A 4 kg blocks A is placed on the top of 8 kg block B which rests on a smooth table .A just slips on B when a force of 12 N is applied on B to make Both A and B move together is (4x) N Find the value of x.

Answer

To solve the problem, we need to determine the value of \( x \) in the expression \( 4x \) N, which represents the force required to make both blocks A and B move together.

### Step-by-Step Solution:

1. **Identify the masses of the blocks:**
- Mass of block A, \( m_A = 4 \, \text{kg} \)
- Mass of block B, \( m_B = 8 \, \text{kg} \)

2. **Understand the condition for slipping:**
- Block A slips on block B when a force of \( 12 \, \text{N} \) is applied to block B. This force represents the maximum static friction force that can act between the two blocks.

3. **Calculate the acceleration of block A:**
- The force causing block A to slip is the maximum static friction force, which is \( 12 \, \text{N} \).
- Using Newton's second law, \( F = m \cdot a \), we can find the acceleration \( a_A \) of block A:
\[ a_A = \frac{F}{m_A} = \frac{12 \, \text{N}}{4 \, \text{kg}} = 3 \, \text{m/s}^2 \]

4. **Determine the total force required to move both blocks together:**
- When both blocks A and B move together, they will have the same acceleration. The total mass of the system (A + B) is:
\[ m_{\text{total}} = m_A + m_B = 4 \, \text{kg} + 8 \, \text{kg} = 12 \, \text{kg} \]
- The acceleration of both blocks when a force \( F \) is applied can be expressed as:
\[ F = m_{\text{total}} \cdot a \]
- Substituting the values:
\[ F = 12 \, \text{kg} \cdot 3 \, \text{m/s}^2 = 36 \, \text{N} \]

5. **Relate the total force to \( 4x \):**
- According to the problem, the total force \( F \) is also given as \( 4x \):
\[ 4x = 36 \, \text{N} \]
- To find \( x \), we solve for \( x \):
\[ x = \frac{36 \, \text{N}}{4} = 9 \]

### Final Answer:
The value of \( x \) is \( 9 \).