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. $\mu =2tan\theta$
• B. $\mu =tan\theta$
• C. $\mu =\frac{1}{tan\theta }$
• D. $\mu =\frac{2}{tan\theta }$

Answer 1

Answer: (A)

Given,
The upper half of an inclined plane of inclination $\theta$ is perfectly smooth while lower half is rough.
Let's consider the length of the inclined plane as $2l$ and the height of the top of the inclined plane from the ground as $h=2l\sin \theta$ , as shown below

By Work-Energy Theorem,
Work done by all the forces is equal to the change in kinetic energy of the block.
i.e.,
$W_{\text{gravity}}+W_{\text{friction}}=\Delta K.E.$
$\Rightarrow mgh+-\mu mg\cos \theta l=0$
$\Rightarrow mg2l\sin \theta -\mu mgl\cos \theta =0$
$\Rightarrow 2mgl\sin \theta =\mu mgl\cos \theta$
$\Rightarrow \mu =2\tan \theta$