
The maximum force acting is, Fmax=(14000+2000)N
=16000N
The formula for power is given by, P=Fmaxv.
It is given, v=3ms−1.
⇒P=16000N×3ms−1=48000W=48kW
⇒W=48kW
JEE Main 2023 — Physics Mechanics
If the maximum load carried by an elevator is 1400kg (600kg-Passengers+800kg-elevator) , which is moving up with a uniform speed of 3ms−1 and the frictional force acting on it is 2000N, then the maximum power used by the motor is _____________kW.(g=10ms−2)
Held on 10 Apr 2023 · Verified 6 Jul 2026.
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