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Step-by-Step Solution
Step 1: Identify the physical situation
A block starts from rest at the top of an inclined plane of length $l$ and inclination $\phi$. The upper half of this inclined plane (length $\tfrac{l}{2}$) is smooth (no friction), and the lower half ($\tfrac{l}{2}$) is rough with friction coefficient $\mu$. The block comes to rest again at the bottom of the incline.
Step 2: State the work-energy theorem
The work-energy theorem says that the total work done on the block is equal to its change in kinetic energy:
$$
W_{\text{total}} = \Delta K.
$$
Since the block starts and ends at rest, $\Delta K = 0$, implying that the net work done on the block must be zero.
Step 3: Calculate the work done by gravity
The component of gravity acting along the incline is $mg \sin \phi$. Over the total length $l$ of the incline, the work done by gravity is
$$
W_{g} = (mg \sin \phi) \times l.
$$
Step 4: Calculate the work done by friction
Friction acts only on the rough part of the incline (length $\tfrac{l}{2}$). The frictional force on the block is
$$
f = \mu \, mg \cos \phi.
$$
Since friction opposes the motion, the work done by friction is negative:
$$
W_{\text{friction}} = - ( \mu \, mg \cos \phi ) \times \frac{l}{2}.
$$
Step 5: Apply net work = 0 condition
According to the work-energy theorem,
$$
W_{g} + W_{\text{friction}} = 0.
$$
Substitute the expressions for $W_{g}$ and $W_{\text{friction}}$:
$$
mg \, l \sin \phi - \left( \mu \, mg \cos \phi \right) \frac{l}{2} = 0.
$$
Step 6: Solve for the friction coefficient $\mu$
Simplify the above equation by canceling out common factors $mg$ and $l$, we get:
$$
\sin \phi - \frac{\mu \cos \phi}{2} = 0
\quad \Longrightarrow \quad
\frac{\mu \cos \phi}{2} = \sin \phi
\quad \Longrightarrow \quad
\mu = 2 \, \tan \phi.
$$
Step 7: Final Answer
The coefficient of friction for the lower half of the inclined plane must be
$$
\mu = 2 \tan \phi.
$$
