Two blocks having masses of 400 g and 200 g are connected by a...

By the principle of conservation of mechanical energy, the gain in the total kinetic energy of the system is equal to the net loss in its potential energy.

Let $m_1 = 400 \text{ g} = 0.4 \text{ kg}$ and $m_2 = 200 \text{ g} = 0.2 \text{ kg}$.

When the heavier block $m_1$ falls through a distance $h = 50 \text{ cm} = 0.5 \text{ m}$, the lighter block $m_2$ rises by the same distance $h$.

The net loss in potential energy of the system is given by:

$\Delta U = m_1 g h - m_2 g h = (m_1 - m_2) g h$

Substituting the given values and taking $g = 9.8 \text{ m/s}^2$:

$\Delta U = (0.4 - 0.2) \times 9.8 \times 0.5$

$\Delta U = 0.2 \times 9.8 \times 0.5 = 0.98 \text{ J}$

Since the system starts from rest, the total kinetic energy of the system (including the translational kinetic energy of both blocks and the rotational kinetic energy of the pulley) is equal to the net loss in potential energy.

Therefore, the kinetic energy of the system is $0.98 \text{ J}$.