Question

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 ${30}^{\circ }$, the box starts to slip and slides $4.0\text{m}$ down the plank in $4.0\text{s}$. The coeffficient of static and kinetic friction between the box and the plank will respectively be

• A. $0.4$ and $0.3$
• B. $0.6$ and $0.6$
• C. $0.6$ and $0.5$
• D. $0.5$ and $0.6$

Answer 1

The correct option is C $0.6$ and $0.5$

Static friction
${\mu }_s=\tan \theta$ [$\theta$-At which it starts to slide]
${\mu }_s=\tan {30}^{\circ }=\frac{1}{\sqrt{3}}=0.6$
Now as the body slips down, acceleration $=g\sin \theta -{\mu }_{k}g\cos \theta$
$=g\sin {30}^{\circ }-{\mu }_{k}g\cos {30}^{\circ }$
$s=ut+\frac{1}{2}a{t}^2$
$4=0+\frac{1}{2}[g\left(\frac{1}{2}\right)-\frac{\mu \times g\sqrt{3}}{2}]t^2$
$4=\frac{1}{2}[\frac{g}{2}-\frac{{\mu }_{k}g\sqrt{3}}{2}]16$
$\frac{1}{2}=\frac{1}{2}\left(g\right)[1-\sqrt{3}{\mu }_k]$
$1-\sqrt{3}{\mu }_k=\frac{1}{10}$
$1-\sqrt{3}{\mu }_k=0.1$
${\mu }_k=\frac{0.9}{\sqrt{3}}=0.52$