Question

A piece of ice starting from rest slides down a rough ${45}^{\circ }$ incline in twice the time it takes to slide down a frictionless ${45}^{\circ }$ incline. What is the coefficient of friction between the ice and incline?

• A. $0.25$
• B. $0.50$
• C. $0.75$
• D. $0.40$

Answer 1

The correct option is C $0.75$

In both the case, the initial speed $u$ and the distance $s$ travelled is
According to formula: $s=ut+\frac{1}{2}a{t}^2$
$a{t}^2$ is equal (as $u=0$ in both the cases).
Let $a_1$ be the acceleration of the ice block on rough incline and $t_1$ be the time taken to slide down this rough plane. Similarly, let $a_2$ and $t_2$ be the similar quantities in case of frictionless inclined plane.
Given that $t_1=2t_2$
Further, $\frac{a_1}{a_2}=\frac{t_2^2}{t_1^2}$ ($\because a{t}^2=$ constant)
$\Rightarrow \frac{a_1}{a_2}=\frac{t_2^2}{4t_2^2}=\frac{1}{4}$ ... (1)
But $a_1=g(\sin \theta -\mu \cos \theta )$ ... (2)
and $a_2=g\sin \theta$ ... (3)
$\therefore \frac{g(\sin \theta -\mu \cos \theta )}{g\sin \theta }=\frac{1}{4}$
$\Rightarrow 4\sin \theta -4\mu \cos \theta =\sin \theta$
$\Rightarrow 3\sin \theta =4\mu \cos \theta$
$\therefore \mu =\frac{3}{4}\tan \theta =\frac{3}{4}\tan {45}^{\circ }\Rightarrow \mu =\frac{3}{4}=0.75$
Hence, the correct answer is option (c).