Question 232: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

First, let's find the total weight of the elevator and passengers:
Total weight = (mass of passengers + mass of elevator) ÃÂ g
Total weight = (600 kg + 800 kg) ÃÂ 10 m/sÃÂ²
Total weight = 1400 kg ÃÂ 10 m/sÃÂ² = 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 ÃÂ Velocity
Here, the force is the tension in the cable, and the velocity is the speed of the elevator:
Power = 16,000 N ÃÂ 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.

Work, Energy and Power & Circular Motion

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