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A particle of mass m is moving in a hori...

A particle of mass m is moving in a horizontal circle of radius r under centripetal force given by `(- k//r^(2))` where k is a constant . Then the total energy of the particle is _________ .

A

`-k//2r`

B

`k//r`

C

`k//2r`

D

`-k//r`

Text Solution

Verified by Experts

The correct Answer is:
A
`(mv^(2))/(r) = (k)/(r^(2))`
K.E. `= (1)/(2)mv^(2)= (k)/(2r)`
P.E. `=-int_(oo)^(r)Fdr = int_(oo)^(r)(k)/(r^(2))dr`
= `-(k)/(r )`
Total energy E = KE +PE
`= (k)/(2r) -(k)/( r) = -(k)/(2r)`
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