Determine the acceleration of the 500 g block as shown in the...

Let $a$ be the common acceleration of the system and $T_1, T_2$ be the tensions in the strings.

Assuming the $500\text{ g}$ block moves downward, the equations of motion for the three masses are:

For the $500\text{ g}$ block:
$500g - T_2 = 500a$

For the $100\text{ g}$ block on the incline:
T_2 - T_1 - 100g \sin 30^{}^{{}^{\circ}} = 100a

For the $50\text{ g}$ block:
$T_1 - 50g = 50a$

Adding the three equations, we get:
(500g - T_2) + (T_2 - T_1 - 100g \sin 30^{}^{{}^{\circ}}) + (T_1 - 50g) = (500 + 100 + 50)a

$500g - 100g \times \dfrac{1}{2} - 50g = 650a$

$500g - 50g - 50g = 650a$

$400g = 650a$

$a = \dfrac{400g}{650} = \dfrac{8g}{13}$

Since $a$ is positive, the $500\text{ g}$ block moves downward with an acceleration of $\dfrac{8g}{13}$.