To solve this problem, we need to analyze the forces and the motion in the system, considering the tension in the wire, the acceleration of the system, and the forces acting on the pulley.

Step 1: Define Variables and Given Information

1. Mass of the box, $m_1 = 12.0 \, \text{kg}$
2. Mass of the hanging weight, $m_2 = 5.00 \, \text{kg}$
3. Mass of the pulley, $M = 2.30 \, \text{kg}$
4. Diameter of the pulley, $D = 0.500 \, \text{m}$, so radius $R = 0.250 \, \text{m}$

Since the surface is frictionless, and the pulley and wire are also frictionless, we will use Newton’s second law for both the box and the hanging weight, and rotational dynamics for the pulley.

Step 2: Set Up Equations of Motion

Box (horizontal motion)

The box has mass $m_1$ and experiences tension $T_{\text{horizontal}}$ to the right. This tension will cause it to accelerate along the x-axis: $m_1 \cdot a = T_{\text{horizontal}}$

Hanging Weight (vertical motion)

The hanging weight experiences gravitational force $m_2 g$ downward and tension $T_{\text{vertical}}$ upward. This force causes it to accelerate downward: $m_2 g - T_{\text{vertical}} = m_2 \cdot a$

Pulley (rotational motion)

The pulley has a rotational inertia $I = \frac{1}{2} M R^2$ and experiences a torque due to the tensions in the wire segments on each side (horizontal and vertical). The net torque on the pulley is given by: $\tau = I \cdot \alpha$ where $\alpha = \frac{a}{R}$ is the angular acceleration of the pulley. The torque generated by the tensions is: $T_{\text{vertical}} \cdot R - T_{\text{horizontal}} \cdot R = I \cdot \alpha$ Substituting $I = \frac{1}{2} M R^2$ and $\alpha = \frac{a}{R}$, we get: $T_{\text{vertical}} - T_{\text{horizontal}} = \frac{1}{2} M \cdot a$

Step 3: Solve for Tensions and Acceleration

1. Find the acceleration $a$

Combine the three equations: $m_1 a = T_{\text{horizontal}}$ $m_2 g - T_{\text{vertical}} = m_2 a$ $T_{\text{vertical}} - T_{\text{horizontal}} = \frac{1}{2} M a$ Substitute $T_{\text{horizontal}} = m_1 a$ from the first equation and rearrange the third equation to solve for $T_{\text{vertical}}$: $T_{\text{vertical}} = m_1 a + \frac{1}{2} M a$ Substitute this into the second equation: $m_2 g - \left( m_1 a + \frac{1}{2} M a \right) = m_2 a$ Solve for $a$: $m_2 g = \left( m_1 + m_2 + \frac{1}{2} M \right) a$ $a = \frac{m_2 g}{m_1 + m_2 + \frac{1}{2} M}$ Substitute the values: $a = \frac{(5.00)(9.8)}{12.0 + 5.00 + \frac{1}{2} (2.30)}$ $a = \frac{49}{13.15} \approx 3.725 \, \text{m/s}^2$

2. Find the Tensions $T_{\text{horizontal}}$ and $T_{\text{vertical}}$

• Horizontal tension $T_{\text{horizontal}}$: $T_{\text{horizontal}} = m_1 \cdot a = (12.0)(3.725) \approx 44.7 \, \text{N}$

• Vertical tension $T_{\text{vertical}}$: $T_{\text{vertical}} = m_2 g - m_2 a = (5.00)(9.8) - (5.00)(3.725)$ [ T_{\text{vertical}} \approx 49 - 18.625 =