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Question

A body of mass 1 kg starts moving from rest at t=0 in a circular path of radius 8 m. Its kinetic energy varies with time as k=2t2 J, then the magnitude of centripetal acceleration (in m/s2) at t=2 s is:

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Solution

Given, Kinetic energy k=2t2
On differentiating, we get
dkdt=4t
dk=4tdt=F.ds
4tdt=aTds (since mass is 1 kg)
Integrating,
t04tdt=v0vdv
4t22+c=v22
Using the given condition v=0 at t=0, we get c=0
v=2t
At t=2 sv=4 m/s
So, centripetal acceleration
ac=v2R=428=2 m/s2

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