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 $\dfrac{I}{1000}$. The nearest integer to $I$ is ______.
Numerical value type. Enter your answer.
Answer: 346
Going up: retardation $a_1 = g(\sin\theta + \mu\cos\theta)$; coming down: acceleration $a_2 = g(\sin\theta - \mu\cos\theta)$. Over the same distance $d$:
$$v_0^{2} = 2a_1d,\qquad \frac{v_0^{2}}{4} = 2a_2d \Rightarrow a_1 = 4a_2$$
$\sin\theta + \mu\cos\theta = 4\sin\theta - 4\mu\cos\theta \Rightarrow \mu = \dfrac35\tan30^\circ = \dfrac{3}{5\sqrt3} \approx 0.346$.
So $I = 346$.
Solution by Sreeraj P, M.Sc Physics