How Much Work Is Done Stretching a Spring?

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The discussion focuses on calculating the work done when stretching a spring with a force constant of 450 N/m. The work for stretching the spring from its relaxed position to an extension of 10 cm is calculated as 2.25 J. An additional 10 cm stretch requires 6.75 J, totaling 9 J for the full extension of 20 cm. The return of the spring to its relaxed position involves oscillation, indicating that energy is conserved but not stored in a static state.

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[SOLVED] Stretching things out a bit, I

Can anyone help me with this problem?

You have a spring with force constant 450 N/m. How much work do you do in
(a) stretching the spring from its relaxed position to an extension of 10 cm.
(b) stretching it an additional 10 cm.
(c) allowing it to return from an extension of 20 cm to its relaxed position?

is this right?
a) W = 1/2(450)(.1^2) = 2.25J

b) W = 1/2(450)(.2^2) - 1/2(450)(.1^2) = 6.75J

c) W = 1/2(450)(.2^2) - 0 = 9J
 
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a) and b) right
c)W=9 j,when you make spring to stay at rest in his relaxed position.But if you just release spring without giving any energy it'll return to his relaxed position,but it'll not stop there.The object will oscillate around relaxed position.
 
Thank you so much azatkgz.
 

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