The text describes a physics problem involving three ballot boxes connected by cords, with one cord passing over a pulley.
The masses of the boxes are given as mA = 34 kg, mB = 43 kg, and mC = 10 kg.
The pulley is described as having negligible friction and mass.
The system is released from rest.
The question asks for the tension in the cord between B and C when released.
It also asks for the distance box A moves in the first 0.250 seconds, assuming it does not reach the pulley.

To solve the problem, we'll approach it in two parts:

1. **Find the tension in the cord between boxes B and C.**
2. **Determine the distance box A moves in the first 0.250 seconds.**

### 1. Finding the Tension in the Cord between B and C

First, let's analyze the forces acting on the system.

- \( m_A = 34 \, \text{kg} \)
- \( m_B = 43 \, \text{kg} \)
- \( m_C = 10 \, \text{kg} \)

Let's assume the acceleration of the system is \( a \) (which is the same for all boxes since they are connected by cords). The tension in the cord between B and C is \( T_{BC} \).

For each box, applying Newton's second law:

- For \( m_A \):
  \[
  m_A a = T_{AB} - m_A g \quad \text{(Equation 1)}
  \]

- For \( m_B \):
  \[
  m_B a = T_{BC} - T_{AB} \quad \text{(Equation 2)}
  \]

- For \( m_C \):
  \[
  m_C a = m_C g - T_{BC} \quad \text{(Equation 3)}
  \]

Since the pulley is frictionless and massless, the tensions in the ropes can be related.

Let's combine the equations to eliminate the tensions and find the acceleration \( a \) of the system.

### Adding the three equations:

\[
m_A a + m_B a + m_C a = m_C g - m_A g
\]
\[
a(m_A + m_B + m_C) = m_C g - m_A g
\]
\[
a = \frac{(m_C - m_A)g}{m_A + m_B + m_C}
\]

Substituting the given values:

\[
a = \frac{(10 - 34) \times 9.8}{34 + 43 + 10} = \frac{-24 \times 9.8}{87} \approx -2.70 \, \text{m/s}^2
\]

(The negative sign indicates that the acceleration is in the opposite direction to what we assumed.)

Now, substitute \( a \) into Equation 3 to find \( T_{BC} \):

\[
T_{BC} = m_C g - m_C a
\]

Substitute the values:

\[
T_{BC} = 10 \times 9.8 - 10 \times (-2.70) = 98 + 27 = 125 \, \text{N}
\]

So, the tension \( T_{BC} \) is **125 N**.

### 2. Determining the Distance Box A Moves in 0.250 Seconds

The box A starts from rest, so we use the equation of motion:

\[
s = ut + \frac{1}{2} a t^2
\]

Since initial velocity \( u = 0 \):

\[
s = \frac{1}{2} a t^2
\]

Substitute \( a = 2.70 \, \text{m/s}^2 \) and \( t = 0.250 \, \text{seconds} \):

\[
s = \frac{1}{2} \times 2.70 \times (0.250)^2 = 0.0844 \, \text{meters}
\]

So, the distance moved by box A is approximately **0.0844 meters** (or **8.44 cm**).

### Summary of Results

1. **Tension in the cord between B and C**: \( T_{BC} = 125 \, \text{N} \)
2. **Distance moved by box A in the first 0.250 seconds**: \( 0.0844 \, \text{m} \)

Would you like more details or have any questions?

Here are five relative questions for further exploration:
1. How would the tension change if the mass of box C was increased?
2. What would happen if the pulley had friction?
3. How would the problem change if all the boxes were on a rough surface?
4. What if box A reaches the pulley? How would you calculate the new tension and acceleration?
5. How does the gravitational acceleration affect the system if it's on a different planet?

**Tip:** Always consider all forces acting on each object separately to write down accurate equations of motion.

The text describes a physics problem involving three ballot boxes connected by cords, with one cord passing over a pulley. The masses of the boxes are given as mA = 34 kg, mB = 43 kg, and mC = 10 kg. The pulley is described as having negligible friction and mass. The system is released from rest. The question asks for the tension in the cord between B and C when released. It also asks for the distance box A moves in the first 0.250 seconds, assuming it does not reach the pulley.

Solve a physics problem involving three masses connected by cords over a pulley. Learn how to calculate the tension in the cord between two boxes and determine the distance one box moves within 0.250 seconds using Newton's Second Law and kinematic equations. Suitable for Grades 10-12 physics students.

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