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Spring New mass-Please Help

  1. Nov 3, 2006 #1
    A spring with K=450N/m hangs vertically. You attach 202kg block to it and allow the mass to fall. What is the maximum distance the block will fall before it begins moving upward?


    =21.56 N

    F=+/- kx

    Is this the max distance I calcul;ated or the new equilibrium distance? Can u give me a hint.
  2. jcsd
  3. Nov 3, 2006 #2


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    Staff Emeritus
    Science Advisor

    The equilibrium position will be where the force equals the weight:
    450x= 2.2(9.8), measured below the initial position. Since the mass gains momentum as it falls, it will have its maximum velocity when it passes there, a distance x below it's initial position. It will continue to fall until the spring overcomes that momentum. Because of the symmetry of the situation, that distance below equilibrium will also be x. The "maximum distance the block will fall before it begins moving upward" will be 2x where x satisfies 450x= 2.2(9.8).
  4. Apr 10, 2011 #3
    HallsofIvy, I don't understand what you are saying.
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