07_InstSolManual_PDF-1 - 7-1WORK ANDENERGY7Answers to...

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Unformatted text preview: 7-1WORK ANDENERGY7Answers to Multiple Choice Problems1. B2. D3. A4. C5. C6. C7. A, D8. B, C, E9. A, D, E10. B, D11. B, D12. B, D13. B, D14. A15. CSolutions to Problems7.1. Set Up:Assume the fsherman is holding the pole straight out in Front oF him so that the pole and fshing line areroughly parallel to the water. Thus may be used in the relation Solve:Refect:Becauseall oF the Force that the fsherman applies through the line is used to perForm work on the fsh.7.2. Set Up:Use On the way up, the displacement is upward and the gravity Force isdownward, so On the way down, both the displacement and Force are downward, so Solve:On the way up:On the way down:Refect:When the Force and displacement are in opposite directions, the work done is negative.7.3. Set Up:Use with Solve:Refect:Since the relatively small angle oF allows the boat to apply approximately 97% oF the180 N Force to pulling the skier.7.4 Set Up:Use In part (a),In part (b), solve For Solve: (a)(b)and Refect: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:or 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,Wn51ncosf2s51ns2 1cos90250.0 Jf 590.Wf51Ffcosf2s510.600 N2 1cos1802 11.50 m25 20.900 Jf 5180.Wstudent512.40 N2 1cos2 11.50 m253.60 Jf 5.Wnet5Wstudent1Wgrav1Wn1Wf.1Fcosf2s.W5Fis5fFi,FFi5F.f 563.9cosf 5W/1Fs2565.0 J/3 18.50 N2 117.4 m2 450.439W51Fcosf2s518.50 N2 1cos2 117.4 m25148 Jf.f 5.W5Fis51Fcosf2s.15.0cos15.0<0.97,W51Fcosf2s51180 N2 1cos15.02 1300.0 m255.223104Jf 515.0.W5Fis51Fcosf2sW51mgcos2s515.8031022kg2 19.80 m/s22 1cos2 16.17 m253.51 JW51mgcos1802s515.8031022kg2 19.80 m/s22 1cos1802 16.17 m25 23.51 Jf 5.f 5180.W5Fis51Fcosf2s.f 5,W5Fis51Fcosf2s53 125.0 N2 1cos2 4 112.0 m25300 JW5Fis51Fcosf2s.f 5(d)Since gravity acts downward and perpendicular to the tabletop,(e)Refect: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 nis perpendicular to the surface of the ramp while the displacement is parallel to the surface ofthe ramp, so and (d)(e)Now and Refect:In part (b), gravity does negative work since the gravity force acts downward and the carton moves upward....
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07_InstSolManual_PDF-1 - 7-1WORK ANDENERGY7Answers to...

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