The particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration `a_(c)` is varying with time t as `a_(c)` = `k^(2)` , Where k is a constant .The power delivered to particle by the forces actingon it is

Explanation

In general for a particle moving with constant speed, the power delivered by the total (centripetal force is zero, the reason being, force is always radial and velocity is always tangerial, the angle between then being `90^@`.
However, if the particle has some tangential acceleration too, then the power delivered by all the forces will not be zero, since in this case, tangential forces, besides radial forces, also act. Therefore, it is wise, to first known, whether the velocity (and hence KE) of the particle change with time or remains constant)
If v be the instantaneous velocity, then
`(mv^2)/(r)=k^2rt^2impliesv=(krt)/(sqrtm)`
Obviously, the velocity depends on time. Therefore, the power delivered is not zero.
Now, if `a_t` be the tangerial acceleration,
Then `a_t=(dv)/(dt)=(kr)/(sqrtm)`
Therefore, tangerial force `F_t=ma_t=sqrtm(kr)`
Now, power P delivered is given by `P=(vecF_r+vecF_t)*vecv`
`vecF_r*vecv+vecF_t*vecv` `[:' vecF_r is _|_ t o vecv]`
`=vecF_t*vecv` [`:' vecF_t` is along the same direction as `vecv`]
`=sqrtm(kr)((krt)/(sqrtm))`
`P=k^2r^2t`
Alternatively: `E=1/2mv^2=1/2k^2r^2t^2`
`P=(dE)/(dt)=k^2r^2t^2`