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IF a particle of mass m is moving i...

IF a particle of mass m is moving in a horizontal circle of radius r with a centripetal force `(-(K)/(r^(2)))` , then its total energy is

A

`(K ) /(2r)`

B

`-(K)/(r )`

C

`-(2k) /(r ) `

D

`-(4K )/(r ) `

Text Solution

Verified by Experts

The correct Answer is:
A
Centripetal force (F ) `=(mv^(2))/( r ) = (K )/(r^(2))("given ")`
` therefore mv^(2) =(K )/(r ) therefore K.E =(1)/(2) mv^(2) =(K )/(2r) `
and `P.E ` = work done `= int F dr `
` =int (K)/(r^(2))dr=-(K )/(r )`
`therefore ` Total energy `=K.E +P.E =( K)/(2r) -(K )/(r ) =-(K )/(2r )`
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