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Transcript
Chapter 5: Newton’s Laws of Motion Answers and Solutions 1. No, it is not possible for a stationary object to have one force acting on it. If it did, it would accelerate and no longer remain stationary. However, it is possible for a stationary object to have two forces acting on it as long as the net force is zero. Furthermore, an object with one force acting on it can momentarily be at rest, such as when a ball thrown upward gets to the top of its flight. 2. Picture the Problem: A shopping cart accelerates in the forward direction due to a pushing force. Strategy: Use Newton’s second law to find the acceleration of the cart, then use the acceleration to find the distance traveled in 2.50 s. F (10.1 N ) Solution: 1. Find a using Newton’s second law: a= = = 0.821 m/s 2 m 12.3 kg ( Δx = 12 a t 2 = 2. Solve the position-time equation for the distance traveled: 1 2 ) (0.821 m/s ) (2.50 s) 2 2 = 2.57 m Insight: Doubling the applied force would double the acceleration and also double the distance traveled. 3. Picture the Problem: This is a follow-up question to Guided Example 5.4. Moe, Larry, and Curly push on a 752-kg boat that floats next to a dock. They each exert an 85.0-N force parallel to the dock, and the boat has an acceleration of 0.530 m/s2. Strategy: Use Newton’s second law to find the mass of the boat. The net force is the vector sum of the three forces. F 3 (85.0 N ) Solution: Solve Newton’s second law for mass: m= = = 481 kg a 0.530 m/s 2 Σ ( ) Insight: If Larry, Moe, and Curly were pushing in different directions, the net force on the boat would have been smaller, and so would have been the acceleration. 4. Picture the Problem: A force acts on a car of known mass and causes it to accelerate. Strategy: Use Newton’s second law to find the force required to produce the given acceleration. F = m a = (1500 kg ) 3.8 m/s 2 = 5700 N = 5.7 kN Solution: Solve Newton’s second law for the force: Σ ( ) Insight: No information is given about the direction of the acceleration, but we know from Newton’s second law that the direction of the net force must be the same as the direction of the acceleration. The force is reported in kilonewtons to emphasize there are only two significant figures in the answer. 5. Picture the Problem: You exert a horizontal force that accelerates both your little sister and the sled. Strategy: Apply Newton’s second law to the sled plus sister combination, and solve for the mass of your sister. F 120 N = = 48 kg a 2.5 m/s 2 Solution: 1. Use Newton’s second law to find the total mass of the sled plus sister combination: mtotal = 2. Find the mass of your sister by subtracting: msister = mtotal – msled = 48 kg – 7.4 kg = 41 kg Insight: A more massive sister would decrease the acceleration of the sled if the pulling force remained the same. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5–1 Chapter 5: Newton’s Laws of Motion 6. Pearson Physics by James S. Walker Picture the Problem: This is a follow-up question to Guided Example 5.5. Two groups of canoeists meet in the middle of a lake. After a brief visit, a person in canoe 1 pushes on canoe 2 with a force of 46 N to separate the canoes. The mass of canoe 1 and its occupants is m1 = 150 kg, and the initial mass of canoe 2 and its occupants is m2 = 250 kg. Strategy: Use Newton’s third law to determine the magnitude of the force on canoe 2, and Newton’s second law to find the acceleration of canoe 2 if its mass is increased to 25,000 kg. Solution: 1. Newton’s third law states that both canoe 1 and canoe 2 experience the same magnitude force, but in opposite directions. If the mass of canoe 2 were increased, its acceleration would decrease according to Newton’s second law because the magnitude of the force would not change. We conclude that the acceleration of canoe 2 will decrease if its mass is increased. 2. Use Newton’s second law to find the acceleration of a massive ship that replaces canoe 2: a= ΣF = m 46 N = 0.0018 m/s 2 25, 000 kg Insight: As predicted, the 0.0018 m/s2 acceleration we found when canoe 2 was replaced by a whip with a mass of 25,000 kg is much smaller than the 0.18 m/s2 acceleration experienced by the 250-kg canoe 2. 7. (a) The two forces that act on a book at rest in your hand are the force of gravity and the upward force your hand exerts on the book. (b) Yes, because the book is not accelerating we conclude there is no net force on the book, and the hand force must be equal and opposite to the gravitational force. (c) No, the two forces are not an action-reaction pair because both forces act upon the book. Action-reaction force pairs always work on two different objects. 8. Picture the Problem: The parent and child push off each other while on ice skates. The child is pushed in the positive x direction and the parent is pushed in the negative x direction. Strategy: The situation is governed by Newton’s third law, where the force Fparent exerted on the parent by the child is equal and opposite to the force Fchild exerted on the child by the parent. Solution: Set the magnitudes of the forces equal to find the acceleration of the parent using Newton’s second law: Fchild = -Fparent mchild achild = mparent aparent aparent = æ19 kg ö÷ mchild ÷÷(2.6 m/s 2 ) = 0.70 m/s 2 achild = ççç mparent è 71 kg ÷ø Insight: If the child had the same mass as the parent, they would each experience the same magnitude force and the same magnitude acceleration, but in opposite directions. 9. A resting object cannot move unless it accelerates, and it will not accelerate unless it is acted upon by a nonzero net force. Neither will a moving object change its speed or direction unless it is acted upon by a nonzero net force. 10. When the net force acting on an object triples in magnitude, the acceleration will also triple in magnitude. 11. Suppose you push on a loaded railroad car and it does not move. You might be tempted to argue that it does not move because the force you exert on the railroad car is equal and opposite to the force the railroad car exerts on you. While this is true by Newton’s third law, the two forces cannot cancel each other because they act on different objects; one is a force on the railroad car and the other is a force on you. The railroad car does not move because the force of static friction on the railroad car balances out the force you exert on the railroad car. The two forces cancel because they are equal and opposite, but they are not an action-reaction pair. 12. (a) A shopping cart will remain at rest if you do not push it because there is no net force on it that can accelerate it. (b) When you push on the cart, the vector sum of forces on the cart is no longer zero, and the cart will accelerate according to Newton’s second law. (c) According to Newton’s second law, the more massive the cart, the smaller its acceleration, so that the product of mass and acceleration always equals the magnitude of the net force. The acceleration is linearly proportional to the force and inversely proportional to the mass. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5–2 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 13. The astronaut should push the otherwise useless jet pack away from him, in the opposite direction from the spaceship. As a result, the reaction force exerted on him by the pack will accelerate him toward the spaceship. 14. Yes, it is possible. While the acceleration of an object is always in the direction of the net force, the velocity of an object does not need to be in the direction of the acceleration. For instance, when a soccer ball is kicked downfield it follows a parabolic trajectory. The net force on the ball is due to gravity alone, and its acceleration is always downward, but the ball moves along the parabolic path, not downward. 15. Picture the Problem: Two force vectors are depicted as acting on the tire of a front wheel drive car. Strategy: The net force on the car is in the forward direction. Therefore, the external force responsible for accelerating the car (the force of friction) must exert a force in the forward direction. Meanwhile, the engine tries to rotate the front tire clockwise, which exerts a backward force on the ground. Solution: 1. (a) Because the external force on the car is in the forward direction in order to accelerate it, and because the engine rotates the front tire clockwise, which exerts a backward force on the ground, we conclude that F1 is the force exerted by the car on the ground and F2 is the force exerted by the ground on the car. 2. (b) the force exerted by the car on the ground and the force exerted by the ground on the car are an action-reaction pair of forces. Therefore, the expression (2) F2 = −F1 is the correct relationship between the two forces. Insight: Friction between the tires and the ground is the force responsible for propelling the car, not the engine force. Without friction the tires spin due to the engine force but the car goes nowhere. 16. Picture the Problem: A baseball is accelerated by a net force. Strategy: Use Newton’s second law to find the force required to accelerate a ball of known mass. Solution: Apply Newton’s second law: Σ F = m a = (0.15 kg) (12 m/s ) = 1.8 N 2 Insight: This is a pretty small force, equivalent to 1.8 N × 1 lb/4.45 N = 0.40 lb = 6.5 ounces of force. 17. Picture the Problem: An airplane is accelerated horizontally in the direction opposite its motion in order to slow it down from its initial landing speed. Strategy: Use Newton’s second law to find the acceleration of the airplane. – 4.30 × 105 N F = −1.23 m/s 2 a= = Solution: Solve Newton’s second law for a : m 3.50 × 105 kg ( ) ( ) Insight: The typical landing speed of a Boeing 747-200 is 71.9 m/s (161 mi/h) and it has a specified landing roll distance of 2,121 m, requiring an average landing acceleration of −1.22 m/s2. 18. Picture the Problem: The free-body diagram for the car and the trailer is shown at right. The diagram assumes there is no friction. F1 −F1 F2 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 F1 exerted on it by the car, and it has the same acceleration (1.85 m/s2) as the car. For the car there are two forces acting on it, the engine F2 and the trailer −F1 . Apply Newton’s second and third laws as appropriate to find the requested forces. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5–3 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker Solution: 1. (a) Write Newton’s second law for the trailer: Σ F = F = m a = (560 kg ) (1.85 m/s ) = 1.0 kN 2. (b) Newton’s third law states that the force the trailer exerts on the car is equal and opposite to the force the car exerts on the trailer: − F1 = −1.0 kN in the backward direction 3. (c) Write Newton’s second law for the car: Σ F = M a = (1400 kg) (1.85 m/s ) = 2.6 kN 2 1 in the forward direction 2 in the forward direction Insight: The engine force F2 must be 3.6 kN because it must not only balance the 1.0-kN force from the trailer but also accelerate the car in the forward direction, requiring an additional 2.6 kN of force. 19. Picture the Problem: This is a follow-up question to Guided Example 5.7. The fire alarm goes off and a 97-kg firefighter slides down a pole. Strategy: Use Newton’s second law to find the acceleration of the firefighter. Solution: Solve Newton’s second law for a y : ay = ΣF y m = Fpole − mg m = 650 N − 9.81 m s 2 = −3.11 m/s 2 97 kg Insight: The negative sign indicates that the firefighter still accelerates downward, but at a much lower rate than the −9.81 m/s2 acceleration he would have in a free fall. It is also less than the −4.2 m/s2 he has when the pole force is only 540 N, as described in Guided Example 5.7. 20. Picture the Problem: A suitcase is accelerated straight upward by an applied force. Strategy: There are two forces acting on the suitcase, the applied force F acting upward and the force of gravity W acting downward. Use Newton’s second law together with the known acceleration to determine the mass and weight of the suitcase. Let upward be the positive direction. ΣF Solution: 1. (a) Write out Newton’s second law: y m= = F − mg = ma F = ma + mg = m ( a + g ) F 105 N = = 9.99 kg a + g 0.705 m/s 2 + 9.81 m/s 2 ( ) W = mg = (9.99 kg ) 9.81 m/s 2 = 98.0 N 2. (b) Now use the definition of the weight: Insight: Note that mass and weight have different units. A 98.0-N suitcase weighs about 22 pounds. 21. Picture the Problem: A spring is stretched by an applied force. Strategy: Use Hooke’s law to find the spring constant. F =kx Solution: Solve Hooke’s law for k: k= F 1.2 N = = 19 N/m x 0.064 m Insight: This is a fairly weak spring, where 19 N (about 4.2 lb) of force would stretch it an entire meter. 22. Picture the Problem: A spring is stretched by an applied force. Strategy: Use Hooke’s law to find the stretch distance. Solution: Solve Hooke’s law for x: F =kx x= F 2.3 N = = 0.12 m = 12 cm k 19 N/m Insight: Compare this answer with the previous question, where roughly half the force stretches the spring by about half as much distance. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5–4 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 23. The steps to drawing a free-body diagram are: (1) draw vectors to represent all of the forces that act on the object, (2) choose a coordinate system, (3) resolve each force into components, and (4) apply Newton’s second law to each coordinate direction separately. 24. The weight of an object with mass m is calculated from the equation W = mg. 25. If an object is at rest, and remains at rest, it is not accelerating, and you can conclude that the vector sum of all the forces that act on it are zero. If the object is only momentarily at rest, such as at the top of the flight of a baseball that is thrown straight upward, there must be a net force on the object because it is accelerating. 26. Hooke’s law states that the distance a spring stretches is linearly proportional to the amount of force that is exerted on it. The constant of proportionality is the spring constant k: F = kx. 27. In order to decrease the sag angle in a clothesline it is necessary to increase the tension. In every case there must be an upward component to the tension force in order to balance the weight of the clothesline and the objects hanging from it. That upward component can be provided by a force that is inclined at a small angle as long as the magnitude of the force is very large. 28. Picture the Problem: A 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 of the chair acting upward and the force of gravity Wchild acting downward. There are three forces acting on the chair: the normal force N of the floor acting upward, the weight of the child acting downward, and the force of gravity Wchair acting downward. Write Newton’s second law in the vertical direction for each case and then use the equations to find N. Solution: 1. (a) Write out Newton’s Second Law in the vertical direction and solve for N: ΣF y (a) child N ( N = mg = (9.3 kg ) 9.81 m/s 2 ΣF y N = N − mg = 0 ) Wchild = 91 N 2. (b) Write out Newton’s Second Law in the vertical direction for the chair and solve for N: (b) chair Wchild + Wchair = N − mchild g − mchair g = 0 N = ( mchild + mchair ) g ( = (9.3 + 3.7 kg ) 9.81 m/s 2 ) N = 130 N = 0.13 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 need only support the weight of the child. 29. Picture the Problem: A spring is stretched by an applied force. Strategy: Use Hooke’s law to find the required force. Solution: Use Hooke’s law to find F: F = k x = ( 62 N/m )( 0.19 m ) = 12 N Insight: A 12 N force is equivalent to about 2.5 lb, so this spring is fairly easy to stretch. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5–5 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 30. Picture the Problem: The free-body diagram for the satellite is shown at right. Strategy: The force F1 from one astronaut adds as a vector to the force F2 from the second astronaut to produce the net force F, whose angle θfrom the x axis is known. Use the addition of vectors together with the trigonometric tangent function to find the magnitude of F2 . Solution: The angle is determined by the tangent function: tan θ = F2 F1 F2 = F1 tan θ = ( 42 N ) tan15° = 11 N Insight: Once the two astronaut forces are known, we can find the magnitude of the net force (the vector sum of the two forces) by calculating F = F12 + F22 = 422 + 112 = 43 N . 31. Picture the Problem: This is a follow-up question to Guided Example 5.11. A salt shaker with a mass of 50.0 g slides along a table, slowing down due to a coefficient of kinetic friction μ k = 0.120. Strategy: There is only one force on the salt shaker that changes its horizontal motion: the friction force. Use the weight of the salt shaker and the coefficient of friction to determine the magnitude of the friction force on the salt shaker. Then use the force, together with Newton’s second law, to find the acceleration. ( ) Solution: 1. Find the magnitude of the friction force: f k = μk mg = ( 0.120)( 0.0500 kg ) 9.81 m s 2 = 0.0589 N 2. Apply Newton’s second law to find a: ΣF = f k = ma a= fk 0.0589 N = = 1.18 m/s 2 m 0.0500 kg Insight: A higher coefficient of kinetic friction in this follow-up question resulted in a higher acceleration when compared with Guided Example 5.11. Note that a = f k m = μ k mg m = μ k g . 32. Picture the Problem: A baseball player slides in a straight line and comes to rest due to the frictional force. Strategy: Find the acceleration of the player using Newton’s second law, and insert the result into the velocity-distance equation to find the slide distance. Let the initial velocity v i point in the positive direction. Solution: 1. Find the acceleration of the player using Newton’s second law: 2. Use the velocity-distance equation to find the slide distance: ΣF x = f k = ma a= μ mg fk =− k = − μk g m m 0 − vi2 v 2 − vi2 ( 4.0 m/s) = = = 1.8 m 2a −2μ k g 2 ( 0.46) 9.81 m/s 2 2 Δx = ( ) Insight: If the player were running faster, say, 8.0 m/s, a µk of 0.46 would result in a slide distance of 7.1 m, a quarter of the 27 m between the bases! He might very well overshoot the base and be tagged out. 33. Picture the Problem: This is a follow-up question to Guided Example 5.12. Sand poured slowly into a pail exerts sufficient force (11 N) on a string to cause a jack-o-lantern to slide horizontally. Strategy: The pumpkin will begin sliding when the weight of the sand matches the maximum static friction force. Solution: 1. (a) If the coefficient of static friction is reduced from 0.61 to 0.55, the maximum static friction force will decrease. We conclude that the weight of the pail and sand necessary to start the pumpkin moving in this case is less than 11 N. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5–6 Chapter 5: Newton’s Laws of Motion 2. (b) Set the weight of the pail equal to the maximum static friction force: Pearson Physics by James S. Walker ( ) Wpail = μs mg = ( 0.55)(1.8 kg ) 9.81 m s 2 = 9.7 N Insight: Because the friction force scales linearly with μs , a 10% reduction in μs (from 0.61 to 0.55) resulted in a 10% reduction in the required weight of the pail (from 11 N to 9.7 N). 34. Picture the Problem: A book slides in a straight line across the top of the tabletop. A larger force is needed to start the book sliding, and a smaller force is needed to keep it sliding with constant speed. 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 µk can be determined. Solution: 1. When the book begins sliding, the applied force equals the maximum static friction force: F = fs = μs mg 2. When the book is sliding at constant speed, the applied force equals the kinetic friction force: Fapp = f k = μ k mg app Fapp μs = mg μk = Fapp mg 2.25 N = (1.80 kg ) (9.81 m/s 2 ) = 1.50 N = 0.127 (1.80 kg ) (9.81 m/s 2 ) = 0.0849 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. 35. Stacking a second book on top of the first book will double the normal force exerted by the tabletop, because that normal force must now support the weight of two books. Doubling the normal force will double the force of kinetic friction because kinetic friction varies linearly with the normal force. 36. Skidding to a stop actually requires a longer distance than does coming to a stop without skidding. This is because a rolling wheel harnesses the coefficient of static friction (the point of contact is momentarily at rest) whereas a skidding tire is affected by sliding friction. The coefficient of static friction is usually higher than the coefficient of sliding friction. Therefore, a rolling wheel can stop the bicycle with a larger force and in a shorter distance. 37. Static friction involves surfaces that are not moving with respect to each other, and it tries to hold objects in place. Sliding friction involves surfaces that slip past each other, and it describes the force on a moving object. 38. (a) Friction is undesirable for the rotation of a bicycle tire around its axle and for the motion of an ice skate across the surface of a rink. (b) Friction is helpful for walking without slipping, and for the mechanism that grips paper while feeding it through a printer or photocopier. 39. Picture the Problem: A book slides horizontally across a table at constant speed due to a pushing force. Strategy: The pushing force must balance the sliding friction force. Solution: Set the pushing force equal to the sliding friction force: ΣF = F push − fk = 0 ( Fpush = f k = μk N = μ k mg = ( 0.11)( 0.27 kg ) 9.81 m/s 2 ) = 0.29 N Insight: This force is pretty small, equivalent to about 1.0 oz. A larger force would be needed to overcome the maximum force of static friction and start the book sliding. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5–7 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 40. Picture the Problem: A book slides horizontally across a table at constant speed due to a pushing force. Strategy: Use Newton’s second law to find the required pushing force that will both balance the sliding friction force and accelerate the book. Solution: Use Newton’s second law to find the required pushing force: ΣF = F push − fk = m a Fpush = f k + m a = μk N + m a = μ k mg + m a ( ) ( = ( 0.21)( 0.39 kg ) 9.81 m/s 2 + (0.39 kg ) 0.18 m/s 2 ) Fpush = 0.803 + 0.070 = 0.87 N Insight: This force is pretty small, equivalent to about 3.1 oz. Note that the pushing force does two things: it balances (or “overcomes”) friction and it accelerates the book. 41. Picture the Problem: A flowerpot begins sliding horizontally when a pushing force exceeds the maximum force of static friction. Strategy: Find the maximum static friction force from the given values of mass and coefficient of friction. Solution: Set the pushing force equal to the maximum static friction force: ΣF = F push − fs, max = 0 ( Fpush = fs, max = μs N = μs mg = ( 0.64)(14 kg ) 9.81 m/s 2 ) = 88 N Insight: This sizeable force (equivalent to 20 lb) is a consequence of the large mass (the 14-kg flowerpot weighs 31 lb) and the substantial coefficient of static friction. 42. Picture the Problem: A coin slides across a countertop with an initial speed, but friction gradually brings it to a stop. Strategy: Use the velocity-position equation to determine the acceleration of the coin, and then use Newton’s second law to find the friction force and then the coefficient of kinetic friction. Solution: 1. (a) Solve the velocityposition equation for acceleration: vf2 = vi2 + 2 a Δ x vf2 − vi2 0 − (1.8 m/s ) = = −1.9 m/s 2 2Δ x 2 ( 0.86 m ) 2 a= 2. Write Newton’s second law in the x direction: ΣF 3. The normal force must balance the weight: ΣF 4. Substitute the normal force into the expression from step 2 and solve for μk: x = − f k = ma − μ k N = ma y = N − mg = 0 μk = − N = mg −1.9 m/s 2 ma ma a =− =− =− = 0.19 N mg g 9.81 m/s 2 5. (b) The result of part (a) is independent of the mass of the coin. Although two coins would produce a normal force, and thus a friction force, that is twice as large as one coin, they would also have twice the inertia. Their acceleration would be the same, so the two coins would come to rest in the same distance as before. Insight: The result of part (b) is similar to gravity: massive objects experience a greater gravitational force, but they also have a larger inertia, and their acceleration is the same as less massive objects. 43. A force is a result of some physical interaction between two objects, but a net force is the vector sum of all forces acting on a single object. 44. Inertia is the property of an object that resists any change in its motion. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5–8 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 45. When you hit the brakes of your car, a force on the car changes its motion by slowing it down. Your body’s inertia resists that change in motion, attempting to maintain its speed in the forward direction. It is this inertia that makes your body move toward the front of the car while it is slowing down. 46. (a) According to Newton’s second law, if the net force Fnet = m a doubles, while the mass remains the same, the acceleration also doubles. (b) According to Newton’s second law, if the mass m in Fnet = m a doubles, while the net force remains the same, the acceleration will be cut in half. 47. If the tablecloth is pulled rapidly, it can exert a force on the place settings for only a very short time. In this brief time, the objects on the table accelerate, but only slightly. Therefore, the objects may have barely moved by the time the tablecloth is completely removed. 48. As the dog shakes its body, it starts water in its fur moving in one direction. When it then begins to shake its body in the opposite direction, much of the water continues in the same direction because of the law of inertia (Newton’s First Law). As a result, water leaves the fur with each reversal in direction. 49. Picture the Problem: A small car collides with a large truck. Strategy: Consider Newton’s laws of motion when answering the conceptual question. Solution: 1. (a) Newton’s third law states that when the car exerts a force on the truck, the truck exerts an equal and opposite force on the car. We conclude that the magnitude of the force experienced by the car is equal to the magnitude of the force experienced by the truck. 2. (b) The best explanation is A. Action-reaction forces always have equal magnitude. Statements B and C are false. Insight: The force has a larger effect (produces a larger acceleration) on the smaller car due to its smaller inertia. 50. Picture the Problem: A small car collides with a large truck. Strategy: Consider Newton’s Laws of motion when answering the conceptual question. Solution: 1. (a) Newton’s third law states that when the car exerts a force on the truck, the truck exerts an equal and opposite force on the car. However, the car’s mass is less than the truck’s mass, so by Newton’s second law an equal magnitude force on a smaller mass will produce a larger acceleration. We conclude that the acceleration experienced by the car is greater than the acceleration experienced by the truck 2. (b) The best explanation is B. Both vehicles experience the same magnitude of force, therefore the lightweight car experiences the greater acceleration. Statements A and C are both false because the forces are of equal magnitude. Insight: If the car’s mass were half the mass of the truck, its acceleration would be twice that of the truck. 51. While you do exert an upward force on the Earth (of about 700 N), the mass of the Earth is so large that its acceleration is indiscernible. During a 2.0 s freefall dive, you will fall almost 20 meters while the Earth rises up a distance of Δy = 12 a t 2 = 1 2 (700 N 6.0 × 10 24 ) kg ( 2.0 s ) = 2.3 × 10 −22 m , a distance a million times smaller than the nucleus of a 2 single atom. 52. Picture the Problem: Three identical pucks with different initial velocities are each acted upon by the same force. Strategy: Consider Newton’s second law when answering the conceptual question. Solution: Newton’s second law states that the acceleration of an object must always point in the same direction as the net force on it. Because the net forces on each puck are identical, so are their accelerations, regardless of the magnitude or direction of the initial velocity. The ranking of the accelerations is therefore A = B = C. Insight: The accelerations of pucks A and B are identical even though puck A is slowing down at first and puck B is speeding up. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5–9 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 53. Picture the Problem: A golf cart is accelerated by a net force. Strategy: Use Newton’s second law to find the required net force that will accelerate the cart by the stated amount. Solution: Use Newton’s second law to find the required net force: Σ F = m a = (540 kg) (2.5 m/s ) = 1350 N = 1.4 kN 2 Insight: The largest (empty) golf cart available from one particular manufacturer weighs 798 lb, equivalent to a mass of 362 kg. This 540 kg golf cart must be weighed down by passengers and gear! 54. Picture the Problem: A backpack is accelerated by a net force. Strategy: Use Newton’s second law to find the mass of a backpack that is accelerated by a net force. Solution: Use Newton’s second law to find the mass: Σ F = ma m= ΣF = a 23 N = 6.1 kg 3.8 m/s 2 Insight: A 6.1-kg backpack would have a weight of 60 N or about 13.5 lb. 55. Picture the Problem: A rainbow trout is accelerated by a net force. Strategy: Use Newton’s second law to find the acceleration of a rainbow trout due to the net force. Solution: Use Newton’s second law to find the acceleration: Σ F = ma a= Σ F = 3.4 N = 2.6 m/s m 2 1.3 kg Insight: A 2.6-kg rainbow trout that experienced the same net force would accelerate at half this rate. F1 56. Picture the Problem: The free-body diagrams for the truck and the trailer is shown at right. The diagram assumes there is no friction. −F1 F2 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 F1 exerted on it by the truck, and the trailer has the same acceleration (1.16 m/s2) as the truck. F = F1 = m a = ( 620 kg ) 1.16 m/s 2 = 720 N Solution: Write Newton’s second law for the trailer: in the forward direction Insight: The engine force F2 must be 2.8 kN because it must not only balance the 0.72-kN force from the trailer but also ( Σ ( ) ) accelerate the truck in the forward direction, requiring an additional ma = (1800 kg ) 1.16 m/s 2 = 2.1 kg of force. F1 57. Picture the Problem: The free-body diagrams for the car and the trailer is shown at right. The diagram assumes there is no friction. −F1 F2 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 truck there are two forces acting on it, the engine F2 and the trailer − F1 . Apply Newton’s second law to find the net force on the truck. F = M a = (1800 kg ) 1.16 m/s 2 = 2.1 kN Solution: Write Newton’s second law for the truck: Σ ( ) in the forward direction Insight: The engine force F2 must be 2.8 kN in the forward direction because it must not only balance the 0.72-kN force from the trailer but also accelerate the car in the forward direction, requiring an additional 2.1 kN of force. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 10 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 58. Picture the Problem: A parent and a child push off each other while on ice skates. The child is pushed in the positive x direction and the parent is pushed in the negative x direction. Strategy: By Newton’s third law the force Fparent exerted on the parent by the child is equal and opposite to the force Fchild exerted on the child by the parent. However, Newton’s second law indicates that the accelerations of the parent and the child are inversely proportional to their masses. Solution: 1. (a) The magnitude of the force experienced by the child is the same as the magnitude of the force experienced by the parent, but they point in opposite directions. 2. (b) The acceleration of the child is more than the acceleration of the parent. The child experiences the same force as the parent but has less mass, so she experiences the larger acceleration. Insight: If the child had the same mass as the parent, they would each experience the same magnitude force and the same magnitude acceleration, but in opposite directions. 59. Picture the Problem: A parent and a child push off each other while on ice skates. The child is pushed in the positive x direction and the parent is pushed in the negative x direction. Strategy: By Newton’s third law the force Fparent exerted on the parent by the child is equal and opposite to the force Fchild exerted on the child by the parent. However, Newton’s second law indicates that the accelerations of the parent and the child are inversely proportional to their masses. Fchild = -Fparent Solution. Set the magnitudes of the forces equal to find the acceleration of the parent mchild achild = mparent aparent using Newton’s second law: æ 21 kg ö÷ m ÷(0.31 m/s 2 ) = 0.072 m/s 2 aparent = child achild = çç çè 91 kg ÷ø÷ mparent Insight: If the child were older and had the same mass as the parent, they would each experience the same magnitude force and the same magnitude acceleration, but in opposite directions. 60. Picture the Problem: A jet is accelerated in a straight line by the force of a catapult. Strategy: Use the change in velocity to find the acceleration of the jet, and then apply Newton’s second law to find the mass of the jet. Solution: 1. Find the acceleration of the jet: aav = 2. Apply Newton’s second law to find the mass of the jet: m= Dv æç 250 km/h – 0 öæ ÷÷çç1000 m öæ ÷÷çç 1 h ö÷÷ = 34.7 m/s 2 =ç ÷ ç ç 2.00 s øè km ÷øèç 3600 s ø÷ Dt è F 9.35 × 105 N = = 2.69 × 10 4 kg a 34.7 m/s 2 Insight: This mass corresponds to a weight of about 30 English tons (27 metric tons), and is approximately correct for the takeoff mass of the F-14 Tomcat fighter jet. A formidable weapon! 61. Picture the Problem: A car is accelerated horizontally in the direction opposite its motion in order to slow it down from 16.0 m/s to 9.50 m/s. Strategy: Use Newton’s second law and the definition of acceleration to determine the net force on the car as it slows down. Then use the average velocity equation to find the distance traveled by the car as it slows down. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 11 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker Solution: 1. (a) Use Newton’s second law and the definition of acceleration to find the net force on the car: (9.50 − 16.0 m/s) = −5100 N Δv = (950 kg ) F = ma = m Δt 1.20 s = 5.1 kN opposite to the direction of motion 2. (b) Use the average velocity equation to find the distance traveled by the car as it slows down: Δx = 1 2 (v0 + v ) Δt = 12 (16.0 + 9.50 m/s)(1.20 s) = 15.3 m Insight: We must consider “950 kg” as having only two significant figures because the zero is ambiguous. That limits the net force to two significant figures, even though the acceleration a = (9.50 − 16.0 m/s ) 1.20 s = −5.42 m/s 2 has three significant figures. 62. Picture the Problem: A firefighter slides down a pole, slowly accelerating until he lands at the floor below. Strategy: There are two forces acting on the firefighter, gravity and the upward friction from the pole. Those two forces add together to produce the acceleration of the firefighter. Find the acceleration of the firefighter using the velocityposition equation, and then use Newton’s second law to find the force due to the friction from the pole. Let upward be the positive direction. v y2 − vi,2 y Solution: 1. Find the acceleration of the firefighter using the velocity-position equation: ay = 2. Find the mass of the firefighter: W = mg 3. Use Newton’s second law to find Fpole : ΣF 2Δy y 2 − 4.2 m/s ) − 02 ( = 2 ( −3.3 m ) m= = −2.7 m/s 2 W 782 N = = 79.7 kg g 9.81 m/s 2 = Fpole − W = ma ( ) Fpole = ma + W = ( 79.7 kg ) −2.7 m/s 2 + 782 N = 570 N = 0.57 kN Insight: If the landing speed of the firefighter were 2.1 m/s instead of 4.2 m/s, the upward friction force would be larger, but not twice as much. In fact, the acceleration would slow to −0.668 m/s2 and the pole force would increase to 729 N. 63. Picture the Problem: Two people push together in the same direction on an object and it accelerates at a rate given by a1. The same two people push in opposite directions with the same amount of force each, and the object accelerates at a rate given by a2. Strategy: Use Newton’s second law to write two equations representing each of the two cases described in the problem. Solve the two equations for the two unknowns F1 and F2 . Let both people push in the positive direction in the first case, but let person 2 push in the negative direction in the second case. Solution: 1. Write Newton’s second law for each of the two cases: ΣF ΣF x = F1 + F2 = ma 1 x = F1 − F2 = ma2 2. Solve the second equation for F2 : F2 = F1 − ma2 3. Substitute the result into the first equation: F1 + ( F1 − ma2 ) = ma 1 ( 2 F1 = m a 1 + a2 4. Now use that result in the equation from step 2: ) F2 = éê 12 m (a 1 + a2 )ùú - ma2 = ë û F1 = 1 2 1 2 ( m a 1 + a2 ) m (a 1 - a2 ) Insight: If it turns out that F2 was the larger of the two forces, then a2 would have a negative value. 64. Because force is equal to mass multiplied by acceleration, and because Earth’s gravity accelerates all masses near its surface at the same rate (namely, g = 9.81 m/s2 downward), the gravitational force that Earth exerts on an object is equal to mg. We call that force the weight of an object. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 12 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 65. A sketch that shows all the forces acting on an object is referred to as a free-body diagram. Once all of the forces are drawn on a free-body diagram, a coordinate system is chosen and each force is resolved into components. At this point, Newton’s second law can be applied to each coordinate direction separately. 66. The normal force exerted by a horizontal road on a car must balance the car’s weight to ensure the car does not accelerate along the vertical direction. It must therefore be directed upward and have magnitude mg. 67. Your apparent weight in an elevator is related to your acceleration. While the elevator is moving at constant speed you do not feel any of the effects of apparent weight. You only experience them when the elevator is either speeding up or slowing down, in other words, when it is accelerating. 68. Hooke’s law indicates that the amount of stretch by a spring is linearly proportional to the stretching force. Therefore, if the force exerted on a spring is doubled, the stretch increases by a factor of two. 69. Tension forces occur whenever a string, rope, or other object is stretched. Examples include a clothesline, a guitar string, or the force along the rope that pulls a water skier. 70. An object is in equilibrium whenever its velocity (speed and direction) is constant, whether its speed be zero or 1000 m/s. 71. A clothesline has a finite mass, and so the tension in the line must have an upward component to oppose the downward force of gravity. Therefore, even an empty clothesline sags in a manner similar to when a weight is hanging from it. 72. When the magnitude of the upward friction force exerted on the girl by the rope equals the magnitude of her weight, her acceleration will be zero because the net force on her is zero. As a result, she moves with constant velocity. 73. The brick is at rest, which means that the net force on it is zero. There are two upward forces, the spring force and the normal force, and one downward force, the weight of the brick. Because the two upward forces are balanced by the weight, we conclude that the normal force acting on the brick is less than the brick’s weight. 74. Picture the Problem: You jump out of an airplane and open your parachute after an extended period of free fall. Strategy: Consider Newton’s second law when answering the conceptual question. Solution: 1. (a) In order to slow your descent you must accelerate in the upward direction and there must therefore be an upward net force on your body. Considering the upward force of the parachute and the downward force of gravity as vectors, we conclude that the parachute force must be greater than your weight to produce a net upward force. 2. (b) The best explanation is C. To decelerate a skydiver in free fall, the net force acting on the skydiver must be upward. Statements A and B are both false. Insight: An object will slow down whenever its velocity and acceleration vectors point in opposite directions. 75. Picture the Problem: The displayed free-body diagrams depict the forces exerted on identical hockey pucks. Strategy: Add the forces as vectors to find the net force. The largest net force will produce the largest acceleration according to Newton’s second law. Solution: The net force for case A is ( 7 − 5 N ) = 2 N. For case B its magnitude is 32 + 32 N 2 = 4.24 N. For case C we must add the components: Cx = 3 N + (3 N ) cos 45° = 5.12 N C y = (3 N ) sin 45° = 2.12 N C = 5.122 + 2.122 N 2 = 5.54 N Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 13 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker Finally, the net force in case D is 3 N. Therefore, the ranking of the accelerations, from smallest to greatest, is: A < D < B < C. Insight: The question could be answered without any calculation. First we note that the net force in case A is 2 N, which is less than the 3-N net force in case D. Second, both cases B and C have net forces larger than 3 N. Third, we note that the two 3-N forces in case C are more nearly in the same direction, as compared with the two 3-N forces in case B. It follows that the puck in case C has the larger acceleration. 76. Picture the Problem: The free-body diagram for this problem is shown at right. Strategy: Use the free-body diagram to determine the net force F = Fastronaut + W on Σ Fastronaut the rock, then apply Newton’s second law to find the acceleration of the rock. Let upward be the positive direction. F = ( 46.2 N ) + ( − 40.0 N ) = 6.2 N upward Solution: 1. Find the net force: W Σ 2. Now apply Newton’s second law to find a : a= ΣF = m 6.2 N = 1.2 m/s 2 upward 5.00 kg Insight: If the astronaut were to exert less than 40.0 N of upward force on the rock, it would accelerate downward. 77. Picture the Problem: The free-body diagram of the potatoes is shown at right: Strategy: There are two forces acting on the potatoes: the normal force N of the shopping cart acting upward and the force of gravity W acting downward. Solution: 1. (a) The free-body diagram for the potatoes is shown at right. 2. (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. Note also that for a real sack of potatoes in a shopping cart, a very small static friction force (between the cart and the potatoes) in the forward direction balances the air friction force in the backward direction. The net force is still zero, but there are some small horizontal forces involved. 78. Picture the Problem: An elevator accelerates up and down, changing your apparent weight Wa. A free-body diagram of the situation is depicted at right. Strategy: There are two forces acting on you: the applied force F = Wa of the scale acting upward and the force of gravity W acting downward. The force Wa represents your apparent weight because it is both the force the scale exerts on you and the force you exert on the scale. Use Newton’s second law together with the known force Wa to determine the acceleration a. Solution: 1. (a) The direction of acceleration is upward. An upward acceleration results in an apparent weight greater than the actual weight. 2. (b) Use Newton’s second law together with the known forces to determine the acceleration a. ΣF y = Wa − W = ma a= Wa − W Wa − W 730 − 610 N = = 9.81 m/s 2 = 1.9 m/s 2 m W g 610 N ( ) Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 14 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 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 traveling upward, or slowing down if it is traveling 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. 79. Picture the Problem: A spring is stretched by an applied force. Strategy: Use Hooke’s law to find the required force. F = k x = (55 N/m )( 0.12 m ) = 6.6 N Solution: Use Hooke’s law to find F: Insight: A 6.6 N force is equivalent to about 1.5 lb, so this spring is fairly easy to stretch. 80. Picture the Problem: A spring is compressed by an applied force. Strategy: Solve Hooke’s law to find the compression distance. F =kx Solution: Solve Hooke’s law for x: x= F 4.3 N = = 0.061 m = 6.1 cm k 71 N/m Insight: A 4.3 N force is equivalent to about 1.0 lb. Note that Hooke’s law applies to both the stretching (as in the previous question) and the compressing of a spring. 81. Picture the Problem: A spring is stretched by an applied force. Strategy: Use Hooke’s law to find the spring constant. F =kx Solution: Solve Hooke’s law for k: k= F 4.8 N = = 51 N/m x 0.095 m Insight: The 4.8 N force is equivalent to about 1.1 lb and it stretches the spring by 3.7 in. 82. Picture the Problem: A thread from a spider web acts like a spring and stretches from the weight of the spider. Strategy: Use Hooke’s law to find the stretch distance, taking the force to be the spider’s weight mg. Solution: Solve Hooke’s law for x: Insight: The spider’s weight is very small, about F =kx ( ) 2 F mg ( 0.00026 kg ) 9.81 m/s x= = = = 0.36 mm k k 7.1 N/m 1 oz, but it still stretches the spider web thread by a tiny amount. 100 83. Picture the Problem: A steel wire is stretched by an applied force. Strategy: Use Hooke’s law to find the spring constant. Solution: Solve Hooke’s law for k: F =kx k= F 360 N = = 3.3 × 105 N/m = 0.33 MN/m x 0.0011 m Insight: The tendency for materials like steel to stretch when subjected to a tension is often characterized by a coefficient called Young’s modulus, but the behavior is very much like a spring constant, and the spring constant model works just fine for the situation described in this problem. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 15 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 84. Picture the Problem: A child pulls a sled along an icy surface that is slippery enough that friction can be ignored. The pulling force is inclined above the direction of motion. Strategy: Only the horizontal component of the force will accelerate the sled. Write Newton’s second law along the horizontal direction to find the acceleration of the sled, and then use the acceleration to find the velocity. Solution: 1. Find the horizontal component of force: Fx = F cos θ = ( 6.20 N ) cos (35.0°) = 5.08 N 2. Use Newton’s second law together with the horizontal force to determine the acceleration a. ΣF 3. Use the velocity-time equation to find vf: vf = vi + a t = 0 + 1.10 m/s 2 (1.15 s ) = 1.27 m/s x = Fx = ma a= ( Fx 5.08 N = = 1.10 m/s 2 m 4.60 kg ) Insight: If there were substantial friction the problem would become much more difficult. The vertical component of the pulling force reduces the normal force and thus changes the amount of friction f k = μk N which, in turn, affects the horizontal motion. 85. 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 x direction (parallel to a ) in order to find the acceleration of the sled. Solution: Write Newton’s second law in the x direction: ΣF x = 2 F cos 35° − 57 N = ( msled + mchild ) ax ax = 2 F cos 35° − 57 N msled + mchild ax = 2 (55 N ) cos 35° − 57 N = 1.5 m/s 2 19 + 3.7 kg Insight: Some of the force exerted by the teenagers is exerted in the y direction and cancels out. Only the x components of the forces accelerate the sled. 86. Picture the Problem: A ship accelerates up and down as it is carried by the waves. Meanwhile, a crew member on the ship stands on a bathroom scale. As the ship accelerates the apparent weight of the crew member changes. Strategy: There are two forces acting on the crew member: the applied force F of the scale acting upward and the force of gravity W acting downward. The force F represents the apparent weight of the crew member because that is both the force the scale exerts on the crew member and the force the crew member exerts on the scale. Use Newton’s second law together with the known force F acceleration to determine the acceleration a. The maximum upward acceleration corresponds to the largest apparent weight. Solution: 1. (a) Use Newton’s Second Law together with the known forces to determine the acceleration a. 2. (b) Repeat (a) for the smallest apparent weight: ΣF y = F -W = ma a= a= F -W F – W æç 910 N – 750 N ö÷ 2 2 = =ç ÷÷ø(9.81 m/s ) = 2.1 m/s çè 750 N m W g æ 560 N – 750 N ö÷ F -W 2 2 g = çç ÷÷(9.81 m/s ) = -2.5 m/s çè ø W 750 N Insight: A similar effect occurs when you stand in an elevator and are accelerated either upward or downward. Apparent weight can be observed in any frame of reference that is accelerating. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 16 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 87. Picture the Problem: Two astronauts push on a satellite to accelerate it. The freebody diagram for the satellite is shown at right. Strategy: The force F1 from one astronaut adds as a vector to the force F2 from the second astronaut to produce the net force Fnet . Use a ruler and a protractor to add the vectors head-to-tail, preserving their length and direction. Then use the magnitude of Fnet to find the mass of the satellite by applying Newton’s second law. Solution: 1. (a) Use a ruler and a protractor to construct a vector diagram similar to the one shown at the right. Measure the length of Fnet to scale to find its magnitude is around 100 N. Measure the angle θto find it is about 35°. The net force on the satellite is thus 100 N at 35° above the positive x axis. 2. (b) Apply Newton’s second law to find the mass of the satellite: ΣF F2 Fnet = Fnet = ma m= Fnet 100 N = = 830 kg a 0.12 m/s 2 θ F1 Insight: A careful addition of vector components along the x axis and y axis reveals that the magnitude of the vector sum is 100.8 N at 33.5° above the x axis. 88. Picture the Problem: Three astronauts push on a 680-kg satellite to accelerate it. The free-body diagram for the satellite is shown at right. Strategy: The force F1 from one astronaut adds as a vector to the forces F2 and F3 from the second and third astronauts, respectively, to produce the net force Fnet . Use a ruler and a protractor to add the vectors head-to-tail, preserving their length and direction. Then use the magnitude of Fnet to find the acceleration of the satellite by applying Newton’s second law. Solution: 1. (a) Use a ruler and a protractor to construct a vector diagram similar to the one shown at the right. Measure the length of Fnet to scale to find its magnitude is around 90 N. Measure the angle θto find it is around 20°. The net force on the satellite is thus 90 N at 20° above the positive x axis. 2. (b) Apply Newton’s second law to find the acceleration of the satellite: ΣF F3 Fnet 63° θ F1 F2 = Fnet = ma a= Fnet 90 N = = 0.13 m/s 2 m 680 kg Insight: A careful addition of vector components along the x axis and y axis reveals that the magnitude of the vector sum is 91.0 N at 19.9° above the x axis. 89. Picture the Problem: The figure at right depicts the normal force on a suitcase as a function of the angle θthat the pulling force makes with the horizontal. Strategy: The curves in the figure provide specific angles at which the normal force is either zero or mg 2 . Use Newton’s second law and the given information to determine the magnitude of the applied force F in each case. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 17 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker Solution: 1. (a) Write out Newton’s second law in the vertical direction and solve for F: ΣF 2. Substitute N = mg 2 for the case θ= 90°: F= mg − mg 2 = sin 90° 3. (b) Substitute N = 0 for the case θ= 90°: F= mg − 0 = mg sin 90° 4. (c) Substitute N = 0 for the case θ= 30°: F= mg − 0 = 2mg sin 30° y = F sin θ + N − mg = 0 1 2 F mg − N sin θ mg Insight: As the applied force F becomes larger, the normal force becomes smaller because the applied force supports more of the weight of the suitcase. If an applied force F = 2mg is exerted at an angle greater than 30°, the suitcase will accelerate upward. 90. (a) Kinetic and static friction are both forces that are proportional to the normal force between the two surfaces involved. The two forces both oppose the direction of motion (or the direction of the pulling force in the case of static friction). (b) Kinetic and static friction are different because kinetic friction has a constant value, but the static friction force ranges from zero to a maximum value. When the maximum value is exceeded the object begins to slide. 91. A stuck car needs friction between the tires and the icy road in order to get unstuck. One way to accomplish this is to increase the normal force between the tires and the road. Because the normal force depends upon the amount of weight on the car’s axle, sitting on the trunk of the car will increase the weight on the rear axle and therefore the normal force and the friction force will each increase. 92. 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. 93. Picture the Problem: A box remains at rest in each situation depicted in the figure. Strategy: The magnitude of the static friction force is equal and opposite to the magnitude of the horizontal component of the net force exerted upon the box. This is due to Newton’s second law and the fact that the box is not accelerating; the net force on the box must be zero. Solution: Box 1 has the largest horizontal component of force acting upon it, so the static friction must be largest in that case. Box 3 has the smallest horizontal component of force, so static friction must be smallest in that case. We conclude that the ranking of the static friction forces is box 3 < box 2 < box 1. Insight: The maximum static friction force depends upon the coefficient of static friction and the magnitude of the normal force. If we were to rank the maximum static friction forces we would note that box 2 has the largest magnitude normal force (due to F1 and its weight) and boxes 1 and 3 have similar normal forces, so box 1 = box 3 < box 2. 94. Picture the Problem: Two bricks are pushed across a rough surface. In case 1 they are placed end-to-end and in case 2 they are stacked one on top of the other. Strategy: Use the mathematical relationship between the force of kinetic friction and the normal force to answer the conceptual question. Solution: 1. (a) The force of kinetic friction is proportional to the normal force between the bricks and the tabletop. In case 2 the normal force will be twice the normal force of a single brick, but in case 1 there are two bricks and therefore Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 18 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker two friction forces. Another way to view the situation is to say there are two bricks of the same mass but different shapes. However, the friction force does not depend upon shape, only upon the magnitude of the normal force. We conclude that the force of kinetic friction in case 1 is equal to the force of kinetic friction in case 2. 2. (b) The best explanation is B. The normal force is the same in the two cases, and friction is independent of surface area. Statements A and C are both false. Insight: If we were to stand the bricks up on their small ends, the friction forces would still remain the same. 95. Picture the Problem: Two drivers stop suddenly to avoid a deer. Driver 1 stops by locking up his brakes and screeching to a halt; driver 2 stops by applying her brakes just to the verge of locking, so that the wheels continue to turn until her car comes to a complete stop. Strategy: Use Newton’s second law together with the difference between kinetic and static friction to answer the conceptual question. Solution: 1. (a) Driver 1 stops his car using kinetic friction as the car tires skid, but driver 2 uses static friction because the point of contact between a rolling tire and the road is instantaneously at rest. The maximum magnitude of the static friction force is larger than the magnitude of the kinetic friction force, so driver 2 has a larger braking force to utilize and experiences a larger braking acceleration. We conclude that the stopping distance of driver 1 is greater than the stopping distance of driver 2 2. (b) The best explanation is C. Locked-up brakes result in sliding (kinetic) friction, which is less than rolling (static) friction. Statements A and B are both false. Insight: Race cars have specialized tires with very high coefficients of friction to increase the amount of force they can generate for rounding turns at high speeds. They also greatly decrease the braking distance. 96. Picture the Problem: A baseball player slides in a straight line and comes to rest due to the frictional force. Strategy: Find the acceleration of the player using Newton’s second law. Solution: Find the acceleration of the player using Newton’s second law: ΣF x = f k = ma a= fk μ mg =− k = − μ k g = − ( 0.46) 9.81 m/s 2 = − 4.5 m/s 2 m m ( ) Insight: Note that the acceleration is independent of the baseball player’s mass and his initial speed. It depends only on the coefficient of kinetic friction and the acceleration of gravity. The negative sign indicates the acceleration is opposite the direction of motion. 97. Picture the Problem: A baseball player slides in a straight line and comes to rest due to the frictional force. Strategy: Find the acceleration of the player using Newton’s second law. Solution: 1. (a) The acceleration of the player, as found in the previous question, is independent of the player’s mass. We conclude that if the mass of the player is doubled, the player’s acceleration while sliding will stay the same. 2. (b) Find the acceleration of the player using Newton’s second law: ΣF x = f k = ma a= μ mg fk =− k = − μ k g = − ( 0.46) 9.81 m/s 2 = − 4.5 m/s 2 m m ( ) Insight: Note that the acceleration is independent of both the baseball player’s mass and his initial speed. It depends only on the coefficient of kinetic friction and the acceleration of gravity. 98. Picture the Problem: A baseball player slides in a straight line and comes to rest due to the frictional force. Strategy: Find the acceleration of the player using Newton’s second law. Solution: 1. (a) The acceleration of the player, as found in the previous question, is independent of the player’s initial velocity. We conclude that if the speed of the player is doubled, the acceleration while sliding will stay the same. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 19 Chapter 5: Newton’s Laws of Motion 2. (b) Find the acceleration of the player using Newton’s second law: Pearson Physics by James S. Walker ΣF x = f k = ma a= μ mg fk =− k = − μ k g = − ( 0.46) 9.81 m/s 2 = − 4.5 m/s 2 m m ( ) Insight: Note that the acceleration is independent of both the baseball player’s mass and his initial speed. It depends only on the coefficient of kinetic friction and the acceleration of gravity. 99. Picture the Problem: A car 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 car needed to accelerate it at 12 m/s2. Use Newton’s second law to find the minimum coefficient of static friction. Solution: 1. Write Newton’s second law for the car: ΣF = fs = μs mg = ma 2. Solve the equation for µs: μs = a 12 m/s 2 = = 1.2 g 9.81 m/s 2 x 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. 100. Picture the Problem: A sprinter accelerates in a straight line due to the static friction between her shoes and the track. Strategy: The static friction between her shoes and the track provides the forward force needed to accelerate the sprinter. First find her acceleration from the velocity-position equation, and then use Newton’s second law to find the minimum coefficient of static friction. Solution: 1. (a) Find the sprinter’s acceleration from the velocity-position equation: v 2 − v02 v 2 − 02 (13 m/s ) a= = = = 3.8 m/s 2 2 Δx 2 Δx 2 ( 22 m ) 2. Write Newton’s second law for the sprinter: ΣF 3. Solve the equation for µs: μs = 2 x = fs = μs mg = ma a 3.8 m/s 2 = = 0.39 g 9.81 m/s 2 Insight: A larger coefficient of friction could produce a larger acceleration, although the sprinter will likely find the limitations of the human body exceed the limitations of the shoes! 13 m/s is a very fast pace, as world class sprinters average about 10 m/s over a 100 m distance (though they must start from rest, so their final speed is faster than 10 m/s). 101. 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 maximum acceleration of the car, and then use the velocity-time equation to find the smallest amount of time in which the car can accelerate to 15 m/s. Solution: 1. (a) Write Newton’s second law for the coffee cup: ΣF 2. Solve the equation for a: a = μs g = ( 0.24) 9.81 m/s 2 = 2.35 m/s 2 = 2.4 m/s 2 x = fs = μs mg = ma ( ) v − v0 15 m/s − 0 = = 6.4 s a 2.35 m/s 2 Insight: If the person owned a Ferrari 575M Maranello capable of going from zero to 60 mi/h in 4.2 seconds (6.4 m/s2), 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! 3. (b) Use the velocity-time equation to find the minimum time: t= Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 20 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 102. Picture the Problem: The free-body diagram of the block is shown at right. Strategy: Choose coordinate axes with x pointing parallel to the incline and down, and y pointing parallel to N . Write Newton’s second law in the y and x directions and combine the equations to solve for θ. Solution: 1. Write Newton’s second law in the y direction: ΣF = N − mg cos θ = 0 2. Write Newton’s second law in the x direction: ΣF = mg sin θ − f s = ma y N = mg cos θ x ma = mg sin θ − μs N 3. Substitute the expression for N from step 1 and solve for θ: 0 = mg sin θ − μs ( mg cos θ ) μs cos θ = sin θ θ = tan −1 ( μs ) = tan −1 ( 0.25) = 14° Insight: This approach can also be inverted to find the coefficient of friction μs = tan θ ; simply vary the incline angle and record the minimum angle required to overcome static friction and get the block sliding. 103. Picture the Problem: A book slides in a straight line across a 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 µk can be determined. Solution: 1. When the book begins sliding, the applied force equals the maximum static friction force: Fapp = fs = μs mg 2. When the book is sliding at constant speed, the applied force equals the kinetic friction force: Fapp = f k = μ k mg μs = μk = Fapp mg Fapp mg = = 2.05 N (1.92 kg ) (9.81 m/s 2 ) 1.03 N (1.92 kg ) (9.81 m/s 2 ) = 0.109 = 0.0547 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. 104. It is true that if Mr. Ed pulls on the wagon, the wagon pulls back on him with an equal and opposite force (Newton’s third law). However, these two forces do not cancel because they act on different objects; one is a force on the horse, the other is a force on the cart. If you want to know about the motion of the cart, you must consider the forces on the cart only, and ignore all other forces. 105. The whole brick has twice the inertia (it is harder to accelerate) but it also experiences twice the force due to gravity. As a result, these two effects (more inertia, more force) exactly cancel, and the free fall acceleration is independent of mass. 106. The whole brick has twice the force of gravity on it, but it also has twice the inertia (it is harder to accelerate). As a result, these two effects (more force, more inertia) exactly cancel, and the free fall acceleration is independent of mass. 107. Although all objects have zero weight in orbit, they still retain their inertia, or the resistance to change in motion, due to their mass. If an astronaut pushes on two objects with the same amount of force, the more massive object will have a smaller acceleration. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 21 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 108. Picture the Problem: An elevator accelerates up and down, changing your apparent weight Wa. A free-body diagram of the situation is depicted at right. Strategy: There are two forces acting on you: the applied force F = Wa of the scale acting upward and the force of gravity W acting downward. The force Wa represents your apparent weight because it is both the force the scale exerts on you and the force you exert on the scale. Use Newton’s second law together with the known force Wa acceleration to determine the acceleration a. Solution: 1. The direction of acceleration is downward. A downward acceleration results in an apparent weight less than the actual weight. 2. Use Newton’s second law together with the known forces to determine the magnitude of the acceleration a. ∑F y = Wa − W = ma a = Wa − W W −W 480 − 540 N = a = (9.81 m/s2 ) = 1.1 m/s2 540 N m W g 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. 109. Picture the Problem: Various objects either accelerate or move with constant velocity. Strategy: Consider Newton’s second law as a vector equation when answering the conceptual questions. Solution: 1. (a) “A car accelerating northward from a stoplight.” The acceleration vector points northward and therefore the net force points northward, due to the vector nature of Newton’s second law. 2. (b) “A car traveling southward and slowing down.” In this case the acceleration points opposite to the direction of motion, and so the acceleration and the net force point northward. 3. (c) “A car traveling westward with constant speed.” In this case the acceleration and the net force are zero. 4. (d) “A skydiver parachuting downward with constant speed.” In this case the acceleration and the net force are zero. 5. (e) “A baseball during its flight from pitcher to catcher (ignoring air resistance).” In this case the acceleration points downward (due to gravity) and therefore the net force points downward. Insight: The acceleration of an object must always point in the same direction as the net force on it. 110. Picture the Problem: A spring in a cushion of a chair compresses from the force applied to it. Strategy: Use Hooke’s law to find the compression distance. Solution: Solve Hooke’s law for x: F =kx x= F 51 N = = 0.088 m = 8.8 cm k 580 N/m Insight: This is a pretty weak spring, as it compresses a distance equivalent to 3.5 in when subjected to a relatively small force (51 N is the same as 11 lb). 111. Picture the Problem: A branch is bent by the weight of a chimpanzee. Strategy: Use Hooke’s law to find the spring constant. Solution: Solve Hooke’s law for k: F =kx k= ( ) 2 F mg (22 kg ) 9.81 m/s = = = 1660 N/m = 1.7 kN/m x x 0.13 m Insight: The 216 N force is equivalent to about 49 lb and it bends the branch by 5.1 inches. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 22 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 112. Each time an astronaut exerts a force on a baseball, the baseball exerts an equal and opposite force on the astronaut because of Newton’s third law. Without any friction to inhibit motion, the astronaut will move backward as the baseball goes forward toward the other astronaut. When the second astronaut catches the ball, she exerts a force on the baseball, but the ball exerts an equal and opposite force on her, accelerating her backward (in the original direction of the ball’s motion). As the game of catch progresses, the astronauts will move farther and farther apart and with increasing speed. 113. Picture the Problem: A parachutist is moving straight downward at 3.85 m/s, slowing down and coming to rest over a distance of 0.750 m as she lands on the ground. Strategy: Use velocity-position equation to determine the acceleration of the parachutist, and then use Newton’s second law to find the net force on her. v y2 = v02y + 2a Δy Solution: 1. (a) Use the velocity-position equation to find the acceleration of the parachutist: 2 2 v y − v0 y 0 − ( −3.85 m/s ) a= = = 9.88 m/s 2 upward 2 Δy 2(0 − 0.750 m) Fground − mg = ma ∑ F = Fground + W = ma 2 2. Use Newton’s second law to find the force exerted by the ground: Fground = ma + mg = (42.0 kg) (9.88 m/s 2 + 9.81 m/s 2 ) = 827 N upward 3. (b) If the parachutist comes to rest in a shorter distance, the acceleration a will be greater than 9.88 m/s2 and the force exerted by the ground will therefore be greater than that in part (a). Insight: There solution assumes the parachute lines are slack, so that the parachute exerts no force during the landing. 114. The force the road exerts on the wheels of the drag-racing car is a reaction to the force the wheels exert on the road. The engine turns the wheels and provides the necessary force on the road. 115. Picture the Problem: A samara (the fruit of a maple tree) falls from a tree at constant speed. Strategy: Because the samara is falling with constant speed, we conclude that the net force on it is zero. Use Newton’s second law to determine the force the air exerts on the fruit. Solution: 1. (a) Use Newton’s second law ∑ F = Fair + W = 0 to find the force of air resistance: Fair = − W = − (− mg ) = (0.00121 kg ) 9.81 m/s 2 = 0.0119 N ( ) 2. (b) If the constant speed of descent is greater than 1.1 m/s, the acceleration still remains zero, so that the force of air friction remains the same as in part (a). Insight: The force of air friction depends upon speed and the shape of the object that is passing through the air. A fruit with a more streamlined shape will fall at a greater and greater speed until Fair increases to 0.0119 N and the net force becomes zero once again. At that point the fruit is said to have achieved its terminal velocity. 116. Picture the Problem: The tension in a helicopter cable lifts two men straight upward. Strategy: Write Newton’s second law in the vertical direction for the men and solve for the cable tension. Solution: 1. (a) The tension in the cable is greater than the combined weight of the men because it must not only support the weight of the men but also accelerate them upward. 2. (b) Write Newton’s second law in the y direction and solve for T: ∑F y = T − mg = ma ( ) T = m (a + g ) = (172 kg ) 1.10 + 9.81 m/s 2 = 1880 N = 1.88 kN Insight: The tension is larger than the 1.69 kN weight of the men. To accelerate them at 9.81 m/s2 upward, the cable must exert a force equal to twice the weight of the men, or about 3.37 kN. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 23 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 117. Picture the Problem: The force of the gecko’s foot pad acts straight upward in order to support the animal’s weight. Strategy: Use the definition of weight to find the number of geckos that could be suspended by a single foot. Then use an estimate of 700 N for a person’s weight and a foot size of 25 cm × 8 cm to find an estimate of the force per square centimeter the human body exerts on the soles of the feet. F = nW = nmg Solution: 1. (a) Use the definition of weight to find the number of geckos that F 11 N n= = = 4.5 geckos or 4 geckos could be suspended from one gecko foot: mg ( 0.250 kg ) 9.81 m/s 2 ( 2. (b) Find the force per square centimeter exerted on the soles of a 700-N person’s two feet: ) F 700 N = = 1.8 N/cm 2 A 2 × ( 25 cm × 8 cm ) Insight: If humans had feet like geckos, they could walk on ceilings, too, and even hang by one foot! In fact, a single human-sized gecko foot could support the weight of three humans that each weigh 700 N. 118. Picture the Problem: The two workers pull on a raft in the directions indicated by the figure at the right. Strategy: Place the x-axis 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: ∑F y = − (130 N ) sin 34° + F sin 45° = 0 F = (130 N ) sin 34° = 100 N = 0.10 kN sin 45° Insight: The second worker doesn’t have to pull as hard as the first because a larger component of that force is pulling in the y direction. However, the second worker’s force in the forward direction (73 N) is not as large as the first worker’s force (110 N). 119. Picture the Problem: A hot air balloon is floating a few feet above the ground with zero net force on it. Then a late arrival climbs in and the balloon accelerates downward. Strategy: Apply Newton’s second law to find the mass of the late arrival. Since the balloon was initially neutrally buoyant, we know that the buoyant force upward equals the mass of the balloon and other occupants, and those forces cancel. Therefore the weight of the late arrival is the only unbalanced force that produces the acceleration of the balloon. Let downward be the positive direction. Solution: Write Newton’s second law in the vertical direction for the balloon, and solve for the mass of the late arrival mLA : ∑F y = Wlate arrival = mLA g = (M + mLA ) a mLA ( g − a ) = M a mLA = ( ) 2 M a (1220 kg ) 0.56 m/s = = 74 kg g −a 9.81 − 0.56 m/s 2 Insight: Because the mass of the basket and occupants has changed by 74/1220 = 6.1%, the buoyant force created by the hot air of the balloon must be increased by more than 6.1% in order to accelerate the balloon upward. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 24 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 120. Picture the Problem: You push a box horizontally across a rough surface. Strategy: Write Newton’s second law in the horizontal direction for the box in each of the two cases. There will be two equations and two unknowns, the mass and μ k . Subtract the two equations to solve for the unknowns. Solution: 1. (a) Write Newton’s second law for each case specified. The friction force is f = μk N = μ k mg . Subtract the first equation from the second, and solve for the mass: ∑F ∑F x x ( ) = − μ mg + 81 N = m (0.75 m/s ) 0 + (81 − 75 N ) = m (0.75 − 0.50 m/s ) = − μ k mg + 75 N = m 0.50 m/s 2 2 k 2 6N = m = 24 kg 0.25 m/s 2 2. (b) Now solve the first equation for μ k : ( 75 N − m 0.50 m/s 2 mg )=μ k = ( 75 N − ( 24 kg ) 0.50 m/s 2 ( 24 kg ) (9.81 m/s 2 ) ) = 0.27 Insight: We bent the rules for significant figures a bit in step one, where the subtraction of 81 − 75 N = 6 N leaves us with only one significant figure, but we kept two in order to avoid rounding error in step 2. 121. Answers will vary. Students should research friction and prepare a report that explores the advantages and disadvantages of friction. Friction is useful when you do not want two surfaces to slide relative to each other, such as when a tire helps a car navigate a turn, or when you want a child to stay in a snow sled when you pull it. Friction should be minimized when you want to surfaces to slide freely, such as when a snow sled travels down a hill or when a hockey puck slides across an ice rink. 122. Answers will vary. Students should research static and kinetic friction and prepare a report that evaluates the coefficients for a variety of different surfaces. You will find that there are not any cases where the coefficient of static friction is less than the coefficient of kinetic friction, but there are some cases where they equal. 123. According to Newton’s second law, the net force on an object determines its acceleration, not its speed. Acceleration is the rate of change of speed, and is independent of speed. When you push on a heavy box, the box pushes back on you with an equal and opposite force, consistent with Newton’s third law. 124. Picture the Problem: The driver moves forward with the initial speed of the car and then comes to rest when acted upon by the force of the seat belt. Strategy: Find the acceleration of the driver using the velocity-position equation and then use Newton’s second law to find the net force on the driver. If we take the initial velocity v i to be in the positive direction, the acceleration and force will each be negative. The approximate value can also simply be read from the provided graph. Solution: 1. Use the velocity-position equation to find the acceleration of the driver: v2 v 2 − vi2 02 − vi2 a= = =− i 2 Δx 2 Δx 2 Δx 2. Apply Newton’s second law to find the net force. The calculated answer is choice C. 2 65.0 kg )(18.0 m/s ) mvi2 ( = −1.05 × 104 N ∑ F = ma = − 2 Δx = − 2 (1.00 m ) Insight: The negative sign means that the force and the acceleration act in the backward direction. The stopping force for this 18.0 m/s (40.3 mi/h) crash is 1.2 tons distributed over the contact area between the seat belt and the driver’s body. Car crashes are dangerous! Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 25 Chapter 5: Newton’s Laws of Motion Pearson Physics by James S. Walker 125. Picture the Problem: The driver moves forward with the initial speed of the car and then comes to rest when acted upon by the force of the hard steering wheel. Strategy: Find the acceleration of the driver using the velocity-position equation and then use Newton’s second law to find the net force on the driver. If we take the initial velocity v i to be in the positive direction, the acceleration and force will each be negative. The approximate value can also simply be read from the provided graph. Solution: 1. Use the velocity-position equation to find the acceleration of the driver: v2 v 2 − vi2 02 − vi2 = =− i a= 2 Δx 2 Δx 2 Δx 2. Apply Newton’s second law to find the net force. The calculated answer is choice C. mvi2 (65.0 kg )(18.0 m/s)2 = −2.11 × 105 N = m = − = − F a ∑ 2 Δx 2 (0.0500 m ) Insight: The negative sign means that the force and the acceleration act in the backward direction. The safety devices in the previous problem cut the stopping force (and the injury risk) by a factor of 20. 126. Picture the Problem: The driver moves forward with the initial speed of the car and then comes to rest when acted upon by the force of the seat belt. Strategy: Find the acceleration of the driver using the velocity-position equation and then use Newton’s second law to find the net force on the driver. If we take the initial velocity v i to be in the positive direction, the acceleration and force will each be negative. The approximate value can also simply be read from the provided graph. Solution: 1. Use the velocity-position equation to find the acceleration of the driver: v2 v 2 − vi2 02 − vi2 = =− i a= 2 Δx 2 Δx 2 Δx 2. Apply Newton’s second law to find the net force. The calculated answer is choice D. 2 65.0 kg )(36.0 m/s ) mvi2 ( = − 4.21 × 104 N ∑ F = ma = − 2 Δx = − 2 (1.00 m ) Insight: The stopping force for this 36.0 m/s (80.5 mi/h) crash is 4.7 tons distributed over the contact area between the seat belt and the driver’s body. This is four times more force than endured in the 18.0-m/s crash of the first problem presented in this series. Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 – 26