Answer:
87.5 kJ
Step-by-step explanation:
The potential energy stored in a stretch spring is given by the expression below
PE = Work = force * distance
So:

This then simplifies to:

where k is the spring constant= 140 N/m
Given that x= 25 m
Substituting our data into the expression for P.E stored we have

Hence the energy stored in the spring is 87.5kJ