Question

A body slides down an inclined plane of inclination $\theta$.
The coefficient of friction down the plane varies as $\mu =\alpha x.$ Here, $\alpha$ is a positive constant and x is the distance moved by the body down the plane. The kinetic energy (K) versus distance (x) graph will be as

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Initially downward acceleration of body is g sin $\theta$. At distance x, the acceleration is a = g sin $\theta -\alpha$ xg cos $\theta ,$ i.e., net acceleration gradually decreases and it becomes zero at, $x=tan\theta /\alpha ,$
For $x<tan\theta /\alpha ,a=gsin\theta -\alpha xgcos\theta$
$v\left(dv/dx\right)=gsin\theta -\alpha xgcos\theta$
or ${\int }_0^{v}vdv={\int }_0^x(gsin\theta -\alpha xgcos\theta )dx$
or $\frac{v^2}{2}=gxsin\theta -\frac{g\alpha x^{2}cos\theta }{2}$
or $K=\frac{1}{2}m{v}^2=m(gxsin\theta -\frac{a{x}^{2}gcos\theta }{2})$
i.e., K versus x graph will be a parabola till K becomes constant.
Hence, the correct answer is option (a).