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Work done in a spring

  1. Mar 19, 2006 #1
    Hi all...

    I'm confused with the following question. The questions in this book are just terribly unclear..

    "For the pressure gauge in Figure 6.4, what work is done by the air on the spring?"


    Is it asking the magnitude of the work? I think it's impossible to find the magnitude of the work since the force constant of the spring is not given...

    Having said that, maybe it's asking the "type of work". If so, how should I answer this question? Is it acceptable to just say: the work done on the spring is equivalent to the elastic potential energy (W = Ee = (kx^2)/2).
    Last edited: Mar 19, 2006
  2. jcsd
  3. Mar 19, 2006 #2
    For a constant pressure process, the work done is equal to:

    [tex] W=Pdv=F_{spring}dx=\frac{1}{2}kx^2 [/tex]
  4. Mar 20, 2006 #3
    Thanks cyrusabdollahi.

    I never heard of Pdv before. But I did some research on google. That formula is used to calculate the work done during a change of volume of a gas. It could make sense to use this formula. But why is it a constant pressure process? I thought that the pressure would have to increase to push the spring?

    But let's assume the formula should work. How should I word my answer? Would this answer be acceptable:

    "For the pressure gauge in Figure 6.4, what work is done by the air on the spring?"

    As the volume of the air increases in the pressure gauge, the air pushes the spring to move 2.0 cm. This process causes the air to do work on the spring. The work done by the air on the spring, can be expressed as the following:

    W = P (delta)v = F(spring) dx = 1/2 kx^2, where x = 0.02 m.
  5. Mar 20, 2006 #4
    It is constant because the tire pressure is constant. It will push the spring and compress it until the force of the spring resisting compression is equal and opposite to the force of the gas pushing on the spring.

    <I hope im not giving you poor advice >
    Last edited: Mar 20, 2006
  6. Mar 20, 2006 #5

    Doc Al

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    Staff: Mentor

    While it makes sense that the gas pressure is essentially constant, the spring force would not be. I'd think that the work done by the gas would be twice the energy needed to compress the spring. (The extra energy would be wasted via friction.)
  7. Mar 20, 2006 #6
    Of course, I did not mean for it to be that way. The spring force will vary linearly from 0 initially until it is equal to the pressure force, P*A. I was sloppy. :frown:

    Hmmmm. I assumed that this was frictionless. You always know more than I, can you explain that please :smile:.

    Perhaps I should have wrote:

    [tex] PdV=F(x)dx [/tex]
    Last edited: Mar 20, 2006
  8. Mar 20, 2006 #7
    hmmm?? I'm losing you guys.. So how should I answer this question?
  9. Mar 20, 2006 #8

    Doc Al

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    Staff: Mentor

    Exactly. The spring force will vary, but the pressure force will not. Call the pressure force [itex]F_p[/itex]. When the spring comes to equilibrium, its displacement (D) will satisfy [itex]F_p = k D[/itex]. So the work done by the pressure force will equal [itex]F_p D = k D^2[/itex], while the energy stored in the spring will equal half that.

    If it was frictionless, the plunger would oscillate about the equilibrium point.

    This seems exactly like hanging a mass from a vertical spring. If you attach the mass and let it fall from the unstretched position, the work done by gravity will be twice the elastic PE when the spring is at its equilibrium position. Of course, that extra energy will be the KE of the mass. If there were nothing to damp the motion, it will keep oscillating.

    (We should get a pressure gauge and play with it. See if what I'm saying is true. :smile: )
  10. Mar 20, 2006 #9
    Yes, you are 100% correct :tongue:. I had the 1/2 relationship but never spotted the missing 1/2 energy.

    We could ignore the energy stored in the mass of the plunger since the mass would be insignificantly small. Then it would act like a critically damped osciallator as you say. And all the energy would be lost to the friction. Also, I would venture to say that you are putting energy into the air at the lower pressure that surrounds the spring as you compress it in the process as well.
    Last edited: Mar 20, 2006
  11. Jan 28, 2007 #10
    I was having a heck of a time with this question as well. Seeing as neither the force applied (pressure) nor the spring constant (k) are given (and I don't see how it could be determined without physical tests), I just opted to use the spring constant from the support question (640 N/m), noted that on the question, and moved on.
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