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Unformatted text preview: CHAPTER 7 WORK, ENERGY, AND POWER ActivPhysics can help with these problems: Activity 5.1 Section 7-1: Work Problem 1. How much work do you do as you exert a 75-N force to push a shopping cart through a 12-m-long supermarket aisle? Solution If the force is constant and parallel to the displace- ment, W =F r= F r = (75 N)(12 m) = 900 J . Problem 2. If the coefficient of kinetic friction is 0.21, how much work do you do when you slide a 50-kg box at constant speed across a 4.8-m-wide room? Solution If you push parallel to a level floor, the applied force equals the frictional force (since the acceleration is zero), and the normal force equals the weight. The applied force is constant and parallel to the displacement, so W a = F a r = f k r = k N r = k mg r = 0 . 21(50 kg)(9 . 8 m / s 2 )(4 . 8 m) = 494 J . Problem 3. A crane lifts a 650-kg beam vertically upward 23 m, then swings it eastward 18 m. How much work does the crane do? Neglect friction, and assume the beam moves with constant speed? Solution Lifting the beam at constant speed, the crane exerts a constant force vertically upward and equal in magnitude to the weight of the beam. During the horizontal swing, the force is the same, but is perpendicular to the displacement. The work done is F r = ( mg ) ( y + x ) = mg y = (650 kg) (9 . 8 m / s 2 )(23 m) = 147 kJ . Problem 4. You lift a 45-kg barbell from the ground to a height of 2.5 m. (a) How much work do you do on the barbell? (b) You hold the barbell aloft for 2.0 min. How much work do you do on the barbell during this time? (c) You lower the barbell to the ground. Now how much work do you do on it? Solution (a) If we assume the barbell is lifted at a constant velocity by a vertical applied force equal to the weight and positive upward parallel to the displacement, then W app = mg y = (45 kg)(9 . 8 m / s 2 )(2 . 5 m) = 1 . 10 kJ . (b) Just holding the weight stationary, you must still exert an applied force of mg to balance gravity, but the displacement through which the force acts is zero. Hence the work done on the barbell is also zero. (Actually, individual muscle fibers are continually contracting even though the overall muscles are stationary, so internal work is being done in your muscles and you feel tired just holding a weight.) (c) When the weight is lowered at constant velocity, the upward applied force, mg , is opposite to the displacement downward, y = 2 . 5 m; hence the work done on the barbell is negative, W app = 1 . 10 kJ . Problem 5. The worlds highest waterfall, the Cherun-Meru in Venezuela, has a total drop of 980 m. How much work does gravity do on a cubic meter of water dropping down the Cherun-Meru?...
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- Fall '09