A block is suspended by an ideal spring of force constant force F and if maximum displacement of block from its initial mean position of rest is `x_(0)` then

Options

A. increase in energy stored in spring is `kx_(0)^(2)`
B. `x_(0)=(3F)/(2k)`
C. `x_(0)=(2F)/(k)`
D. work done by applied force F is `Fx_(0)`

Answer

To solve the problem, we need to analyze the situation of a block suspended by an ideal spring and derive the maximum displacement \( x_0 \) from its mean position.

### Step-by-step Solution:

1. **Understanding the Forces**:
The block is subjected to two forces: the gravitational force \( mg \) acting downward and the spring force \( kx \) acting upward, where \( k \) is the spring constant and \( x \) is the displacement from the mean position.

2. **Equilibrium Condition**:
At maximum displacement \( x_0 \), the block is in equilibrium. Therefore, the downward force due to gravity is balanced by the upward spring force:
\[ mg = kx_0 \]

3. **Solving for Maximum Displacement**:
Rearranging the equation gives:
\[ x_0 = \frac{mg}{k} \]

4. **Work Done by the Applied Force**:
The work done \( W \) by the applied force \( F \) to stretch the spring to the maximum displacement \( x_0 \) can be calculated using the formula for work done against a spring:
\[ W = F \cdot x_0 \]

5. **Potential Energy in the Spring**:
The potential energy stored in the spring at maximum displacement \( x_0 \) is given by:
\[ PE = \frac{1}{2} k x_0^2 \]

6. **Substituting \( x_0 \)**:
Substitute \( x_0 = \frac{mg}{k} \) into the potential energy equation:
\[ PE = \frac{1}{2} k \left(\frac{mg}{k}\right)^2 = \frac{1}{2} \frac{m^2 g^2}{k} \]

7. **Final Expressions**:
- Maximum displacement: \( x_0 = \frac{mg}{k} \)
- Work done by applied force: \( W = F \cdot x_0 \)
- Potential energy stored in the spring: \( PE = \frac{1}{2} \frac{m^2 g^2}{k} \)

### Conclusion:
The maximum displacement \( x_0 \) of the block from its initial mean position is given by:
\[ x_0 = \frac{mg}{k} \]