If a spring has a constant of 9 (kg)/s^29kgs2, how much work will it take to extend the spring by 22 cm 22cm?

1 Answer
Feb 29, 2016

The work done in extending the spring a distance, e, is given by the area under the force-extension graph for the spring. This equals:
W=1/2*F*eW=12Fe
for a spring that obeys Hooke's law.

The relationship between force and extension is given by F=k*eF=ke
If you substitute the above into the first equation you arrive at:
W=1/2*k*e^2W=12ke2
Hence:
W= 1/2*9*0.22^2=0.22JW=1290.222=0.22J