The position (x) of a particle of mass 1 kg moving along X-axis at time t is given by `(x=1/2 t^(2))` metre. Find the work done by force acting on it in time interval from `t=0` to `t=3 s`.

Options

A. 4.2 J
B. 5.4 J
C. 4.6 J
D. 4.5 J

Answer

D

Explanation

`x=1/2 t^(2)`
`implies v=(dx)/(dt)=1/2 (2t)=t`
`:.` At `t=0, v_(i)=0 implies` At `t=3 s, v_(f)=3 ms^(-1)`
According to work -energy theorem,
`W=Delta K=K_(f)-K_(i)`
`=1/2 mv_(f)^(2)-1/2 mv_(i)^(2)=1/2xx1xx3^(2)=4.5 J`