Question

A body of mass $m$ is launched up a rough inclined plane of angle ${45}^{\circ }$ with the horizontal. If the time of ascent is half the time of descent, the coefficient of friction is

• A. $\frac{2}{5}$
• B. $\frac{3}{5}$
• C. $\frac{3}{4}$
• D. $\frac{5}{8}$

Answer 1

The correct option is B $\frac{3}{5}$

Up the plane:
$s=ut-\frac{1}{2}a{t}^2$ ($v$ is the final velocity) and
$a=-g(\sin \theta +\mu \cos \theta )$
$s=\frac{1}{2}g(\sin \theta +\mu \cos \theta )t_1^2$;

Down the plane: $s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s=\frac{1}{2}g(\sin \theta -\mu \cos \theta )t_2^2$
Given: $t_1=\frac{t_2}{2};\Rightarrow \frac{1}{2}g(\sin \theta +\mu \cos \theta )t_1^2=\frac{1}{2}g(\sin \theta -\mu \cos \theta )t_2^2$
$\Rightarrow (\sin \theta +\mu \cos \theta )=(\sin \theta -\mu \cos \theta )4$
Substituting $\theta ={45}^{\circ }$
$\Rightarrow 1+\mu =(1-\mu )4$
$\Rightarrow \mu =\frac{3}{5}$