If the maximum load carried by an elevator is \(1400\,kg\) (\(600\,kg\) - Passenger \(+800\,kg\) - elevator), which is moving up with a uniform speed of \(3\,ms ^{-1}\) and the frictional force acting on it is \(2000\,N\), then the maximum 10.Power used by the motor is \(...........\,kW \left( g =10\,m / s ^2\right)\).

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Answer & Solution

Correct Answer

(C) \(48\)

Step-by-step Solution

Detailed explanation

\(P_{\max }=F_{\max } \times V\)

\(F_{\max }=1400\,g +\text { friction }\)

\(=14000+2000=16000\)

\(P_{\max }=16000 \times 3=48000\,W =48\,KW\)

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