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 plane is given by

• A. $2\tan \theta$
• B. $\tan \theta$
• C. $2\cot \theta$
• D. $2\cos \theta$

Answer 1

The correct option is A $2\tan \theta$

The given condition is shown in the figure as

First of all we have to calculate the acceleration of the block in both the regions i.e. smooth ($a_S$) and rough ($a_R$)
Force acting down the plane in smooth region is $F=mg\sin \theta$
$\therefore$ acceleration $a_S=\frac{mg\sin \theta }{m}=g\sin \theta$
Similarly, force acting down the plane in rough region is
$F=mg\sin \theta -f=mg\sin \theta -\mu mg\cos \theta$
$\therefore$ acceleration $a_R=\frac{mg(\sin \theta -\mu \cos \theta )}{m}=g(\sin \theta -\mu \cos \theta )$

Now, applying equation of motion in both the regions:

1. For smooth plane
Let the velocity gained by block at the end of smooth surface be $v$
So, from equation we have
$v^2-u^2=2a_{S}S$
$v^2-(0{)}^2=2(g\sin \theta )S$
$v^2=2gS\sin \theta$

2. For rough plane
The initial velocity of the block at the rough surface will be the velocity gained by block at the end of smooth surface.
Let the final velocity of the block at the end of the rough surface be $v_1$
then, from the equation we have
$v_1^2-v^2=2a_{R}S$
$\Rightarrow 0^2-v^2=2g(\sin \theta -\mu \cos \theta )S$
$\Rightarrow -v^2=2g(\sin \theta -\mu \cos \theta )S$
$\Rightarrow -2gS\sin \theta =2gS(\sin \theta -\mu \cos \theta )$
$\Rightarrow -\sin \theta =\sin \theta -\mu \cos \theta$
$\Rightarrow -2\sin \theta =-\mu \cos \theta$
$\Rightarrow \mu =\frac{2\sin \theta }{\cos \theta }=2\tan \theta$