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.