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227_Physics ProblemsTechnical Physics

# 227_Physics ProblemsTechnical Physics - Chapter 8*P8.29 229...

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Chapter 8 229 *P8.29 As the locomotive moves up the hill at constant speed, its output power goes into internal energy plus gravitational energy of the locomotive-Earth system: P t mgy f r mg r f r = + = + sin θ P = + mgv fv f f sin θ As the locomotive moves on level track, P = fv i 1 000 27 hp 746 W 1 hp m s F H G I K J = f b g f = × 2 76 10 4 . N Then also 746 000 160 000 9 8 5 2 76 10 4 W kg m s m 100 m N 2 = F H G I K J + × b g e j e j . . v v f f v f = × = 746 000 10 7 04 5 W 1.06 N m s . P8.30 We shall take the zero level of gravitational potential energy to be at the lowest level reached by the diver under the water, and consider the energy change from when the diver started to fall until he came to rest. E mv mv mgy mgy f d mg y y f d f mg y y d f i f i k i f k k i f = + = ° = − = = + = 1 2 1 2 180 0 0 70 0 9 80 10 0 5 00 5 00 2 06 2 2 cos . . . . . . d i d i b g e j a f kg m s m m m kN 2 P8.31 U K E U K i i f f + + = + mech : m gh fh m v m v 2 1 2 2 2 1 2 1 2 = + f n m g = = µ µ 1 m gh m gh
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