Find the work done.

A particle of mass m is moving in a circle of radius R such that its centripetal acceleration 'a' varies with time 't' as a = K t 2 t^2 , where K is a constant . Find the work done on the particle in first 't' seconds .

(mKR t 2 t^2 )/2 2mKR t 2 t^2 (mKR t 2 t^2 )/(3R) mKR t 2 t^2

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1 solution

Work-Energy theorem gives work done =change in kinetic energy in the first t seconds =(1/2)m(v^2)=(1/2)mkR(t^2)

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