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) Numerical answer type. Work it out, then open the answer.

Answer: 48. At uniform speed F = mg + f = 1400 × 10 + 2000 = 16000 N. P = Fv = 16000 × 3 = 48000 W So P = 48 kW.

If the maximum load carried by an elevator is 1400 kg ( 600 kg -… | JEE Main 2023 (10 Apr, Shift 2)

Q20JEE Main 2023 · 10 APR · SHIFT 2Medium

Power

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 __________ $\text{kW}\ \left(g = 10\text{ m/s}^2\right)$

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 48

At uniform speed $F = mg + f = 1400 \times 10 + 2000 = 16000\text{ N}$ .
$P = Fv = 16000 \times 3 = 48000\text{ W}$
So $P = 48\text{ kW}$ .

Report a mistake

Power question from JEE Main 2023 (10 Apr, Shift 2)

More questions on Power