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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 ms−1. The frictional force opposing the motion is 4000 N. What is minimum power delivered by the motor to the elevator?

A
22 kW
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B
44 kW
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C
66 kW
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D
88 kW
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Solution

The correct option is B 44 kW
Here, m=1800 kg
Frictional force, f=4000 N
Uniform speed, v=2 ms1
Downward force on elevator is
F=mg+f
=(1800 kg×10 ms2)+4000 N
=22000 N
The motor must supply enough power to balance this force. Hence,
P=Fv=(20000 N)(2 ms1)
=44000 W=44×103
=44 kW.

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