When a load of 10 kg is hung from the wire, then extension of 2m is produced. Then work done by restoring force is

Options

A. 200 J
B. 100 J
C. 50 J
D. 25 J

Explanation

To solve the problem, we need to find the work done by the restoring force when a load of 10 kg is hung from a wire, causing an extension of 2 m.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Load (mass) = 10 kg
- Extension (elongation) = 2 m
- Acceleration due to gravity (g) = 9.8 m/s² (approximately 10 m/s² for simplicity)

2. **Calculate the Weight of the Load:**
The weight (force due to gravity) can be calculated using the formula:
\[ F = m \cdot g \]
Where:
- \( m = 10 \, \text{kg} \)
- \( g \approx 10 \, \text{m/s}^2 \)

Thus,
\[ F = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N} \]

3. **Determine the Maximum Stretching Force:**
The maximum stretching force (which is equal to the weight of the load when it is fully extended) is:
\[ F_{\text{max}} = 100 \, \text{N} \]

4. **Calculate the Work Done by the Restoring Force:**
The work done by the restoring force when the wire is stretched can be calculated using the formula:
\[ W = \frac{1}{2} \times F_{\text{max}} \times \text{elongation} \]
Substituting the values:
\[ W = \frac{1}{2} \times 100 \, \text{N} \times 2 \, \text{m} \]
\[ W = \frac{1}{2} \times 200 \, \text{N m} = 100 \, \text{J} \]

5. **Final Answer:**
The work done by the restoring force is:
\[ W = 100 \, \text{J} \]