Question

A block of mass $m$ slides down an inclined plane of slope angle $\theta$ with constant velocity. Now, same block is projected up an inclined plane with an initial velocity $u$. If the coefficient of kinetic friction between the block and the plane is $\mu$, the distance up to which the block will rise up the plane, before coming to rest is :-

• A. $\frac{u^2\mu }{2gsin\theta }$
• B. $\frac{u^2\mu }{2gcos\theta }$
• C. $\frac{u^2}{4gsin\theta }$
• D. $\frac{u^2}{4gcos\theta }$

Answer 1

Answer: (C)

$\frac{u^2}{4gsin\theta }$

When block is coming down the plane, it has constant velocity which means total force on it is balanced.

$\mu mgcos\theta =mgsin\theta$ ........(1)

When it is projected up the plane net force acting downwards

$ma=-\mu mgcos\theta -mgsin\theta$

Total downward acceleration using equation (1)
$a=-gsin\theta -gsin\theta =-2gsin\theta$
$\therefore v^2-u^2=2as$
$\Rightarrow 0-u^2=2\times (-2gsin\theta )\times S$
$\Rightarrow S=\frac{u^2}{4gsin\theta }$