A force acts on a 2   kg object so that its position is given as a function of time as x = 3 t 2 + 5 .…

Here, the position of the object is $x=3t^2+5$, therefore the velocity of the object will be,

$v=\frac{dx}{dt}=\frac{d\left(3t^2+5\right)}{dt}$

$v=6t+0$

From the work-energy theorem, all the work done by the force acting on it will be equal to the change in its kinetic energy, i.e.,

$W=KE\left(t=5 \mathrm{s}\right)-KE\left(t=0 \mathrm{s}\right)$
$W=\frac{1}{2}\times 2\times {\left(6\times 5\right)}^2-\frac{1}{2}\times 2\times {\left(6\times 0\right)}^2$
$W=900 \mathrm{J}$.