Question

A block of mass M is released from point P on a rough inclined plane with inclination angle $\theta$, shown in the figure below. The coefficient of friction is $\mu$. if $\mu <\tan \theta ,$ then the time taken by the block to reach another point Q on the inclined plane, where PQ= s, is

• A. $\sqrt{\frac{2s}{g\cos \theta (\tan \theta -\mu )}}$
• B. $\sqrt{\frac{2s}{g\cos \theta (\tan \theta +\mu )}}$
• C. $\sqrt{\frac{2s}{g\sin \theta (\tan \theta -\mu }}$
• D. $\sqrt{\frac{2s}{g\sin \theta (\tan \theta +\mu }}$

Answer 1

The correct option is A $\sqrt{\frac{2s}{g\cos \theta (\tan \theta -\mu )}}$

FDB of block

$N=mgcos\theta$
From Newton's second law
$mg\sin \theta -\mu mg\cos \theta =ma$
$a=g\sin \theta -\mu g\cos \theta$
$S=ut+\frac{1}{2}a{t}^2$
(as initial velocity, u = 0)

$S=0+\frac{1}{2}(g\sin \theta -\mu g\cos \theta )t^2$

$t=\sqrt{\frac{2s}{g\cos \theta [\tan \theta -\mu ]}}$