	The two barges shown here are coupled by a cable of negligible mass. The mass of the front barge is  2.00×103kg
  and the mass of the rear barge is  3.00×103kg.
  A tugboat pulls the front barge with a horizontal force of magnitude  20.0×103N,
  and the frictional forces of the water on the front and rear barges are  8.00×103N
  and  10.0×103N,
  respectively. Find the horizontal acceleration of the barges and the tension in the connecting cable.

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?

---

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.

The two barges shown here are coupled by a cable of negligible mass. The mass of the front barge is 2.00×10^3 kg and the mass of the rear barge is 3.00×10^3 kg. A tugboat pulls the front barge with a horizontal force of magnitude 20.0×10^3 N, and the frictional forces of the water on the front and rear barges are 8.00×10^3 N and 10.0×10^3 N, respectively. Find the horizontal acceleration of the barges and the tension in the connecting cable.

Determine the horizontal acceleration and tension in the connecting cable between two barges pulled by a tugboat with forces of 20.0×10^3 N, 8.00×10^3 N, and 10.0×10^3 N applied. This solution covers Newton's Second Law of Motion and force analysis, suitable for Grades 11-12.

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