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Spring Force Sign Question

  1. Sep 22, 2013 #1
    I am sorry if I am doing anything wrong. I'm new to this site and physics in general. Please feel free to correct me if I'm making some stupid mistake in formatting.

    1. The first figure gives spring force Fx versus position x for the spring–block arrangement of the second figure. The scale is set by Fs = 160 N. We release the block at x = 12.0 cm. How much work does the spring do on the block when the block moves from xi =+9.0 cm to (a)x=+6.0 cm, (b)x=-6.0 cm, (c)x=-9.0 cm, and (d)x=-10.0 cm?


    2. Relevant equations
    I am probably missing an equation here, but here are the ones I used:

    F = kx
    Work = -.5k(Δx)^2 ​

    3.
    Okay, first I converted everything to meters, which just ended up moving the decimal two places left.

    Second, I found k. (Force is Kg*m/(s*s), x is meters, so k is in Kg/(s*s), which is kinda a weird unit, but whatever)

    F = kx
    k = F/x
    k = 160/.02
    k = 8000​

    Third, I plugged in the numbers for the work equation.

    Work = -.5k Δx
    -.5k = -.5*8000 = -4000
    Δx = xf - xi
    Work = -4000(xf - .09)^2 ​

    Lastly, I plugged in the four ending points. (x's are in meters)

    Axf = .06-----> Wa = -3.6 Joules
    Bxf = -.06----> Wb = -90 Joules
    Cxf = -.09----> Wc = -129.6 Joules
    Dxf = -.1-----> Wd = -144.4 Joules​



    Okay, as I was writing this up, I found out that I had made a stupid mistake in my calculations (I didn't square delta x in the work equations). However, I'm on my last shot on this online homework, and I want to get it RIGHT. So are these values correct? Or should they be positive? (I thought Joules were always positive, but I don't think I missed a minus anywhere...)
    Thank you for your help!
     
  2. jcsd
  3. Sep 22, 2013 #2

    Doc Al

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    Staff: Mentor

    Careful! That's the work done by the spring when it is stretched a distance Δx from its unstretched position, not for an arbitrary Δx.

    From F = ma, you should recognize Kg*m/(s*s) as equivalent to a Newton.

    OK.

    No. Using your equation, you want ΔW.

    You'll need to redo your calculations.

    Joule is a unit of work/energy. Work can be positive or negative.
     
  4. Sep 22, 2013 #3
    Thanks! Could you tell me why, exactly, I would be solving for the change in work? I know you mentioned this:
    But how do I know that I'm not supposed to do what I did?


    Additionally, how do you solve for the change in work? (I know that ΔW is Wfinal - Winitial, but what x's make up those works? Would it be that the Winitial Δx would be .12-.9? Or am I wrong, again?)
     
    Last edited: Sep 22, 2013
  5. Sep 22, 2013 #4

    Doc Al

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    Staff: Mentor

    I think you'll have an easier time if you think in terms of the elastic potential energy stored in the spring. That is given by:
    PE = .5kx2, where x is the displacement from equilibrium.

    The work done by the spring in moving from one position to another will equal -ΔPE. (If the PE increases, that means the spring has done negative work.)

    Note that Δ(y2) ≠ (Δy)2.
     
  6. Sep 22, 2013 #5
    ...We haven't gone over potential energy yet; so I'm a bit clueless on what you just said there. Sorry about that.
     
  7. Sep 22, 2013 #6

    Doc Al

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    Staff: Mentor

    No worries. In that case you'll just have to use your formula, but with the caveat that I explained in my earlier post.

    For example, you'll first have to calculate the work done by the spring when stretched to +9.0 cm. Then compare that to the work done when stretched to +6.0 cm. And so on.

    A most important tip is:
     
  8. Sep 22, 2013 #7
    Ahh okay thank you very much!
     
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