Answer:
E = 2.6411 J
Explanation:
we know that elastic potential energy could be calculated by:
[tex]E = \frac{1}{2}kx^2[/tex]
where k is the constant of the spring and x is the deform of the spring.
Now x can be calculated as:
x = 3.52m-2.54m
x = 0.98 m
Finally, replacing the data, we get:
[tex]E = \frac{1}{2}(5.5 N/m)(0.98m)^2[/tex]
E = 2.6411 J
