Quick QuizzesAnswers to Even Numbered Conceptual QuestionsAnswers to Even Numbered ProblemsProblem SolutionsChapter 5 Energy Quick Quizzes 1. (c). The work done by the force is ()cosWFxθ=∆ , where θ is the angle between the direction of the force and the direction of the displacement (positive x-direction). Thus, the work has its largest positive value in (c) where 0θ=° , the work done in (a) is zero since 90θ=°, the work done in (d) is negative since 90 270θ°< < °, and the work done is most negative in (b) where 180θ=°. 2. (d). All three balls have the same speed the moment they hit the ground because all start with the same kinetic energy and undergo the same change in gravitational potential energy. 3. (c). They both start from rest, so the initial kinetic energy is zero for each of them. They have the same mass and start from the same height, so they have the same initial potential energy. Since neither spends energy overcoming friction, all of their original potential energy will be converted into kinetic energy as they move downward. Thus, they will have equal kinetic energies when they reach the ground. 4. (c). The decrease in mechanical energy of the system is fk∆x. This has a smaller value on the tilted surface for two reasons: (1) the force of kinetic friction fk is smaller because the normal force is smaller, and (2) the displacement ∆x is smaller because a component of the gravitational force is pulling on the book in the direction opposite to its velocity. 143144 CHAPTER 5 Answers to Even Numbered Conceptual Questions 2. (a) The chicken does positive work on the ground. (b) No work is done. (c) The crane does positive work on the bucket. (d) The force of gravity does negative work on the bucket. (e) The leg muscles do negative work on the individual. 4. (a) Kinetic energy is always positive. Mass and speed squared are both positive. (b) Gravitational potential energy can be negative when the object is lower than the chosen reference level. 6. (a) Kinetic energy is proportional to the speed squared. Doubling the speed makes the object’s kinetic energy four times larger. (b) If the total work done on an object in some process is zero, its speed must be the same at the final point as it was at the initial point. 8. The total energy of the bowling ball is conserved. Because the ball initially has gravitational potential energy mgh and no kinetic energy, it will again have zero kinetic energy when it returns to its original position. Air resistance and friction at the support will cause the ball to come back to a point slightly below its initial position. On the other hand, if anyone gives a forward push to the ball anywhere along its path, the demonstrator will have to duck. 10. (a) The effects are the same except for such features as having to overcome air resistance outside. (b) The person must lift his body slightly with each step on the tilted treadmill. Thus, the effect is that of running uphill. 12. Both the force of kinetic friction exerted on the sled by the snow and the resistance force exerted on the moving sled by the air will do negative work on the sled. Since the sled is maintaining constant velocity, some towing agent must do an equal amount of positive work, so the net work done on the sled is zero. 14. The kinetic energy is converted to internal energy within the brake pads of the car, the roadway, and the tires. 16. Work is actually performed by the thigh bone (the femur) on the hips as the torso moves upwards a distance h. The force on the torso G is approximately the same as the normal force (since the legs are relatively light and are not moving much), and the work done by G minus the work done by gravity is equal to the change in kinetic energy of the torso. torsoFotorsFhFtorsourvurFtorsourEnergy 145 At full extension the torso would continue upwards, leaving the legs behind on the ground (!), except that the torso now does work on the legs, increasing their speed (and decreasing the torso speed) so that both move upwards together. Note: An alternative way to think about problems that involve internal motions of an object is to note that the net work done on an object is equal to the net force times the displacement of the center of mass. Using this idea, the effect of throwing the arms upwards during the extension phase is accounted for by noting that the position of the center of mass is higher on the body with the arms extended, so that total displacement of the center of mass is greater. 18. The normal force is always perpendicular to the surface and the motion is generally parallel to the surface. Thus, in most circumstances, the normal force is perpendicular to the displacement and does no work. The force of static friction does no work because there is no displacement of the object relative to the surface in a static situation. 20. Before the jump, the system (vaulter plus pole) has kinetic energy. After the vaulter leaves the ground and the pole is bent, the system has less kinetic energy, but the gravitational potential energy of the system increases and some energy is stored as elastic potential energy in the bent pole. When the vaulter reaches the top of the vault, the kinetic energy is at its minimum, the gravitational potential energy is at its maximum, and no energy is stored in the pole, which is now straight. Some mechanical energy is lost due to air resistance and the frictional force between the pole and the ground during the ascent.146 CHAPTER 5 Answers to Even Numbered Problems 2. 30.6 m 4. 1.6 kJ 6. (a) 900 J (b) 0.383 8. (a) 31.9 J (b) 0 (c) 0 (d) 31.9 J 10. 160 m s 12. (a) –168 J (b) –184 J (c) 500 J (d) 148 J (e) 5.64 m s 14. 90.0 J 16. (a) 1.2 J (b) 5.0 m s (c) 6.3 J 18. 2.0 m 20. 0.5 m 22. (a) 0.768 m (b) 51.68 10 J×24. 26.5 m s 26. 5.1 m 28. (a) 9.90 m s (b) 7.67 m s 30. (a) 5.94 m s, 7.67 m sBCvv== (b) 147 J 32. 5.11 m s 34. 61 m 36. (a) 9.90 m s (b) 11.8 J 38. (a) No, ()sinkkgFfmµθ=−( (b) )cos , skFfWFx W inkmgF xθµθ−−16.9 JkfW=−==4.23 N, 47.9 J, kFfW== (c) normal force, gravitational force, and vertical component of applied force (d) 40. 77 m sEnergy 147 42. (a) 2.29 m s (b) 15.6 J 44. ()24sin 15hθ+ 46. 1.5 m (measured along the incline) 48. (a) 21 kJ (b) 0.92 hp 50. 2.9 m s 52. 194 m 54. (a) 7.92 hp (b) 14.9 hp 56. (a) 7.50 J (b) 15.0 J (c) 7.50 J (d)