A block starts moving up an inclined plane of inclination $30^{\circ}$ with an initial velocity of $v_{0}$. It comes back to its initial
position with velocity $\frac{v_{0}}{2}$. The value of the coefficient of kinetic friction between the block and the inclined plane is
close to $\frac{I}{1000}$. The nearest integer to $I$ is
(346)
Acceleration of block while moving up an inclined plane,
$a_{1}=g \sin \theta+\mu g \cos \theta$
$\Rightarrow a_{1}=g \sin 30^{\circ}+\mu g \cos 30^{\circ}$
$=\frac{g}{2}+\frac{\mu g \sqrt{3}}{2}$ ..(i) $\left(\because \theta=30^{\circ}\right)$
Using $v^{2}-u^{2}=2 a(s)$
$\Rightarrow v_{0}^{2}-0^{2}=2 a_{1}(s)$
$\Rightarrow v_{0}^{2}-2 a_{1}(s)=0$ $(\because u=0)$
$\Rightarrow s=\frac{v_{0}^{2}}{a_{1}}$ ...(ii)
Acceleration while moving down an inclined plane
$a_{2}=g \sin \theta-\mu g \cos \theta$
$\Rightarrow a_{2}=g \sin 30^{\circ}-\mu g \cos 30^{\circ}$
$\Rightarrow a_{2}=\frac{g}{2}-\frac{\mu \sqrt{3}}{2} g$ ...(3)
Using again $v^{2}-u^{2}=2 a s$ for downward motion
Equating equation (ii) and (iv)
$\frac{v_{0}^{2}}{a_{1}}=\frac{v_{0}^{2}}{4 a_{2}} \Rightarrow a_{1}=4 a_{2}$
$\Rightarrow \frac{g}{2}+\frac{\mu g \sqrt{3}}{2}=4\left(\frac{g}{2}-\frac{\mu \sqrt{3}}{2}\right)$
$\Rightarrow 5+5 \sqrt{3} \mu=4(5-5 \sqrt{3} \mu)$ (Substituting, $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )
$\Rightarrow 5+5 \sqrt{3} \mu=20-20 \sqrt{3} \mu \Rightarrow 25 \sqrt{3} \mu=15$
$\Rightarrow \mu=\frac{\sqrt{3}}{5}=0.346=\frac{346}{1000}$
So, $\frac{I}{1000}=\frac{346}{1000}$