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Spring constant of molecule

  1. Mar 14, 2008 #1
    1. The problem statement, all variables and given/known data
    A molecule of DNA lies along a straight line. It is 1.12 μm long. The ends of the molecule become singly ionized; negative on one end, positive on the other. The helical molecule acts as a spring and compresses .4% upon becoming charged.
    The Coulomb constant is 8.99 X 109 N*m2/C2.
    Determine the effective spring constant of the molecule. Take into account the compressed length when calculating the distance between the ends of the molecule. Answer in units N/m.


    2. Relevant equations
    Fe= ke * ( q1q2 / d^2 )



    3. The attempt at a solution
    d = 1.2*10-6 - (1.2*10-6 * .004) = 1.1952 μm
    Fe = ke * ( q1q2 / d2 ) = (8.99*109) * (( 1.6*10-19 )2) / (1.1952*10-6)2 = 2014 N
    Fs = kx
    x = d
    Fe = Fs ---> 2014 = (1.1952*10-6) k ---> k = 1.685*109 N/m
     
  2. jcsd
  3. Mar 16, 2008 #2

    Doc Al

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    Staff: Mentor

    Is the distance 1.2 or 1.12 μm?
    In Hooke's law (F = kx), x is the amount of compression from the uncompressed/unstretched position. So x does not equal d!
     
  4. Mar 16, 2008 #3
    distance is 1.12μm , so
    d=(1.12E-9) - (1.2E-9 * .004) =1.11552E-9

    & i totally butchered the rest. I'm not sure what its supposed to look like
    Fe=8.99E+9 * ????? / 1.11552E-9

    Then
    Fs=kx
    Fs*x =K
     
  5. Mar 16, 2008 #4

    Doc Al

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    Staff: Mentor

    Good.

    You had the right idea before:
    [tex]F_e = k_e q^2/d^2[/tex]

    F = kx
    so, k = F/x.

    What's x? (By what distance is the "spring" compressed?)
     
  6. Feb 25, 2009 #5
    This guy is right. x does not equal d. x equals (1.12 μm*.004)
     
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