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 .

Answer 1

When the block is moving up, $mg\sin \theta$ and friction acts down the incline,

Therefore, Acceleration of block while moving up an inclined plane,
$a_1=g\sin \theta +\mu g\cos \theta$

$\Rightarrow a_1=g\sin {30}^{\circ }+\mu g\cos {30}^{\circ }$

$\Rightarrow a_1=\frac{g}{2}+\frac{\mu g\sqrt{3}}{2}$

Using $v^2=u^2+2a\left(s\right)$

Here,$v=0,u=v_0\text{and}{a}_1\to -ve\left(\text{downwards}\right)$
$\Rightarrow v_0^2-2a_1\left(s\right)=0$
$\Rightarrow s=\frac{v_0^2}{a_1}--\left(1\right)$

While the block is sliding down, friction acts up the incline,

Acceleration while moving down an inclined plane
$a_2=g\sin \theta -\mu g\cos \theta$

$\Rightarrow a_2=g\sin {30}^{\circ }-\mu g\cos {30}^{\circ }$

$\Rightarrow a_2=\frac{g}{2}-\frac{\mu \sqrt{3}}{2}g$
Again, Using $v^2-u^2=2as$ for downward motion,
Here, $u=0,v=\frac{v_0}{2}$

$\Rightarrow {\left(\frac{v_0}{2}\right)}^2=2a_2\left(s\right)\Rightarrow s=\frac{v_0^2}{4a_2}---\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, $a_1=4a_2$

$\Rightarrow \frac{g}{2}+\frac{\mu g\sqrt{3}}{2}=4(\frac{g}{2}-\frac{\mu \sqrt{3}g}{2})$

$\Rightarrow 1+\mu \sqrt{3}=4-4\mu \sqrt{3}$

$\Rightarrow \mu =\frac{\sqrt{3}}{5}=0.346=\frac{346}{1000}$

So, $I=346$

Hence, $346$ is the correct answer.