Question

A small mass slides down an inclined plane of inclination $\theta$ with the horizontal. The coefficient friction is $\mu ={\mu }_{0}x$ where $x$ is the distance through which the mass slides down and ${\mu }_0$ is a positive constant. Then the distance covered by the mass before it stops 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 $\frac{2}{{\mu }_0}\tan \theta$

$F_{net}=mg\sin \theta -\mu mg\cos \theta$
$=mg\sin \theta -{\mu }_{0}xg\cos \theta$
$a=\frac{F_{net}}{m}=g\sin \theta -{\mu }_{0}xg\cos \theta$
$\therefore v.\frac{dv}{dx}=g\sin \theta -{\mu }_{0}xg\cos \theta$
$or{\int }_0^{0}vdv={\int }_0^{x_m}(g\sin \theta -{\mu }_{0}xg\cos \theta )dx$
$\text{Solving this equation we get,}$
$x_m=\frac{2}{{\mu }_0}\tan \theta$