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

Explanation

From work - energy theorem `W = Delta KE`
`x = (t^(3))/3, "Velocity" v = (dx)/(dt) = t^(2)`
At ` t = 0, v_(1) = 0 , At t = 2s, v_(2) = 4m/s`
`W = 1/2m(v_(2)^(2) - v_(1)^(2)) = 1/2 xx 2 (4^(2) - 0)= 16 J`