PYQ NEET- Work Energy And Power L-7

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

Answer:#

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

Total weight $=($ mass of passengers + mass of elevator $) \times \mathrm{g}$

Total weight $=(600 \mathrm{~kg}+800 \mathrm{~kg}) \times 10 \mathrm{~m} / \mathrm{s}^2$

Total weight $=1400 \mathrm{~kg} \times 10 \mathrm{~m} / \mathrm{s}^2=14,000 \mathrm{~N}$

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 $=14,000 \mathrm{~N}+2,000 \mathrm{~N}=16,000 \mathrm{~N}$

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

Power $=$ Force $\times$ Velocity

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

Power $=16,000 \mathrm{~N} \times 3 \mathrm{~m} / \mathrm{s}=48,000 \mathrm{~W}$

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

Power $=48,000 \mathrm{~W} / 1,000=48 \mathrm{~kW}$

So, the maximum power used by the motor is $48 \mathrm{~kW}$.