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.

Let $m_1 = 2 \text{ kg}$ and $m_2 = 1.5 \text{ kg}$.
Let $g = 10 \text{ m/s}^2$.
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 ($F_{net} = m_2 a$):

$$T - m_2 g = m_2 a \quad (2)$$

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 $a$:

$$a = \frac{(m_1 - m_2) g}{m_1 + m_2}$$

$$a = \frac{(2 \text{ kg} - 1.5 \text{ kg}) \times 10 \text{ m/s}^2}{2 \text{ kg} + 1.5 \text{ kg}}$$ $$a = \frac{(0.5 \text{ kg}) \times 10 \text{ m/s}^2}{3.5 \text{ kg}}$$ $$a = \frac{5 \text{ N}}{3.5 \text{ kg}}$$ $$a = \frac{5}{3.5} \text{ m/s}^2$$ $$a = \frac{50}{35} \text{ m/s}^2$$ $$a = \frac{10}{7} \text{ m/s}^2$$

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