The position of a particle of mass 4 g, acted upon by a constant force is given by `x=4t^(2)+t`, where x is in matre and t in second. The work done during the first 2 s is

Options

A. 128 mJ
B. 512 mJ
C. 576 mJ
D. 144 mJ

Answer

C

Explanation

The position of a particle
`x=ut^(2)+t`
`v=(dx)/(dt)=8t`
`a=(d^(2)x)/(dt)=8`
The distance in first 2 s
`x=4t^(2)+t=4xx(2)^(2)+2`
`x=18m`
The work done during the first 2 s
`W=F.x`
`=m.a.x`
`=4xx10^(-3)xx8xx18`
`=576xx10^(-3)J`
`W=576mJ`