As shown in the figure, a spring is fixed at the bottom end of an...

Let the spring constant be $k$ and the coefficient of friction be $\mu$.

During the downward motion, the block starts from rest and comes to a momentary halt after compressing the spring. The total distance moved down the incline is $s_1 = 4.8 + 0.2 = 5.0 \text{ m}$.

Applying the work-energy theorem for the downward journey:
$W_{\text{gravity}} + W_{\text{friction}} + W_{\text{spring}} = \Delta K = 0$
$mg s_1 \sin 37^\circ - \mu mg s_1 \cos 37^\circ - \dfrac{1}{2}kx^2 = 0$
$\dfrac{1}{2}kx^2 = mg s_1 (\sin 37^\circ - \mu \cos 37^\circ)$ --- (1)

During the upward rebound, the block starts from rest at maximum compression and moves a distance $s_2 = 1.0 \text{ m}$ up the incline before coming to rest.

Applying the work-energy theorem for the upward journey:
$W_{\text{gravity}} + W_{\text{friction}} + W_{\text{spring}} = \Delta K = 0$
$-mg s_2 \sin 37^\circ - \mu mg s_2 \cos 37^\circ + \dfrac{1}{2}kx^2 = 0$
$\dfrac{1}{2}kx^2 = mg s_2 (\sin 37^\circ + \mu \cos 37^\circ)$ --- (2)

Equating (1) and (2):
$mg s_1 (\sin 37^\circ - \mu \cos 37^\circ) = mg s_2 (\sin 37^\circ + \mu \cos 37^\circ)$
$s_1 (\sin 37^\circ - \mu \cos 37^\circ) = s_2 (\sin 37^\circ + \mu \cos 37^\circ)$

Substituting $s_1 = 5.0 \text{ m}$, $s_2 = 1.0 \text{ m}$, $\sin 37^\circ = 0.6$, and $\cos 37^\circ = 0.8$:
$5.0 (0.6 - 0.8\mu) = 1.0 (0.6 + 0.8\mu)$
$3.0 - 4.0\mu = 0.6 + 0.8\mu$
$2.4 = 4.8\mu$
$\mu = \dfrac{2.4}{4.8} = 0.5$

Answer: $0.5$