Help with Elasticity Questions | girlInterrupt

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SUMMARY

This discussion focuses on the elasticity of DNA, specifically calculating the change in length and elastic energy when a force is applied. The user, girlInterrupt, successfully calculated the change in length (5 nm) and the elastic energy (4.5 x 10^-25 J) for a DNA strand of 580 nm length with a spring constant of 3.6 x 10^-17 N/nm. For part c, the user determined that folding the DNA strand in half increases its stiffness, leading to a reduced stretch of 1.25 x 10^-9 m when the same force is applied. The discussion emphasizes the relationship between force, spring constant, and the geometry of the DNA strand.

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  • Understanding of elastic properties and Hooke's Law
  • Familiarity with the concepts of spring constant and elastic energy
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  • Ability to perform unit conversions (e.g., nm to meters)
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girlinterrupt
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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.

Homework Statement



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

Homework Equations



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

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
 
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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
 
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 therefore its stretch is 4 times less. Good work!
 

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