Question

The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme activities, such as sprinting, it can be subjected to forces as high as 13.0 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is 78.1 mm^2, its average length is 25.0 cm, and its average Young's modulus is 1474 MPa.

A)How much tensile stress is required to stretch this muscle by 5.00% of its length?
I first changed the units on the numbers:
78.1mm^2= .0781 m^2
25cm= .25m
1474MPa= 1.474*10^9 Pa
5% length= .0125 m

I tried using this equation:
F= YA/L *(delta)L

I got the answer 5755970, and they want the answer in Pa.


B)If we model the tendon as a spring, what is its force constant?

F=kx
Force from above part/.25m=k ??

C)If a 75.0 kg sprinter exerts a force of 13.0 times his weight on his Achilles tendon, by how much will it stretch?

Answer 1

You erred on changing square mm to square meters. There are 10^6 mmsquared in 1m^2, not 1000.

b) yes, force corrected.
c) if it is linear, then do the math F=kx

Answer 2

I'm still getting A wrong....

F=(1.474*10^9)(7.81*10^-5)/(.25) *.0125

=5755.95 Pa

Answer 3

achilles

Answer 4

A) 7.7*10^7 Pa

Answer 5

k= A(Y)/L

MPa=Pa(1/1000)^2(mPa)

K=4.4*10^5