Work done by a spring when stretched

  • Thread starter bobby3280
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  • #1
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I have a simple problem that i can't seem to get right.

An ideal spring has a spring constant k = 22.0 N/m. What is the amount of work that must be done to stretch the spring 0.70 m from its relaxed length?

I tried W = (-)1/2 k x^2
= - (.5)(22.0) (.7^2)
= -5.39
but this isn't correct any suggestions.

Thanks
 

Answers and Replies

  • #2
Doc Al
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Why do you have a minus sign?
 
  • #3
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Thanks, I had the minus sign becuase that is how the formula is in the book.
 
  • #4
radou
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bobby3280 said:
Thanks, I had the minus sign becuase that is how the formula is in the book.

Since this is a conservative system, the work done from point 1 to 2 equals the change of potential energy, i.e. W = V2 - V1. In your case V1 = 0 (the spring is relaxed), so W = V2, hence, there should not be any minus sign involved in the book.
 
  • #5
Doc Al
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bobby3280 said:
Thanks, I had the minus sign becuase that is how the formula is in the book.
The formula for what? The energy stored in a spring equals 1/2kx^2 (no minus sign).
 
  • #6
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Um there is usually a minus sign isnt there? Because that equation is found by taking the intergral of Hooke's Law.

| means sigma for this because I cant find how to type it lol....


F=-Kx
W=|(fx)(dx)
W=|(-kx)(dx)
W=|-K(x^2/2)(dx)
W=-1/2Kx^2

I think that is right. And it makes sense because the work done by a spring is negative because it opposes an objects displacement.
 
  • #7
Doc Al
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Realize that the work done by the spring is the negative of the work required to stretch it (which was the question in this thread, despite the title). The work required to stretch (or compress) a spring is positive; the energy stored in a stretched (or compressed) spring is positive.
 
  • #8
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Oh, well that makes more sense. Didn't realize what was being asked.
 

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