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Question

An elevator which can carry a maximum load of 1800kg (elevator+passengers)is moving up with a constant speed of 2m/s. The frictional force opposing the motion is 4000N. Determine the minimum power delivered by the motor to the elevator in watt and horsepower .

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Solution

Dear Student,
1.weight in the elevator,w=mgw=1800×9.8=17640Ntotal force acting downward=weight+frictional forceF=w+fF=17640+4000F=21640Npower delivered by motor=force×velocityP=F×vP=21640×2P=43280 Watts

We know that 1 horse power = 746 Watt

so Power = 43280/746 = 58.01 Hp

Regards

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