Question

An object is placed on the surface of a smooth inclined plane of inclination $\theta .$ It takes time $t$ to reach the bottom. If the same object is allowed to slide down a rough inclined plane of inclination $\theta ,$ it takes time $nt$ to reach the bottom where $n$ is the number greater than one. The coefficient of friction $\mu$ is given by

• A. $\mu =\tan \theta \left[1-\frac{1}{n^2}\right]$
• B. $\mu =\cot \theta \left[1-\frac{1}{n^2}\right]$
• C. $\mu =\tan \theta {\left[1-\frac{1}{n^2}\right]}^{1/2}$
• D. $\mu =\cot \theta {\left[1-\frac{1}{n^2}\right]}^{1/2}$

Answer 1

Answer: (A)

On smooth inclined plane, the acceleration of body down the plane $=g\sin \theta .$ For rough inclined plane, the downward acceleration of body $=g\sin \theta -\mu g\cos$ $\theta$. Let $l$ be the length of inclined plane, then
$l=\frac{1}{2}(g\sin \theta )t^2\text{ and }l=\frac{1}{2}(g\sin \theta -\mu g\cos \theta )(nt{)}^2$
$\therefore \frac{1}{2}(g\sin \theta )t^2=\frac{1}{2}(g\sin \theta -\mu g\cos \theta )(nt{)}^2$
or $\sin \theta =(\sin \theta -\mu \cos \theta )n^2=n^2\sin \theta -\mu n^2\cos \theta$
or $\mu n^2\cos \theta =\left(n^2-1\right)\sin \theta$
or $\mu =\frac{\left(n^2-1\right)\sin \theta }{n^2\cos \theta }=\tan \theta \left(1-\frac{1}{n^2}\right)$