If the maximum load carried by an elevator is 1400   kg ( 600   kg -Passengers   + 800  …

If the maximum load carried by an elevator is 1400 kg (600 kg-Passengers +800 kg-elevator) , which is moving up with a uniform speed of 3 m s-1 and the frictional force acting on it is 2000 N, then the maximum power used by the motor is _____________kW. g=10 m s-2

Solution

 

The maximum force acting is, Fmax=14000+2000 N

= 16000 N

The formula for power is given by, P=Fmaxv.

It is given, v=3 m s-1.

P = 16000 N × 3 m s-1 = 48000 W=48 kW

W = 48 kW

Asked in: JEE Main 2023 (10 Apr Shift 2)

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