Respuesta :
Answer:
Part(a): The value of the spring constant is [tex]3.11 \times 10^{2}~Kg~s^{-2}[/tex].
Part(b): The work done by the variable force that stretches the collagen is [tex]1.5 \times 10^{-6}~J[/tex].
Explanation:
Part(a):
If '[tex]k[/tex]' be the force constant and if due the application of a force '[tex]F[/tex]' on the collagen '[tex]\Delta l[/tex]' be it's increase in length, then from Hook's law
[tex]F = k~\Delta l....................................................................(I)[/tex]
Also, Young's modulus of a material is given by
[tex]Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)[/tex]
where '[tex]A[/tex]' is the area of the material and '[tex]l[/tex]' is the length.
Comparing equation ([tex]I[/tex]) and ([tex]II[/tex]) we can write
[tex]&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}[/tex]
Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.
Substituting the given values in equation ([tex]III[/tex]), we have
[tex]k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}[/tex]
Part(b):
We know the amount of work done ([tex]W[/tex]) on the collagen is stored as a potential energy ([tex]U[/tex]) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as
[tex]W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)[/tex]
Substituting all the values, we can write
[tex]W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J[/tex]
