Question

A given object takes $n$ times more time to slide down a $45^{\circ }$ rough inclined plane as it takes to slide down a perfectly smooth $45^{\circ }$ incline. The coefficient of kinetic friction between the object and the incline is :

• A. $\frac{1}{1-n^2}$
• B. $1-\frac{1}{n^2}$
• C. $\sqrt{1-\frac{1}{n^2}}$
• D. $\sqrt{\frac{1}{1-n^2}}$

Answer 1

Answer: (B)

Consider the following figures:

We know that for a body moving with constant acceleration, the kinematics equation is given as $s=ut+\frac{1}{2}a{t}^2$.
Initial velocity, $u=0\Rightarrow s=\frac{1}{2}a{t}^2$
$\Rightarrow 2s=a{t}^2$
$\Rightarrow t=\sqrt{\frac{2s}{a}}$
$\Rightarrow t\propto \frac{1}{\sqrt{a}}$
Now for smooth inclined plane $a_s=g\sin \theta$
For rough inclined plane $a_r=g\sin \theta -g\mu \cos \theta$
Also, time taken to travel down the smooth inclined plane $t_s=t$ and time taken to travel down the rough inclined plane $t_r=nt.$
Therefore, $\frac{t_s}{t_r}=\sqrt{\frac{a_t}{a_s}}$
$\Rightarrow \frac{t_s^2}{t_r^2}=\frac{a_t}{a_s}$
$\Rightarrow \frac{t_s^2}{t_r^2}=\frac{g\sin \theta -g\mu \cos \theta }{g\sin \theta }=1-\mu \tan \theta$
$\Rightarrow \frac{t^2}{n^2{t}^2}=1-\mu \tan \theta$
$\Rightarrow \frac{1}{n^2}=1-\mu \tan \theta$
$\Rightarrow \mu \tan \theta =1-\frac{1}{n^2}$
$\Rightarrow \mu =\left(1-\frac{1}{n^2}\right)\tan \theta$
Now we know $\theta =45^{\circ }\Rightarrow \tan 45^{\circ }=1$
Therefore,
$\mu =1-\frac{1}{n^2}$