Question

The upper half of an inclined plane of inclination $\theta$ is perfectly smooth while the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom if the coefficient of friction between the block and lower half of the plane is equal to:
\[\begin{align}
  & \text{A}\text{. }\mu =\dfrac{1}{\tan \theta } \\
 & \text{B}\text{. }\mu =\dfrac{2}{\tan \theta } \\
 & \text{C}\text{. }\mu \text{= 2 tan }\theta \\
 & \text{D}\text{. }\mu \text{= tan }\theta \\
\end{align}\]

Answer 1

Hint: To find the coefficient of friction, consider the work done by the body. Since friction is only due to rough surface, we can equate it to the work done due to gravity to balance the forces and this will give us the required friction.

Formula used:
$W=W_{f}+W_{g}=0$
$W_{f} =-\mu mg cos\theta l$
$W_{g}=mg(2l)sin\theta$


Complete step by step answer:
Clearly, since the plane is inclined by an angle $\theta$, the plane will provide an opposing force called the friction, which opposes the motion of the block on the plane. Let us take the free body diagram:

Consider the length of the plane as \[2l\]. Let the initial velocity of the block, before it moves as $u=0$, and the final velocity of the block when it reaches the bottom as $v=0$
Then the work done by the moving block is the change in its kinetic energy: $W.D=\Delta K.E=\dfrac{1}{2}(mu^{2}-mv^{2})=0$
Then the total work done by the block, is $W=W_{f}+W_{g}=0$, where $W_{f}$ is the work done by the friction and $W_{g}$ is the work done by the gravity.
Then, the work done by the smooth surface is zero, whereas the work done due to rough surface of distance $l$ is given by the free body diagram $W_{f} =-\mu mg cos\theta l+0=-\mu mg cos\theta l$ where,$cos \theta$ the force component along the slope is, $\mu$ is the coefficient of friction.
Similarly, the work done due to gravity is given as $W_{g}=mg(2l)sin\theta$, where $sin \theta$ is the force component in the downward direction. Here we take \[2l\] as the gravity acts on the body throughout the motion.
Then, we have $-\mu mgcos\theta l+mg(2l)sin \theta=0$
Or $-\mu cos\theta l=-2l sin\theta$
Or $\mu=2 tan\theta$
Hence the coefficient of friction is $\mu=2 tan\theta$.
Thus the answer is D.

Note:
The work done by the smooth surface of distance $l$ is zero, whereas the work done the due rough surface of distance $l$ is given by the free body diagram $W_{f} =-\mu mg cos\theta l+0=-\mu mg cos\theta l$, where negative sign indicates opposition to the motion, also, the work done due to gravity is given as $W_{g}=mg(2l)sin\theta$. We can equate it to the work done due to gravity, this will give us friction.