A point mass of `0.5` kg is moving along `x-` axis as `x=t^(2)+2t` , where, `x` is in meters and `t` is in seconds. Find the work done (in J) by all the forces acting on the body during the time interval `[0,2s]`.

Options

A. `6J`
B. `7J`
C. `8J`
D. `9J`

Answer

C

Explanation

Method
`I:` `W_(all)=DeltaKE`
`=(1)/(2)m(v_(2)^(2)-v_(0)^(2))`
`v=(dx)/(dt)=2t+2=2(t+1)`
`v_(2)=6m//s,v_(0)=2m//s`
`W_(all)=(1)/(2)(1/2)(6^(2)-2^(2))=8J`
Method `II :`
`x=t^(2)+2t`
`a=(d^(2)x)/(dt^(2))=2m//s^(2)=` constant
so `F=ma=1N=` constant
so, `W=Fd=(1)(8-0)=8J`