Relative Extension of Collagen Fiber: %0.8

AI Thread Summary
The discussion revolves around calculating the percentage of relative extension of collagen fiber under stress. The user initially applies the formula F/A = E Δl/l but struggles to achieve the expected result of 0.8%. A key error identified is the incorrect conversion of the cross-sectional area from mm² to m² and a miscalculation in the modulus of elasticity. After correcting these mistakes, the user expresses gratitude for the assistance received. The conversation highlights the importance of careful unit conversion in physics calculations.
edorf
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1. Collagen fiber is stressed with 12 N force. The cross-sectional area of the fiber is 3 mm2, its coefficient of elasticity is 500 MPa. Give the percentage of relative extension.

I think i have found the correct equation but i can't seem to find the rigth answer which is suposed to be 0.8% but i always end up with 8*10^-4..

Homework Equations



F/A=E Δl/l

The Attempt at a Solution



12N/(0,003m^2 * 5*10^6 Pa)

and then I get a result of 8*10^-4, and I am stuck.. I believe there is something fundamental I have missed but I can't seem to find it.. Thanks in advance!
 
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welcome to pf!

hi edorf! welcome to pf! :smile:

(try using the X2 tag just above the Reply box :wink:)
edorf said:
The cross-sectional area of the fiber is 3 mm2, its coefficient of elasticity is 500 MPa. Give the percentage of relative extension.

12N/(0,003m^2 * 5*10^6 Pa)

erm … 3 mm2 isn't .003 m2, is it? :redface:

(and you left the 00 out of 500)
 
ahhhhh...! Thank you so much!:D I guess that's what happends when you try to solve problems for 8h straight :P
 
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