Question 170 of 210

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 $\dfrac{{v}_{0}}{2}.$ The value of the coefficient of kinetic friction between the block and the inclined plane is close to $\dfrac{1}{1000},$ The nearest integer to $I$ is :

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Answer: 346

$AtoB$
${a}_{1}=g\sin 30^{\circ}+\mu g\cos 30^{\circ}$
$=\dfrac{g}{2}+\dfrac{\mu g\sqrt{3}}{2};g=10m{s}^{2}$
${v}_{0}^{2}-2{a}_{1}\left(s\right)=0$
$s=\dfrac{{v}_{0}^{2}}{{a}_{1}}$ .....(i)
$BtoA$
${a}_{2}=\dfrac{g}{2}-\dfrac{\mu \sqrt{3}}{2}g$
${\left(\dfrac{{V}_{0}}{2}\right)}^{2}=2{a}_{2}\left(s\right)$
$s=\dfrac{{V}_{0}^{2}}{4{a}_{2}}$ .....(ii)
From equation (i) and (ii)
$\dfrac{{V}_{0}^{2}}{{a}_{1}}=\dfrac{{V}_{0}^{2}}{4{a}_{2}}$
$\Rightarrow {a}_{1}=4{a}_{2}$
$\Rightarrow 5+5\sqrt{3}\mu =4\left\{5-5\sqrt{3}\mu \right\}$
$\Rightarrow 5+5\sqrt{3}\mu \Rightarrow 25\sqrt{3}\mu =15\Rightarrow \mu =\dfrac{\sqrt{3}}{5}=0.346=\dfrac{346}{1000}$
So, $\dfrac{1}{1000}=\dfrac{346}{1000}$