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07_InstSolManual_PDF-1

# 07_InstSolManual_PDF-1 - WORK AND ENERGY 7 6 C 7 A D 8 B C...

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7-1 W ORK AND E NERGY 7 Answers to Multiple Choice Problems 1. B 2. D 3. A 4. C 5. C 6. C 7. A, D 8. B, C, E 9. A, D, E 10. B, D 11. B, D 12. B, D 13. B, D 14. A 15. C Solutions to Problems 7.1. Set Up: Assume the fisherman is holding the pole straight out in front of him so that the pole and fishing line are roughly parallel to the water. Thus may be used in the relation Solve: Reflect: Because all of the force that the fisherman applies through the line is used to perform work on the fish. 7.2. Set Up: Use On the way up, the displacement is upward and the gravity force is downward, so On the way down, both the displacement and force are downward, so Solve: On the way up: On the way down: Reflect: When the force and displacement are in opposite directions, the work done is negative. 7.3. Set Up: Use with Solve: Reflect: Since the relatively small angle of allows the boat to apply approximately 97% of the 180 N force to pulling the skier. 7.4 Set Up: Use In part (a), In part (b), solve for Solve: (a) (b) and Reflect: More work is done in part (a) where In part (b), because the box is pulled at an angle other than zero, and less work is done even though the force and displacement are the same. 7.5. Set Up: For parts (a) through (d), identify the appropriate value of and use the relation In part (e), apply the relation Solve: (a) Since you are applying a horizontal force, Thus, (b) The friction force acts in the horizontal direction, opposite to the motion, so (c) Since the normal force acts upward and perpendicular to the tabletop, W n 5 1 n cos f 2 s 5 1 ns 21 cos 90 ° 2 5 0.0 J f 5 90 ° . W f 5 1 F f cos f 2 s 5 1 0.600 N 21 cos 180 ° 21 1.50 m 2 5 2 0.900 J f 5 180 ° . W student 5 1 2.40 N 21 cos 0 ° 21 1.50 m 2 5 3.60 J f 5 0 ° . W net 5 W student 1 W grav 1 W n 1 W f . 1 F cos f 2 s . W 5 F i s 5 f F i , F F i 5 F . f 5 63.9 ° cos f 5 W / 1 Fs 2 5 65.0 J / 31 8.50 N 21 17.4 m 24 5 0.439 W 5 1 F cos f 2 s 5 1 8.50 N 21 cos 0 ° 21 17.4 m 2 5 148 J f . f 5 0 ° . W 5 F i s 5 1 F cos f 2 s . 15.0 ° cos 15.0 ° < 0.97, W 5 1 F cos f 2 s 5 1 180 N 21 cos 15.0 ° 21 300.0 m 2 5 5.22 3 10 4 J f 5 15.0 ° . W 5 F i s 5 1 F cos f 2 s W 5 1 mg cos 0 ° 2 s 5 1 5.80 3 10 2 2 kg 21 9.80 m / s 2 21 cos 0 ° 21 6.17 m 2 5 3.51 J W 5 1 mg cos 180 ° 2 s 5 1 5.80 3 10 2 2 kg 21 9.80 m / s 2 21 cos 180 ° 21 6.17 m 2 5 2 3.51 J f 5 0 ° . f 5 180 ° . W 5 F i s 5 1 F cos f 2 s . f 5 0 ° , W 5 F i s 5 1 F cos f 2 s 5 31 25.0 N 21 cos 0 ° 241 12.0 m 2 5 300 J W 5 F i s 5 1 F cos f 2 s . f 5 0 °

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(d) Since gravity acts downward and perpendicular to the tabletop, (e) Reflect: Whenever a force acts perpendicular to the direction of motion, its contribution to the net work is zero. 7.6. Set Up: Use Calculate the work done by each force. In each case, identify the angle In part (d), the net work is the algebraic sum of the work done by each force. Solve: (a) Since the force exerted by the rope and the displacement are in the same direction, and (b) Gravity is downward and the displacement is at above the horizontal, so (c) The normal force n is perpendicular to the surface of the ramp while the displacement is parallel to the surface of the ramp, so and (d) (e) Now and Reflect: In part (b), gravity does negative work since the gravity force acts downward and the carton moves upward.
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