A particle of mass `m` is moving is a horizontal circle of radius `x` under a centripetal force equal to `- (kv^(2))` where it is constant The total energy of the particle is

Options

A. `-(k) /(r )`
B. `-(k)/(2r )`
C. `(k)/(2r)`
D. `(2k)/(r )`

Answer

B

Explanation

Since the particle is moving in horizontal circle exectripetal force
`F = (mv^(2))/(r ) = (k)/(y^(2))…..(i)`
`(mv^(2)) = (k)/(r )`
kinetic enrgy of the particle
`K = (1)/(2) mv^(2) = (k)/(2r)(Using (i))`
As `F = (-dU)/(dv)`
`:.` Potential energy
`U=-underset(infty)overset(r)intFdr=- underset(infty)overset(r)int r^(-r)dr=((-k)/(r^(2)))dr=Kdr=(-k)/(r)`
`:.` Total `"energy "=K+U=(k)/(2r)-(k)/(r)=(-k)/(2r)`.