Question

A body of mass $m$ is launched up on a rough inclined plane making an angle of ${30}^{\circ }$ with the horizontal. The coefficient of friction between the body and plane is $\frac{\sqrt{x}}{5}$ if the time of ascent is half of the time of descent. The value of $x$ is ______.

Answer 1

Given:
$t_a=\frac{1}{2}{t}_d$

Let's assume, the length of inclined plane is $s$.

$s=\frac{1}{2}a{t}^2\Rightarrow t=\sqrt{\frac{2s}{a}}$

So from given condition,

$\sqrt{\frac{2s}{a_a}}=\frac{1}{2}\sqrt{\frac{2s}{a_d}}.........\left(1\right)$


Here, friction force, $f=\mu mg\cos \theta$

$a_a=g\sin \theta +\mu g\cos \theta$

$a_a=\frac{g}{2}+\frac{\sqrt{3}}{2}\mu g(\because \theta ={30}^{\circ })$

$a_d=g\sin \theta -\mu g\cos \theta$

$a_d=\frac{g}{2}-\frac{\sqrt{3}}{2}\mu g$

using the above values of $a_a$ and $a_d$, and putting in equation $\left(1\right)$, we will get

$\mu =\frac{\sqrt{3}}{5}$

$\therefore \text{Correct answer :}3$