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January 26, 2015

January 26, 2015

Posted by **Brianna** on Wednesday, July 14, 2010 at 10:30am.

A cylindrically shaped piece of collagen is being stretched by a force that increases from 0 to 3.0 x 10^-2 N. The length and radius of the collagen are respectively 2.5 and 0.091 cm and Young’s molulus is 3.1 x 10^6 N/m^2.

(A) If the stretching obeys Hooke’s Law, what is the spring constant for collagen?

(B) How much work is done by the variable force that stretches the collagen?

- Physics -
**bobpursley**, Wednesday, July 14, 2010 at 11:15amforce= Y (deltaL/L)area

so, if hookes law is F=k deltax

then force/deltax= k=Y*area/length

calculate k.

work done? 1/2 k x^2

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