Observe the setup illustrated in the figure. The system is released...

Let $a_1$ and $v_1$ be the acceleration and velocity of the $1.0\text{ kg}$ block, and $a_2$ and $v_2$ be the acceleration and velocity of the $4.0\text{ kg}$ block.

From the constraint of the movable pulley, when the $1.0\text{ kg}$ block descends by a distance $h$, the $4.0\text{ kg}$ block moves by a distance $2h$. Thus, $v_2 = 2v_1$ and $a_2 = 2a_1$.

Given that the $1.0\text{ kg}$ block descends by $h = 1\text{ m}$ and attains a speed of $v_1 = 0.3\text{ m/s}$, we can find its acceleration using the kinematic equation:
$v_1^2 = u^2 + 2a_1 h$
$(0.3)^2 = 0 + 2a_1(1)$
$0.09 = 2a_1 \Rightarrow a_1 = 0.045\text{ m/s}^2$

The acceleration of the $4.0\text{ kg}$ block is:
$a_2 = 2a_1 = 2(0.045) = 0.09\text{ m/s}^2$

Let $T$ be the tension in the string. The $1.0\text{ kg}$ block is supported by two segments of the string, so the upward force is $2T$. Writing the equation of motion for the $1.0\text{ kg}$ block ($m = 1.0\text{ kg}$):
$mg - 2T = m a_1$
$1.0(10) - 2T = 1.0(0.045)$
$10 - 2T = 0.045 \Rightarrow 2T = 9.955 \Rightarrow T = 4.9775\text{ N}$

Writing the equation of motion for the $4.0\text{ kg}$ block ($M = 4.0\text{ kg}$), which is subjected to the tension $T$ and kinetic friction $f_k = \mu_k Mg$:
$T - \mu_k Mg = M a_2$
$4.9775 - \mu_k(4.0)(10) = 4.0(0.09)$
$4.9775 - 40\mu_k = 0.36$
$40\mu_k = 4.6175$
$\mu_k = \dfrac{4.6175}{40} \approx 0.1154$

Rounding to two decimal places, we get $\mu_k \approx 0.12$, which matches option (4).