Question

A smooth block is released from rest on an incline plane of inclination $\theta$ and length $l$. The time taken to slide down the rough incline is $\alpha$ times the time taken to slide down the smooth incline. The coefficient of friction is

• A. ${\mu }_k=\frac{{\alpha }^2-1}{{\alpha }^2\tan \theta }$
• B. ${\mu }_k=\frac{{\alpha }^2-1}{{\alpha }^2\cot \theta }$
• C. ${\mu }_k=\sqrt{\frac{{\alpha }^2-1}{{\alpha }^2\tan \theta }}$
• D. ${\mu }_k=\sqrt{\frac{{\alpha }^2-1}{{\alpha }^2\cot \theta }}$

Answer 1

The correct option is B ${\mu }_k=\frac{{\alpha }^2-1}{{\alpha }^2\cot \theta }$

When the surface is smooth, acceleration along inclined plane,

$a=g\sin \theta \text{(constant)}$

Length of the incline $=l$

Using $s=ut+\frac{1}{2}a{t}^2$, we can write that,

$l=(0\times t)+\frac{1}{2}\times \left(g\sin \theta \right)t_1^2$

$\Rightarrow t_1=\sqrt{\frac{2l}{g\sin \theta }}......\left(1\right)$

When the surface is rough, acceleration along the incline is given by,

$a=g\sin \theta -\mu g\cos \theta$

where $\mu$ is the coefficient of friction,

Again using, $s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow l=(0\times t)+\frac{1}{2}(g\sin \theta -\mu g\cos \theta )t_2^2$

$\Rightarrow t_2=\sqrt{\frac{2l}{g\sin \theta -\mu g\cos \theta }}.......\left(2\right)$

From the data given in the question ,

$t_2=\alpha t_1$

From $\left(1\right)$ and $\left(2\right)$ ,

$\sqrt{\frac{2l}{g\sin \theta -\mu g\cos \theta }}=\alpha \sqrt{\frac{2l}{g\sin \theta }}$

$\Rightarrow \frac{1}{g\sin \theta -\mu g\cos \theta }={\alpha }^2\frac{1}{g\sin \theta }$

$\Rightarrow {\alpha }^2=\frac{1}{1-\mu \cot \theta }$

$\Rightarrow {\alpha }^2-\mu {\alpha }^2\cot \theta =1$

$\Rightarrow \mu =\frac{{\alpha }^2-1}{{\alpha }^2\cot \theta }$

Since block is moving, so we can write that,

${\mu }_k=\frac{{\alpha }^2-1}{{\alpha }^2\cot \theta }$

Thus, option (b) is the correct answer.