A body of mass `3 kg` is under a force , which causes a displacement in it is given by `S = (t^(3))/(3)` (in metres). Find the work done by the force in first `2` seconds.

Options

A. 2 J
B. 3.8 J
C. 5.2 J
D. 24 J

Answer

D

Explanation

`S=(t^(3))/(3)therefore dS=t^(2)dt`
`a=(d^(2)S)/(dt^(2))=(d^(2))/(dt^(2))[(t^(3))/(3)]=2t m//s^(2)`
Now work done by the force `W= int_(0)^(2)F.ds = int_(0)^(2)ma.dS`
`int_(0)^(2)3xx2t xx t^(2)dt = int_(0)^(2)6t^(3)dt=(3)/(2)[t^(4)]_(0)^(2)=24 J`