Animated Solution for Physics - Work, Energy, and Power: A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to (Take, $1\text{ HP}=746\text{ W},g=10{\text{ms}}^{-2}$)

Imagine you are standing inside a fully loaded elevator. As it ascends, it's not just fighting against gravity; it's also battling the frictional forces of the guide rails and pulleys. The motor at the top is the unsung hero, constantly pulling the cable to keep you moving.

In this problem, we are given a $60\text{ HP}$ motor lifting a $2000\text{ kg}$ elevator against a $4000\text{ N}$ frictional force. The key phrase here is "speed of the elevator at full load". When an elevator reaches its operating speed, it moves at a constant velocity. This means the net acceleration is zero, and by Newton's First Law, the net force must also be zero!

Since the net force is zero, the upward force provided by the motor must perfectly balance the downward forces. What are these downward forces? 1. The weight of the elevator: $W=mg$ 2. The frictional force: $f$

So, the force the motor needs to exert is:

Let's plug in the numbers. The mass $m$ is $2000\text{ kg}$, and $g$ is $10{\text{ ms}}^{-2}$.

Now, we need to relate this force to the power of the motor. The mechanical power $P$ delivered by a force $F$ moving an object at velocity $v$ is given by:

But wait! The power is given in Horsepower (HP). We must convert this to the standard SI unit, Watts (W). We are given that $1\text{ HP}=746\text{ W}$.

Now, we substitute everything into our power equation to find the velocity $v$:

Let's simplify the fraction. We can cancel a $60$ from the numerator and denominator:

Looking at our options, the closest value is $1.9{\text{ ms}}^{-1}$. This is a classic JEE problem that tests your ability to combine Newton's laws with the concept of mechanical power, while also keeping an eye on unit conversions!