Chapter 7: Work and Kinetic EnergyAnswers to Even-Numbered Conceptual Questions2. False. Any force acting on an object can do work. The work done by different forces may add to produce agreater net work, or they may cancel to some extent. It follows that the net work done on an object can bethought of in the following two equivalent ways: (i) The sum of the works done by each individual force; or(ii) the work done by the net force.4. If the net work done on an object is zero, it follows that its change in kinetic energy is also zero. Therefore, itsspeed remains the same.6. Frictional forces do negative work whenever they act in a direction that opposes the motion. For example,friction does negative work when you push a box across the floor, or when you stop your car.8. The fact that the ski boat’s velocity is constant means that its kinetic energy is also constant. Therefore, thenet work done on the boat is zero. It follows that the net force acting on the boat does no work. (In fact, thenet force acting on the boat is zero, since its velocity is constant.)10. No. What we can conclude, however, is that the net force acting on the object is zero.12. No. Power depends both on the amount of work done by the engine, and the amount of time during whichthe work is performed. For example, if engine 2 does its work in less than half the time of engine 1, it canproduce more power even if the amount of work the engines do is the same.Solutions to Problems and Conceptual Exercises1. Picture the Problem: The International Space Station orbits the Earth in an approximately circular orbit at a height of above the Earth’s surface.Strategy: According to equation 7-3, the work done on an object is positive if the force and the displacement are along the same direction, but zero if the force is perpendicular to the displacement.Solution: The gravitational force exerted by the Earth is radial (toward the center of the circular orbit), but the displacement is tangential, perpendicular to the force. We conclude that the work done by the Earth on the space stationis zero.Insight: If the Earth were to do any work on the space station, its energy would change. Of course the rockets that lifted the parts of the space station into space did do work on them, increasing their energy sufficiently to achieve orbit.2. Picture the Problem: A pendulum bob swings from point I to point II along the circular arc indicated in the figure at right.Strategy: Apply equation 7-3, which says that the work done on an object is positive if the force and the displacement are along the same direction, but zero if the force is perpendicular to the displacement.Solution: 1. (a) As the pendulum bob swings from point I to point II, the force of gravity points downward and a component of the displacement is downward. Therefore, the work done on the bob by gravity is positive.2. (b) As the pendulum bob swings, the force exerted by the string is radial (toward the pivot point) but the displacementis tangential, perpendicular to the force. We conclude that the work done on the bob by the string is zero.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.7 – 1Chapter 7: Work and Kinetic Energy James S. Walker, Physics, 4th EditionInsight: The work done by the Earth is negative if the bob swings from point II to point III because a component of the displacement is upward but the force is downward.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.7 – 2Chapter 7: Work and Kinetic Energy James S. Walker, Physics, 4th Edition3. Picture the Problem: A pendulum bob swings from point II to point III along the circular arc indicated in the figure at right.Strategy: Apply equation 7-3, which says that the work done on an object is positive if the force and the displacement are along the same direction, but zero if the force is perpendicular to the displacement.Solution: 1. (a) As the pendulum bob swings from point II to point III, the force of gravity points downward and a component of the displacement is upward. Therefore, the work done on the bob by gravity is negative.2. (b) As the pendulum bob swings, the force exerted by the string is radial (toward the pivot point) but the displacementis tangential, perpendicular to the force. We conclude that the work done on the bob by the string is zero.Insight: The work done by the Earth is positive if the bob swings from point I to point II because a component of the displacement is downward and the force is downward.4. Picture the Problem: The farmhand pushes the hay horizontally.Strategy: Multiply the force by the distance because in this case the two point along the same direction.Solution: Apply equation 7-1 directly: W =Fd = 88 N( )3.9 m( )=340 J = 0.34 kJInsight: The 26-kg mass is unneeded information unless we needed to know the amount of friction or the acceleration of the bale.5. Picture the Problem: The children lift the bucket vertically.Strategy: Multiply the force by the distance because in this case the two point along the same direction.Solution: Apply equation 7-1: W =Fd =mgdNow solve for m: m =Wgd=201 J9.81 m/s2( )4.70 m( )= 4.36 kgInsight: The applied force equals the weight as long as the bucket does not accelerate.6. Picture the Problem: The pumpkin is lifted vertically then carried horizontally.Strategy: Multiply the force by the distance because during the lift the two point along the same direction.Solution: 1. (a) Apply equation 7-1 directly: W =mgd= 3.2 kg( )9.81 m/s2( )1.2 m( )= 38 J2. (b) The force is perpendicular to the displacement so W = 0.Insight: You can still get tired carrying a pumpkin horizontally even though you’re doing no work!7. Picture the Problem: The suitcase is pushed horizontally.Strategy: Determine the applied force and solve for d.Solution: Solve equation 7-1 for d: d =Wfk=Wμkmg=642 J0.272( )71.5 kg( )9.81 m/s2( )= 3.37 mInsight: The applied force equals the friction force as long as the suitcase does not accelerate.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is