A particle of mass m moves in a horizontal ci | vedclass

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(D) For a particle moving in a circular path,the centripetal force is provided by the given force: $mv^2/r = k/r^2$. From this,the kinetic energy $(K.

Vedclass · Accepted Answer

$-k/(2r)$

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(D) For a particle moving in a circular path,the centripetal force is provided by the given force: $mv^2/r = k/r^2$. From this,the kinetic energy $(K.E.)$ is: $K.E. = 1/2 mv^2 = k/(2r)$. The potential energy $(U)$ is calculated by integrating the force: $U = -\int_{\infty}^{r} F \, dr = -\int_{\infty}^{r} (-k/r^2) \, dr = -k/r$. The total energy $(E)$ is the sum of kinetic and potential energy: $E = K.E. + U = k/(2r) - k/r = -k/(2r)$. The negative sign indicates that the particle is in a bound state.

$A$ particle of mass $m$ moves in a horizontal circle of radius $r$ under the influence of a centripetal force given by $F = -k/r^2$,where $k$ is a constant. Calculate the total energy of the particle.

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