#### A block starts moving up an inclined plane of inclination $30^{\circ}$ with an initial velocity of $\nu _{0}$. It comes back to its initial position with velocity $\frac{\nu _{0}}{2}.$ 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 _______. Option: 1 346 Option: 2 12345 Option: 3 7689 Option: 4 12341

Let the length of the inclined plane as S (as shown in the figure)

For upward motion

\begin{aligned} &\text { from work energy theorem }\\ &W_{f}=k_{f}-k_{i} \end{aligned}

So

$\frac{1}{2} m v_{0}^{2}=\left(m g \frac{S}{2}+\mu m g \frac{\sqrt{3}}{2} S\right)....(1)$

For downward motion

\begin{aligned} &\text { from work energy theorem }\\ &W_{f}=k_{f}-k_{i} \end{aligned}

$\frac{1}{2} \frac{m v_{0}^{2}}{4}=\left(m g \frac{S}{2}-\mu m g \frac{\sqrt{3}}{2} \cdot S\right).......(2)$

From equation (1) and (2)

we get

$\begin{array}{l} 4=\frac{(1+\mu \sqrt{3})}{(1-\mu \sqrt{3})} \\ \\ \Rightarrow 4-4 \mu \sqrt{3}=1+\mu \sqrt{3} \\ \Rightarrow 3=5 \mu \sqrt{3} \\ \\ \therefore \mu=\frac{\sqrt{3}}{5} \approx \frac{346.41}{1000} \\ I=346 \end{array}$