A force act on a `30 gm` particle in such a way that the position of the particle as a function of time is given by `x = 3 t - 4 t^(2) + t^(3)`, where `x` is in metros and `t` is in seconds. The work done during the first `4` second is

Options

A. `5.28 J`
B. `450 mJ`
C. `490 mJ`
D. `530 mJ`

Answer

A

Explanation

`v = (dx)/(dt) = 3 - 8t + 3t^(2)`
`:. v_(0) = 3 m//s and v_(4) = 19 m//s`
`W = (1)/(2) m (v_(4)^(2) - v_(0)^(2))` (According to work energy theorem)
`= (1)/(2) xx 0.03 xx (19^(2) - 3^(2)) = 5.28 J`