Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Vertical spring problem

  1. Jul 18, 2010 #1
    1. The problem statement, all variables and given/known data

    A spring with a spring constant of 450 N/m hangs vertically. You attach a 2.2 kg block to it and allow the mass to fall. What is the maximum distance the block will fall before it begins moving upward?

    2. Relevant equations

    I'm not exactly sure how the conservation of energy equation would look for this. So, when the mass pulls the spring down to its maximum displacement, there is only elastic potential energy in the isolated spring system. Now I'm not sure what other point to use this information in reference to.

    If it's the equilibrium position that I'm also supposed to look at, then I'm not sure what forms of energy are present at that point.

    3. The attempt at a solution
  2. jcsd
  3. Jul 18, 2010 #2
    Wait nevermind ... I googled it. The equilibrium position no longer equals x = 0 for a vertical spring, and the formula becomes h = mg/k.
    Whoops lol.
  4. Jul 18, 2010 #3
    Does the problem make sense though?
  5. Jul 18, 2010 #4
    Erm, from what I understand... A mass of 2.2kg is attached to a spring while it is at the equilibrium position. Gravity makes the spring stretch downwards until it reaches a maximum point where there is only elastic potential energy, and then it starts to move back up again.

    I'm still a bit confused about what energy is acting where in terms of the "isolated system".

    And this formula, h = mg/k, actually isn't giving me the correct answer. ...So I guess it doesn't make sense lol. :/
  6. Jul 18, 2010 #5
    Okay I think I got it!

    At the equilibrium position, only gravitational potential energy is present since the mass-spring system has not moved yet at that instant and since the spring is vertical.

    The mass pulls the spring down to its maximum displacement, and at that point only elastic potential energy is present.


    Eg = Ee
    m * g * h = 0.5 * k * x^2

    At the equilibrium position, the height of the mass should be equal to the maximum displacement after the mass falls. So, x = h:

    m * g * x = 0.5 * k * x^2
    0 = 0.5*450*x^2 - 2.2*9.8*x
    x= 0 and 0.095m

    Woot right answer :). Is my reasoning all correct?
  7. Jul 19, 2010 #6
    Maybe this will help with the concepts.


    Woot? Am I helping an Aggie?

    \m/ thats a hookem horns sign... :)

    Your reasoning using energy looks correct without doing the math myself.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook