Under the action of a force, a `2 kg` body moves such that its position x as a function of time is given by `x =(t^(3))/(3)` where x is in metre and t in second. The work done by the force in the first two seconds is .

Options

A. 1600 J
B. 160 J
C. 16 J
D. 1.6 J

Answer

C

Explanation

(c ) Speed of the body, `v=(dx)/(dt)=(d)/(dt)((t^(3))/(3))=t^(2)`
At`" "` t=0, v=0,
At`" " t=0 s, v=4 ms^(-1)`
From work energy theorem,
W=change in kinetic energy
`K_(f)-K_(i)`
`=(1)/(2)m(v_(f)^(2)-v_(i)^(2))=(1)/(2)xx2xx(16-0)=16J`