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Question

An elevator can carry a maximum load of 1800 kg (elevator + passengers ) is moving up with a constant speed of 2 m/s. The frictional force opposing the motion is 400 N. Determine the minimum power delivered by

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Solution

Given that,

Mass of elevator + passengers, M=1800Kg

Speed, v=2m/s

Friction force, f=400N

Power delivered, P=force×velocity.........(1)

Force acting downward,

F=mg+f

F=1800×10+4000

F=22000N

Put the value of F in equation (1)

P=22000×2

P=44000Watts


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