A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches ${30}^0$ the box starts to slip and slides $4.0 m$down the plank in $4.0 \mathrm{s}$. The coefficients of static and kinetic friction between the box and the plank will be, respectively

Let ${\mu }_s$ and ${\mu }_k$ be the coefficienls of static and kinetic friction between the box and the plank respectively.When the angle of inclination $\theta$ reaches ${30}^0$, the block just slides,

${\mu }_s= \tan \theta = \tan {30}^0 = \frac{1}{\sqrt{3}} = 0.6$

If $a$ is the acceleration produced in the block, then 

$ma = mg \sin \theta -f_k$

(where $f_k$ is the force of kinetic friction) 

$$\begin{gathered} = mg \sin \theta - {\mu }_{k}N \\ = mg \sin \theta -{\mu }_{k}mg \cos \theta \\ a =g(\sin \theta -{\mu }_k\cos \theta ) \\ As, g= 10 m{s}^{-2} and \theta = {30}^0 \\ a = (10 m{s}^{-2})(\sin {30}^0 - {\mu }_k \cos {30}^0) \end{gathered}$$

If $s$ is the distance travelled by the block in time $t$, then 

$s = \frac{1}{2}a{t}^2 (as u = 0) or a = \frac{2s}{t^2}$

But $s = 4.0 m and t = 4.0 s \left(given\right)$

$a = \frac{2(4.0 m)}{{\left(4.0s\right)}^2} = \frac{1}{2}m{s}^{-2}$

Substituting this value of $a$ in eqn $\left(i\right)$, we get ,

$$\begin{gathered} \frac{1}{2}m{s}^{-2} = (10 m{s}^{-2})\left(\frac{1}{2}-{\mu }_k\frac{\sqrt{3}}{2}\right) \\ \frac{1}{10} = 1-\sqrt{3}{\mu }_k or, \sqrt{3}{\mu }_k =1-\frac{1}{10} = \frac{9}{10} = 0.9 \\ {\mu }_k =\frac{ 0.9}{\sqrt{3}} = 0.5 \end{gathered}$$