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JEE MAIN - Physics (2023 - 10th April Evening Shift - No. 23)

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 __________ $$\mathrm{kW}\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$$
Answer
48

Explanation

First, let's find the total weight of the elevator and passengers:

Total weight = (mass of passengers + mass of elevator) × g

Total weight = (600 kg + 800 kg) × 10 m/s²

Total weight = 1400 kg × 10 m/s² = 14,000 N

Now, we need to calculate the total force acting on the elevator as it moves upwards. Since the elevator is moving at a constant speed, the net force acting on it is zero. Therefore, the tension in the cable must balance the total weight and frictional force:

Tension = Total weight + Frictional force Tension = 14,000 N + 2,000 N = 16,000 N

The power used by the motor can be calculated using the formula:

Power = Force × Velocity

Here, the force is the tension in the cable, and the velocity is the speed of the elevator:

Power = 16,000 N × 3 m/s = 48,000 W

To convert the power to kilowatts, divide by 1,000:

Power = 48,000 W / 1,000 = 48 kW

So, the maximum power used by the motor is 48 kW.

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