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Spring, weight and potential energy

  1. Jan 30, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle is connected to a spring at rest. Because of weight, the mass moves a distance [itex]\Delta x[/itex]. Calculate the value of the elastic constant [itex]k[/itex]


    2. Relevant equations
    [itex]U_{PE} = \frac{1}{2}k\Delta x^2[/itex]
    [itex]U = mgh[/itex]
    [itex]F = ma[/itex] (in general)
    [itex]F = kx[/itex] (Hooke's Law)

    3. The attempt at a solution
    I've tried to solve this problem from two points of view, but the result has been different in each one.

    First, I've thought that, when the mass is at rest after being connected to the spring and got down to the new equilibrium position:
    [itex]mg = k\Delta x \Longrightarrow \boxed{k = \frac{mg}{\Delta x}}[/itex]

    However, from the point of view of energies:
    [itex]mgh_0 = mgh + \frac{1}{2}k\Delta x^2[/itex] (I use [itex]\Delta x[/itex] because the initial position is 0)

    As [itex]h_0 - h = \Delta x[/itex], I've got:
    [itex]mg\Delta x = \frac{1}{2}k\Delta x^2 \Longrightarrow \boxed{k = \frac{2mg}{\Delta x}}[/itex]

    Obviously there is a wrong factor of 2.

    What's the problem here? Where is the mistake?

    Thanks!
     
  2. jcsd
  3. Jan 30, 2014 #2

    BvU

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    Energy case is for the situation you let go at h=0. When it is at delta x, it's still moving. Until it is a 2 delta x, when it hangs still again. But now the spring is pulling harder and up she goes!
    In other words, before there is rest, some energy has to be dissipated! You now know how much.
     
  4. Jan 30, 2014 #3
    Intuitively I understand what you say, but I don't see how to write it analytically.

    I mean, as we are in a conservative field, we only have to consider initial and final states. How should I have written the main equation ([itex]mgh_0 = mgh + \frac{1}{2}k\Delta x^2[/itex]) to include what you say?

    Thanks!
     
  5. Jan 30, 2014 #4

    BvU

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    That would be a little difficult, precisely because it describes an isolated case. It says there is energy left over, that has to go somewhere to get the "in rest" situation. E.g. in overcoming the air resistance which heats up the air a little.
     
  6. Jan 30, 2014 #5
    Okay. So the dissipated energy has a value of [itex]\frac{1}{2}k\Delta x^2[/itex], hasn't it?

    What I don't understand is why, if we are in a conservative field, we can't just consider initial and final states. Could you explain me the reason?

    Thank you.
     
  7. Jan 30, 2014 #6

    BvU

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    You can, actually. It's just that it isn't a steady state. mg (h-h0) is converted into spring energy plus kinetic energy. If there is no friction, the oscillation goes on forever.

    Is there a risk that we are misinterpreting the OP ? My idea was Δx is measured after the thing has come to rest.
     
  8. Jan 30, 2014 #7
    No misinterpretation. Now I see all you're trying to explain. For some reason I thought there wasn't friction. It's obvious that, if there isn't it, I can't apply energy equations "at rest" situation, simply because there isn't that situation. Am I right?

    So, when I study energies at the beginning ([itex]h = 0[/itex]) and at the end ([itex]h = \Delta x[/itex], when entire system is at rest) I should include that dissipated energy.

    If [itex]TE_0[/itex] and [itex]PE_0[/itex] are the total mechanical energy and the potential energy at the beginning respectively, [itex]TE[/itex] and [itex]PE[/itex] are the same energies but at the end, [itex]SP[/itex] is the spring energy at the end, and [itex]Q[/itex] is the dissipated energy (say heat), then:
    [itex]TE_0 = PE_0[/itex]
    [itex]TE = SP + PE[/itex]
    [itex]TE_0 = TE + Q[/itex]

    Do I agree?

    Thank you!
     
  9. Jan 30, 2014 #8

    BvU

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    Well, there is of course an "at rest" situation, it's just that it doesn't have the same energy as the situation where you attach the weight to the spring.

    Yes. in fact, you can probably do that yourself: Suppose you exercise a force to keep the weight in position h0, then gently lower the weight to position h. If you do it slow enough, the force is mg - kx where x runs from 0 to h and h = mg/k. The weight does work, namely ∫ F(x) dx where the integral goes from 0 to h. F(x) = mg - kx. The integral (you can do it) is [itex]\frac{1}{2}k h^2[/itex]. Looks familiar ?
     
  10. Jan 30, 2014 #9
    When I say that there isn't an "at rest" situation I'm talking about the frictionless case. If there isn't friction, the spring will experiment an non-stop simple harmonic motion, won't it?


    Thank you! I've solved that integral and I've got that result. Amazing!
     
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