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Step-by-Step Solution
1. Understanding the Problem
We have an inclined plane of total length 2l, making an angle $\theta$ with the horizontal. The upper half (length $l$) is frictionless, and the lower half (length $l$) has a coefficient of friction $\mu$. A block starts from rest at the top and comes to rest again at the bottom. We need to find $\mu$ in terms of $\theta$.
2. Energy Considerations in the Frictionless Portion (Upper Half)
Since the block starts from rest at the top and the upper half is smooth (no friction), the entire loss in potential energy converts into kinetic energy at the midpoint:
Length of the frictionless portion = $l$.
Drop in height for the upper half = $l \,\sin\theta$ (vertical component of the inclined plane length).
Loss of potential energy from top to midpoint = $mg\,l\,\sin\theta$.
This equals the kinetic energy gained at the midpoint:
\[
\frac{1}{2}mv_{\text{mid}}^2 = mg\,l\,\sin\theta.
\]
Hence,
\[
v_{\text{mid}}^2 = 2g\,l\,\sin\theta.
\]
3. Energy Considerations in the Frictional Portion (Lower Half)
The block enters the lower half with speed $v_{\text{mid}}$ and finally comes to rest at the bottom. During this lower half, two changes in energy occur:
It loses additional potential energy as it descends another vertical height of $l \,\sin\theta$.
Friction does negative work against the motion over distance $l$.
Total mechanical energy available at the start of the lower half (midpoint) is:
Kinetic energy at midpoint = $mg\,l\,\sin\theta.$
Additional potential energy to be lost from midpoint to bottom = $mg\,l\,\sin\theta.$
Hence, total energy to be dissipated by friction from midpoint to bottom is
\[
mg\,l\,\sin\theta + mg\,l\,\sin\theta = 2\,mg\,l\,\sin\theta.
\]
4. Work Done by Friction
The frictional force $f$ on the block in the lower half is given by
\[
f = \mu\,mg\,\cos\theta.
\]
Since the friction acts over a distance $l$ down the plane, the total work done against friction is
\[
f \times l = \mu\,mg\,\cos\theta \times l = \mu\,mg\,l\,\cos\theta.
\]
Because the block comes to rest at the bottom, all of its available mechanical energy in the lower half is dissipated by friction:
\[
2\,mg\,l\,\sin\theta = \mu\,mg\,l\,\cos\theta.
\]
5. Solving for $\mu$
Canceling common factors $mg\,l$ from both sides, we get:
\[
2\,\sin\theta = \mu\,\cos\theta
\quad \Longrightarrow \quad
\mu = \frac{2\,\sin\theta}{\cos\theta} = 2\,\tan\theta.
\]
6. Final Answer
The required coefficient of friction is
\[
\mu = 2\,\tan\theta.
\]
