Under the action of a force, a 2kg 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.

Under the action of a force, a 2kg body moves such that its position x

1.	Under the action of a force, a 2kg 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.
Answer» Solution :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) XX 2(4^(2) - 0) = 16 J`