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))`
Acceleration of block A and B.

Explanation

To solve the problem, we need to analyze the forces acting on both blocks and apply Newton's laws of motion.

### Given:
- Mass of block A (m1) = 50 kg
- Mass of block B (m2) = 1 kg
- Horizontal force (F) applied on block A = 10 N
- Acceleration due to gravity (g) = 10 m/s²

### Step 1: Identify the forces acting on block A (50 kg)
The only horizontal force acting on block A is the applied force of 10 N. Since the surface is smooth (frictionless), the only force acting in the vertical direction is the gravitational force (weight) and the normal force from block B.

**Weight of block A (W1)**:
\[ W_1 = m_1 \cdot g = 50 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 500 \, \text{N} \]

### Step 2: Identify the forces acting on block B (1 kg)
Block B only experiences the normal force from block A (N2) and its own weight.

**Weight of block B (W2)**:
\[ W_2 = m_2 \cdot g = 1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 10 \, \text{N} \]

### Step 3: Apply Newton's second law to block A
For block A, we can write the equation of motion in the horizontal direction:
\[ F = m_1 \cdot a_A \]
Where \( a_A \) is the acceleration of block A.

Substituting the values:
\[ 10 \, \text{N} = 50 \, \text{kg} \cdot a_A \]

### Step 4: Solve for acceleration of block A
Rearranging the equation gives:
\[ a_A = \frac{10 \, \text{N}}{50 \, \text{kg}} = 0.2 \, \text{m/s}^2 \]

### Step 5: Apply Newton's second law to block B
Since block B is resting on block A and there is no horizontal force acting on it (due to the smooth surface), the acceleration of block B will be the same as that of block A.

Thus,
\[ a_B = a_A = 0.2 \, \text{m/s}^2 \]

### Final Results:
- Acceleration of block A (50 kg) = 0.2 m/s²
- Acceleration of block B (1 kg) = 0.2 m/s²