A body at rest slides down a 30^ inclined plane. The time taken by...

When a plane is inclined to the horizontal at an angle $\theta$, which is greater than the angle of repose, then the body placed on the inclined plane slides down with an acceleration a. From figure, $R=m g \cos \theta$.....(i) Net force on the body down the inclined plane $f=m g \sin \theta-F$....(ii) i.e. $\quad f=m a=m g \sin \theta-\mu R \quad\left[\because \mu=\frac{F}{R}\right]$$\begin{aligned} \therefore \quad m a & =m g \sin \theta-\mu m g \cos \theta \quad \text{ [from Eq. (i)] } \\ & =m g(\sin \theta-\mu \cos \theta) \\ \Rightarrow \quad a & =g(\sin \theta-\mu \cos \theta)\end{aligned}$$$Time taken by the body to slide down the plane$$\begin{aligned} & t_1=\sqrt{\frac{2 s}{a}}=\sqrt{\frac{2 s}{(g \sin \theta-\mu g \cos \theta)}} \\ & t_2=\sqrt{\frac{2 s}{g \sin \theta}} \quad \text{ [in absence of friction] }\end{aligned}\because \quad t_1=2 t_2 \Rightarrow \quad t_1^2=4 t_2^2$ [given]\begin{aligned} & \left|\sqrt{\frac{2 s}{(g \sin \theta-\mu g \cos \theta)}}\right|^2=4 \times\left|\sqrt{\frac{2 s}{g \sin \theta}}\right|^2 \\ \Rightarrow & \frac{2 s}{g(\sin \theta-\mu \cos \theta)}=4 \times \frac{2 s}{g \sin \theta} \\ \Rightarrow & \quad \sin \theta=4 \sin \theta-4 \mu \cos \theta \\ \Rightarrow & \mu=\frac{3}{4} \tan \theta=\frac{3}{4} \tan 30^{\circ}=0.43\end{aligned}$$