How far must this spring be compressed to store 35.0 J of potential energy?
2 Answers

PE = (1/2) kx^2
=> 35 = (1/2) * 82 * x^2
=> x ^2 = 70/82
=> x = 0.924 m.
[Of course, the spring must be sufficiently long to be able to compress this length.]

The energy E = ((k * x) * (x))/ 2 (The area under the Force vs. Extension graph)
35 = (82 * x^2)/2
x = SQR(70/82) = 0.9239 meters
Assuming the spring is linear and large enough
Source(s): Jim Dept. of Physics Imperial College London