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Spring and work

  1. Mar 16, 2006 #1

    I am messing something up here. I think it's my logic.

    Given k=2.5 * 10^3 N/m

    Find how much mass would have to be suspended from the vertical spring to move it 6.0 cm

    I get W=1/2kx^2=.5(2500)*(.06m)^2=4.5J
    I think that's right.

    Then I do
    W=F*d=m*a*d or m=W/a*d=4.5/(9.8m/s^2 / .06 m)=7.65 kg
    But that's not the answer.

  2. jcsd
  3. Mar 16, 2006 #2

    Doc Al

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    Staff: Mentor

    Look up Hooke's law and the meaning of k.

    Note: I assume the question is asking for the mass that would--if gently added to the spring, not dropped from a height--stretch the spring by the given amount.
    Last edited: Mar 16, 2006
  4. Mar 16, 2006 #3


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    Staff Emeritus
    Science Advisor
    Gold Member

    Why are you using work, it would be much easier to use one of Hooke's laws;
    [tex]F = - kx[/tex]

    Can you arrive at the correct answer using this?

    [Edit] It seems Doc Al got there before me, damn he's so quick! :mad: [Edit]
  5. Mar 16, 2006 #4
    Oh, so

    m*9.8=2500 * .06=150/9.8= 15 kg

    I got it. Man, I sure went abot that the completely wrong way.

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