A light cord passes over a friction-free pulley. Each end of the cord has a mass attached, one mass of 25 kg, and the other 12 kg. Calculate: 4.2.1 The acceleration of the masses. 4.2.2 The tension in the cord.

Let $T$ be the tension in the cord and $a$ be the acceleration of the system. Since $m_1 > m_2$, $m_1$ will accelerate downwards and $m_2$ will accelerate upwards.

For $m_1$ (downwards positive):

$$m_1 g - T = m_1 a \quad (1)$$

For $m_2$ (upwards positive):

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

Add equation (1) and equation (2):

$$(m_1 g - T) + (T - m_2 g) = m_1 a + m_2 a$$ $$m_1 g - m_2 g = (m_1 + m_2) a$$

Factor out $g$ on the left and $a$ on the right:

$$g(m_1 - m_2) = a(m_1 + m_2)$$

Solve for $a$:

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

Substitute the given values:

$$a = \frac{9.81 \text{ m/s}^2 (25 \text{ kg} - 12 \text{ kg})}{25 \text{ kg} + 12 \text{ kg}}$$ $$a = \frac{9.81 \text{ m/s}^2 (13 \text{ kg})}{37 \text{ kg}}$$ $$a = \frac{127.53}{37} \text{ m/s}^2$$ $$a \approx 3.446756 \text{ m/s}^2$$

Rounding to three significant figures:

$$\boxed{\text{3.45 m/s}^2}$$

4.2.2 The tension in the cord.

Using equation (2):

$$T - m_2 g = m_2 a$$ $$T = m_2 a + m_2 g$$ $$T = m_2 (a + g)$$

Substitute the values:

$$T = 12 \text{ kg} (3.446756 \text{ m/s}^2 + 9.81 \text{ m/s}^2)$$ $$T = 12 \text{ kg} (13.256756 \text{ m/s}^2)$$ $$T \approx 159.081 \text{ N}$$

Rounding to three significant figures:

$$\boxed{\text{159 N}}$$