An elevator of mass 500 kg is to be lifted up at a constant velocity of 0.4 m/s. What should be the minimum horse power of the motor to be used? (Take g= 10 m/s² and 1 hp = 750 watts):

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8/3 hp

4/3 hp

2000 hp

75 hp

answer is A.

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Detailed Solution

P=F⋅v=mgv=500×10×0.4W = 2000 W=2000750hp=83hp

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