A block of mass 50 kg is kept on another block of mass 1 kg as shown in figure. A horizontal force of 10 N is applied on the 50kg block. (All surface are smooth). Find
`(g = 10 m//s^(2))`
Force exerted by B on A.

Explanation

To solve the problem of finding the force exerted by block B on block A, we will follow these steps:

### Step 1: Identify the forces acting on the blocks
- Block A (1 kg) is on top of block B (50 kg).
- A horizontal force of 10 N is applied to block B.
- The gravitational force acting on block A is \( m_1 \cdot g = 1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 10 \, \text{N} \).
- The gravitational force acting on block B is \( m_2 \cdot g = 50 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 500 \, \text{N} \).

### Step 2: Determine the acceleration of block B
Using Newton's second law, the net force acting on block B can be calculated:
\[ F = m \cdot a \]
For block B:
\[ 10 \, \text{N} = 50 \, \text{kg} \cdot a_B \]
Solving for \( a_B \):
\[ a_B = \frac{10 \, \text{N}}{50 \, \text{kg}} = 0.2 \, \text{m/s}^2 \]

### Step 3: Determine the force exerted by block B on block A
The force exerted by block B on block A (denoted as \( N_2 \)) can be found using the weight of block A and the acceleration of block B. Since block A is not accelerating horizontally (it is at rest relative to block B), the normal force \( N_2 \) must balance the weight of block A and provide the necessary force to accelerate it:
\[ N_2 = m_1 \cdot g + m_1 \cdot a_B \]
Substituting the values:
\[ N_2 = 1 \, \text{kg} \cdot 10 \, \text{m/s}^2 + 1 \, \text{kg} \cdot 0.2 \, \text{m/s}^2 \]
\[ N_2 = 10 \, \text{N} + 0.2 \, \text{N} = 10.2 \, \text{N} \]

### Step 4: Conclusion
The force exerted by block B on block A is \( 10.2 \, \text{N} \).