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Hookes law

  1. Nov 23, 2009 #1
    1. The problem statement, all variables and given/known data
    A mass of 2.3 kg is dropped from a height of 4.82 meters above a vertical spring anchored at its lower end to the floor. If the spring constant is 20 N/cm, how far, to the nearest tenth of a cm, is the spring compressed?
    answer is 34.1

    2. Relevant equations
    PE = mgh
    KE = (1/2)m(v^2)
    Hookes Law= (1/2)K(x^2)



    3. The attempt at a solution
    mgh = (1/2)K(x^2)
     
    Last edited: Nov 23, 2009
  2. jcsd
  3. Nov 23, 2009 #2

    Pengwuino

    User Avatar
    Gold Member

    There you go, that is how you solve the equation. The actual solution, however, is wrong. Did you keep your units?
     
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