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A 500 kg load is lifted to a height of 20 meters by means of a pulley block which has an efficiency of 92%. Determine: a) The

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Answer #1

given

mass m = 500 kg

height h = 20 m

efficiency = 0.92

a) work out put w = mgh

w = 500*9.81*20 = 98100 J

wout = 0.92*98100 = 90252 J

b) work lost = wlost = 98100-98100*0.92 = 7848 J

c) power p = 90252/2*60 = 752.1 W

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