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 $2\tan \theta$


For smooth plane
Applying equation of motion

$v^2-u^2=2as$, here $a=g\sin \theta$
$v^2-(0{)}^2=2(g\sin \theta )s$ $\Rightarrow v^2=2g\sin \theta \times s$

For rough plane $($ here the initial velocity will be final velocity in the smooth plane $)$
Acceleration of block is $a=g(\sin \theta -\mu \cos \theta )$

$v^2-u^2=2as$ $(v=0,u^2=2g\sin \theta \times s)$

Putting the values we get
$-2g\sin \theta \times s=2g(\sin \theta -\mu \cos \theta )\times s$

$2\sin \theta =\mu \cos \theta$ $\Rightarrow \mu =2\tan \theta$