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Homework Help: Determine the effective spring constant of the molecule

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data
    A molecule of DNA is 2.17 micrometers long. The ends of the molecule becom singly ionized: negative on one end, positive on the other. The helical molecule acts like a spring and compresses 1.00% upon becoming charged. Determine the effective spring constant of the molecule.


    2. Relevant equations
    I need a spring constant equation... Is this what I use?
    F=kx

    3. The attempt at a solution
    The distance is going to be .01(2.17), but I'm not sure where to go from there. Could you please point me in the right direction? Thanks!
     
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  3. Jan 14, 2010 #2

    ideasrule

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    Yes, F=kx is what you use.

    What's the force compressing the molecule? Remember that the molecule is singly ionized.
     
  4. Jan 14, 2010 #3

    Matterwave

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    Hooke's law gives F=-kx (note the negative).

    Since you already know x in this situation, all you need is F. Where do you think the force of compression comes from, and what is that force?
     
  5. Jan 14, 2010 #4
    WEll... Could I use Coulomb's Law?
    I know it is compressing because the two charges would be attracted to eachother and want to come closer... but I'm not really sure how to get the force from that.
     
  6. Jan 14, 2010 #5

    Matterwave

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    Use Coulomb's law. F=-kQ1Q2/r^2 (note, this k is not the spring constant k, but a constant for Coulomb's law)
     
  7. Jan 14, 2010 #6
    but it doesn't say how much my charges are ionized by... should I just say they are + and - 1 ?
     
  8. Jan 14, 2010 #7
    And is r going to be 2.17 micrometers?
     
  9. Jan 14, 2010 #8

    Matterwave

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    It does say that they are "singly ionized". Which means your charges are +1e and -1e. Your r is actually going to be .99(2.17) micrometers, since that's the final position. The force of the "spring" is the force applied in the final position, so the force of the charges should be the force applied in the final position as well, although I don't think this will affect your calculations much.
     
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