A force acts on a 3 g particle in such a way that position of particle as a function of time is given by `x=3t - 4t^2 + t^3` where x is in metre and is in sec. The work done during first 4 s is 

Options

A. 570 mJ
B. 450 mJ
C. 490 mJ
D. 528 mJ

Answer

D

Explanation

`x = 3t - 4t^2 + t^3`
`v = (dx)/(dt) = (3 - 8t + 3t^2) , a = (d^2 x)/(dt^2) = (-8 + 6t)`
`W = intFdx = int m(d^2x)/(dt^2) ((dx)/(dt)) dt`
`= int_(0)^(4) (3 xx 10^(-3)) (6t - 8)(3t^2 - 8t + 3) dt`
`= 3 xx 10^(-3) int_(0)^(4) (18 t^3 - 48t^2 + 18t - 24t^2 + 64t - 24)dt`
`= 3 xx 10^(-3) [(18t^4)/(4) - (72 t^3)/(3) + (82t^2)/(2) - 24 t]_(0)^(4)`
`= 528 xx 10^(-3) ` Joule = 528 mJ .