A) \[-\frac{k}{2r}\]
B) \[-\frac{k}{r}\]
C) \[-\frac{2k}{r}\]
D) \[-\frac{4k}{r}\]
Correct Answer: A
Solution :
\[\frac{m{{v}^{2}}}{r}=\frac{k}{{{r}^{2}}}\] Þ \[m{{v}^{2}}=\frac{k}{r}\]\ K.E.= \[\frac{1}{2}m{{v}^{2}}=\frac{k}{2r}\] P.E.\[=\int{F\,dr}\]\[=\int{{}}\frac{k}{{{r}^{2}}}dr=-\frac{k}{r}\] \ Total energy = K.E. + P.E. \[=\frac{k}{2r}-\frac{k}{r}=-\frac{k}{2r}\]You need to login to perform this action.
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