A body of mass 20 kg is resting on a table. A light string is attached to it and passed over a smooth pulley at the edge of the table and a mass 3kg is attached at the free end. What is the acceleration of the system of the coefficient of kinetic friction between the mass and the table is `0.04` ? What is the tension in the string ?

Explanation

To solve the problem step by step, we will analyze the forces acting on both masses and apply Newton's second law.

### Step 1: Identify the forces acting on the system
- For the 20 kg mass on the table:
- Weight (downward): \( W_1 = m_1 \cdot g = 20 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 196 \, \text{N} \)
- Normal force (upward): \( N = W_1 = 196 \, \text{N} \)
- Frictional force (opposing motion): \( F_k = \mu \cdot N = 0.04 \cdot 196 \, \text{N} = 7.84 \, \text{N} \)

- For the 3 kg mass hanging off the table:
- Weight (downward): \( W_2 = m_2 \cdot g = 3 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 29.4 \, \text{N} \)

### Step 2: Write the equations of motion
- For the 3 kg mass (downward motion):
\[ m_2 g - T = m_2 a \quad \text{(1)} \]
Substituting the values:
\[ 29.4 - T = 3a \]

- For the 20 kg mass (horizontal motion):
\[ T - F_k = m_1 a \quad \text{(2)} \]
Substituting the values:
\[ T - 7.84 = 20a \]

### Step 3: Solve the equations simultaneously
From equation (1):
\[ T = 29.4 - 3a \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ (29.4 - 3a) - 7.84 = 20a \]
Simplifying:
\[ 29.4 - 7.84 = 20a + 3a \]
\[ 21.56 = 23a \]
\[ a = \frac{21.56}{23} \approx 0.94 \, \text{m/s}^2 \]

### Step 4: Find the tension in the string
Substituting the value of \( a \) back into equation (3):
\[ T = 29.4 - 3(0.94) \]
Calculating:
\[ T = 29.4 - 2.82 = 26.58 \, \text{N} \]

### Final Answers
- The acceleration of the system is approximately \( 0.94 \, \text{m/s}^2 \).
- The tension in the string is approximately \( 26.58 \, \text{N} \).

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