Question

A body of mass $m$ is released from the top of a rough inclined plane length $l$. If the frictional force is $f$, and the height of body from ground is $h$ then the velocity of the body of the bottom in ${ms}^{-1}$ will be :-

• A. $\sqrt{\frac{2}{m}(mgh-fl)}$
• B. $2gh-f/l$
• C. $\sqrt{\frac{2}{m}}gh$
• D. $zero$

Answer 1

The correct option is A $\sqrt{\frac{2}{m}(mgh-fl)}$

Here,$\sin \theta =\frac{h}{l}$
from FBD, $ma=mg\sin \theta -f\Rightarrow a=\frac{gh}{l}-\frac{f}{m}$
by using, $v^2-u^2=2aS$,
$v^2-0=2(\frac{gh}{l}-\frac{f}{m})l$
$v^2=\frac{2}{m}(mgh-fl)$
$\therefore v=\sqrt{\frac{2}{m}(mgh-fl)}$