A block starts moving up an inclined plane of inclination
Question:

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

Solution:

(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, $\frac{I}{1000}=\frac{346}{1000}$

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