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To solve **Problem 14**, let's analyze the given information and calculate the acceleration and tension in the cord.

### Problem Statement
Given:
- \( m_A = 10 \, \text{kg} \) (mass of block A)
- \( m_B = 5 \, \text{kg} \) (mass of block B)
- Coefficient of kinetic friction \( \mu_k = 0.2 \) (for block A on the surface)
  
We need to find:
1. The **acceleration** of the system.
2. The **tension** in the cord.

**Answers provided for verification**:
- Acceleration \( a = 1.96 \, \text{m/s}^2 \)
- Tension \( T = 39.2 \, \text{N} \)

### Solution

This is a typical problem involving two blocks connected by a cord over a frictionless pulley. Block A is on a horizontal surface, and block B is hanging vertically.

#### Step 1: Draw Free-Body Diagrams
For block A:
- The forces acting are:
  - Tension \( T \) to the right.
  - Frictional force \( f_k = \mu_k \cdot m_A \cdot g \) to the left.
  
For block B:
- The forces acting are:
  - Tension \( T \) upward.
  - Weight \( m_B \cdot g \) downward.

#### Step 2: Set Up Equations of Motion

**For Block A (horizontal):**
Using Newton's second law:
\[
T - f_k = m_A \cdot a
\]
where \( f_k = \mu_k \cdot m_A \cdot g \).

Substitute the values:
\[
f_k = 0.2 \times 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N}
\]
Thus, the equation becomes:
\[
T - 19.6 = 10 \cdot a
\]
or
\[
T = 10a + 19.6
\]

**For Block B (vertical):**
Using Newton's second law:
\[
m_B \cdot g - T = m_B \cdot a
\]
Substitute values:
\[
5 \times 9.8 - T = 5 \cdot a
\]
or
\[
49 - T = 5a
\]
Thus,
\[
T = 49 - 5a
\]

#### Step 3: Solve the Equations

Now, we have two equations:
1. \( T = 10a + 19.6 \)
2. \( T = 49 - 5a \)

Set them equal to each other:
\[
10a + 19.6 = 49 - 5a
\]
Combine terms:
\[
15a = 29.4
\]
\[
a = \frac{29.4}{15} = 1.96 \, \text{m/s}^2
\]

#### Step 4: Find the Tension

Now, substitute \( a = 1.96 \) back into one of the equations for \( T \). Using \( T = 10a + 19.6 \):
\[
T = 10 \cdot 1.96 + 19.6
\]
\[
T = 19.6 + 19.6 = 39.2 \, \text{N}
\]

### Final Answers
- **Acceleration** \( a = 1.96 \, \text{m/s}^2 \)
- **Tension** \( T = 39.2 \, \text{N} \)

These match the provided answers.

Would you like a further breakdown of any step or additional clarification?

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### Related Questions
1. How would the acceleration change if the mass of block B increased?
2. What effect does increasing the friction coefficient have on the system's acceleration?
3. If the pulley has mass, how would you account for it in the equations?
4. How would the tension vary if there were an incline for block A?
5. Can we solve for the acceleration and tension if \( \mu_k \) is unknown?

### Tip
Always start by analyzing each body's forces individually and apply Newton's second law to set up equations.

Solve Problem 14 to find the acceleration and tension in a two-block pulley system with kinetic friction. Given masses (10 kg and 5 kg) and a friction coefficient of 0.2, learn how to apply Newton's laws to determine acceleration (1.96 m/s²) and tension (39.2 N). Suitable for students in Grades 11-12.

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