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 times a person’s weight. According to one set of experiments, the average area of the Achilles tendon is 78.1 mm², its average length is 25 cm, and its average Young’s modulus is 1474 MPa. How much tensile stress is required to stretch this muscle by 5.0% of its length?

To calculate the tensile stress required to stretch the Achilles tendon by 5% of its length, we can use Hooke's Law, which states that stress (σ) is equal to the applied force divided by the cross-sectional area (A) of the tendon.

First, let's calculate the change in length (ΔL) of the tendon:

ΔL = (5% / 100) * length of the tendon
= (0.05) * 25 cm
= 1.25 cm

Next, let's calculate the initial cross-sectional area (A) of the tendon:

A = 78.1 mm²

Now, we can calculate the tensile stress (σ) using Hooke's Law:

σ = F / A

To find the force (F), we use the Young's modulus (E), which describes the stiffness of the material:

F = E * A * ΔL / length of the tendon

Substituting the given values:

F = (1474 MPa) * (78.1 mm²) * (1.25 cm) / (25 cm)

Let's convert the units accordingly to ensure consistency:

1 MPa = 1 N / mm²
1 cm = 10 mm

F = (1474 N / mm²) * (78.1 mm²) * (1.25 cm) / (25 cm)
= (1474 N / mm²) * (78.1) * (1.25) / 25
≈ 91.2 N

Therefore, the tensile stress required to stretch the Achilles tendon by 5% of its length is approximately 91.2 N.