A block of mass 'm' is connected to another block of mass 'M' by a spring (massless) of spring constant 'k' The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched Then a constant force 'F' starts acting on the block of mass 'M' to pull it. Find the force on the block of mass 'm' :

Options

A. `(mf)/(M)`
B. `((M + m)F)/(m)`
C. `(mF)/((m + M))`
D. `(MF)/((m + M))`

Explanation

To solve the problem, we will analyze the forces acting on both blocks and apply Newton's second law of motion.

### Step-by-Step Solution:

1. **Identify the System**:
We have two blocks:
- Block of mass \( m \) (let's call it Block 1)
- Block of mass \( M \) (let's call it Block 2)
These blocks are connected by a massless spring of spring constant \( k \) and are on a smooth horizontal plane.

2. **Understanding the Forces**:
A constant force \( F \) is applied to Block 2 (mass \( M \)). This force will cause both blocks to accelerate. The spring will exert a force (tension \( T \)) on Block 1.

3. **Applying Newton's Second Law**:
- For Block 2 (mass \( M \)):
\[ F - T = M \cdot a \quad \text{(Equation 1)} \]
- For Block 1 (mass \( m \)):
\[ T = m \cdot a \quad \text{(Equation 2)} \]

4. **Expressing Acceleration**:
From Equation 1, we can express the tension \( T \):
\[ T = F - M \cdot a \]

5. **Substituting Tension in Equation 2**:
Now, substitute the expression for \( T \) from Equation 1 into Equation 2:
\[ F - M \cdot a = m \cdot a \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ F = m \cdot a + M \cdot a \]
\[ F = (m + M) \cdot a \]

7. **Finding Acceleration**:
Now, we can solve for acceleration \( a \):
\[ a = \frac{F}{m + M} \]

8. **Finding the Force on Block 1**:
Now, substitute \( a \) back into Equation 2 to find the force \( T \) (the force on Block 1):
\[ T = m \cdot a = m \cdot \frac{F}{m + M} \]
\[ T = \frac{mF}{m + M} \]

### Final Answer:
The force on the block of mass \( m \) is:
\[ T = \frac{mF}{m + M} \]