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 $3 \mathrm{~ms}^{-1}$ and the frictional force acting on it is 2000 N, then the maximum power used by the motor is ______ $\mathrm{kW}$ ($\mathrm{g}=10 \mathrm{~m}/\mathrm{s}^{2}$).

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Work, Power & Energy

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 power used by the motor is ______

kW

(

g = 10 m / s^{2}

Answer: 48

Solution

As the elevator rises at constant speed its acceleration is zero, so the motor force must balance both the total weight and the frictional drag:

F = m g + f = 1400 \times 10 + 2000 = 16000 N

. The power delivered is

P = F v = 16000 \times 3 = 48000 W

. The maximum power used by the motor is

48 kW

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Integer answer

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