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Conceptual Problem with Springs and Energy

  1. Oct 30, 2006 #1
    Here is the problem, it is simple:

    An air glider is attached to a massless spring, which is compressed 0.18 m from it's relaxed position. Find the velocity of the glider at its original position, and at 0.25 m. Spring constant is 10 N/m, mass of glider is 0.15 kg.

    I have solved this problem, but I am having trouble understanding the potential energy issues for the distance of 0.25 m. The first is easy:

    kx^2/2 = mv^2/2

    The second is also not difficult:

    kx^2/2 = mv^2/2 + k(0.25-0.18)^2/2

    But I do not really "get" this second equation. I understand that the massless spring can't have any kinetic energy, since k = mv^2/2, and it is massless, then k-0 when the spring is at its original position, and I think that would be the highest speed.

    But as it continues past that point, it must build up a potential energy. But I don't know how to think about this. There must be some energy somewhere that is being converted into potential energy, but if there is truly no kinetic energy, where is the potential energy of the spring coming from? I can't believe it is coming from the glider, that makes no sense, but there is nothing else that is doing anything here.

    If someone could help me understand this, I would be so grateful.

  2. jcsd
  3. Oct 30, 2006 #2
    Just a thought: Is it that the intial potential energy gets converted into potential energy in the opposite direction? But doesn't it have to go through an intermediate stage first, Us -> Ks -> -Us ?

    Thanks again,
  4. Oct 30, 2006 #3

    Doc Al

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    Staff: Mentor

    The spring stores PE when compressed a distance X as well as when stretched a distance X. But spring PE is always >= 0 (the zero point is the unstretched position). (Energy doesn't have direction--it's a scalar.)

    Yes, PE gets converted first to KE, then back to PE. (But not negative PE!)
  5. Oct 30, 2006 #4


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    Staff: Mentor

    Is the spring attached so that the glider is not free past the uncompressed point of the spring? Normally a spring-launched glider would just fly off past that point, so the velocity would be the same as at the x=0 point (neglecting air resistance). But if the spring is fully attached to the (un-)glider, then it will start to rob KE from the glider and convert it back to PE. Just calculate how much PE is in the spring at 0.25m, and adjust the glider's KE accordingly.
  6. Oct 30, 2006 #5
    Thank you both. Yes, energy is a scalar. That must be my mantra for now.

    And thank you, Berkeman. That must be what is happening. Yes, the glider is attached to the spring, and that makes perfect sense.

    Thank you both so much.
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