A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches $30^\circ$, the box starts to slip and slides $4.0 \text{ m}$ down the plank in $4.0 \text{ s}$. The coefficients of static and kinetic friction between the box and the plank will be, respectively:

Coefficient of Static Friction ($\mu_s$): The box just begins to slip when the angle of inclination reaches the angle of repose, $\theta = 30^\circ$. At this point, the maximum static friction balances the component of weight down the plane ($mg \sin\theta = \mu_s mg \cos\theta$). $\mu_s = \tan\theta = \tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$ Rounding to one decimal place, $\mu_s \approx 0.6$. (Reference: NCERT Class 11, Physics Part I, Sec 5.9, Example 4.8).

Coefficient of Kinetic Friction ($\mu_k$): Once the box starts sliding, it covers a distance $s = 4.0 \text{ m}$ in time $t = 4.0 \text{ s}$ starting from rest ($u = 0$). Using the kinematic equation $s = ut + \frac{1}{2}at^2$ [NCERT Class 11, Physics Part I, Sec 3.6, Eq 2.9b]: $4.0 = 0 + \frac{1}{2}a(4.0)^2 \implies 4.0 = 8a \implies a = 0.5 \text{ m/s}^2$

Dynamic Equation: The net force acting down the plane is gravity minus kinetic friction: $ma = mg \sin\theta - \mu_k mg \cos\theta$ $a = g(\sin 30^\circ - \mu_k \cos 30^\circ)$ Taking $g = 10 \text{ m/s}^2$: $0.5 = 10(0.5 - \mu_k \times 0.866)$ $0.05 = 0.5 - 0.866 \mu_k$ $0.866 \mu_k = 0.45$ $\mu_k = \frac{0.45}{0.866} \approx 0.519$ Rounding to one decimal place, $\mu_k \approx 0.5$.