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
3 comments:
i think these equations are easy but get harder once you have to apply them to equations we previously learned. there are a few examples like that on the homework that i dont get but hopefully we'll go over it in class tomorrow
when is the lab due that we did last week because you gave no specific due date? i think it was called work and power lab.
Bryan - I'd like to get the lab by Thursday - We'll get into a more regular routine next semester...
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