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Elasticity Questions

  1. Sep 11, 2008 #1
    Hi, I need some help regarding an elasticity question I am attempting.

    I have done the first two questions and need someone to check over what I've done.

    I also need some direction on how to attempt part c, as I'm feeling a bit clueless on where to begin tackling this.

    1. The problem statement, all variables and given/known data

    Q. The elastic properties of DNA mean that it can be stretched, bent and twisted into different shapes which affect how DNA binds to other molecules. If we have a strand of DNA 580nm long with an spring constant of 3.6×10−17N/nm and apply a stretching force of 1.8 × 10−16N then calculate the following:

    a) The change in length of the DNA strand.

    b) The amount of elastic energy that is contained in the stretched strand of DNA.

    c) The change in length of the DNA strand if it were folded in half when the force was applied (i.e. an equivalent piece of DNA with half the length but twice the cross-sectional area of the first strand).

    What is known:
    F = 1.8 X 10^-16
    k = 3.6 X 10-17 N/nm = to meters: 3.6 x 10-8 N/m
    L = 580nm = 5 x 10^-7m

    2. Relevant equations

    a) Change in L = F/K
    b) U = kx^2 / 2
    c) Not sure!

    3. The attempt at a solution

    a)

    Change in L = F/K
    = (1.8 x 10^-16) / (3.6 x 10 -8)
    = 5 x 10-9
    which is equivalent to: 5nm

    b)

    U = kx^2 / 2
    = 3.6 x 10 -8 x (5 x 10-9)^2 / 2
    = 4.5 x 10-25

    c)

    I'm not sure how to work this out as the question has no reference to the area of the rod.. do I need to work out the area first.. any help is appreciated!


    Thanks!
    girlInterrupt
     
  2. jcsd
  3. Sep 11, 2008 #2
    I've just had a go at part c, see if this makes sense...

    Note: for simplicity sake, k1, L1 and A1 refer to the original values in the equations and k2, L2 and A2 refer to part c where the length of DNA is folded in half.

    My Solution:

    k1 = AY/L

    Therefore:
    k2 = A2Y / L2

    = 2 x A1 Y / (1/2) L1

    = 4 X A1 Y / L1

    = 4k1

    = 4 x 3.6 x 10^-8
    = 1.44 x 10^-7

    So a change in length with the original force will equal:

    = F / k
    = (1.8 x 10^-16) / (1.44 x 10^-7)
    = 1.25 x 10 ^-9

    Does that look right?

    Thanks,
    girlInterrupt
     
  4. Sep 11, 2008 #3

    PhanthomJay

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    Homework Helper
    Gold Member

    I didn't check your numbers, but your solution is correct. The only correction is that K = AE/L, where E is the elasticity modulus of the material in units of force per length squared, which gives you the proper units of K in units of force/length. But since E is common to both cases, you don't need to know it. The folded strand is 4 times stiffer than the unfolded strand, and therfore its stretch is 4 times less. Good work!
     
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