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Inconsistent calculations with Hooke's Law/Elastic Potential Energy

  1. Dec 10, 2006 #1
    1. The problem statement, all variables and given/known data

    My question is in regards to a relatively simple problem:

    A massless, ideal spring with a spring constant k hangs from the ceiling; it is originally neither compressed nor stretched. We attach a block of mass m to the spring. In terms of the mass of the block, the spring constant k, and the acceleration due to gravity, g, how far does the spring stretch from it's equilibrium position?

    While this question is not terribly complicated, I am confused because I feel that I should be able to do it two ways and get the same answer: calculations using 1) force (Hooke's Law) and 2) Gravitational Potential Energy/Elastic Potential Energy should presumably give the same result for the displacement from equilibrium (x) of the spring. But they don't (at least when I do it). I will demonstrate what I have done below.

    2. Relevant equations

    In order to do this Problem using net force and Hooke's Law we only need two equations:

    F(net) = ma
    F(spring) = -kx

    In order to do this Problem using conservation of energy we only need three:

    Total Energy(initial) = Total Energy (final)
    U(spring) = (1/2)kx^2
    U(gravity) = mgh

    3. The attempt at a solution

    Via net force equations:

    F(net) = ma = 0 because the block is suspended from the spring. This means that

    F(spring) - F(gravity) = 0

    so F(spring) = F(gravity)

    so kx = mg
    so x = mg/k

    Via Energy equations:

    Energy(initial) = mgh(initial) = Energy(final) = U(spring) + mgh(final)

    Rearranging we get:

    mg(h(initial)-h(final)) = (1/2)kx^2

    h(initial)-h(final) = x (the displacement) so:

    mgx = (1/2)kx^2

    Solving this we get:

    x= 2mg/k

    This is clearly different than the original answer.

    So my first question is: Why are these two calculations different? It seems like they should give me the same value for x.

    And my second question is: how can I add an addtional term into my energy equation to get the same answer as the force method? This should be fixable.

    I know this is a long post, but any clarification on this topic would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Dec 11, 2006 #2

    OlderDan

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    Your force calculation is telling you the equilibrium position of the mass hanging on the spring. Your energy calculation is telling you the total distance the mass falls from where the spring is relaxed to its lowest position. Your energy calculation should be 2 times your force calculation. See if you can think why this is true. No additional calculations are needed. What will happen to the mass if it is released from rest at the point where it is attached to the relaxed spring?
     
  4. Dec 16, 2006 #3
    With all due respect Dan, I don't find your input very helpful (although now having figured out the answer on my own I can see what you were driving at), but I did want to post to clarify things. In particular, after I posted this question I thought about it (as I fell asleep that night) and realized what my problem was: the two sets of calculations actually describe two different situations. The force calculations describe a situation in which the mass is SLOWLY lowered to an equilibrium position and the energy calculations describe a situation in which the mass is just dropped. In the former case I have added in an extra non-conservative force (i.e. my hand to keep the mass from simply dropping). This means that to bring the energy calculations in line I have to add an extra term for the WORK done by my hand. I was stuck for a little while on why the amount of work done by my hand would be (1/2)mgx (which it has to be in order to work out mathematically), but then I realize that at equilibrium my hand must supply a full mg of force (to balance gravity) while at the bottom the force will be zero (all of the job of keep the mass in equilibrium is taken over by the spring). Because the relationship between force and distance is linear (F=-kx), the average force over the distance (x) will be (1/2) mg. This means that the overall work done will be (1/2)mgx. If we insert that into the equation I gave above everything works out swimmingly and the energy and force calculations give the same answer. =)
     
  5. Dec 16, 2006 #4

    OlderDan

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    With all due respect freethinker, you have totally changed the problem from what you originally posted.

    The answer to your first question is that one of your original calculations was a force calculation and the second was an energy calculation in which the values that you represented by the single variable x in both equations in fact correspond to two different points in the motion. The results were not contradictory, except in your attempt to equate two x values that represent different positions.

    Changing your first calculation by introducing an additional force is changing the problem, and introduces an unnecessary assumption in the calculation of the equilibrium position. The implication of your statement "The force calculations describe a situation in which the mass is SLOWLY lowered to an equilibrium position", that the mass must be slowly lowered to make the force calculation valid is NOT true. The force provided by the spring is a function only of the elongation of the spring and it is completely independent of any other forces that are acting. The equilibrium position of the mass hanging on the spring is the same no matter how the mass arrives at that position. Why do you think it has to be at rest when it gets there? In the original problem statement, the mass would be moving with maximum speed when it gets to equilibrium, which is the point exactly in the middle of the range of motion of a harmonic oscillator, and that is why your "first x" is exactly half of your "second x".

    You can't and you have not. You resolved your contradiction by turning one problem into two different problems that have two different answers, and solved the problem of finding the equilibrium position using a valid, but needlessly complex energy calculation. At equilibrium, mg = kx, and that is all you need to find the equilibrium position.
     
    Last edited: Dec 16, 2006
  6. Jun 1, 2009 #5
    older dan, thanks a heap, those two lines has helped me a heap and my class a week before the mid year exams.
     
  7. Jan 17, 2012 #6
    I thought about this for a while. I realize now what is going on. I know this is an ancient post, but I read alot of the old ones, so hopefully this will clarify things for some readers.

    At equilibrium (i.e. when the block is at rest and gravity is balanced by the spring) the stretch of the spring is x.

    To properly do an energy analysis, one must consider dropping the mass from the relaxed point of the spring and watch mother nature do her bit. Naturally, the mass will fall through the equilibrium position, and on down to double the x from the static position mentioned earlier. The total displacement in the energy analysis is therefore 2x.

    If one equates gravitational potential with elastic potential, BOTH sides must receive this new displacement of 2x and the equation resolves to the same equation as in the static case. I believe freethinker only put 1x on both sides, which causes the factor of two confusion.

    mg(2x)=(0.5)k(2x)^2
    2mgx=(0.5)k(4)x^2
    mgx=kx^2
    mg=kx

    Same as in the equilibrium analysis... no need for the magic hand to slow the decent.
     
  8. Feb 13, 2012 #7
    Thanks jtebb. I just wanted to let you know that at least you helped someone. I was doing my homework and I was thinking, why is the potential energy in a spring half that of its gravitational potential? You explained it great. Thanks!
     
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