To solve this, let’s break down the problem into steps, first determining the horizontal acceleration of the system, then calculating the tension in the cable connecting the two barges.

Step 1: Horizontal Acceleration of the System

1. Total mass of the system: $m_{\text{total}} = m_{\text{front}} + m_{\text{rear}} = 2.00 \times 10^3 \, \text{kg} + 3.00 \times 10^3 \, \text{kg} = 5.00 \times 10^3 \, \text{kg}$

2. Total horizontal force acting on the system: The tugboat exerts a pulling force of $F_{\text{tug}} = 20.0 \times 10^3 \, \text{N}$. However, we must subtract the frictional forces from both barges: $F_{\text{net}} = F_{\text{tug}} - (f_{\text{front}} + f_{\text{rear}})$ $F_{\text{net}} = 20.0 \times 10^3 \, \text{N} - (8.00 \times 10^3 \, \text{N} + 10.0 \times 10^3 \, \text{N}) = 20.0 \times 10^3 \, \text{N} - 18.0 \times 10^3 \, \text{N} = 2.0 \times 10^3 \, \text{N}$

3. Calculate acceleration: Using Newton's second law, $F = m \cdot a$: $a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{2.0 \times 10^3 \, \text{N}}{5.0 \times 10^3 \, \text{kg}} = 0.4 \, \text{m/s}^2$

   So, the horizontal acceleration $a$ of the barges is: $a = 0.4 \, \text{m/s}^2$

Step 2: Tension in the Cable

To find the tension in the cable connecting the two barges, consider only the forces acting on the rear barge:

1. Force needed to accelerate the rear barge: Since the rear barge has a mass of $m_{\text{rear}} = 3.00 \times 10^3 \, \text{kg}$ and it accelerates at $a = 0.4 \, \text{m/s}^2$: $F_{\text{required}} = m_{\text{rear}} \cdot a = 3.00 \times 10^3 \, \text{kg} \times 0.4 \, \text{m/s}^2 = 1.2 \times 10^3 \, \text{N}$

2. Determine the tension: The tension in the cable, $T$, must overcome the frictional force acting on the rear barge in addition to providing the force required to accelerate it. So: $T = F_{\text{required}} + f_{\text{rear}}$ $T = 1.2 \times 10^3 \, \text{N} + 10.0 \times 10^3 \, \text{N} = 11.2 \times 10^3 \, \text{N}$

Thus, the tension in the connecting cable is: $T = 11.2 \times 10^3 \, \text{N}$

Final Answers

• Horizontal acceleration of the barges: $0.4 \, \text{m/s}^2$
• Tension in the cable: $11.2 \times 10^3 \, \text{N}$

Would you like a deeper explanation on any part of the solution?

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Here are five related questions to deepen understanding:

1. How would the acceleration change if the frictional force on each barge doubled?
2. What would happen to the tension if the rear barge's mass increased by 50%?
3. How would the system's behavior change if the cable had a significant mass?
4. What if the tugboat pulled with a different angle, not purely horizontal—how would that affect the acceleration?
5. How does the solution change if the frictional forces depend on the velocity of the barges?

Tip: When dealing with systems of objects, breaking down the forces acting on each component can clarify the net forces and accelerations involved.