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Homework Help: Vertical mass less spring problem

  1. Apr 29, 2010 #1
    A 1.00 kg block is compressed a distance of x = 0.130 m against a vertical massless spring with spring constant 280 N/m. The block is released from rest. How far above the spring's equilibrium point does the block travel?

    I used conservation of energy and got
    0.5kx^2 + 0.5mv^2 + mgh = 0.5kX^2 + 0.5mV^2 + mgH
    0.5(280)(0.13^2) + 0 + 0 = 0 + 0 + (1)(9.8)H
    2.366 = 9.8H
    H= 0.2414

    This is marked wrong so i thn subtracted the initial 0.13 and it was still marked wrong.
    Can anyone tell me why and how to do it? Please?
  2. jcsd
  3. Apr 29, 2010 #2


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    The 0.24 m that you found represents the distance from maximum spring compression to maximum height. As you suspected, you need to subtract from it the distance to the "spring's equilibrium point". Is that 0.13 m? I think not because it is the overall compression of the spring. When the 1 kg mass is placed on the spring and the system is at rest (equilibrium) that's the equilibrium point. So by what distance is the spring compressed when you place the 1 kg mass on it and you just let it sit there?
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