Q 11-05-102JEE MainJEE Main 2023 (10 Apr, Shift 2)Easy
If the maximum load carried by an elevator is $1400\ \text{kg}$ ($600\ \text{kg}$ passengers $+\ 800\ \text{kg}$ elevator), which is moving up with a uniform speed of $3\ \text{m s}^{-1}$ and the frictional force acting on it is $2000\ \text{N}$, then the maximum power used by the motor is ______ kW. ($g=10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 48
At uniform speed the motor force balances weight plus friction: $F=14000+2000=16000\ \text{N}$.
$P=Fv=16000\times3=48000\ \text{W}=48\ \text{kW}$.
Solution by Sreeraj P, M.Sc Physics