A spring is fixed at the bottom end of an incline of inclination...

Let the coefficient of kinetic friction between the block and the incline be $\mu$.

During the downward motion, the block travels a total distance $s_1 = 4.8 + 0.2 = 5 \text{ m}$ before coming to a momentary halt. Applying the work-energy theorem from the initial position to the point of maximum compression:

$W_{\text{gravity}} + W_{\text{friction}} + W_{\text{spring}} = \Delta K$

mg s_1 \sin 37^{}^{{}^{\circ}} - \mu mg s_1 \cos 37^{}^{{}^{\circ}} - \dfrac{1}{2} k x^2 = 0

Substituting $m = 2 \text{ kg}$, $g = 10 \text{ m/s}^2$, $s_1 = 5 \text{ m}$, $x = 0.2 \text{ m}$, \sin 37^{}^{{}^{\circ}} = \dfrac{3}{5}, and \cos 37^{}^{{}^{\circ}} = \dfrac{4}{5}:

$2 \times 10 \times 5 \times \dfrac{3}{5} - \mu \times 2 \times 10 \times 5 \times \dfrac{4}{5} = \dfrac{1}{2} k (0.2)^2$

$60 - 80\mu = 0.02k$

During the upward rebound, the block travels a distance $s_2 = 1 \text{ m}$ up the incline from the point of maximum compression. Applying the work-energy theorem for the upward journey:

-mg s_2 \sin 37^{}^{{}^{\circ}} - \mu mg s_2 \cos 37^{}^{{}^{\circ}} + \dfrac{1}{2} k x^2 = 0

Substituting the values with $s_2 = 1 \text{ m}$:

$-2 \times 10 \times 1 \times \dfrac{3}{5} - \mu \times 2 \times 10 \times 1 \times \dfrac{4}{5} + 0.02k = 0$

$-12 - 16\mu + 0.02k = 0$

$0.02k = 12 + 16\mu$

Equating the two expressions for $0.02k$:

$60 - 80\mu = 12 + 16\mu$

$48 = 96\mu \Rightarrow \mu = 0.5$

Substituting $\mu = 0.5$ into the equation for $k$:

$0.02k = 12 + 16(0.5)$

$0.02k = 20$

$k = \dfrac{20}{0.02} = 1000 \text{ N/m}$