A body of mass `6kg` is under a force which causes displacement in it given by `S= (t^(2))/(4)` metres where `t` is time . The work done by the force in `2` sec is

Options

A. `12 J`
B. `9 J`
C. `6J`
D. `3J`

Answer

D

Explanation

`s = (t^(2))/(4) :. Dx = (1)/(2) dt`
`F = ma = (md^(2)s)/(dt^(2)) = (6d^(2))/(dt^(2)) [(t^(2))/(4)] = 3N`
Now
`W = int _(0)^(2) F ds = int_(0)^(2) 3 (t)/(2)dt = (3)/(2) [(t^(2))/(2)]_(0)^(2) = (3)/(4) [(2)^(2) - (0)^(2)] = 3 J`