A spring of spring constant `5 xx 10^(3) N//m` is stretched initially by 5 cm from the upstretched position. The work required to further stretch the spring by another 5 cm is .

Options

A. `6.25 N-m`
B. `12/50 N-m`
C. `18.75 N-m`
D. `25.00 N-m`

Explanation

To find the work required to further stretch a spring that has already been stretched, we can use the formula for the potential energy stored in a spring, which is given by:

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

where:
- \( PE \) is the potential energy,
- \( k \) is the spring constant,
- \( x \) is the displacement from the equilibrium position.

### Step-by-Step Solution:

1. **Identify the given values:**
- Spring constant, \( k = 5 \times 10^3 \, \text{N/m} \)
- Initial stretch, \( x_i = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Additional stretch, \( x_f = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Total stretch after additional stretch, \( x_f = x_i + 0.05 = 0.05 + 0.05 = 0.1 \, \text{m} \)

2. **Calculate the initial potential energy (PE_initial):**
\[ PE_i = \frac{1}{2} k x_i^2 = \frac{1}{2} (5 \times 10^3) (0.05)^2 \]
\[ PE_i = \frac{1}{2} (5 \times 10^3) (0.0025) = \frac{1}{2} (12.5) = 6.25 \, \text{J} \]

3. **Calculate the final potential energy (PE_final):**
\[ PE_f = \frac{1}{2} k x_f^2 = \frac{1}{2} (5 \times 10^3) (0.1)^2 \]
\[ PE_f = \frac{1}{2} (5 \times 10^3) (0.01) = \frac{1}{2} (50) = 25 \, \text{J} \]

4. **Calculate the work done (W):**
The work done to stretch the spring further is the difference between the final and initial potential energy:
\[ W = PE_f - PE_i = 25 \, \text{J} - 6.25 \, \text{J} = 18.75 \, \text{J} \]

### Final Answer:
The work required to further stretch the spring by another 5 cm is **18.75 J**.