Answer:
0.815 mm
Explanation:
The rod in made of Kevlar 49, so it has an Young's modulus of
E = 125 GPa
The stiffness of a rod is given by:
k = E * A / L
k = E * π/4 * d^2 / L
So:
k = 125*10^9 * π/4 * 0.01^2 / 0.1 = 98.17 MN/m
Of the pulling forces only one is considered because when you pull on something there is always another force on the other side of equal magnitude and opposite direction to maintain equilibrium.
Hooke's law:
Δl = P/k
Δl = 80*10^3 / 98.17*10^6 = 0.000815 m = 0.815 mm
