A 2kg mass on a smooth inclined plane is connected over a smooth pulley to a 3kg mass. Find the acceleration of the system.

Step 1: Identify the given masses and angle. Let $m_1 = 2 \text{ kg}$ be the mass on the inclined plane. Let $m_2 = 3 \text{ kg}$ be the hanging mass. The angle of inclination is $\theta = 30^\circ$. The plane is smooth, meaning there is no friction. The pulley is also smooth, meaning it is massless and frictionless.

Step 2: Determine the forces acting on each mass. For mass $m_2$ (hanging mass): The forces are its weight $m_2 g$ acting downwards and the tension $T$ in the string acting upwards. Since $m_2$ is heavier than $m_1$ (considering the component of $m_1$'s weight along the incline), $m_2$ will accelerate downwards. Applying Newton's second law ($F_{net} = m_2 a$): $m_2 g - T = m_2 a \quad (1)$

For mass $m_1$ (on the inclined plane): The forces acting along the incline are the tension $T$ acting upwards along the incline and the component of its weight $m_1 g \sin\theta$ acting downwards along the incline. Since $m_2$ accelerates downwards, $m_1$ will accelerate upwards along the incline. Applying Newton's second law ($F_{net} = m_1 a$): $T - m_1 g \sin\theta = m_1 a \quad (2)$

Step 3: Solve the system of equations for acceleration $a$. Add equation (1) and equation (2) to eliminate $T$: $(m_2 g - T) + (T - m_1 g \sin\theta) = m_2 a + m_1 a$ $m_2 g - m_1 g \sin\theta = (m_1 + m_2) a$ Factor out $g$ on the left side and $a$ on the right side: $(m_2 - m_1 \sin\theta) g = (m_1 + m_2) a$ Solve for $a$: $a = \frac{(m_2 - m_1 \sin\theta) g}{m_1 + m_2}$

Step 4: Substitute the given values into the equation for $a$. We know that $\sin(30^\circ) = \frac{1}{2}$. $a = \frac{(3 kg - 2 kg \times \sin(30^\circ)) g}{2 kg + 3 kg}$ $a = \frac{(3 - 2 \times \frac{1}{2}) g}{5}$ $a = \frac{(3 - 1) g}{5}$ $a = \frac{2g}{5}$

The final answer is $\boxed{\frac{2g}{5}}$.