**Skip mo na to.** A box with mass M=M= 14.3 [kg] is placed on a rough surface with coefficient of kinetic friction of 0.245. The mass is attached to two balls m1m1 and m2=m2= 10.6 [kg] via two pieces of rope and the system starts sliding as shown. The rope slides around the pulley mechanism frictionlessly. Given that the first rope exerts a tension force of 112 [N] on both the box and m1m1, what is the mass of m1m1?

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.

A box with mass M= 14.3 [kg] is placed on a rough surface with 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 as shown. The rope slides around the pulley mechanism frictionlessly. Given that the first rope exerts a tension force of 112 [N] on both the box and m1, what is the mass of m1?

Solve for the mass of m1 in a system involving a box, pulley, and two masses. With a given tension of 112 N, friction coefficient of 0.245, and mass of 14.3 kg, we apply Newton's second law, kinetic friction, and force equations. Ideal for students in Grades 10-12.

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