Question
The upper half of an inclined plane of inclination $$\theta $$ is perfectly smooth while 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
A.
$$\mu = \frac{1}{{\tan \theta }}$$
B.
$$\mu = \frac{2}{{\tan \theta }}$$
C.
$$\mu = 2\tan \theta $$
D.
$$\mu = \tan \theta $$
Answer :
$$\mu = 2\tan \theta $$
Solution :
Net work done by the block in going from top to bottom of the inclined plane, must be equal to the work done by frictional force.
The block may be stationary, when
$$\eqalign{ & mg\sin \theta \cdot L = \mu \,mg\cos \theta \frac{L}{2} \cr & {\text{or}}\,\mu = \frac{{mg\,\sin \theta \cdot L}}{{mg\,\cos \theta \cdot \frac{L}{2}}} = 2\frac{{\sin \theta }}{{\cos \theta }} = 2\,\tan \theta \cr & \mu = 2\tan \theta \cr} $$
Net work done by the block in going from top to bottom of the inclined plane, must be equal to the work done by frictional force.
The block may be stationary, when
$$\eqalign{ & mg\sin \theta \cdot L = \mu \,mg\cos \theta \frac{L}{2} \cr & {\text{or}}\,\mu = \frac{{mg\,\sin \theta \cdot L}}{{mg\,\cos \theta \cdot \frac{L}{2}}} = 2\frac{{\sin \theta }}{{\cos \theta }} = 2\,\tan \theta \cr & \mu = 2\tan \theta \cr} $$
