A spring,which is initially is its unstretched condition,is first stretched by a length 5cm and then again by a further length 5cm.The work done in the first case is `W_(1)` , and in the second case is `W_(2)` .Then the work required to stetch it further by another 5 cm is (`K = 5xx10^(3)`)

Options

A. 18.75 J
B. 25.00 J
C. 6.25 J
D. 12.50 J

Explanation

To solve the problem, we will calculate the work done in stretching the spring in two different scenarios using Hooke's Law. The work done on a spring is given by the formula:

\[ W = \frac{1}{2} k x^2 \]

where:
- \( W \) is the work done,
- \( k \) is the spring constant,
- \( x \) is the displacement from the equilibrium position.

### Step 1: Calculate \( W_1 \) (Work done to stretch the spring by 5 cm)

1. **Convert the displacement from cm to meters**:
\[ x_1 = 5 \text{ cm} = 0.05 \text{ m} \]

2. **Use the formula for work done**:
\[ W_1 = \frac{1}{2} k x_1^2 \]
Given \( k = 5 \times 10^3 \, \text{N/m} \):
\[ W_1 = \frac{1}{2} \times 5 \times 10^3 \times (0.05)^2 \]

3. **Calculate \( W_1 \)**:
\[ W_1 = \frac{1}{2} \times 5 \times 10^3 \times 0.0025 \]
\[ W_1 = \frac{1}{2} \times 5 \times 10^3 \times 2.5 \times 10^{-3} \]
\[ W_1 = \frac{1}{2} \times 12.5 = 6.25 \, \text{J} \]

### Step 2: Calculate \( W_2 \) (Work done to stretch the spring from 5 cm to 10 cm)

1. **Calculate the total displacement for \( W_2 \)**:
\[ x_2 = 10 \text{ cm} = 0.1 \text{ m} \]

2. **Use the formula for work done for the total stretch**:
\[ W_2 = \frac{1}{2} k x_2^2 \]
\[ W_2 = \frac{1}{2} \times 5 \times 10^3 \times (0.1)^2 \]

3. **Calculate \( W_2 \)**:
\[ W_2 = \frac{1}{2} \times 5 \times 10^3 \times 0.01 \]
\[ W_2 = \frac{1}{2} \times 5 \times 10^3 \times 10^{-2} \]
\[ W_2 = \frac{1}{2} \times 50 = 25 \, \text{J} \]

### Step 3: Calculate the work done to stretch the spring from 10 cm to 15 cm

1. **Calculate the work done for the stretch from 10 cm to 15 cm**:
\[ x_3 = 15 \text{ cm} = 0.15 \text{ m} \]

2. **Use the formula for work done for the total stretch**:
\[ W_3 = \frac{1}{2} k x_3^2 \]
\[ W_3 = \frac{1}{2} \times 5 \times 10^3 \times (0.15)^2 \]

3. **Calculate \( W_3 \)**:
\[ W_3 = \frac{1}{2} \times 5 \times 10^3 \times 0.0225 \]
\[ W_3 = \frac{1}{2} \times 5 \times 10^3 \times 22.5 \times 10^{-3} \]
\[ W_3 = \frac{1}{2} \times 112.5 = 56.25 \, \text{J} \]

### Final Calculation of Work Required to Stretch from 10 cm to 15 cm

To find the work done to stretch from 10 cm to 15 cm, we subtract the work done to stretch to 10 cm from the work done to stretch to 15 cm:

\[ W_{10 \text{ to } 15} = W_3 - W_2 = 56.25 - 25 = 31.25 \, \text{J} \]

### Summary of Results

- \( W_1 = 6.25 \, \text{J} \)
- \( W_2 = 25 \, \text{J} \)
- Work required to stretch from 10 cm to 15 cm is \( 31.25 \, \text{J} \).