Hookes law- force of the spring=(constant)(x)
Elastic potential energy- PEs=(1/2)kx*2
practice problem:
How much force is needed to stretch the spring .8m using a constant of .6N/m
Fs=kx
Fs=(.6N/m)(.8m)
Fs=.48N
How much work is done on the spring?
W=Et=changePE+changeKE+changeQ
W=changePE
W=(1/2)kx*2
W=(1/2)(.6N/m)(.8m)*2
W=.192J
Doing work on a spring changes its potential energy