Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • A particle is moving in a circular path of radius a under the action of an attractive potential U=  . Its total energy is :

IMG_20190128_102517.jpg

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 :

Answers (1)

best_answer

@Shilpa

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

 

 

Posted by

avinash.dongre

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

MBBS seat count crosses 1.29 lakh; 19 states get new medical colleges in one year
anam-khan

Karnataka NEET UG 2025: Stray vacancy seat allotment result declared; 107 allotted MBBS, BDS seats
anam-khan

NEET PG 2025 Seat Allotment (OUT) LIVE: MCC round 2 counselling result on mcc.nic.in; admission process

Latest Question