A body of mass `3 kg` is under a constant force which causes a displacement `s` metre in it, given by the relation `s=(1)/(3)t^(2)`, where `t` is in seconds. Work done by the force in 2 seconds is

Options

A. `19/5 J`
B. `5/19 J`
C. `3/8 J`
D. `8/3 J`

Answer

`(4)`

Explanation

`s = (t^(2))/3, ds /dt = (2t)/3, (d^(2)s)/(dt^(2)) = 2/3 `
Work done, W` = int F ds = int m (d^(2)s)/(dt^(2)) ds` = `int m( d^(2)s)/(dt^(2)) (ds)/(dt)dt = int _(0)^(2) 3 xx 2/3 xx (2t)/3 dt = 4/3 int_(0)^(2) t dt`
` = 4/3 int _(0)^(2) tdt = 4/3|(t^(2))/(2)|_(0)^(2) = 4/3 xx 2 = 8/3 J`