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Spring Question

  1. Oct 23, 2005 #1
    If I held a block of mass m directly above a vertical spring with constant k and dropped it, I need to find out how far down it would go before rebounding to the equilibrium position. I don't have a specific problem; I'm just trying to learn the concept.

    I know that the magnitude of the force on the block due to the spring is |F|=kx. The force on the spring due to the block is |F|=mg. I am able to find the equilibrium position by equating those two, but how can I determine how far past the equilibrium position the block will go if it is dropped? I cannot find anything about this in my book or online. Could someone give me a hint on how to begin with the information I have above? I'd appreciate it.

    Thank you.
     
  2. jcsd
  3. Oct 23, 2005 #2

    Päällikkö

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    [tex]W = \int_A^B \mathbf{F} \cdot d\mathbf{s}[/tex]
    Does this sound gibberish to you?
     
  4. Oct 23, 2005 #3
    I know what it is -- line integral -- and how to use them (in most problems), but I cannot see how it's applied here. Are you suggesting something like the following?:

    [tex]\text{mgx}=\text{k}\int_0^xs\,ds[/tex]

    Then I could say:

    [tex]x=\frac{2\text{mg}}{\text{k}}[/tex]

    Is this reasoning / answer alright?

    Thanks again for your help.
     
    Last edited: Oct 23, 2005
  5. Oct 23, 2005 #4
    I suggest using the law of conservation of energy. but that will require the distance the block from the spring initially.
     
  6. Oct 23, 2005 #5
    I basically mean zero distance from the block to the spring -- as if someone was holding the block above the spring, barely making contact with it.
     
  7. Oct 23, 2005 #6
    Anyways, I partitioned the distance into x1 (the distance from the release point to the equilibrium point) and x2 (the distance from the equilibrium point to the point where the block begins to rebound back to the eq. point).

    [tex]{v_1}^2=2a\Delta x_1\implies {v_1}^2=\frac{2mg^2}{k}[/tex]

    For the above step, I solved for x1 and replaced it into the left equation.

    [tex]{v_1}^2=2a\Delta x_2\implies\frac{2mg^2}{k}=x_2\left(\frac{k\left(x_1+x_2\right)}{m}-g\right)[/tex]

    I can get x2 from this, but it's much longer than what I expected. Does anyone know a better way?
     
  8. Oct 23, 2005 #7
    Hmmm... I actually tried using something else.

    [tex]F=kx-mg[/tex]

    [tex]W=\int_{0}^{x}\left(ks-mg\right)\,ds=\frac{1}{2}kx^2-mgx[/tex]

    But since both the initial and final velocities are zero, it is clear that [itex]\Delta K=0[/itex].

    [tex]\frac{1}{2}kx^2-mgx=0\implies x=\frac{2mg}{k}[/tex]

    ...which is what I came up with earlier.

    Is this alright?

    Thank you very much.
     
  9. Oct 24, 2005 #8
    conservation of energy

    Initial energy of the system = 0 (taking the initial level of the block as the reference level to calculate the gravitational potential energy)
    Energy of the system when the block is at the lowest position =
    1/2*kx^2+mg(-x)
    Initial energy of the system = energy of the system at that position
    x=2mg/k
     
  10. Oct 25, 2005 #9

    VietDao29

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    Yup. This is correct.
    The work done by its weight is equal with the opposite of the work done by the tension of the string.
    This is also correct.
    This is wrong since the acceleration is not constant. It's not 'g'.
    Viet Dao,
     
    Last edited: Oct 25, 2005
  11. Oct 25, 2005 #10
    Yes, I understand that now. I don't know what I was thinking. Anyways, thank you all very much for your help.
     
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