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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 a centripetal force equal to `(-K//r^(2))`, where k is a constant. The total energy of the particle is -

A

`-k/r`

B

`-k/(2r)`

C

`k/(2r)`

D

`(2k)/r`

Text Solution

Verified by Experts

The correct Answer is:
B
Since the particle is moving in horizontal circle, centripetal force,
`F=(mv^2)/r=k/r^2,mv^2=k/r` ...(i)
Kinetic energy of the particle,
`K=1/2mv^2 =k/(2r)` (Using (i))
As `F=(-dU)/(dr)`
`therefore` Potential energy,
`U= - underset(oo)oversetrintFdr=-undersetoooversetrint((-k)/r^2)dr =kundersetoooversetrintr^(-2) dr=(-k)/r`
`therefore` Total energy = K+U = `k/(2r)-k/r =(-k)/(2r)`
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