By what amount does the 52-cm-long femur of an 73 kg runner compress at this moment? The cross-section area of the bone of the femur can be taken as 5.2*10-4m2 and its Young's modulus is 1.6*1010N/m2.

4.55*10^-5m

Explanation:

Young modulus of the material is equal to the ratio of the tensile stress to tensile strain of the elastic material.

Young modulus = Tensile stress/Tensile strain

Tensile stress = Force/cross sectional area

Give mass = 73kg

Force = mg = 73*10 = 730N

Cross sectional area = 5.2*10^-4m²

Tensile stress = 730/5.2*10^-4

Tensile stress = 1.4*10^6N/m

Strain = extension/original length

Given original length = 52cm = 0.52m

Tensile strain = extension (e) / 0.52

Substituting the values given into the young modulus formula we have;

1.6*10^10 = 1.4*10^6/{e/0.52}

1.6*10^10 = 1.4*10^6*0.52/e

e = 1.4*10^6*0.52/1.6*10^10

e = 7.28*10^5/1.6*10^10

e = 4.55*10^-5m

This shows that the femur compresses by 4.55*10^-5m