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 / s²)

Work, Energy and Power: JEE Main 2023 Physics question

JEE Main 2023 · 10 Apr, Evening shift · Q73

JEE Main 202310 Apr, Evening shiftPhysicsHardWork, Energy and Power70× since 2002

If the maximum load carried by an elevator is

1400 \mathrm{~kg}

(

600 \mathrm{~kg}

- Passengers + 800

\mathrm{kg}

- elevator), which is moving up with a uniform speed of

3 \mathrm{~m} \mathrm{~s}^{-1}

and the frictional force acting on it is

2000 \mathrm{~N}

, then the maximum power used by the motor is __________

\mathrm{kW}\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2023 Work, Energy and Power Question with Answer (10 Apr Evening shift, Q73) | JEE Arc