Animated Solution for Physics - Work, Energy, and Power: The upper half of an inclined plane with inclination $\varphi$ is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is given by

The problem presents a fascinating scenario: a block sliding down an inclined plane, but the plane has a split personality. The upper half is perfectly smooth, offering no resistance, while the lower half is rough, acting as a brake. The block starts from rest at the top and, surprisingly, comes to rest exactly at the bottom. Our mission is to find the coefficient of friction for that rough lower half.

Imagine you are standing at the top of this incline. You let go of the block. On the smooth upper half, gravity pulls it down, and it accelerates freely. It gains kinetic energy. But as soon as it hits the rough lower half, friction kicks in. This friction is strong enough to not only stop the acceleration but to decelerate the block, bringing it to a complete halt just as it reaches the bottom.

This means the block's initial velocity $u$ is zero, and its final velocity $v$ is also zero. Consequently, the total change in kinetic energy $\Delta K$ is exactly zero.

Work-Energy Theorem

Whenever you see a problem involving changes in speed over a distance, the Work-Energy Theorem should be your go-to tool. It states that the net work done by all forces acting on an object equals its change in kinetic energy:

Since $\Delta K=0$, the net work done must be zero. Now, let's break down the forces doing the work.

There are three main forces acting on the block:

1. Gravity ($mg$): This force acts downwards. Its component along the incline is $mg\sin \varphi$. This component acts in the direction of motion over the entire length $l$ of the incline. Therefore, the work done by gravity is positive:

2. Normal Force ($N$): This force acts perpendicular to the surface. Since the block doesn't move perpendicular to the surface, the normal force does zero work.

3. Kinetic Friction ($f_k$): This force only acts on the rough lower half, which has a length of $l/2$. The frictional force is given by $\mu N$, where $N=mg\cos \varphi$. Since friction opposes motion, its work is negative:

Now, we plug these individual work components back into our Work-Energy equation:

Notice the beauty of this equation. The mass $m$, the acceleration due to gravity $g$, and the total length $l$ are present in both terms. We can divide the entire equation by $mgl$, making them vanish! This tells us a profound physical truth: the result is completely independent of how heavy the block is or how long the incline is.

Finally, rearranging to solve for the coefficient of friction $\mu$:

And there we have it! The coefficient of friction must be exactly $2\tan \varphi$ to bring the block to a perfect stop at the bottom.