# 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 = (EA

k = (E*πR

E is the Young's modulus

πR

L

Δx = F / k = 0.0418 / 315.18 = 0.00013 m = 1.3x10

F= - kΔx = (EA

_{0}/L_{0})Δx = Δxk = (E*πR

^{2}/ L_{0}) = (3.1x10^{6}* 3.1415* (0.0934x10^{-2})^{2}/ 2.69x10^{-2}) = 315.82 N/mE is the Young's modulus

πR

^{2}is the original cross-sectional area through which the force is applied;L

_{0}is the original length of the object.Δx = F / k = 0.0418 / 315.18 = 0.00013 m = 1.3x10

^{-4}mA = kx^{2}/2 = 315.18* (1.3x10^{-4})2/2 = 2.66 x 10^{-6 }J.Need a fast expert's response?

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