Respuesta :
Answer:
0.156 mm
Explanation:
Here is the complete question
The normal force of the ground on the floor can reach three times a runner's body weight when the foot strikes the pavement. By what amount does the 52-cm-long femur of an 85 kg runner compress at this moment? The cross-section area of the bone of the femur can be taken as 5.2 × 10⁻⁴ m² and its Young's modulus is 1.6 × 10¹⁰ N/m²
The Young's modulus of the bone Y = stress/strain = σ/ε = F/A ÷ ΔL/L = FL/AΔL where F = force on bone = 3mg(since it is 3 times his weight) where m = mass of runner = 85 kg and g = acceleration due to gravity = 9.8 m/s². L = length of femur = 52 cm = 0.52 m, A = cross-sectional area of femur = 5.2 × 10⁻⁴ m² and ΔL = compression of femur.
Making ΔL subject of the formula,
ΔL = FL/AY
ΔL = 3mgL/AY
Substituting the values of the variables into the equation, we have
ΔL = 3mgL/AY
ΔL = 3 × 85 kg × 9.8 m/s² × 0.52 m/(5.2 × 10⁻⁴ m² × 1.6 × 10¹⁰ N/m²)
ΔL = 1299.48 kgm²/s² ÷ 8.32 × 10⁻⁶ N
ΔL = 156.1875 × 10⁻⁶ m
ΔL = 0.1561875 × 10⁻³ m
ΔL = 0.1561875 mm
ΔL ≅ 0.156 mm
The amount does the 52-cm long femur of 85 kg is 0.156 mm.
Calculation of the amount:
Since
The Young's modulus of the bone Y should be
= stress/strain
= σ/ε
So,
= F/A ÷ ΔL/L
here F = force on bone = 3mg
m = mass of runner = 85 kg
and g = acceleration due to gravity = 9.8 m/s²
L = length of femur = 52 cm = 0.52 m,
A = cross-sectional area of femur = 5.2 × 10⁻⁴ m²
and ΔL = compression of femur.
Now
ΔL = FL/AY
ΔL = 3mgL/AY
Now
ΔL = 3mgL/AY
= 3 × 85 kg × 9.8 m/s² × 0.52 m/(5.2 × 10⁻⁴ m² × 1.6 × 10¹⁰ N/m²)
= 1299.48 kgm²/s² ÷ 8.32 × 10⁻⁶ N
= 156.1875 × 10⁻⁶ m
= 0.1561875 × 10⁻³ m
= 0.1561875 mm
= 0.156 mm
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