200_Physics ProblemsTechnical Physics

# 200_Physics ProblemsTechnical Physics - 202 P7.35 Energy...

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202 Energy and Energy Transfer P7.35 v i = 200 . m s µ k = 0100 . KfxW K ik f −+ = other : 1 2 0 2 mv f x −= 1 2 2 mv mg x = x v g i k == = 2 2 2 2 0 100 9 80 204 . .. . ms m bg af a f Section 7.8 Power *P7.36 P av kg m s s W = = × = W t K t mv t f ∆∆ 2 2 3 2 0875 0620 221 10 801 . ej P7.37 Power = W t = mgh t 700 10 0 800 875 N m s W a f . . P7.38 A 1 300-kg car speeds up from rest to 55.0 mi/h = 24.6 m/s in 15.0 s. The output work of the engine is equal to its final kinetic energy, 1 2 1 300 24 6 390 2 kg m s kJ b g . = with power = 390 000 10 4 J 15.0 s W ~ around 30 horsepower. P7.39 (a) WK =∆ , but K = 0 because he moves at constant speed. The skier rises a vertical distance of 60 0 30 0 30 0 .s i n . . m m °= . Thus, WW g in 2 kg m s m J kJ =− = = × = 700 98 300 206 10 206 4 . . . . (b) The time to travel 60.0 m at a constant speed of 2.00 m/s is 30.0 s. Thus, input J 30.0 s W h p × W t 686 0 919 4 . P7.40 (a) The distance moved upward in the first 3.00 s is yv
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