Question

Starting from rest, a body slides down a ${45}^{\circ }$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of kinetic friction between the body and the inclined plane is

• A. $0.33$
• B. $0.25$
• C. $0.75$
• D. $0.80$

Answer 1

The correct option is C $0.75$

Acceleration without friction $a=g\sin \theta$
Acceleration with friction $a_f=g\sin \theta -\mu g\cos \theta$
Now time taken $t=\sqrt{\frac{2h}{a}}$ where $h$ is the vertical distance down the inclined plane, which doesn't change.

Thus, $\frac{T}{T_f}=\sqrt{\frac{a_f}{a}}$
$=\sqrt{\frac{g\sin \theta -\mu g\cos \theta }{g\sin \theta }}$
$=\sqrt{1-\mu \cot \theta }$

Given $T_f=2T$
$\Rightarrow \frac{1}{2}=\sqrt{1-\mu \cot \theta }$
$\Rightarrow \frac{1}{4}=1-\mu \cot \theta$
$\Rightarrow 1-\frac{1}{4}=\mu \cot \theta$
$\frac{3}{4}=\mu \cot {45}^{\circ }=\mu$
$\therefore \mu =\frac{3}{4}=0.75$