Two masses 2kg and 1.5kg are tied to the ends of a light inextensible string which passes over a smooth light pulley fixed to the ceiling. Find the acceleration of the 2kg mass.

Step 1: Identify the given values and set up equations for each mass. Let $m_1 = 2 \text{ kg}$ and $m_2 = 1.5 \text{ kg}$. Let $g = 10 m/s^2$. Let $T$ be the tension in the string and $a$ be the acceleration of the system. Since $m_1 > m_2$, the mass $m_1$ will accelerate downwards and $m_2$ will accelerate upwards.

For mass $m_1$ (2 kg), applying Newton's second law ($F_{net} = m_1 a$): $m_1 g - T = m_1 a \quad (1)$

For mass $m_2$ (1.5 kg), applying Newton's second law ($F_{net} = m_2 a$): $T - m_2 g = m_2 a \quad (2)$

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

Step 3: Substitute the given values into the equation for $a$. $a = \frac{(2 kg - 1.5 kg) \times 10 m/s^2}{2 kg + 1.5 kg}$ $a = \frac{(0.5 kg) \times 10 m/s^2}{3.5 kg}$ $a = \frac{5 N}{3.5 kg}$ $a = \frac{5}{3.5} m/s^2$ $a = \frac{50}{35} m/s^2$ $a = \frac{10}{7} m/s^2$

The acceleration of the 2kg mass is $\frac{10}{7} m/s^2$ downwards.

The final answer is $\boxed{\frac{10}{7} m/s^2}$.