## In This post you will get answer of Q.

If potential energy is given by $U=frac{a}{r^{2}}-frac{b}{r}$. Then find out maximum force. (given a = 2, b = 4)

Solution:

## $F=-frac{du}{dr}=-left[-frac{2a}{r^{ 3}}+frac{b}{r^{2}}right]$

$=frac{2a}{r^{3}}-frac{b}{r^{2}}$

$frac{df}{dr}=-frac{6a}{r^{4}}+frac{2b}{r^{3}}=0$

$Rightarrowquadquad frac{6a}{r}=2b$

$Rightarrowquadquad F_{max}=frac{2times2}{left(frac{3}{2}right)^{3}}-frac{4}{left(frac{3}{2}right)^{2}}$

$quadquadquadquad=frac{4times8}{27}-frac{4times4}{9}=frac{32-16times3}{27}=-frac{16}{27}$