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. `(5)/(19) J`
B. `(3)/(8) J`
C. `(8)/(3) J`
D. `(19)/(5) J`

Explanation

Work done by the force
`= Force xx Displacement`
or `W = F xx s` (i)
But from Newton's 2nd law , we have
Force = Mass xx Acceleration
i.e., `F = ma` (ii)
Hence, from Eqs (i) and (ii), we get
`W = mas = m ((d^(2)s)/(dt^(2))) s` (ii)
`(because . a = (d^(2)s)/(dt^(2)))`
Now, we have `s = (1)/(3) t^(2)`
` (d^(2)s)/(dt^(2)) = (d)/(dt) [(d)/(dt) ((1)/(3) t^(2))]`
`= (d)/(dt) xx ((1)/(3) t) = (2)/(3) (d)/(dt) = (2)/(3)`
hence, Eq. (iii) becomes
`W = (2)/(3) ms = (2)/(3) m xx (1)/(3) t^(2) = (2)/(9) mt^(2)`
We have given `m = 3 kg, t= 2s`
`:. W = (2)/(3) xx 3 xx (2)^(2) = (8)/(3) J`