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. For part (a), the work done to stretch the spring from its relaxed position to an extension of 10 cm is correctly calculated as 2.25 J. In part (b), the additional work for stretching it another 10 cm is calculated to be 6.75 J. Part (c) addresses the work done when allowing the spring to return from an extension of 20 cm to its relaxed position, which totals 9 J. The calculations confirm that while the spring returns to its relaxed position, it will oscillate around that point if released without external energy.
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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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