227_Physics ProblemsTechnical Physics

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

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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 tm g yf rm g r f r =+ = + ∆∆ sin θ mgv fv ff sin As the locomotive moves on level track, = fv i 1000 27 hp 746 W 1 hp ms F H G I K J = f bg f 276 10 4 . N Then also 746 000 160 000 9 8 5 4 W k g m s m 100 m N 2 = F H G I K J ej e j .. vv v f = × = 746 000 10 704 5 W 1.06 N . 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. Em v m v m g y m g y f d mg y y f d f mg y y d fif i k if k k =−+ = ° −− = = = + = 1 2 1 2 180 00 70 0 9 80 10 0 5 00 500 206 22 cos . . di af kg m s m m m kN 2 P8.31 UK E U K ii f f ++ = + mech : mg h f h mv 21 2 2 2 1 2 1 2 −= + fnm g == µµ 1 h h m m v 1
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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