If the anterior cruciate ligament (ACL) in an athlete's knee has length 1.0cm, breaking point 1.9×10^8 Pa, and Young's Modulus 6.0×10^8 Pa, how far must it be stretched from its relaxed length to tear it?
PLEASE HELP!
My answer is 0.0032m, but I am not sure if this is correct!
To determine the distance the ACL must be stretched to tear it, we need to use the concept of stress and strain, and the formula for Young's modulus.
Young's modulus (Y) is a measure of the stiffness of a material and relates stress (σ) to strain (ε) through the equation: Y = σ/ε.
In this case, we know the breaking point of the ACL (σ) is 1.9×10^8 Pa, and the Young's modulus (Y) is 6.0×10^8 Pa. We need to find the strain (ε) that corresponds to the breaking point.
Rearranging the equation, we can solve for strain (ε): ε = σ/Y.
ε = 1.9×10^8 Pa / 6.0×10^8 Pa = 0.3167
So, the strain (ε) corresponding to the breaking point of the ACL is approximately 0.3167.
Now, we know that strain (ε) is defined as the change in length (ΔL) divided by the original length (L). Rearranging the equation, we can solve for the change in length (ΔL): ΔL = ε * L.
Given that the original length of the ACL (L) is 1.0 cm (or 0.01 m), we can calculate the change in length:
ΔL = 0.3167 * 0.01 m = 0.003167 m or approximately 0.0032 m.
So, the ACL must be stretched approximately 0.0032 meters (or 3.2 mm) from its relaxed length to tear it.
Therefore, your answer of 0.0032 m is correct!