A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases f

Question

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to 3.06 10-2 N. The length and radius of the collagen are, respectively, 2.7 and 0.093 cm, and Young’s modulus is 3.10 106 N/m2.
(a) If the stretching obeys Hooke’s law, what is the spring constant k for collagen?
(b) How much work is done by the variable force that stretches the collagen?

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Nguyệt Ánh 6 months 2021-08-31T05:17:39+00:00 1 Answers 0 views 0

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  1. Cherry
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    2021-08-31T05:18:46+00:00
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    Answer:

    Part(a): The value of the spring constant is 3.11 \times 10^{2}~Kg~s^{-2}.

    Part(b): The work done by the variable force that stretches the collagen is 1.5 \times 10^{-6}~J.

    Explanation:

    Part(a):

    If ‘k‘ be the force constant and if due the application of a force ‘F‘ on the collagen ‘\Delta l‘ be it’s increase in length, then from Hook’s law

    F = k~\Delta l....................................................................(I)

    Also, Young’s modulus of a material is given by

    Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)

    where ‘A‘ is the area of the material and ‘l‘ is the length.

    Comparing equation (I) and (II) we can write

    && Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}

    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 (III), we have

    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}

    Part(b):

    We know the amount of work done (W) on the collagen is stored as a potential energy (U) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as

    W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)

    Substituting all the values, we can write

    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

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