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Spring questions

  1. Jul 26, 2011 #1
    1. The problem statement, all variables and given/known data
    1. An unstretched spring with spring constant 20 N/cm is suspended from the ceiling. A 4.72-kg mass is attached to the spring and let fall. To the nearest tenth of a centimeter, how far does it stretch the spring?

    2. An unstretched spring with spring constant 5.9 N/cm is suspended from the ceiling. A 4-kg mass is attached to the spring and slowly let down. To the nearest tenth of a centimeter, how far does it stretch the spring?


    2. Relevant equations

    F=-kx
    F=ma

    3. The attempt at a solution
    1.
    ma=kx
    (4.72 kg)(9.81 m/s^2)=(2000 N/m)(x)
    x=2.3 cm
    Wrong. The answer is 4.6 cm.

    2.
    ma=kx
    (4 kg)(9.81 m/s^2)=(590 N/m)(x)
    x=6.7 cm
    Correct. Even though I used the same steps, this one is correct and the other is wrong. Am I missing a crucial difference between "let fall" and "slowly let down"?

    Thanks in advance!
     
  2. jcsd
  3. Jul 26, 2011 #2
    yes you are. All bodies in motion have kinetic energy equal to mv2/2
    Because one is let down (V=0), and the other is dropped (V=/=0)

    In other words, you assumed both were static equilibrium problems, when in reality only the second one is.

    Hope that helps :)
     
  4. Jul 26, 2011 #3
    Thank you, that makes sense.

    How does this fit into the equations though, since I am using forces and this relates to energy?
     
  5. Jul 26, 2011 #4
    Simply remember the definition of Work, W=F*d or W=integrate(F,d,d1,d2)
    If you're in calc based physics, you should be able to derive the equation, if not, then find the equation for PE of a spring on your equation sheet.
     
  6. Jul 26, 2011 #5
    Oh, I see.

    Just for future reference, this is the solution:

    Potential energy due to gravity is mgh; h in this case is x.

    mgx=1/2kx^2
    mg=1/2kx
    (4.72 kg)(9.81 m/s^2)=1/2(2000 N/m)(x)
    x=4.6 cm

    Thanks again for the help!
     
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