Question
The upper half of an inclined plane of inclination $$\theta $$ is perfectly smooth while tower 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
A.
$$\mu = \frac{2}{{\tan \theta }}$$
B.
$$\mu = 2\tan \theta $$
C.
$$\mu = \tan \theta $$
D.
$$\mu = \frac{1}{{\tan \theta }}$$
Answer :
$$\mu = 2\tan \theta $$
Solution :
For upper half of inclined plane
$${v^2} = {u^2} + 2a\frac{S}{2} = 2\left( {g\sin \theta } \right)\frac{S}{2} = gS\sin \theta $$
For lower half of inclined plane
$$\eqalign{ & 0 = {u^2} + 2g\left( {\sin \theta - \mu \cos \theta } \right)\frac{S}{2} \cr & \Rightarrow - gS\sin \theta = gS\left( {\sin \theta - \mu \cos \theta } \right) \cr & \Rightarrow 2\sin \theta = \mu \cos \theta \cr & \Rightarrow \mu = \frac{{2\sin \theta }}{{\cos \theta }} = 2\tan \theta \cr} $$
For upper half of inclined plane
$${v^2} = {u^2} + 2a\frac{S}{2} = 2\left( {g\sin \theta } \right)\frac{S}{2} = gS\sin \theta $$
For lower half of inclined plane
$$\eqalign{ & 0 = {u^2} + 2g\left( {\sin \theta - \mu \cos \theta } \right)\frac{S}{2} \cr & \Rightarrow - gS\sin \theta = gS\left( {\sin \theta - \mu \cos \theta } \right) \cr & \Rightarrow 2\sin \theta = \mu \cos \theta \cr & \Rightarrow \mu = \frac{{2\sin \theta }}{{\cos \theta }} = 2\tan \theta \cr} $$
