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Hooke's law for Springs

  1. Feb 23, 2008 #1
    A spring with spring constant k=340 N/m is used to weigh a 6.7 KG fish. How does the spring stretch?

    I used Hooke's law (F=-Kx), but ended up having a negtive distance x=-0.2m. Is this expected? in the problem they say the spring stretches...this is confusing me...thank you for your help guys in solving this problem
     
  2. jcsd
  3. Feb 23, 2008 #2
    I think it depends on how you define your coordinate system. Say downward is the negative direction and the spring at equilibrium is at x=0. Then when you add the weight the spring stretches downward and x is at some negative value, so [tex]\Delta x[/tex] is negative. Force on the weight is opposite the displacement in the positive direction.

    On the other hand, if you choose up to be the negative direction, then your change in x is positive and the force on the weight is in the negative direction.

    Either way, the spring stretches the same distance, the negative sign would just tell you which direction the end of the spring moves (and whether negative is up or down depends on your coordinate system)
     
  4. Feb 23, 2008 #3
    fs =-kx

    During compression delta x is negative.

    you are correct.negative sign just indicates the direction of motion of spring.
     
  5. Feb 23, 2008 #4
    Thanks guys!!!!! this was so helpful
     
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