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Work of a spring

  1. Oct 9, 2004 #1
    Hello everyone!
    The problem I have is really easy, but something is confusing me.

    A spring with spring constant 24 N/m is stretched 0.24 m from its equilibrium position. How much work must be done to stretch it an additional 0.072 m?

    I used the formula : W = (1/2)*k(Xf^2 - Xi^2), (it came out negative) but the answer I get is not the correct one. For Xf I used 0.24+0.072 since that's the distance from the equilibrium position.
    Can you please give me a hint?
     
  2. jcsd
  3. Oct 9, 2004 #2

    Diane_

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    Homework Helper

    What did you use for xi?
     
  4. Oct 9, 2004 #3
    Hi Diane,
    For X1 I used 0.24 and as I said for Xf i used 0.24+0.072. I think that the work should be negative since the spring goes toward positive x-axis.
    I am not really clear with the work on a spring because the chapter on the book was very brief and that other books I have used are really messy. So I think this should be easy. Thus I just applied the formula I first posted.
     
  5. Oct 10, 2004 #4
    Anyone can give a hint?
     
  6. Oct 10, 2004 #5

    Doc Al

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    Staff: Mentor

    Your method seems correct to me. Why do you say it comes out negative? You are finding the work done to stretch the spring: the force and the displacement are in the same direction, so the work is positive.

    Perhaps you are making an arithmetic mistake.
     
  7. Oct 10, 2004 #6
    Thanks Doc Al,
    I apologize, the work is really positive. That's how I solved it.
    I have to solve the exercise and submit the answer online. I have only 7 chances. So I tried for the first time using the formula and the distances I wrote on the first post. It said "Incorrect"; than I thought maybe work is negative (I knew it was positive) but still I tried once more and I came out
    "Incorrect" again. Then I just considered the first position as 0 and I applied the formula W=-.5 *k*X^2 but I falied again.
    I don't know what is wrong with this!
     
  8. Oct 10, 2004 #7

    Doc Al

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    Staff: Mentor

    Are you stating the problem exactly as given? Are you using proper units in reporting your answer? Show your arithmetic and we can check it over for silly errors.
     
  9. Oct 10, 2004 #8
    Thank you for the time Doc Al,

    I wrote the problem exactly as written on the paper. k is 24 N/m and it is stretched 0.24 m from equilibrium. Then the question is; How much work must be done to stretch it and additional 0.072m? (Answer in J)
    so : W = 0.5*24([0.24+0.072]^2 - 0.24^2)
    W = 12 (0.312^2 - 0.24^2)
    W= 12(0.097344 - 0.0576)
    =12* 0.039744 = 0.476928 J
    This is what I think it should be (which in fact it isn't).
     
  10. Oct 10, 2004 #9

    Doc Al

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    Staff: Mentor

    Your answer looks correct to me. (I'd round it off, though.)
     
  11. Oct 11, 2004 #10
    :cry: I realized that on Friday night the server had failed because I entered the same answer this morning and amazingly it said "Correct". I don't know what happened; probably a server break down.
    Anyway thanks a lot for the help.
     
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