Question

The ratio of time taken by a body to cover an equal distance while sliding down a smooth to rough identical inclined plane of angle $\theta$ is $‘n^{\prime }$. Find the coefficient of friction.

• A. $\frac{\tan \theta }{(1-n^2)}$
• B. $\cot \theta (1-n^2)$
• C. $\sin \theta (1-n^2)$
• D. $\tan \theta (1-n^2)$

Answer 1

The correct option is D $\tan \theta (1-n^2)$

For a smooth inclined plane friction is zero and,
$F=ma\Rightarrow mg\sin \theta =ma$
$a_1=g\sin \theta$
Using the second equation of motion,
$s_1=\frac{1}{2}g\left(\sin \theta \right)t_1^2$


For rough inclined plane:
$F=m{a}_2$
$mg\sin \theta -f=m{a}_2$
Since $f=\mu N=\mu \left(mg\cos \theta \right)$
$mg\sin \theta -\mu \left(mg\cos \theta \right)=m{a}_2$
$\Rightarrow a_2=g(\sin \theta -\mu \cos \theta )$
Hence, $s_2=\frac{1}{2}g(\sin \theta -\mu \cos \theta )t_2^2$; Given: $s_1=s_2$ and $\frac{t_1}{t_2}=n\Rightarrow (1-n^2)\sin \theta =\mu \cos \theta$ $\Rightarrow \mu =\tan \theta (1-n^2)$