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Spring system inside an accelerating box

  1. Dec 29, 2010 #1
    1. The problem statement, all variables and given/known data

    A mass m is resting at equilibrium suspended from a vertical spring of natural length L and spring constant K inside a box.

    The box begins accelerating upward with acceleration a. How much closer does the equilibrium position of the mass move to the bottom of the box?

    2. Relevant equations


    3. The attempt at a solution

    so this is my f=ma statement


    then i solved for x to get x= m(a+g) / k

    but this answer is wrong and the correct answer is

    ma / k
  2. jcsd
  3. Dec 29, 2010 #2


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    Staff: Mentor

    You are looking for the change in the mass' position from the initial equilibrium position, after the box has begun accelerating. What was its initial position?
  4. Dec 29, 2010 #3


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    Staff Emeritus
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    Gold Member

    In other words, what was the equilibrium position of the mass before the box was accelerated?

  5. Dec 29, 2010 #4
    hmm the problem doesnt specify that location.

    Cant we just assume it to be at location 0
  6. Dec 29, 2010 #5


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    Staff: Mentor

    Sure. You can assign your zero position to that location. Just keep in mind that it isn't the same location as the end of the relaxed, unloaded spring.

    Recapping what's happening:

    1. Box at rest. Spring unloaded, natural length L.
    2. Box at rest. Spring loaded with mass M, settles at equilibrium.
    3. Box accelerating. Spring stretches more, assumes new equilibrium.

    The question is (essentially) asking for the amount of stretching that takes place between items 2 and 3.
  7. Dec 29, 2010 #6

    I think my problem was that i assumed L to be the spring's length when it was loaded. But what you stated makes sense now and it leads to the right answer.

    excellent help. you gave me info but just enough to make me think and understand it

    thank you
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