Under the action of force, 2 kg body moves such that its position x as a function of time t is given by `x=(t^(3))/(3)`, x is in metre and t in second. Calculate the work done by the force in the first 2 second.

Explanation

From work - energy theorem, W = `Delta KE`
`x=t^(3)//3therefore"velocity" " "v=(dx)/(dt)=t^(2)`
`At t = 0, v_(i) = 0^(2) = 0`
`At t = 2, v_(t)=2^(2)=4m//s`
Work done `W = (1)(2) m(v_(1)^(2) - v_(i)^(2))`
`=(1)/(2)xx2(4^(2)-0)=16j`