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°, the box starts to slip and slides 4.0m down the plank in 4.0 s. The coefficients of static and kinetic friction between the box and the plank will be, respectively

 

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

$\therefore {\mu }_s=\tan \theta =\tan 30°=\frac{1}{\sqrt{3}}=0.6$

If $\alpha$ is the acceleration produced in the block, then

$ma=mg\sin \theta -f_k$

$\text{ (where }{f}_k\text{ is force of kinetic friction) }$

$=mg\sin \theta -{\mu }_{k}N \left(\text{ as }{f}_k={\mu }_{k}N\right)$

$=mg\sin \theta -{\mu }_{k}mg\cos \theta \text{ (as }\left.N=mg\cos \theta \right)$

$a=g\left(\sin \theta -{\mu }_k\cos \theta \right)$

$\text{ As }g=10{ \mathrm{ms}}^{-2}\text{and }\theta =30°$

$\therefore a=\left(10{ \mathrm{ms}}^{-2}\right)\left(\sin 30°-{\mu }_k\cos 30°\right)$

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

$s=\frac{1}{2}a{t}^2 \text{ (as }\left.u=0\right)$

$\text{ or }a=\frac{2s}{t^2}$

But  s = 4.0 m and  t = 4.0 s (given)

$\therefore a=\frac{2(4.0 \mathrm{m})}{(4.0 \mathrm{s}{)}^2}=\frac{1}{2}{ \mathrm{ms}}^{-2}$

Substituting this value of $\alpha$ in eqn. (i), we get 

$\frac{1}{2}{ \mathrm{ms}}^{-2}=\left(10{ \mathrm{ms}}^{-2}\right)\left(\frac{1}{2}-{\mu }_k\frac{\sqrt{3}}{2}\right)$

$\frac{1}{10}=1-\sqrt{3}{\mu }_k$

$\text{ or }\sqrt{3}{\mu }_k=1-\frac{1}{10}=\frac{9}{10}=0.9$

${\mu }_k=\frac{0.9}{\sqrt{3}}=0.5$