Question

A small mass slides down an inclined plane of inclination $\theta$ with the horizontal. The co-efficient of friction is $\mu ={\mu }_{0}x$ where '$x$' is the distance through which the mass slides down and '${\mu }_0$' is a constant Then the distance covered by the mass before it stop is :

• A. $\frac{2}{{\mu }_0}tan\theta$
• B. $\frac{4}{{\mu }_0}tan\theta$
• C. $\frac{1}{2{\mu }_0}tan\theta$
• D. $\frac{1}{{\mu }_0}tan\theta$

Answer 1

The correct option is A $\frac{2}{{\mu }_0}tan\theta$

Lets assume, mass = $m$

Contact force on the mass

$N=mgcos\theta$

Force acting along the inclined surface in the direction of mass motion

$F_1=mgsin\theta$

At distance $x$, friction force on the mass which is acting opposite to the direction of mass motion

$f_r={\mu }_{0}mgxcos\theta$

Lets assume that the mass moves $dx$ on the inclined surface.

The mass gets stop as work done by $F_1$, gravitational force, is equal to work done by $f_r$, frictional force.

Work done by $F_1$

$W_{F1}={\int }_0^{x}mgsin\theta dx$

$=mgxsin\theta$

Work done by $f_r$

$W_{fr}={\int }_0^x{\mu }_{0}mgxcos\theta$

$=\frac{1}{2}{\mu }_{0}mg{x}^{2}cos\theta$

Since, $W_{F1}=W_{fr}$

$⟹mgxsin\theta =\frac{1}{2}{\mu }_{0}mg{x}^{2}cos\theta$

$⟹x=\frac{2}{{\mu }_0}tan\theta$

This is the distance covered by the mass before it stop.