Question

A small mass ${}^{\prime }{m}^{\prime }$ slides down an inclined plane of inclination $\theta$ with $\mu ={\mu }_{0}x$ where ${}^{\prime }{x}^{\prime }$ is the distance through which the mass slides down and ${\mu }_0$ is a 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 $\frac{2}{{\mu }_0}tan\theta$

$F_{net}=mg\sin \theta -\mu mg\cos \theta$

$=mg\sin \theta -{\mu }_{0}xmg\cos \theta$

$a=\frac{F_{net}}{m}=g\sin \theta -{\mu }_{0}xg\cos \theta$

$\therefore v\cdot \frac{dv}{dx}=g\sin \theta -{\mu }_{0}xg\cos \theta$

or ${\int }_0^{0}vdv={\int }_o^{x_m}(g\sin \theta -{\mu }_{0}xg\cos \theta )dx$

Solving this equation we get,

$X_m=\frac{2}{{\mu }_0}\tan \theta$.