For shown situation find the maximum elongation in the spring. Neglect friction everywhere. Initiallthe blocks are at rest and spring is unstreched.

A. `4F/3K`
B. `3F/4K`
C. `4F/K`
D. `2F/K`

Options

A. `4F/3K`
B. `3F/4K`
C. `4F/K`
D. `2F/K`

Explanation

To find the maximum elongation in the spring when the blocks are at rest and the spring is unstretched, we can use the concept of reduced mass and the work-energy theorem. Here’s a step-by-step solution:

### Step 1: Understand the System
We have a system of blocks connected by a spring. Initially, the blocks are at rest, and the spring is unstretched. We need to find the maximum elongation of the spring when a force \( F \) is applied.

### Step 2: Define the Reduced Mass
For a system of two masses \( m_1 \) and \( m_2 \), the reduced mass \( \mu \) is given by the formula:
\[ \mu = \frac{m_1 m_2}{m_1 + m_2} \]
In our case, let’s assume we have three blocks of mass \( 2m \), \( 3m \), and \( 6m \). We will consider the two blocks with masses \( 3m \) and \( 6m \) for calculating the reduced mass.

### Step 3: Calculate the Reduced Mass
Using the formula for reduced mass:
\[ \mu = \frac{3m \cdot 6m}{3m + 6m} = \frac{18m^2}{9m} = 2m \]

### Step 4: Apply Newton's Second Law
The net force acting on the system can be expressed as:
\[ F_{\text{net}} = F - F_{\text{spring}} \]
Where \( F_{\text{spring}} \) is the force exerted by the spring when it is elongated by \( x \).

### Step 5: Use Work-Energy Theorem
According to the work-energy theorem, the work done on the system is equal to the change in kinetic energy. In our case, the work done by the applied force \( F \) is equal to the potential energy stored in the spring:
\[ W = \frac{1}{2} k x^2 \]
Where \( k \) is the spring constant and \( x \) is the elongation of the spring.

### Step 6: Set Up the Equation
The work done by the force can also be expressed as:
\[ W = F \cdot d \]
Where \( d \) is the distance moved by the center of mass. Here, we can express the work done as:
\[ F \cdot x = \frac{1}{2} k x^2 \]

### Step 7: Solve for Maximum Elongation
From the forces acting on the blocks, we can set up the equation:
\[ \frac{2F}{3} \cdot x = \frac{1}{2} k x^2 \]
Rearranging gives:
\[ k x^2 - \frac{4F}{3} x = 0 \]
Factoring out \( x \):
\[ x \left( kx - \frac{4F}{3} \right) = 0 \]
Thus, the maximum elongation \( x \) can be found as:
\[ kx = \frac{4F}{3} \implies x = \frac{4F}{3k} \]

### Conclusion
The maximum elongation in the spring is:
\[ x = \frac{4F}{3k} \]
Thus, the correct answer is option A: \( \frac{4F}{3K} \).