A force act on a `30` gm particle in such a way that the position of the particles as a function of time is given by `x = 3 t - 4 t^(2) + t^(3)`, where `x` is in meters 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

`x=3t-4t^2+t^3`
`v=(dx)/(dt)=3-8t+3t^2`
`t=0`, `v_1=3(m)/(s)`
`t=4s`,`v_2=3-8(4)+3(4)^2`
`=3-32+48=19(m)/(s)`
`W_(1rarr2)=K_2-K_1=(1)/(2)m(v_2^2-v_1^2)`
`=(1)/(2)xx(30)/(1000)(19^2-3^2)`
`=0.015xx22xx16=5.28J`