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  • I have a doubt, kindly clarify. - Work Energy and Power - JEE Main-2

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

  • Option 1)

    -\frac{3}{2} \frac{k}{a^{2}}

  • Option 2)

    - \frac{k}{4a^{2}}

  • Option 3)

    \frac{k}{2a^{2}}

  • Option 4)

    Zero

 

Answers (2)

best_answer

U=-\frac{K}{2r^{2}}

\therefore F=\frac{-dU}{dr}=\frac{-K}{r^{3}}

\Rightarrow \frac{K}{r^{3}}=\frac{mv^{2}}{r} or  mv^{2}=\frac{K}{r^{2}}

K.E.=\frac{1}{2}mv^{2}=\frac{K}{2r^{2}}

Total energy = K.E.+P.E.=0

 

Conservative Force -

F=\frac{-dU}{dr}

- wherein

Negative of the rate of change of potential energy with respect to position

 

 


Option 1)

-\frac{3}{2} \frac{k}{a^{2}}

This is incorrect

Option 2)

- \frac{k}{4a^{2}}

This is incorrect

Option 3)

\frac{k}{2a^{2}}

This is correct

Option 4)

Zero

This is incorrect

Posted by

prateek

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