An object of $4\mathrm{kg}$ mass is moving under a force, the relation between time and displacement is $x=\frac{t^3}{3}$ where x is in meters and t is the time in seconds. The following work was completed in the first two seconds:

According to the work-energy theorem, the net work done by forces on an object equals the change in its kinetic energy.

Work done, $W=AKE=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

Given that:

Mass of the object, $m=4\mathrm{kg}$

The relation between time and displacement: $x=\frac{t^3}{3}$

To find work done in two seconds $t=0s$ to $t=2s$

Velocity, $v=\frac{d_x}{dt}=\frac{d\left(t^3\right)}{dt}=\frac{3t^2}{3}=t^2$

Initial velocity at $t=0s,u=0^2=0s$

Final velocity at $t=2s,v=2^2=t^2=4s$

Using the equation: $W=AKE=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

Work done in the first two seconds $=\left(\frac{1}{2}\times 4\times 4^2\right)-\left(\frac{1}{2}\times 4\times 0^2\right)=32J$