Answer:
1575 J
Explanation:
The elastic potential energy of a string is given by:
[tex]E=\frac{1}{2}k \Delta x^2[/tex]
where
k is the spring constant
[tex]\Delta x[/tex] is the stretching/compression of the string with respect to its natural length
For the spring in the problem, we have:
- Spring constant: [tex]k=1400 N/m[/tex]
- Stretching: [tex]\Delta x=2.5 m -1.0 m=1.5 m[/tex]
Therefore, the elastic potential energy stored in the spring is
[tex]E=\frac{1}{2}(1400 N/m)(1.5 m)^2=1575 J[/tex]
