( m A g) = F app + F B on A μ s F N = F N m A g 3

33 F B on A = m A a + μ s F N F app F N = m A g Substitute F N = m A g into the last equation for the horizontal direction. F B on A = m A a + μ s m A g F app = m A (a + μ s g) F app = (4.0 kg){1.4 m/s + (0.5)(9.81 m/s )} N = N B on A = N [backward] Block B exerts a force of N [backward] on block A. 36. Given m = 10.0 kg θ = 30.0 μ k = 0.0 g = 9.81 m/s acceleration of block ( a ) Draw a free-body diagram for the block. Since the block is accelerating downhill, net 0 N parallel to the incline, but net = 0 N perpendicular to the incline. Write equations to find the net force on the block in both directions. direction = N net Fnet + g = F N + Fg direction net = g + fkinetic F net = F g + F fkinetic 33

34 0 = F N + Fg ma = F g + F fkinetic F N = F g Now, F g = mg cos θ F g = mg sin θ and F f = μ kinetic k F N So, F N = ( mg cos θ) ma = mg sin θ + ( μ k F N ) = mg cos θ = mg sin θ μ k F N Substitute F N = mg cos θ into the last equation for the direction. m a = m g sin θ μ k m g cos θ a = g(sin θ μ k cos θ) = (9.81 m/s ){sin 30.0 (0.0)(cos 30.0 ) } = 3. m/s a = 3. m/s [downhill] The acceleration of the block is 3. m/s [downhill]. 37. Given m A = 15 kg m B = 0 kg g = 9.81 m/s [down] magnitude of app = 416 N magnitude of f = 0 N acceleration of each object ( a A and a B ) Since the person is pulling the rope, objects A and B will accelerate up. The rope has a negligible mass. So the tension in the rope is the same on both sides of the pulley. The magnitude of a A is equal to the magnitude of a B. Find the total mass of both objects. m T = m A + m B = 15 kg + 0 kg = 35 kg Choose an equivalent system in terms of m T to analyze the motion. A is equal to the gravitational force acting on m A, and B is equal to the gravitational force acting on m B. Apply Newton s second law to find the net force acting on m T. 34

35 net = app + A + B + f F net = F app + F A + F B + F f = 416 N m A g m B g 0 N = 416 N (m A + m B )g 0 N = 416 N m T g 0 N = 416 N (35 kg)(9.81 m/s ) 0 N = 5.7 kg m s = 5.7 N Use the scalar form of Newton s second law to calculate the magnitude of the acceleration. F net = m T a a = F net m T = 5.7 N 35 kg = kg im 5.7 s 35 kg a A = 1.5 m/s = a B = 1.5 m/s [up] Objects A and B will have an acceleration of 1.5 m/s [up]. 38. (a) Given m A = 6.0 kg m B =.0 kg m C = 4.0 kg μ k = 0.00 g = 9.81 m/s acceleration of object B ( a B ) Since m A > m C, you would expect m A to accelerate down while m C accelerates up. Since object B will accelerate left, choose left to be positive. The strings have negligible mass and do not stretch. So the magnitude of a A is equal to the magnitude of a B, which is also equal to the magnitude of a C. Find the total mass of the system. m T = m A + m B + m C = 6.0 kg +.0 kg kg = 1.0 kg 35

36 Draw a free-body diagram for object B. A is equal to the gravitational force acting on m A, and C is equal to the gravitational force acting on m C. Apply Newton s second law to find the acceleration of m B. horizontal direction vertical direction net h = A + C + f kinetic net = v N + g F net h = F A + F C + F f F kinetic net = F v N + F g m T a B = F A + F C + F f 0 = F kinetic N + ( m B g) = m A g + ( m C g) + ( μ k F N ) = F N m B g = m A g m C g μ k F N F N = m B g = g(m A m C ) μ k F N Substitute F N = m B g into the last equation for the horizontal direction. m T a B = g(m A m C ) μ k m B g a B = m A m C m g m B T m μ k g T = 6.0 kg 4.0 kg 1.0 kg (9.81 m/s ) a B.0 kg (0.00)(9.81 m/s ) 1.0 kg.0 kg = (9.81 m/s ) kg 1.0 (0.00)(9.81 m/s ) = 1.3 m/s = 1.3 m/s [left] Object B will have an acceleration of 1.3 m/s [toward object A]. (b) Given m A = 6.0 kg m B =.0 kg m C = 4.0 kg μ k = 0.00 g = 9.81 m/s a = 1.31 m/s [left] from part (a) B 36

37 tension in each string Draw a free-body diagram for objects A and C. Apply Newton s second law to find the tension in the string between objects A and B. net = A + TA F net = F A + F TA m A a A = m A g + F TA F T A = m A a A m A g = m A (a A g) = (6.0 kg)(1.31 m/s 9.81 m/s ) = 51 N Apply Newton s second law to find the tension in the string between objects B and C. net = C + TC F net = F C + F TC m C a C = m C g + F TC F = m C a C + m C g T C = m C (a C + g) = (4.0 kg)(1.31 m/s m/s ) = 44 N 37

38 and Verify The tension in the string between objects A and B is 51 N, and the tension in the string between objects B and C is 44 N. Both tensions are different because objects A and C have different masses. Also in the case of object A, the tension force opposes the gravitational force on object A, but in the case of object C, the tension force must overcome the gravitational force on object C and also accelerate it. (c) (d) Four action-reaction pairs associated with object B are normal force exerted by the table on object B directed up (action) force exerted by object B on the table directed down (reaction) force exerted by string attached to object A on object B directed left (action) force exerted by object B on string attached to object A directed right (reaction) force exerted by string attached to object C on object B directed right (action) force exerted by object B on string attached to object C directed left (reaction) gravitational force exerted by Earth on object B directed down (action) force exerted by object B on Earth directed up (reaction) From Newton s law of gravitation, F g. r The figure below represents the situation of the problem. 38

39 before after 1 1 F g F g ( r) (10 r) 1 1 r r r r Calculate the factor change of F g = = 1 5 Calculate F g. 1 5 F g = 1 5 (00 N) = 8.00 N The new gravitational force will be 8.00 N [toward Earth s centre]. 40. When each diver steps off the diving tower, they will be in free fall and the only force acting on the divers will be the gravitational force. Since the problem assumes negligible air resistance, both divers will experience the same gravitational acceleration. So the time it takes to reach the water will be the same for both divers. 41. Given m s = 68 t or kg r s = 435 km or m m Earth = kg r Earth = m gravitational field strength at the location of Skylab 1 ( g ) Find the separation distance between Skylab 1 and Earth. r = r s + r Earth = m m = m Use the equation g = GM source r to calculate the magnitude of the gravitational field strength. 39

40 g = Gm Earth r 11 Nm i ( kg ) kg = ( m) = 8.6 N/kg g = 8.6 N/kg [toward Earth s centre] The gravitational field strength at the location of Skylab 1 is 8.6 N/kg [toward Earth s centre]. 4. Given m b = 4.00 kg m Earth = kg r Earth = m m Mars = kg r Mars = m difference in reading on spring scale (ΔF g ) Use the equation g = GM source r to calculate the magnitude of the gravitational field strength on Earth and on Mars. Earth g Earth = Gm Earth (r Earth ) Mars = = N/kg i 4 Nm kg ( m) ( kg ) g Mars = Gm Mars (r Mars ) 11 Nm i ( kg ) kg = ( m) = N/kg 40

41 ΔF g = F g Earth F g Mars = m b g Earth m b g Mars = m b (g Earth g Mars ) = (4.00 kg)(9.783 N/kg N/kg) = 4.3 N The reading on the spring scale will be 4.3 N less on Mars than on Earth. 43. (a) and (b) Given m = 60 kg g = 9.81 m/s [down] (i) a = 0 m/s v = constant (ii) a = 0 m/s v = 0 m/s (iii) a = 4.9 m/s [up] (iv) a = 3.3 m/s [down] (a) reading on the scale (F N ) (b) true weight and apparent weight ( g and w ) Draw a free-body diagram for the student in each situation. (i) (ii) 41

42 (iii) (iv) Use the equation g = m g to find the student s true weight. g = m g F g = (60 kg)( 9.81 m/s ) g = N = N [down] The student is not accelerating left or right. So in the horizontal direction, net = 0 N. For the vertical direction, write an equation to find the net force on the student. (i) and (ii) net = g + N 0 = F g + F N 4

43 F N = F g = ( N) = N N = N [up] Use the equation w = N w to find the student s apparent weight. = N = N [down] (iii) net = g + N F net = F g + F N Apply Newton s second law. ma = F g + F N F N = ma F g = (60 kg)(4.9 m/s ) ( N) = (60 kg)(4.9 m/s ) N N = N Use the equation w = N w = N [up] to find the student s apparent weight. = N = N [down] (iv) net = g + N F net = F g + F N Apply Newton s second law. ma = F g + F N F N = ma F g = (60 kg)(3.3 m/s ) N N = N Use the equation w = N w = N [up] to find the student s apparent weight. = N = N [down] (a) The reading on the scale will be N for situations (i) and (ii), N for situation (iii), and N for situation (iv). (b) The true weight of the student in all situations is N [down]. The apparent weight is N [down] for situations (i) and (ii), 43

44 44. Given m = 60 kg N [down] for situation (iii), and N [down] for situation (iv). air = 00 N [up] g = 9.81 m/s [down] true weight and acceleration of skydiver ( g and a ) Draw a free-body diagram for the skydiver. Use the equation g = m g to find the skydiver s true weight. g = m g F g = (60 kg)(9.81 m/s ) g = N = N [down] The skydiver is not accelerating left or right. So in the horizontal direction, net = 0 N. For the vertical direction, write an equation to find the net force on the skydiver. net = g + air F net = F g + F air Apply Newton s second law. ma = F g + F air = mg + F air 44

45 a = g + 00 N m = 9.81 m/s m 00 kg i s 60 kg = 6.5 m/s a = 6.5 m/s [down] The skydiver has a true weight of N [down] and an acceleration of 6.5 m/s [down]. 45. (a) Given m = 5 kg Δt = 8.0 s d = 300 m g = 9.81 m/s [down] acceleration of boulder ( a ) Calculate the magnitude of the acceleration of the boulder. d = 1 a(δt) a = d (Δt) (300 m) = (8.0 s) = 9.38 m/s a = 9.4 m/s [down] While falling, the boulder has an acceleration of 9.4 m/s [down]. (b) Given m = 5 kg Δt = 8.0 s d = 300 m g = 9.81 m/s [down] a = 9.38 m/s [down] from part (a) air resistance on boulder ( air ) Draw a free-body diagram for the boulder. 45

46 The boulder is not accelerating left or right. So in the horizontal direction, net = 0 N. For the vertical direction, write an equation to find the net force on the boulder. net = g + air F net = F g + F air Apply Newton s second law. ma = mg + F air F air = ma mg = m(a g) = (5 kg)(9.38 m/s 9.81 m/s ) air = 11 N = 11 N [up] The air resistance on the boulder is 11 N [up]. (c) Given m = 5 kg Δt = 8.0 s d = 300 m g = 9.81 m/s [down] a = 9.38 m/s [down] from part (a) air = 10.9 N [up] from part (b) apparent weight of boulder ( w ) The air resistance on the boulder is equivalent to the normal force. N = 10.9 N [up] Use the equation w = N to find the apparent weight of the boulder. w = N = 11 N [down] The boulder has an apparent weight of 11 N [down]. Extensions 46. Given l = 36 cm or 0.36 m m Earth = kg r Earth = m m Moon = kg r Moon = m period of pendulum on Earth and the Moon (T Earth and T Moon ) 46

47 Use the equation g = GM source r to calculate the magnitude of the gravitational acceleration on Earth and on the Moon. Earth g Earth = Gm Earth (r Earth ) 11 Nm i ( kg ) kg = ( m) = m/s T Earth = π l g Earth = π = 1. s 0.36 m m s Moon g Moon = Gm Moon (r Moon ) 11 Nm i ( kg ) kg = ( m) = m/s T Moon = π l g Moon = π 0.36 m m s = 3.0 s The pendulum will have a period of 1. s on Earth and 3.0 s on the Moon. 47. (a) Given m = 150 g or kg = 7.0 m/s [toward net] v i d = 3 m μ k = 0.08 g = 9.81 m/s 47

48 force of kinetic friction on puck ( ) f kinetic Draw a free-body diagram for the puck. Since the puck is accelerating, net 0 N in the horizontal direction, but net = 0 N in the vertical direction. Calculate F f. kinetic F f kinetic = μ k F N = μ k mg = (0.08)(0.150 kg)(9.81 m/s ) f kinetic = 0.1 N = 0.1 N [away from net] The force of kinetic friction on the hockey puck is 0.1 N [away from net]. (b) Given m = 150 g or kg v i d = 3 m μ k = 0.08 g = 9.81 m/s f kinetic = N [away from net] from part (a) acceleration of puck ( a ) = 7.0 m/s [toward net] 48

49 Write equations to find the net force on the puck in both the horizontal and vertical directions. horizontal direction = f kinetic net h F net h = f kinetic ma = F f kinetic F f a = kinetic m vertical direction netv F netv = N kg = N F = 0 m kg i = s kg = 0.78 m/s a = 0.78 m/s [toward net] The acceleration of the puck is 0.78 m/s [toward net]. (c) Given m = 150 g or kg + g = 7.0 m/s [toward net] v i d = 3 m μ k = 0.08 g = 9.81 m/s f kinetic = N [away from net] from part (a) a = m/s [toward net] from part (b) time for puck to stop (Δt) Calculate the time for the puck to stop. a = Δv Δt Δt = Δv a = v f v i a 0 m/s 7.0 m/s = m/s = 8.9 s It will take 8.9 s for the puck to stop. Calculations in the vertical direction are not required in this problem. 49

50 (d) Given m = 150 g or kg 48. = 7.0 m/s [toward net] v i d = 3 m μ k = 0.08 g = 9.81 m/s f kinetic = N [away from net] from part (a) a = m/s [toward net] from part (b) determine if puck will reach the net Calculate the distance the puck travels before it stops. (v f ) = (v i ) + ad 0 = (v i ) + ad d = (v i) a (7.0 m/s) = ( m/s ) = 31 m Since the puck will only travel 31 m, it will not reach the net. Criteria Inertia Objects Newton s 1st Law one object in isolation Newton s nd Law one object experiencing an external non-zero net force Newton s 3rd Law two objects interacting with each other Velocity zero or constant changing zero, constant, or Equation changing net = 0 when Δ v net = m a A on B = B on A = 0 Example gliding while accelerating while satellite orbiting Earth skating riding a bicycle Application seat belts and airbags parachute aircraft propeller 49. Although the effectiveness of these swimsuits is still a matter of debate, those in favour of some standardization might argue that less developed countries cannot afford to keep up with innovations in equipment and would be disadvantaged. Even within any particular country, athletes with less financial means might not be able to afford the best equipment, so the best athletes might not get to represent their 50

51 country. Standardization would also make the comparison of Olympic records more meaningful. Those opposed to standardization of equipment might suggest that faster times, higher jumps, etc., would not be possible and the Olympics may not be as exciting as they could be. Another argument might be that advances in Olympic equipment might become available for the general population to use in everyday activities. 50. Stakeholders and their perspectives can be identified through class discussion. Have students identify those totally in favour of airbags, those in favour with reservation, and those totally against airbags. Stakeholders include the general public, consumer activist groups, medical practitioners, insurance companies, automobile manufacturers, accident investigators, and others. 51. (a) Given m = kg thrust = N [forward] g = 9.81 m/s [down] true weight of rocket ( g ) Use the equation g = m g to find the true weight of the rocket. g = m g F g = ( kg)( 9.81 m/s ) g = N = N [down] The rocket has a true weight of N [down]. (b) Given m = kg thrust = N [forward] g = 9.81 m/s [down] g = N [down] from part (a) net force on rocket at liftoff ( net ) Draw a free-body diagram for the rocket. 51

52 The rocket is not accelerating left or right. So in the horizontal direction, net = 0 N. For the vertical direction, write an equation to find the net force on the rocket. net = thrust + g F net = F thrust + F g = N + ( N) = N N = N net = N [forward] The net force on the rocket at liftoff is N [forward]. (c) Given m = kg thrust = N [forward] g = 9.81 m/s [down] g = N [down] from part (a) net = N [forward] from part (b) initial acceleration of rocket ( a ) The acceleration of the rocket is in the direction of the net force. So use the scalar form of Newton s second law. 5

53 F net = ma a = F net m = N kg = 7 kg im s kg = 3.3 m/s a = 3.3 m/s [forward] The rocket has an acceleration of 3.3 m/s [forward] at liftoff. (d) If the force exerted by the engines remains constant as the fuel burns, the magnitude of the acceleration will increase because the mass of the rocket is decreasing. (e) The first-stage engines are jettisoned after the fuel is consumed to minimize the mass of the rocket. This results in the maximum possible acceleration. 5. (a) Given m p = 70 kg r p to y = 1.0 m m Mars = kg r M to E = m r Earth = m ratio of (F g ) p to (F g ) M 53

54 Suppose your mass is 55 kg. Calculate the separation distance between you and Mars. r M to y = r M to E r Earth = m m = m Calculate F g exerted by Mars on you using Newton s law of gravitation. Gm m Mars (F g ) M = ( rm to y ) y 11 Nm i ( kg )(55 kg ) kg = 11 ( m) = N Calculate F g exerted by the person next to you using Newton s law of gravitation. Gm m p (F g ) p = ( rp to y ) y 11 Nm i (70 kg )(55 kg ) kg = (1.0 m) = N Calculate the ratio of (F g ) p to (F g ) M. (F g ) p (F g ) M = N N = 5.8 The person next to you exerts a gravitational force that is 5.8 times greater than Mars does on you. (b) Since the person next to you exerts a much greater gravitational force on you than Mars does on you, the claim made by astrologers is not valid. 53. This research question will be of particular interest to students interested in medicine or engineering. An artificial hip consists of a stem attached to the femur, a ball (or head) fitted onto the stem, and a cup attached to the acetabulum which provides a smooth gliding surface for the ball. Early materials for artificial joints included glass, Pyrex, ivory, Bakelite, and other plastics but these were discarded because of breakage, wear, bio-incompatibility among other problems. Currently, stems are typically made of alloys of titanium or cobalt-chromium which are considered to be the stongest. Most biocompatible heads are usually made of ceramic or a cobalt- 54

55 chromium alloy. Some acetabular cups consist of a metal shell securely attached to a polyethylene liner. Various types of cement are used to anchor the prosthesis components to bone. Cementless versions have a textured surface to allow for bone cell ingrowth. Some cementless acetabular cups are reinforced with screws. The surfaces of all moving parts are very smooth and strong, and the human body constantly lubricates the joint surfaces. 54. This investigation will involve both science and technology. It nurtures many scientific processes including experimental design and the control of variables. It could also serve as an excellent cross-country skiing field-trip to a local park. Some of the variables involved are temperature, type of snow, snow packing, mass of person, mass of skis, normal force, static and kinetic friction, ski length, ski width, ski base design, and ski waxes. Students could also investigate the advertising and communication strategies used to sell waxes for skis. 55. Gait analysis is the study of how humans walk. In order to start walking, a person leans forward to shift the centre of mass forward slightly. The person exerts an action force backward (static friction) on the ground, one foot at a time. According to Newton s third law, the ground then exerts a reaction force forward on each foot. The reaction force causes the person to experience a non-zero net force and the person accelerates forward. From Newton s second law, the magnitude of the net force is directly proportional to the person s mass and the acceleration desired. In order for a person to stop walking, the process is similar except that the directions are reversed. The person must shift the centre of mass backward slightly. The person exerts an action force forward on the ground, one foot at a time. According to Newton s third law, the ground then exerts a reaction force backward on each foot. The reaction force causes the person to slow down to a stop. The process is similar when a person changes direction. Walking is even more involved than just involving Newton s laws. The chemical energy stored in food is converted to gravitational potential and kinetic energy as the body moves slightly up and down. Also, the legs change the person s velocity. For interesting websites on the topic of gait analysis, follow the links at 55