Question

A block slides down an inclined plane of inclination ${30}^{\circ }$ with a constant acceleration $g/4$ .Coefficient of kinetic friction is given by $\frac{1}{N\sqrt{3}}$. Then, the value of $N$ is

Answer 1

Let the coefficient of kinetic friction be $\mu$.


On applying equilibrium condition along perpendicular to the inclined plane.
$N-mg\cos {30}^{\circ }=0$
$\Rightarrow N=mg\cos {30}^{\circ }...\left(1\right)$
On applying $\sum F=ma$ along the inclined plane,
$mg\sin {30}^{\circ }-f=ma$
where $f=\mu N$ is the kinetic friction force.
$\Rightarrow mg\sin {30}^{\circ }-\mu N=\frac{mg}{4}$
$\because [a=\frac{g}{4}]$
$\Rightarrow mg\sin {30}^{\circ }-\mu mg\cos {30}^{\circ }=\frac{mg}{4}$
$\Rightarrow \frac{1}{2}=\frac{1}{4}+(\mu \times \frac{\sqrt{3}}{2})$
$\Rightarrow \frac{1}{4}=\frac{\sqrt{3}\mu }{2}$
$\Rightarrow \mu =\frac{1}{2\sqrt{3}}$

But $\mu =\frac{1}{N\sqrt{3}}$ [Given]
On comparing, we get $N=2$