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