Question

A small mass slides down an inclined plane of inclination $\theta$ with the horizontal. The coefficient of friction is $\mu ={\mu }_{o}x$ , where $x$ is the distance through which the mass slides down and ${\mu }_o$ 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

Answer: (A)

By equilibrium of block perpendicular to the incline, $N=mgcos\theta$
So kinetic friction is, $f_k=\mu N=\mu mgcos\theta$
Work done by constant force is given as, $W=FScos\theta$
Let distance covered by the mass is $x$ before coming at rest.
So, $W_g=mg\times x\times cos\left(90^{\circ }-\theta \right)=mgxsin\theta$
$W_N=Nxcos90^{\circ }=0$
Work done by variable force is given as $W={\int }_{x_1}^{x_2}Fdx$
So, $W_f={\int }_0^x{f}_{k}dx=-{\int }_0^x\left({\left(\mu \right)}_{0}xmgcos\theta \right)dx$
$\text{ so }{W}_f=-{\left(\mu \right)}_{0}mgcos\theta \left(\frac{x^2}{2}\right)$
By work energy theorem
$W_g+W_N+W_f=\Delta K$
$mgxsin\theta +0-{\left(\mu \right)}_{0}mgcos\theta \left(\frac{x^2}{2}\right)=0$
$x=\frac{2}{{\mu }_0}tan\theta$