Question

The upper half of an inclined plane with inclination $14^{\circ }$ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half $\mu =$____________ . (Take $\tan 14^{\circ }=0.25$ )

Answer 1

Answer: 0.5

As, $v^2-u^2=2as$
In the journey over the upper hall of incline,
$v^2-0=2(g\sin \theta )\frac{s}{2}$
$v^2=g\sin \theta s$
In the journey over the lower hall of incline,
$v^2-u^2=2$ as
$0-gs\sin \theta =2g(\sin \theta -\mu \cos \theta )\frac{s}{2}$
$-\sin \theta =\sin \theta -\mu \cos \theta$
$\therefore \mu =\frac{2\sin \theta }{\cos \theta }$
$\therefore \mu =2\tan \theta$
$=2\tan 14^{\circ }$
$\therefore \mu =2\times 0.25$
$\therefore \mu =0.5$