A block of mass `m` attached in the lower and vertical spring The spring is hung from a calling and force constant value `k` The mass is released from rfest with the spring velocity unstrached The maximum value praduced in the length of the spring will be

Options

A. `Mg//k`
B. `2Mg//k`
C. `4Mg//k`
D. `Mg//2k`

Explanation

To solve the problem of finding the maximum extension of a spring when a mass is attached and released from rest, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a spring with spring constant \( k \) that is hung from the ceiling, and a mass \( m \) is attached to the lower end of the spring. Initially, the spring is unstretched.

2. **Initial Conditions**:
- When the mass is released from rest, the spring is at its natural length, meaning the potential energy of the mass and the spring is zero.

3. **Maximum Extension**:
- Let \( x \) be the maximum extension of the spring from its natural length. At this point, the mass has fallen a distance \( x \).

4. **Energy Conservation**:
- Since there are no external forces acting on the system between the initial and maximum extension points, we can apply the principle of conservation of energy.
- The potential energy lost by the mass when it falls a distance \( x \) is given by:
\[ \text{Potential Energy of mass} = -mgx \]
- The potential energy stored in the spring when it is stretched by \( x \) is:
\[ \text{Potential Energy of spring} = \frac{1}{2} k x^2 \]

5. **Setting Up the Energy Conservation Equation**:
- The loss in potential energy of the mass equals the gain in potential energy of the spring:
\[ mgx = \frac{1}{2} k x^2 \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ kx^2 - 2mgx = 0 \]
- Factoring out \( x \):
\[ x(kx - 2mg) = 0 \]

7. **Solving for \( x \)**:
- This gives us two solutions: \( x = 0 \) (the trivial solution) or:
\[ kx = 2mg \implies x = \frac{2mg}{k} \]

8. **Conclusion**:
- The maximum extension \( x \) of the spring when the mass is released from rest is:
\[ x = \frac{2mg}{k} \]

### Final Answer:
The maximum extension produced in the length of the spring will be \( \frac{2mg}{k} \).

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