A block of mass `m_(1)` rests on a horizontal table. A string tied to the block is passed on a frictionless pulley fixed at the end of the table and to the other end of string is hung another block of mass `m_(2)`. The acceleration of the system is

Options

A. `(m_2g)/(m_1+m_2)`
B. `(m_1g)/(m_1+m_2)`
C. g
D. `(m_2g)/(m_1)`

Answer

To find the acceleration of the system consisting of two blocks, we will analyze the forces acting on each block and apply Newton's second law of motion.

### Step-by-Step Solution:

1. **Identify the System**:
- We have two blocks:
- Block 1 (mass \( m_1 \)) is on a horizontal table.
- Block 2 (mass \( m_2 \)) is hanging off the edge of the table.

2. **Draw Free Body Diagrams**:
- For Block 1 (\( m_1 \)):
- The forces acting on it are:
- Tension \( T \) in the string acting horizontally to the right.
- Normal force \( N \) acting vertically upward.
- Weight \( m_1 g \) acting vertically downward.
- Since there is no vertical motion, we have:
\[ N = m_1 g \]
- In the horizontal direction, applying Newton's second law:
\[ T = m_1 a \quad (1) \]

- For Block 2 (\( m_2 \)):
- The forces acting on it are:
- Weight \( m_2 g \) acting downward.
- Tension \( T \) acting upward.
- Applying Newton's second law in the vertical direction:
\[ m_2 g - T = m_2 a \quad (2) \]

3. **Combine the Equations**:
- From equation (1), we can express \( T \):
\[ T = m_1 a \]
- Substitute \( T \) into equation (2):
\[ m_2 g - m_1 a = m_2 a \]

4. **Rearranging the Equation**:
- Combine the terms involving \( a \):
\[ m_2 g = m_1 a + m_2 a \]
\[ m_2 g = (m_1 + m_2) a \]

5. **Solve for Acceleration \( a \)**:
- Rearranging gives:
\[ a = \frac{m_2 g}{m_1 + m_2} \]

### Final Answer:
The acceleration of the system is:
\[ a = \frac{m_2 g}{m_1 + m_2} \]