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Unformatted text preview: h ic e8 e k# 6 i b iC a aha 6 Problem 12 ( a ): Suppose a person walks about 2 or 3 mph, which is about 1 m/s. Then the energy is 1 2 mv 2 = 38 J. For a good running speed take a 6 minute mile, which is 10 mph, or about 4.4 m/s. This corresponds to a kinetic energy of about 730 J. ( b ): The mass of the electron is given on the inside back cover as m = 9 . 109 10 31 kg. As the speed is v = 2 . 19 10 6 m / s, the kinetic energy is K = 1 2 mv 2 = 2 . 18 10 18 J . ( c ): Suppose your shoulder is about 1.5 m above the floor. The velocity at the end is given by v 2 f = v 2 + 2 gd v f = 5 . 4 m / s . Then the kinetic energy is about 15 J. ( d ): We have K = 1 2 mv 2 v = radicalbigg 2 K m = radicalBigg 2(100 J) 30 kg = 2 . 6 m / s = 5 . 8 mi / hr . Yes, this is reasonable. 6 Problem 17: The total force perpendicular to the slope is N mg cos = 0 N = mg cos . The total force parallel to the slope is F f mg sin . Knowing that F f = k N , the total force F bardbl parallel becomes F bardbl = k mg cos  mg sin . The total distance the box slides is d = h/ sin . The total work done by the parallel forces is W = F bardbl d = mgh k cos + sin sin . This is the change in the kinetic energy. As the box just barely reaches the stranded skier, the final velocity is 0. Thus W = K.E....
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 Summer '08
 Chamberlin
 Energy

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