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Compression of a spring. Hooke's Law vs. Conservation of Energy

  1. Mar 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Lets say I have a spring with a stiffness of k = 250 N/m originally unstretched. I then gently place a 5 kg block on top of the spring. How much does the spring compress?


    2. Relevant equations

    W = mg
    F = -ks
    mgΔh = 0.5k(Δh)2

    3. The attempt at a solution

    Using Hooke's Law I get the following result

    W = 5*9.81 = 49.05 N

    W = F = -kx

    49.05 = 250x meaning x = 0.1962 m

    Using conservation of energy I get this though:

    mgx = 0.5kx2

    0 = 125x2 - 49.05x

    x = 0.3924 m

    So which one is the correct answer? Why is one double the other?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 4, 2014 #2
    The first answer is correct. If you place it gently on the spring, then you need to apply force (and do work) to keep it from accelerating while the spring is compressing. If you don't do this, then the mass will have kinetic energy as it passes through the equilibrium point, and it won't stop until it has traveled twice as far. But this will not be an equilibrium point, and the mass will go back up in the other direction. If you apply the force to keep it from accelerating, then it will only go down to the equilibrium point x = mg/k. The problem with your energy balance was that you omitted the kinetic energy developed by the mass.

    This same question has appeared on Physics Forums many times in the past.

    Chet
     
  4. Mar 4, 2014 #3
    All right, here's my confusion though. If the block is originally at rest then it has no kinetic energy only gravitational potential. Once it starts descending (compressing) it has some kinetic but when it reaches the lowest it can get, for that moment all of its energy is in the form of spring potential. So why wouldn't the initial gravitational potential energy be equal to the spring potential energy?
     
  5. Mar 4, 2014 #4
    The "lowest you can get" is not the same as the position reached when you "gently place it".
    It is actually twice as much.
    You are calculating two different things so you get two different answers.
    When you place it gently the energy of the weight+spring system is not conserved.You have some external force doing work on the system, as was already explained by Chet.
     
  6. Mar 4, 2014 #5
    Ok, I think I'm getting it now. This would be my last counterargument I guess.

    So if the block was released from rest 1 m above the spring then with conservation of energy:

    mg(h+x) = .5kx^2
    49.05 + 49.05x = 125x^2
    x = .8526 m

    I'm assuming this is correct since pretty much the same scenario is in an example in my book.

    So again, if i was to release it from rest when h = 0, then I would go back to mgx = 0.5kx^2. How has anything changed?

    Thank you guys for the help!
     
  7. Mar 4, 2014 #6
    It will be equal to the spring potential energy. But this will not be the equilibrium point. When you get to this point, the spring will have been compressed to twice the distance to the equilibrium point, and the upward force of the spring will be 2mg instead of mg. So the mass will accelerate upward again. The mass will forever oscillate up and down in simple harmonic motion from zero displacement to twice the equilibrium displacement. Whenever it passes through the equilibrium point, its velocity will not be zero; it will pass through a maximum.

    Chet
     
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