Starting from rest, a block slides down an inclined surface that...

Using the second equation of motion:

$s = ut + \dfrac{1}{2}at^2$

Substituting the given values $s = 8 \text{ m}$, $u = 0$, and $t = 2 \text{ s}$:

$8 = 0 + \dfrac{1}{2}a(2)^2$

$8 = 2a \Rightarrow a = 4 \text{ m/s}^2$

The acceleration of a block sliding down a rough inclined plane is given by:

$a = g(\sin\theta - \mu_k \cos\theta)$

Substituting $a = 4 \text{ m/s}^2$, $\theta = 30^\circ$, and taking $g = 10 \text{ m/s}^2$:

$4 = 10$\left(\dfrac{1}{2} - \mu_k \dfrac{\sqrt{3}}{2}\right)<br><br>0.4 = 0.5 - \mu_k $\dfrac{\sqrt{3}}{2}<br><br>$\mu_k $\dfrac{\sqrt{3}}{2} =$0.1$<br><br>$\mu_k = $\dfrac{0.2}{\sqrt{3}} \approx$0.11