Question

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

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

Answer 1

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

Free body diagram:

From free body diagram,
Net force acting downwards parallel to the inclined plane is,

$F_{net}=mg\sin \theta -f_k$

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

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

Thus, $F_{net}=mg\sin \theta -{\mu }_{0}xmg\cos \theta ....\left(1\right)$

From Newton's second law of motion, $F_{net}=ma$

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

$\Rightarrow a=g\sin \theta -{\mu }_{0}xg\cos \theta$

Now, substituting $a=v\frac{dv}{dx}$, we have,

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

Let the body stops at distance $x$.

Since, the intial and final velocities are zero on integrating we get,

${\int }_{v_i=0}^{v_f=0}vdv={\int }_0^x(g\sin \theta -{\mu }_{0}xg\cos \theta )dx$

$\Rightarrow 0=g\sin \theta \times x-{\mu }_{0}g\cos \theta \times \frac{x^2}{2}$

$\Rightarrow \sin \theta ={\mu }_o\cos \theta \times \frac{x}{2}$

$\therefore x=\frac{2}{{\mu }_o}\tan \theta$

Hence, option (a) is the correct answer.