Laws of Motion Part 1

Let’s draw the free body diagram when body slides down on smooth surface For smooth surface, ma = mg sin30º a = g sin30º a = g / 2 Distance covered by the block on the smooth surface in time T, $s=u t+\frac{1}{2} a t^{2}$ $s=0+\frac{1}{2} \left( \frac{g}{2} \right) T^{2}$ $\Rightarrow s=\left( \frac{g}{4} \right) T^{2}$ ...(i) Now, let’s draw the free body diagram when body slides down on rough surface For rough surface, ma = mg sin30º - μ mg cos30º a = g sin30º - μ g Distance covered by the block on the rough surface in time α T, $s=u t+\frac{1}{2} a t^{2}$ $s=0+\frac{1}{2} \left( g \sin 30^{\circ} - \mu g \sin 30\mu \right) t^{2}$ $s=\frac{g}{4} (1-\sqrt{3} \mu) (\alpha T)^{2}$ ...(ii) Distance covered by the block is same for both the case, $\Rightarrow \frac{g}{4} (1-\sqrt{3} \mu) (\alpha T)^{2} = \frac{g}{4} T^{2}$[from Eq. (i) & Eq. (ii)] $\Rightarrow 1-\sqrt{3} \mu = \frac{1}{\alpha^{2}}$ $\Rightarrow \mu = \left( \frac{\alpha^{2}-1}{\alpha^{2}} \right) \frac{1}{\sqrt{3}}$ Comparing with $\left( \frac{\alpha^{2}-1}{\alpha^{2}} \right) \frac{1}{\sqrt{x}}$ The value of the x = 3.