Finding coefficient of friction between block and plane An inclined plane makes angle $\theta$ with horizontal. Upper half plane is perfectly smooth where as lower half is rough. The block of mass m starts sliding from top. If the block again comes to rest at bottom. The coefficient of friction between block and plane.
This means it will first accelerate and the there will be retardation. I am confused as both the thing have come together.
 A: You're right, first it will accelerate (over the smooth surface), and then it must decelerate (over the rough surface, to come to rest again). 
If the plane has length $L$, then it will accelerate (along the plane) over a distance $L/2$, at $g\sin{\theta}$. So we can write, if we call its velocity $v$ when it has gone that distance: $$v^2 = 2g(L/2)\sin{\theta} = gL\sin{\theta}.$$
On the second half of the plane, friction acts, so we can write the block's acceleration $a$ as:$$a = -\mu g \cos{\theta} + g\sin{\theta}.$$
Then, knowing that it stops after a distance $L/2$, we say:
$$0 = v^2 -L(\mu g \cos{\theta} - g\sin{\theta}).$$
Substituting in our expression for $v^2$ from before, we have:
$$gL(\mu  \cos{\theta} - \sin{\theta}) = gL\sin{\theta} \Rightarrow \mu  \cos{\theta} = 2\sin{\theta} \Rightarrow \mu = 2\tan{\theta}.$$
This can also be done using energy (try it!).
A: Hint:
Try to draw FBD (free body diagram)
Acceleration is parallel to slope in upper half portion = $g\sin\theta$
Acceleration in lower half half = $g\sin\theta - \mu g\cos\theta$
