What is the potential energy of the spring when a mass of 65.0kg hangs vertically from it?

Question

A force of 715N stretches a certain spring a distance of 0.155m .Take the free fall acceleration to be = 9.80m/s^2 .

Answer 1

Force F=715 N

extension x =0.155 m

spring or force constant k = 715 /0.166 =4307.23 N/m

When loaded with a mass of 65.0 kg,the force F=mg=65*9.8= 637 N

Potential Energy PE = (1/2)F^2 /k

PE=(1/2)637*637 / 4307.23=47.10 J

Answer 2

A spring stretches according to
F = k*d
where F is the force stretching the spring, k is the spring's constant, and d is the distance stretched from the rest state.

For this spring,
k = 715 / 0.155 N/m

A spring's potential enegy U = (1/2)*k*d^2
Now plug in the data.