Question

Eng e force would (6 points) If the repulsive energy curve follo what increase in repulsive force -109 result from a 5% compressive strain, that is from having r decreased by 10%?

Please show all work and explain in detail.

Thank you very much for your time.

Answer 1

$\\$Energy = $\int $Force x Displacement$ \\\therefore $ Repulsive Energy = $\int $Repulsive Force x dr $\\\therefore A/r^{12} = \int F(r)dr \\\therefore F(r) =\frac{\mathrm{d} (A/r^{12})}{\mathrm{d} r} \\\therefore F(r) = \frac{-12A}{r^{13}}$

$\\$Now we are given 10\% decrease in r$ \\$Let $r_o$ be the initial value$ \\\therefore r_{final} = (1-0.1)r_o = 0.9r_o \\\therefore \Delta F = F(r_{final}) - F(r_o) \\\therefore \Delta F = \frac{-12A}{(0.9r_o)^{13}} + \frac{12A}{r_o^{13}} \\\therefore $ Percentage Change = $\Delta F/(-12A/r_o^{13})*100 \\\therefore $ Percentage Change = $(\frac{(-12A/(0.9r_o)^{13})}{(-12A/r_o^{13})}-1)*100 \\\therefore $ Percentage Change = $(\frac{1}{0.9^{13}}-1)*100 = 293.41\% \\\therefore $ Increase in repulsive force is by 293.41 \% approximately due to applied compressive strain$$