Two blocks are initially in contact sliding down an inclined...

Let the independent accelerations of the $2.0 \text{ kg}$ and $4.0 \text{ kg}$ blocks be $a_1$ and $a_2$ respectively.

For the $2.0 \text{ kg}$ block (which is higher on the incline):
a_1 = g \sin 30^{}^{{}^{\circ}} - \mu_1 g \cos 30^{}^{{}^{\circ}}
$a_1 = 10 \times \dfrac{1}{2} - 0.30 \times 10 \times \dfrac{\sqrt{3}}{2}$
$a_1 = 5 - 1.5 \times 1.732 = 5 - 2.598 = 2.402 \text{ m/s}^2$

For the $4.0 \text{ kg}$ block (which is lower on the incline):
a_2 = g \sin 30^{}^{{}^{\circ}} - \mu_2 g \cos 30^{}^{{}^{\circ}}
$a_2 = 10 \times \dfrac{1}{2} - 0.20 \times 10 \times \dfrac{\sqrt{3}}{2}$
$a_2 = 5 - 1.732 = 3.268 \text{ m/s}^2$

Since $a_2 > a_1$ and the $4.0 \text{ kg}$ block is lower on the incline, the two blocks will separate and move independently.

Therefore, the acceleration of the $2.0 \text{ kg}$ block is approximately $2.4 \text{ m/s}^2$.