If potential energy is given by U = frac{a}{r | vedclass

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(A) The force $F$ is related to potential energy $U$ by the relation $F = -\frac{dU}{dr}$.Given $U = \frac{a}{r^2} - \frac{b}{r}$,we have $F = -\frac{

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$-\frac{16}{27} \ N$

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(A) The force $F$ is related to potential energy $U$ by the relation $F = -\frac{dU}{dr}$.Given $U = \frac{a}{r^2} - \frac{b}{r}$,we have $F = -\frac{d}{dr}(\frac{a}{r^2} - \frac{b}{r}) = -(-\frac{2a}{r^3} + \frac{b}{r^2}) = \frac{2a}{r^3} - \frac{b}{r^2}$.To find the maximum force,we differentiate $F$ with respect to $r$ and set it to zero: $\frac{dF}{dr} = \frac{d}{dr}(\frac{2a}{r^3} - \frac{b}{r^2}) = -\frac{6a}{r^4} + \frac{2b}{r^3} = 0$.This gives $\frac{2b}{r^3} = \frac{6a}{r^4}$,so $r = \frac{3a}{b}$.Substituting $a = 2$ and $b = 4$,we get $r = \frac{3(2)}{4} = 1.5 = \frac{3}{2}$.Now,substitute $r = \frac{3}{2}$ into the expression for $F$:$F = \frac{2(2)}{(3/2)^3} - \frac{4}{(3/2)^2} = \frac{4}{27/8} - \frac{4}{9/4} = \frac{32}{27} - \frac{16}{9} = \frac{32 - 48}{27} = -\frac{16}{27} \ N$.

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

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