A particle of mass m is moving in a horizontal circle of radius r with a centripetal force where k -1 F - k - r 2r- unit- The total energy of the particle isA- -3 k-2 rB- -k-2 rC- - k - rD- 1-5 k - r

The correct option is C f = k/r^2=mv^2/r ⇒ Ke = 1/2 mv^2 = k/2r And potential energy = ∫ Fdr = ∫k/r^2dr = [ k/r ] ...

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Byju's Answer

Standard XII

Physics

Conservative and Non Conservative Forces

A particle of...

Question

A particle of mass m is moving in a horizontal circle of radius r with a centripetal force where k = 1 $F = - \frac{k}{r^{2}}^r$ unit. The total energy of the particle is

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Solution

The correct option is A

$f = \frac{k}{r^{2}} = \frac{m v^{2}}{r} \Rightarrow K e = \frac{1}{2} m v^{2} = \frac{- k}{2 r} And potential energy = - \int F d r = \int \frac{k}{r^{2}} d r = [\frac{- k}{r}]$

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A particle of mass m is moving in a horizontal circle of radius r with a centripetal force where k = 1 F=−kr2^r unit. The total energy of the particle is

The correct option is A f=kr2=mv2r⇒Ke=12mv2=−k2rAnd potential energy =−∫Fdr=∫kr2dr=[−kr]

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The correct option is A

$f = \frac{k}{r^{2}} = \frac{m v^{2}}{r} \Rightarrow K e = \frac{1}{2} m v^{2} = \frac{- k}{2 r} And potential energy = - \int F d r = \int \frac{k}{r^{2}} d r = [\frac{- k}{r}]$