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 given by:

Let the total length of the incline be $S$.Length of smooth part = $S/2$.Length of rough part = $S/2$.Using the Work-Energy Theorem: Total work done = Change in Kinetic Energy.Since the block starts at rest and ends at rest, $\Delta K.E.=0$.Work done by gravity over full length $S$: $W_g=(mg\sin \theta )S$.Work done by friction over the lower half $S/2$: $W_f=-f(S/2)=-(\mu mg\cos \theta )(S/2)$.Setting total work to zero:
$$(mg\sin \theta )S-\mu mg\cos \theta \left(\frac{S}{2}\right)=0$$
$$mg\sin \theta =\mu mg\cos \theta \left(\frac{1}{2}\right)$$
$$\sin \theta =\frac{\mu \cos \theta }{2}$$
$$\mu =2\frac{\sin \theta }{\cos \theta }=2\tan \theta$$