Sure! Let's solve the problem step-by-step:
1. Identify the given information:
- Spring constant ([tex]\(k\)[/tex]): 1400 N/m
- Stretched length of the spring ([tex]\(L_s\)[/tex]): 2.5 m
- Original (unstretched) length of the spring ([tex]\(L_o\)[/tex]): 1.0 m
2. Calculate the deformation (stretch) of the spring:
The deformation ([tex]\(x\)[/tex]) is the difference between the stretched length and the original length:
[tex]\[ x = L_s - L_o \][/tex]
Given:
[tex]\[ L_s = 2.5 \, \text{m} \][/tex]
[tex]\[ L_o = 1.0 \\, \text{m} \][/tex]
Substitute these values:
[tex]\[ x = 2.5 \, \text{m} - 1.0 \, \text{m} \][/tex]
[tex]\[ x = 1.5 \, \text{m} \][/tex]
3. Calculate the elastic potential energy stored in the spring:
The formula to calculate elastic potential energy ([tex]\(E\)[/tex]) stored in a spring is:
[tex]\[ E = \frac{1}{2} k x^2 \][/tex]
Here:
- [tex]\(k\)[/tex] is the spring constant (1400 N/m)
- [tex]\(x\)[/tex] is the deformation (1.5 m)
Substitute the values into the formula:
[tex]\[ E = \frac{1}{2} \times 1400 \, \frac{\text{N}}{\text{m}} \times (1.5 \, \text{m})^2 \][/tex]
[tex]\[ E = \frac{1}{2} \times 1400 \times 2.25 \][/tex]
[tex]\[ E = \frac{1}{2} \times 3150 \][/tex]
[tex]\[ E = 1575 \, \text{J} \][/tex]
4. Conclusion:
The elastic potential energy stored in the spring when it is stretched to a length of 2.5 m is 1575 J.
Therefore, the correct answer is:
[tex]\[ \text{1575 J} \][/tex]
