Question

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

• A. $\frac{\tan \theta }{k}$
• B. $k\tan \theta$
• C. $\frac{\cot \theta }{k}$
• D. $k\cot \theta$

Answer 1

The correct option is A $\frac{\tan \theta }{k}$

The net downward force along the inclined plane on the body at a
distance $x$ is:
$f\left(x\right)=mg\sin \theta -\mu mg\cos \theta$
$=mg(\sin \theta -\mu \cos \theta )$
$=mg(\sin \theta -kx\cos \theta )$
$\therefore f\left(x\right)=0$ at a value of $x=x_0$ given by
$\sin \theta -k{x}_0\cos \theta =0$ $\therefore x_0=\frac{tan\theta }{k}$