Work Energy Power

JEE Main level, single correct, medium

120 of 183

Question 120 of 183

A particle is moving in a circular path of radius a under the action of an attractive potential U=−k2r2.U=-\dfrac{k}{2{r}^{2}}. Its total energy is:

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Answer: (D) Zero

U=−K2r2U=-\dfrac{K}{2{r}^{2}}⇒F=−dUdr=2K2r3=Kr3\Rightarrow F=-\dfrac{dU}{dr}=\dfrac{2K}{2{r}^{3}}=\dfrac{K}{{r}^{3}}⇒mv2r=Kr3\Rightarrow \dfrac{m{v}^{2}}{r}=\dfrac{K}{{r}^{3}}⇒v2=Kmr2\Rightarrow {v}^{2}=\dfrac{K}{m{r}^{2}}
⇒12mv2=K2r2\Rightarrow \dfrac{1}{2}m{v}^{2}=\dfrac{K}{2{r}^{2}}⇒E=K2r2−K2r2=0\Rightarrow E=\dfrac{K}{2{r}^{2}}-\dfrac{K}{2{r}^{2}}=0.

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