Determine the acceleration of the block of mass M for the setup...

Let $T$ be the tension in the string. The block of mass $M$ is pulled up the incline by a single string, so the tension force acting on it is $T$. The movable pulley is supported by two segments of the same string, so the total upward force on it is $2T$.

Let the downward acceleration of the block of mass $2M$ be $a$. By string constraint, if the movable pulley moves downwards by a distance $x$, it requires $2x$ length of string. This string comes from the portion on the incline, meaning the block of mass $M$ must move up the incline by a distance $2x$. Therefore, the acceleration of the block of mass $M$ up the incline is $2a$.

Writing the equation of motion for the block of mass $M$ along the incline:
$T - Mg \sin(30^{{}^{{}^{\circ}}}) = M(2a)$
$T - \dfrac{Mg}{2} = 2Ma$

Writing the equation of motion for the block of mass $2M$ moving downwards:
$2Mg - 2T = (2M)a$
$Mg - T = Ma$

Adding the two equations:
\left(T - \dfrac{Mg}{2}\right) + (Mg - T) = 2Ma + Ma$<br>$\dfrac{Mg}{2} = 3Ma$<br>$a = \dfrac{g}{6}$<br><br>The acceleration of the block of mass$M$is$2a$up the incline:<br>$2a = 2 \left(\dfrac{g}{6}\right) = \dfrac{g}{3}$<br><br>Thus, the acceleration of the block of mass$M$is$\dfrac{g}{3}$up the plane.$