Work Done by String on Block Connected Across Fixed Pulley

To find the work done by the string on the block of mass $m_1 = 0.36\text{ kg}$ during the first second after the system is released from rest, we first determine the acceleration of the system and the tension in the string.

Let:

• $m_1 = 0.36\text{ kg}$
• $m_2 = 0.72\text{ kg}$
• $g = 10\text{ m/s}^2$

Since the pulley is smooth and the string is light and inextensible, both blocks experience the same magnitude of acceleration $a$.

The acceleration $a$ of the system is given by:
$a = \left(\frac{m_2 - m_1}{m_1 + m_2}\right) g$

Substituting the given values:
$a = \left(\frac{0.72 - 0.36}{0.72 + 0.36}\right) \times 10 = \frac{0.36}{1.08} \times 10 = \frac{1}{3} \times 10 = \frac{10}{3}\text{ m/s}^2$

The tension $T$ in the string acting on the block of mass $m_1$ is obtained from its equation of motion (directed upwards):
$T - m_1 g = m_1 a$
$T = m_1 (g + a) = 0.36 \left(10 + \frac{10}{3}\right) = 0.36 \times \frac{40}{3} = 4.8\text{ N}$

The system starts from rest ($u = 0$). The displacement $s$ of the block of mass $m_1$ in the first second ($t = 1\text{ s}$) in the upward direction is:
$s = u t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \times \frac{10}{3} \times (1)^2 = \frac{5}{3}\text{ m}$

Both the tension force $T$ and the displacement $s$ of the block $m_1$ are in the upward direction. Therefore, the work done $W$ by the tension of the string on this block is:
$W = T \cdot s = 4.8\text{ N} \times \frac{5}{3}\text{ m} = 8\text{ J}$

Thus, the work done by the string on the block is $8\text{ J}$.