A body is sliding down a rough inclined plane of angle of inclination $\theta$ for which the coefficient of friction varies with distance x as $\mu (x)=kx$, where k is a constant . Here x is the distance moved by the body down the plane . The net force on the body will be zero at a distance $x_0$ given by :

Question

$\begin{array}{1 1}\frac{\tan \theta}{k}\\k\;tan\;\theta\\\frac{\cot \theta}{k} \\k\;cot \theta\end{array} $

Answer 1

Answer : $\large\frac{\tan \theta}{k}$

The net downward force on the body at a distance x is

http://clay6.com/mpaimg/ph%2021.jpg

$f(x) =mg \sin \theta - \mu mg \cos \theta$

$\qquad= mg( \sin \theta - \mu \cos \theta)$

$\qquad= mg (\sin \theta- kx \cos \theta)$

$\therefore f(x) =0$ at a value of $x =x_0$ given by

$\sin \theta -kx_0 \cos \theta=0$

Which gives $x_0 = \large\frac{\tan \theta}{k}$