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 in covering the same distance, 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:

$F=ma\Rightarrow (mg\sin \theta +f)=m{a}_1$
$mg\sin \theta +\mu mg\cos \theta =m{a}_1$
$a_1=g(\sin \theta +\mu \cos \theta )$
$s_1=ut+\frac{1}{2}a{t}^2$ ($u$ is the initial velocity)
$s_1=\frac{1}{2}g(\sin \theta +\mu \cos \theta )t_1^2$;
Down the plane:

$F=ma\Rightarrow (mg\sin \theta -f)=m{a}_2$
$mg\sin \theta -\mu mg\cos \theta =m{a}_2$
$a_2=g(\sin \theta -\mu \cos \theta )$
$s_2=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s_2=\frac{1}{2}g(\sin \theta -\mu \cos \theta )t_2^2$
Given: $t_1=\frac{t_2}{2};\Rightarrow \frac{1}{2}g(\sin \theta +\mu \cos \theta )t_1^2$ $=\frac{1}{2}g(\sin \theta -\mu \cos \theta )t_2^2$
$\Rightarrow (\sin \theta +\mu \cos \theta )=(\sin \theta -\mu \cos \theta )4\Rightarrow \mu =\frac{3}{5}\tan \theta$
Since $\theta ={45}^{\circ }$
$\mu =\frac{3}{5}\tan {45}^{\circ }=\frac{3}{5}$