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Q. 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_______.

JEE MainJEE Main 2021Laws of Motion

Solution:

$t_{a}=\frac{1}{2} t_{d}$
$\sqrt{\frac{2 s}{a_{a}}}=\frac{1}{2} \sqrt{\frac{2 s}{a_{d}}} \ldots \ldots .$(i)
$a_{a}=g \sin \theta+\mu g \cos \theta $
$=\frac{g}{2}+\frac{\sqrt{3}}{2} \mu g $
$a_{d}=g \sin \theta-\mu g \cos \theta$
$=\frac{g}{2}-\frac{\sqrt{3}}{2} \mu g$
using the above values of $a_{a}$ and $a_{d}$ and putting ineqution (i) we will gate
$\mu=\frac{\sqrt{3}}{5}$