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}=μN=μmgcos\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 $W=\displaystyle \int _{x_{1}}^{x_{2}}Fdx$
So, $W_{f}=\displaystyle \int _{0}^{x}f_{k}dx=-\displaystyle \int _{0}^{x}\left(\left(\mu \right)_{0} x mg cos\theta \right)dx$
$\text{ so }W_{f}=-\left(\mu \right)_{0}mgcos\theta \left(\frac{x^{2}}{2}\right)$
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 $