Asked by Brianna

I think that for energy of a spring I should use (1/2)(k)(x^2) and (1/2)(m)(v^2) but i'm not sure what to do with the equations to solve the problem.

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?

Answers

Answered by bobpursley
force= 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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