In the figure shown, suppose m_2 = 2.0 kg and m_3 = 3.0 kg . What...

Let $T_1$ be the tension in the string connected to $m_1$, and $T_2$ be the tension in the string connecting $m_2$ and $m_3$.

For $m_1$ to remain at rest, the upward tension must balance its weight:
$T_1 = m_1 g$

Since $m_1$ is at rest, the movable pulley is also stationary. Balancing the forces on the massless movable pulley gives:
$T_1 = 2T_2$

With the movable pulley at rest, $m_2$ and $m_3$ form a simple Atwood machine. The tension $T_2$ in this lower string is:
$T_2 = \dfrac{2 m_2 m_3}{m_2 + m_3} g$

Substitute this $T_2$ back into the $T_1$ equation:
$T_1 = 2 \left( \dfrac{2 m_2 m_3}{m_2 + m_3} g \right) = \dfrac{4 m_2 m_3}{m_2 + m_3} g$

Equating the two expressions for $T_1$ gives:
$m_1 g = \dfrac{4 m_2 m_3}{m_2 + m_3} g$

$m_1 = \dfrac{4 m_2 m_3}{m_2 + m_3}$

Substitute the given values $m_2 = 2.0 \text{ kg}$ and $m_3 = 3.0 \text{ kg}$:
$m_1 = \dfrac{4(2.0)(3.0)}{2.0 + 3.0} = \dfrac{24}{5} = 4.8 \text{ kg}$