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 .

Explanation

Method 1: Given that `x=t^3/3`
`:.` Velocity, `v=(dx)/(dt)=t^2impliesdx=t^2dt`
Acceleration, `a=(dv)/(dt)=2t` `:.` Force, `F=ma`
Work done by force,
`W=intFdx=underset0overset2int4t(t^2dt)=4underset0overset2t^3dt`
`=4|t^4/4|_0^2=4/4(2^4-0^4)=16J`
Method 2: By work-energy theorem,
Given `x=t^3/3`
`:.` Velocity `v=(dx)/(dt)=t^2`
At `t=0`, `v_i=0^2=0`
At `t=2s`, `v_f=2^2=4ms^-1`
Work done, `W=1/2m(v_f^2-v_i^2)=1/2xx2xx(4^2-0)=16J`