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)//3 therefore` velocity `v=(dx)/(dt)=t^(2)`
At `t=0, v_(i)=0^(2)=0`, At `t=2, v_(f)=2^(2)=4 m//s`
work done `W=(1)/(2)m(v_(f)^(2)-v_(i)^(2))=(1)/(2)xx2(4^(2)-0)=16 J`