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?
achilles
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
I'm still getting A wrong....
F=(1.474*10^9)(7.81*10^-5)/(.25) *.0125
=5755.95 Pa
A) 7.7*10^7 Pa
k= A(Y)/L
MPa=Pa(1/1000)^2(mPa)
K=4.4*10^5
To answer these questions, we will use the given information and relevant equations. Let's break down each question one by one:
A) To find the tensile stress required to stretch the Achilles tendon by 5% of its length, we can use the formula:
Stress (σ) = Force (F) / Area (A)
But first, we need to determine the force applied. We can use Hooke's Law, which states:
F = Y * (delta L) / L
where F is the force, Y is the Young's modulus, delta L is the change in length, and L is the original length.
Substituting the given values:
Y = 1474 MPa = 1474 * 10^6 Pa (converting MPa to Pa)
delta L = 0.0125 m (5% of 0.25 m)
L = 0.25 m
So, the equation becomes:
F = (1474 * 10^6 Pa) * (0.0125 m) / (0.25 m)
Calculating this, we get F = 7360 * 10^3 N
Now, we can substitute the force and area values into the stress equation:
σ = (7360 * 10^3 N) / (0.0781 m^2)
Calculating this, we get σ = 9.43 * 10^7 Pa (or N/m^2)
Therefore, the tensile stress required to stretch the Achilles tendon by 5% of its length is 9.43 * 10^7 Pa.
B) To find the force constant (k) when modeling the tendon as a spring, we can use the equation:
F = k * x
We already calculated the force (F) in part A as 7360 * 10^3 N, and x is the displacement/stretch of the tendon.
So, the equation becomes:
7360 * 10^3 N = k * x
Since we don't have the displacement value (x), we cannot determine the force constant (k) at this point.
C) To find out how much the Achilles tendon will stretch when a 75.0 kg sprinter exerts a force of 13.0 times his weight on it, we can determine the force applied first.
Weight of the sprinter = mass * acceleration due to gravity
Weight = 75.0 kg * 9.8 m/s^2 = 735 N
Force applied = 13.0 times the weight = 13.0 * 735 N = 9555 N
Now we can use Hooke's Law (F = Y * (delta L) / L) to find the displacement or change in length (delta L):
9555 N = (1474 * 10^6 Pa) * (delta L) / (0.25 m)
Rearranging the equation:
delta L = (9555 N * 0.25 m) / (1474 * 10^6 Pa)
Calculating this, we get delta L = 0.0162 m
Therefore, the Achilles tendon will stretch by approximately 0.0162 m when a 75.0 kg sprinter exerts a force of 13.0 times his weight on it.