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The problem presented involves two blocks connected by a frictionless pulley system. Let’s break down the given information and solve both parts systematically.

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### **Given:**
- \( m_1 = 20 \, \text{kg} \) (mass of block on the horizontal surface)
- \( m_2 = 5 \, \text{kg} \) (mass of hanging block)
- Coefficient of **static friction**, \( \mu_s = 0.4 \)
- Coefficient of **kinetic friction**, \( \mu_k = 0.3 \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)  

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## **(a) Free-body diagram for both masses:**

### Block \( m_1 \) on the horizontal surface:
1. **Weight (downward)**: \( W_1 = m_1 g = 20 \times 9.8 = 196 \, \text{N} \)
2. **Normal force (upward)**: \( N_1 = W_1 = 196 \, \text{N} \)
3. **Friction force (opposing motion)**:  
   \( f_k = \mu_k N_1 = 0.3 \times 196 = 58.8 \, \text{N} \)  
4. **Tension in the rope (to the right)**: \( T \)
  
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### Block \( m_2 \) (hanging):
1. **Weight (downward)**: \( W_2 = m_2 g = 5 \times 9.8 = 49 \, \text{N} \)
2. **Tension in the rope (upward)**: \( T \)

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## **(b) Acceleration of both blocks and tension in the rope**

The two blocks are connected, so their accelerations are the same. Let’s apply **Newton’s second law** to both blocks.

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### **For Block \( m_1 \):**
\[
T - f_k = m_1 a \quad \text{(Equation 1)}
\]
where \( f_k = 58.8 \, \text{N} \).

---

### **For Block \( m_2 \):**
\[
m_2 g - T = m_2 a \quad \text{(Equation 2)}
\]

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### **Solve for \( T \) and \( a \):**

1. From **Equation 2**:
   \[
   T = m_2 g - m_2 a
   \]
   Substituting \( m_2 = 5 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \):
   \[
   T = 49 - 5a
   \]

2. Substitute this value of \( T \) into **Equation 1**:
   \[
   49 - 5a - 58.8 = 20a
   \]
   Simplify:
   \[
   -9.8 = 25a
   \]
   \[
   a = -\frac{9.8}{25} = -0.392 \, \text{m/s}^2
   \]

---

### **Interpretation:**
The negative value indicates that the system is not accelerating in the expected direction—meaning the friction is strong enough to prevent motion, so **the blocks remain at rest**.

To confirm, let’s calculate the static friction force required to keep the system at rest:
\[
f_s = \mu_s N_1 = 0.4 \times 196 = 78.4 \, \text{N}
\]
Since the net force from \( m_2 \) is only \( 49 \, \text{N} \), the static friction is sufficient to prevent motion.

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## **Conclusion:**
1. **The system does not move**, and the acceleration is \( a = 0 \, \text{m/s}^2 \).
2. The tension in the rope is simply the weight of block \( m_2 \):  
   \[ T = 49 \, \text{N}. \]

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Do you want any further clarification or additional steps?  

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### **Five related questions:**
1. How would the acceleration change if \( m_2 \) was heavier?
2. What if the surface was frictionless—how would the system behave?
3. How does the static friction coefficient impact the motion of connected blocks?
4. How can we calculate the work done by friction in this scenario?
5. How would the equations change if the pulley had mass or friction?

---

### **Tip:**  
Always compare the force due to friction with the force driving the system to predict whether the system will move.

A block with a mass m1 = 20 kg is placed on a rough horizontal surface with static and kinetic friction coefficients of µs = 0.4 and µk = 0.3, respectively. It is connected by a rope over a frictionless pulley to another block with a mass m2 = 5 kg hanging vertically. (a) Draw the free-body diagram of both masses. (b) Calculate the acceleration of both blocks and the tension in the rope.

This problem explores a physics scenario with two blocks connected by a pulley, where one block is on a rough surface. Learn how to solve for the system's acceleration and tension in the rope, using concepts such as kinetic friction, static friction, and Newton's second law.

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