A force acts on a 3.0 gm particle in such a way that the position of the particle as a function of time is given by ` x=3t-4t^(2)+t^(3)`, where xx is in metres and t is in seconds. The work done during the first 4 seconds 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`
`{:("at t = 0"," at t = 4 "),(u=3m//s,v=3-32+48=19m//s):}`
According to work energy theorem
`W=Deltak=1/2 mv^2 -1/2"mu"^2`
`implies W=1/(2)0.03 [(19)^2 -(3)^2]=1/2 xx 0.03 xx22xx 16`
`implies W = 5.28 J.`