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Springs / Work

  1. Sep 17, 2012 #1
    What am I missing here?...

    If you place a mass on a spring, the change in potential energy of the mass should equal the stored potential energy in the spring, correct?

    so PE = U

    PE = mgh = Fx (F = force = mg, h = change in spring length x)

    U = 1/2 k*x^2

    Hooke's law: F=-kx, so k=F/x

    U= 1/2 (F/x) * x^2 = 1/2 Fx

    setting PE = U
    Fx = 1/2 Fx

    Clearly I'm doing something wrong...

    (The original problem is the following...)
    5*10^5 Newton force supported by two springs (k_1 = 5.25 E^5, k_2 = 3.6 E^5 N/m)
    The second spring is shorter by 0.5 meters (spring 1 compresses 0.5 meters before the weight hits spring 2). Find the spring compression and find the work done by the springs.

    So I said F = k_1 x_1 + k_2 (x_1 -0.5) and got x_1 = 0.768 meters, and x_2 = 0.268 meters

    So the restoring force F = 5.25*10^5*0.768+3.6*10^5*0.268 = 5*10^5 Newtons (cool, this checks out)

    But for the work, change in PE should equal U

    But the change in PE of the mass (5*10^5 * 0.768 meters) is not equal to the sum of spring potential energies (1/2 k_1 (x_1)^2 + 1/2 k_2 (x_2)^2 )

    Thoughts? Thanks for taking a look!
     
  2. jcsd
  3. Sep 18, 2012 #2
    First, it will be a good idea to write the complete text of the problem.
    The words are important too.

    One of the errors is considering the elastic force a constant, which isn't.
    The "F" in F=mg is not the same force as "F" in F=kx (which depends on x).
     
  4. Sep 18, 2012 #3
    To be at equilibrium, the restoring force (F = kx) should be equal to the gravitational force from the object (F = mg), right?

    And the original problem was stated..

    A 5 * 10^5 Newton block lies on top of a spring with spring constant k_1 = 5.25 * 10^5 N/m.
    Once the first spring compresses 0.5 meters, it hits another spring with k_2 = 3.6 * 10^5 N/m.

    Question 1: How much do the springs compress? (spring 1 compresses 0.768 m and the other spring 0.268 m)

    Question 2: What is the work done by the springs? **This is what I'm having issues with because I think U should = PE, but it doesn't)
     
  5. Sep 18, 2012 #4
    Hello evanave,

    I think this should have been posted in the homework section, so don't be surprised if it gets moved.

    As to the theory behind your question, which is a good one, consider what actually happens.

    Initially the spring is not stretched.

    You hang a mass on it and support the mass in place so it does not move and has zero kinetic energy.

    When you let go the force of gravity acting on the mass pulls it down, stretching the spring.
    During this process the mass looses gravitational energy but acquires kinetic energy and the spring acquires elastic potential energy.

    As a result of the elastic potential energy an increasing restoring force on the mass is developed in the spring, which opposes the force mg due to gravity.

    At the time when the restoring force in the spring is exactly equal to the force due to gravity the mass is still travelling downwards.

    Thus the mass also possesses kinetic energy at this displacement.

    This kinetic energy is transferred to the spring which continues to stretch and the mass slows to rest.

    The restoring force in the spring now exceeds that of gravity so pulls the mass back up.

    So the mass oscillates up and down with simple harmonic motion about the mean position where the restoring force in the spring just balances the gravitational force on the mass.

    Whilst executing this SHM at any one position the gravitational energy lost by the mass will equal the sum of the potential energy in the spring plus the kinetic energy of the mass.

    If you would like to work on this we can take it further.

    Don't forget that the loss of gravitational potential depends upon the height the mass at any instant.
     
    Last edited: Sep 18, 2012
  6. Sep 18, 2012 #5

    sophiecentaur

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    The sums should work out perfectly correctly because GPE equals work done. Is your problem that you are confusing weight force with spring force when extended to the maximum or is it just the numbers that seem to be coming out wrong? The integral of spring force times distance is exactly equal to mgh.
    With the two springs in your example, there would be a difference if the first spring to operate actually reaches a limit (stop) when the second spring kicks in or whether the first spring continues to compress. Do you assume the first spring is operating all the time?
    Are those numbers for compression of the springs your own calculation? Are you calculating the compression for both springs together correctly (the k to use is the sum of the two values for k). I think I'd first calculate force when the second spring kicks in first, then the 'remaining' force would be taken up by both springs compressing, with their combined k. Sorry if I'm stating the obvious but it may help.
     
    Last edited: Sep 18, 2012
  7. Sep 18, 2012 #6

    Doc Al

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    No, not in general. Total mechanical energy will be conserved, but that includes the kinetic energy of the mass.
     
  8. Sep 18, 2012 #7

    sophiecentaur

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    I was assuming that this is a 'statics' problem (a spring balance situation). Could this be resolved, please?
     
  9. Sep 18, 2012 #8
    From the data and the results to part A it looks like they ask for the spring compressions at equilibrium.
    It would make sense to calculate the work done by the springs during their compression from the initial position to the equilibrium position.
    You can simply calculate the work done by each elastic force between x=0 and x=x_equilibrium.
     
  10. Sep 18, 2012 #9

    Doc Al

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    Sounds like a statics problem to me.

    Presumably, the mass is gently lowered onto the springs until they reach their new equilibrium point. The work done by gravity as the mass is lowered is greater than the spring potential energy. (After all, the spring force is always less than the weight of mass until equilibrium is reached.) The "missing" energy is accounted for by whatever did that gentle lowering of the mass.

    If the mass were just dropped, instead of gently being lowered, then the mass would have kinetic energy at the equilibrium point and would continue to oscillate.
     
  11. Sep 18, 2012 #10
    Studiot, I like your explanation (and Doc as well).

    Because this is presented as a statics problem the mass is gently lowered, and some external magical force is slowly lowering it down (removing the energy (kinetic, perhaps)) so that the work done by the springs is less than the change in potential energy.

    U = PE - magically lost energy by slowly lowering the mass onto the spring

    Thanks for the input everyone, I really appreciate it!
     
  12. Sep 18, 2012 #11

    sophiecentaur

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    It suddenly struck me that this problem is analogous to the question of the 'lost energy' when a charged capacitor is connected across an uncharged capacitor. I should have spotted that earlier!
     
  13. Sep 18, 2012 #12
    If the problem is exactly the way you give it here, you don't need to worry about energy conservation (or non-). They just ask to calculate the work done by the springs. You can do this by directly applying the formula for work done by elastic force. You know the spring constant and displacement (from part A) for each spring.

    On the other had, trying to go on a round-about way, you learn some extra stuff.
     
  14. Sep 18, 2012 #13
    Originally only looked at your own work, not at the problem, until it got transferred to homework.

    So I now see that it is about springs in compression not extension. Sorry I should have spotted this earlier.

    There is similar problem in Sears and Zemansky where a weight is dropped onto a spring.

    They ask you to calculate the maximum compression.
     
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