CH_7 - Chapter 7 CHAPTER 7 - Work and Energy 1. The...

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Chapter 7 Page 1 CHAPTER 7 - Work and Energy 1. The displacement is in the direction of the gravitational force, thus W = Fh cos 0° = mgh = (250 kg)(9.80 m/s 2 )(2.80 m) = 6.86 × 10 3 J . 2. The displacement is opposite to the direction of the retarding force, thus W Fx cos 180° = (535 N)(1.25 × 10 3 m)(– 1) = – 6.69 × 10 5 J . 3. Because there is no acceleration, the contact force must have the same magnitude as the weight. The displacement in the direction of this force is the vertical displacement. Thus, W F y = ( mg ) y = (65.0 kg)(9.80 m/s 2 )(20.0 m) = 1.27 × 10 4 J . 4. ( a ) Because there is no acceleration, the horizontal applied force must have the same magnitude as the friction force. Thus, W F x = (230 N)(4.0 m) = 9.2 × 10 2 J . ( b ) Because there is no acceleration, the vertical applied force must have the same magnitude as the weight. Thus, W F y mg y = (1200 N)(4.0 m) = 4.8 × 10 3 J . 5. Because there is no acceleration, from the force diagram we see that F N mg , and F F fr k mg . Thus, W = F x cos 0° = k mg x cos 0° = (0.50)(160 kg)(9.80 m/s 2 )(10.3 m)(1) = 8.1 × 10 3 J . 6. Because the speed is zero before the throw and when the rock reaches the highest point, the positive work of the throw and the (negative) work done by the (downward) weight must add to zero. Thus, W net W throw + mgh cos 180° = 0, or h = – W throw / mg cos 180° = – (80.0 J)/(1.85 kg)(9.80 m/s 2 )(–1) = 4.41 m . 7. 1 J = (1 kg · m/s 2 )(1 m)(1000 g/kg)(100 cm/m) 2 = 1 × 10 7 g · cm/s 2 1 × 10 7 erg . 1 J = (1 N · m)(0.225 lb/N)(3.28 ft/m) = 0.738 ft · lb . 8. The maximum amount of work that the hammer can do is the work that was done by the weight as the hammer fell: W max mgh cos 0° = (2.0 kg)(9.80 m/s 2 )(0.50 m)(1) = 9.8 J . People add their own force to the hammer as it falls in order that additional work is done before the hammer hits the nail, and thus more work can be done on the nail. 9. If we assume the width of a cut is 0.30 m, and we cut the lawn parallel to the long side, the number of cuts required is N = (10.0 m)/(0.30m/cut) = 33 cuts, each 20 m long. Thus the total work (ignoring turn-arounds) is W Fx = (15 N)(33)(20 m) = 1.0 × 10 4 J . m g x y F f fr F N ? x
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Chapter 7 Page 2 10. The minimum work is needed when there is no acceleration. ( a ) From the force diagram, we write Σ F = m a : y -component: F N mg cos = 0; x -component: F min mg sin = 0. For a distance d along the incline, we have W min = F min d cos 0° = mgd sin (1) = (950 kg)(9.80 m/s 2 )(310 m) sin 9.0° 4.5 × 10 5 J . ( b ) When there is friction, we have x -component: F min mg k F N = 0, or F min mg + k mg cos , For a distance d along the incline, we have W min = F min d cos 0° = mgd (sin k cos )(1) = (950 kg)(9.80 m/s 2 )(310 m)(sin 9.0° + 0.25 cos 9.0°) = 1.2 × 10 6 J . 11. We assume that the input force is exerted perpendicular to the lever. When the lever rotates through a small angle , the distance through which the input force acts is ¬ I , and the distance the output force acts is ¬ O . If the output work is equal to the input work, we have W I F I ¬ I W O F O ¬ O , or F O / F I ¬ I / ¬ O .
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This homework help was uploaded on 04/08/2008 for the course PHY 53 taught by Professor Mueller during the Fall '08 term at Duke.

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CH_7 - Chapter 7 CHAPTER 7 - Work and Energy 1. The...

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