Question

A body slides down a rough inclined plane of angle of inclination ${30}^0$ and takes time twice as great as the time taken in slipping down a identical frictionless plane. Find the coefficient of friction between the body and the plane.

• A. $\sqrt{3}/4$
• B. $\sqrt{3}$
• C. 4/3
• D. 3/4

Answer 1

The correct option is A $\sqrt{3}/4$

Let $s$ be the length of the inclined plane.
Force equation along the plane without friction
$ma=mgsin{30}^o$
$a=gsin{30}^o$

Let the time taken by the body to reach the bottom be $t$
$s=\frac{1}{2}gsin{30}^o{t}^2$

Force equation along the plane, with friction is
$ma=mgsin{30}^o-\mu mgcos{30}^o$
$a=g(sin{30}^o-\mu cos{30}^o)$

Let the time taken by the body to reach the bottom of the plane be $t^{\prime }=2t$
$s=\frac{1}{2}g(sin{30}^o-\mu cos{30}^o)t^{\prime 2}$
$s=\frac{1}{2}g(sin{30}^o-\mu cos{30}^o)4t^2$

Equating both the $s$ equations
$\frac{1}{2}g(sin{30}^o-\mu cos{30}^o)4=\frac{1}{2}gsin{30}^o$
$(sin{30}^o-\mu cos{30}^o)4=sin{30}^o$
$3\left(\frac{1}{2}\right)=\mu 2\sqrt{3}$
$\mu =\frac{\sqrt{3}}{4}$