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Answer to Question #1026 in Mechanics | Relativity for Erin

Question #1026
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 0.0418 N. The length and radius of the collagen are, respectively, 2.69 cm and 0.0934 cm, and Young's modulus is 3.1 x 106 N/m2. <br>(a) If the stretching obeys Hooke's law, what is the spring constant k for collagen? <br>(b) How much work is done by the variable force that stretches the collagen?
Expert's answer
According Hooke's law:
F= - kΔx = (EA0/L0)Δx = Δx
k = (E*πR2 / L0) = (3.1x106* 3.1415* (0.0934x10-2)2 / 2.69x10-2) = 315.82 N/m
E is the Young's modulus
πR2 is the original cross-sectional area through which the force is applied;
L0 is the original length of the object.
Δx = F / k = 0.0418 / 315.18 = 0.00013 m = 1.3x10-4 mA = kx2/2 = 315.18* (1.3x10-4 )2/2 = 2.66 x 10-6 J.

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