Question

a spring whose stiffness is 820n/m has a relaxed length of 0.44m. if the length of the spring changes from 0.29m to 0.74m, what is the change in the potential energy of the spring?

Answer 1

At .29 m spring compression is .44 - .29 = -.15 m

At .74 m spring extension is .74 - .44 = .30 m

W = K X2

PE due to compression W1 = 1/2 K (-.15)^2 = .0113 K

PE due to extension W2 = 1/2 K (.30)^2 = .045 K

Total PE = (.0113 + .045) * 820 = 46.2 N