Let's break down the problem into simpler steps using Newton's second law, $F = ma$, where $F$ is the force, $m$ is the mass, and $a$ is the acceleration.

Key information provided:

• Mass of the box, $M = 14.3 \, \text{kg}$.
• Coefficient of kinetic friction, $\mu_k = 0.245$.
• Tension in the first rope, $T_1 = 112 \, \text{N}$.
• Mass of the second ball, $m_2 = 10.6 \, \text{kg}$.

We need to find the mass of $m_1$.

Step 1: Forces on the box

Since the box is sliding on a rough surface, the forces acting on the box in the horizontal direction are:

• Tension $T_1$ pulling the box to the right.
• Kinetic friction force $f_k$, which opposes the motion.

The kinetic friction force is given by: $f_k = \mu_k \cdot N$ where $N$ is the normal force. Since the box is on a horizontal surface, the normal force is equal to the gravitational force on the box: $N = M \cdot g = 14.3 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 140.283 \, \text{N}.$ Thus, the kinetic friction force is: $f_k = 0.245 \cdot 140.283 = 34.859 \, \text{N}.$

Step 2: Net force on the box

The net force acting on the box is the difference between the tension in the rope and the kinetic friction: $F_{\text{net, box}} = T_1 - f_k = 112 \, \text{N} - 34.859 \, \text{N} = 77.141 \, \text{N}.$ Using Newton’s second law, the net force is also equal to the mass of the box multiplied by its acceleration: $F_{\text{net, box}} = M \cdot a.$ Thus, the acceleration of the box is: $a = \frac{F_{\text{net, box}}}{M} = \frac{77.141 \, \text{N}}{14.3 \, \text{kg}} = 5.394 \, \text{m/s}^2.$

Step 3: Forces on $m_1$

Now, consider the forces acting on $m_1$. The forces are:

• The gravitational force on $m_1$, which is $m_1 \cdot g$.
• The tension $T_1$ acting upward.

Using Newton's second law for $m_1$, we have: $m_1 \cdot g - T_1 = m_1 \cdot a.$ Rearranging this equation to solve for $m_1$: $m_1 \cdot g - m_1 \cdot a = T_1$ $m_1 \cdot (g - a) = T_1$ $m_1 = \frac{T_1}{g - a} = \frac{112 \, \text{N}}{9.81 \, \text{m/s}^2 - 5.394 \, \text{m/s}^2}.$ $m_1 = \frac{112}{4.416} = 25.37 \, \text{kg}.$

Thus, the mass of $m_1$ is approximately 25.4 kg.

Would you like further clarification or details?

Here are 5 related questions to explore:

1. How would changing the coefficient of kinetic friction affect the system’s acceleration?
2. What happens to the acceleration if $m_1$ is increased by 10%?
3. How would the system behave if the surface were frictionless?
4. How does the tension in the rope compare with the gravitational force on $m_2$?
5. What would happen if the pulley had friction?

Tip: Always check if the forces are balanced properly when dealing with multiple objects in motion.