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Homework Help: Work and Potential Elastic Energy

  1. Nov 13, 2007 #1
    1. The problem statement, all variables and given/known data
    The following is the data I collected. I set a weight on a spring and measured how far the spring was stretched. The spring was not oscillating, just stretched and left at rest.
    Load (kg):

    Weight (Converted the above load in to N):

    Stretch (m):

    The question is, how do I find the work from the above and how do I find the elastic potential energy.

    2. Relevant equations

    3. The attempt at a solution
    [tex]Work = Fd[/tex]
    [tex]W = (2.94)(0.059) = 0.17346[/tex]

    [tex]F = kx[/tex]
    [tex]k = F/x[/tex]
    [tex]2.94 / 0.059 = 49.8[/tex]
    [tex]U_{elastic} = \frac{1}{2}kx^2[/tex]
    [tex]U_{elastic} = \frac{1}{2}(49.8)(0.059^2) = 0.08673[/tex]

    I'm pretty sure I did something wrong. It also asks me to compare the work done and the elastic potential energy. I thought W = Ue in this case, but the calculations show differently. But I think the calculations are wrong. Any help would be greatly appreciated.
  2. jcsd
  3. Nov 13, 2007 #2
    let's imagine a spring, when the load, F newtons has reached a certain extension, x, such that the force exerted by the spring is equal to F, what would the energy changes be like?

    The work done by the gravity would be Fx. The elastic potential would be 1/2 Fx^2. The other half of the energy lies in the kinetic energy of the load. This position is the equilibrium position of oscillation. It would also be the position whereby the load will remain stationary when damping has taken place.

    At this juncture, we can conclude that the remaining half of the energy has been dissipated into tis surroundings, the kinetic energy into forms of energy like sound.
  4. Nov 13, 2007 #3

    Shooting Star

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    Homework Helper

    Hi odie5533,

    When you gently lower a mass from, say, the height of your head to the ground, where is the work done by gravity going? It is dissipated in your arms. Here, too, the same thing is happening. If you had attached a load to the spring and just let it go, and then measured the extreme stretch while it was oscillating (without friction), then work done would have been equal to elastic PE.
  5. Nov 13, 2007 #4
    Oerg: So the work done by gravity equals the elastic potential? I don't see where the KE takes place, since the mass is not oscillating, v = 0, K = 0.5mv^2 = 0.

    Shooting star:
    [tex]W = U_{elastic} = \frac{1}{2}kx^2[/tex]
    [tex]W = Fd = 2.94 * 0.059 = 0.17346[/tex]
    [tex]0.17346 = \frac{1}{2}kx^2[/tex]
    [tex]k = 99.66[/tex]
    This any better?
  6. Nov 13, 2007 #5
    well, the above example can be described as an oscillating mass whereby damping has occured. The kinetic energy is lost and a stationary load hung from a spring is achieved.

    Or as shooting star has described, it can also be the case whereby the mass is lowered slowly via your hand, in which case, the energy is dissipated through your hand.

    Your initial post was correct, the discrepency between the 2 energy values is the loss of kinetic energy to its surroundings.

    your above post equated the work done with the elastic potential energy which is wrong.
  7. Nov 13, 2007 #6

    Shooting Star

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    If the mass had simply been let drop after being attached to the spring, it would have overshot the current position with max KE and then continued downward for some distance, slowing down all the while, until when again it would reverse velocity and move up to its original position from where it was let go. Without friction, this would be undamped oscillation. The energy would shift between PE due to gravity, elastic PE and KE, the total remaining constant.

    In the 1st eqn, W= Uelastic is not correct. (I will comment on the 2nd later. Time is short now. Pl don’t mind.)

    Please understand that you have taken energy away from the system by not allowing the load to fall naturally after you attached it to the spring. With damping, which is there in the real world, your position of the load now would be where the load ultimately comes to equilibrium after damped oscillation.

    Now, I'm sure you can refine your treatment of the problem.
  8. Nov 13, 2007 #7

    Shooting Star

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    For finding k, use F/extension from your table, as you did in your first post.
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