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This problem can be solved using Newton's second law and the forces acting on the system.

### Given Data:
- \( M = 14.3 \, \text{kg} \) (mass of the box)
- \( m_2 = 10.6 \, \text{kg} \) (mass of the second ball)
- Coefficient of kinetic friction \( \mu_k = 0.245 \)
- Tension force \( T = 112 \, \text{N} \) (in the rope acting on both the box and \( m_1 \))
- We need to find \( m_1 \), the mass of the first ball.

### Step 1: Forces on the Box (Mass \( M \))

For the box \( M \), we have the following forces:
- The tension force \( T = 112 \, \text{N} \) pulling the box to the right.
- The frictional force \( f_k = \mu_k \cdot N \), where \( N = M \cdot g \) is the normal force (since it's on a flat surface).

The frictional force \( f_k \) is:
\[
f_k = \mu_k \cdot M \cdot g = 0.245 \cdot 14.3 \cdot 9.8 \approx 34.2 \, \text{N}
\]

The net force \( F_{\text{net}} \) on the box is the tension minus the frictional force:
\[
F_{\text{net}} = T - f_k = 112 - 34.2 = 77.8 \, \text{N}
\]

Using Newton's second law, the net force on the box is also \( M \cdot a \), where \( a \) is the acceleration:
\[
M \cdot a = 77.8
\]
\[
14.3 \cdot a = 77.8
\]
\[
a \approx 5.44 \, \text{m/s}^2
\]

### Step 2: Forces on \( m_1 \)

For \( m_1 \), the forces acting on it are:
- The tension \( T = 112 \, \text{N} \) acting upward.
- The gravitational force \( m_1 \cdot g \) acting downward.

The net force on \( m_1 \) is:
\[
T - m_1 \cdot g = m_1 \cdot a
\]
Substituting the known values of \( T \), \( g = 9.8 \, \text{m/s}^2 \), and \( a = 5.44 \, \text{m/s}^2 \):
\[
112 - m_1 \cdot 9.8 = m_1 \cdot 5.44
\]
\[
112 = m_1 \cdot (9.8 + 5.44)
\]
\[
112 = m_1 \cdot 15.24
\]
\[
m_1 = \frac{112}{15.24} \approx 7.35 \, \text{kg}
\]

### Step 3: Forces on \( m_2 \)

Similarly, for \( m_2 \), we use the same acceleration \( a \):
\[
m_2 \cdot g - T = m_2 \cdot a
\]
Substitute known values:
\[
10.6 \cdot 9.8 - 112 = 10.6 \cdot 5.44
\]
Solving this equation should yield the same net acceleration to verify consistency.

It looks like the solution needs refinement based on options

A box with mass M = 14.3 [kg] is placed on a rough surface with a coefficient of kinetic friction of 0.245. The mass is attached to two balls m1 and m2 = 10.6 [kg] via two pieces of rope and the system starts sliding. The rope slides around the pulley frictionlessly. Given the first rope exerts a tension force of 112 [N] on both the box and m1, what is the mass of m1?

A detailed step-by-step solution for a physics problem involving a box with mass M = 14.3 kg, kinetic friction (μ_k = 0.245), and a pulley system with masses m1 and m2 = 10.6 kg. Learn how to calculate the mass of m1 using tension, frictional force, and Newton's second law. Suitable for college-level physics students.

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