Answer
As it passes through $x=2.0~m$, the kinetic energy of the block is $~~\frac{7}{3}~J$
Work Step by Step
We can find the work done on the block by the force from $x=0$ to $x=2.0~m$:
$W = \int^{2.0}_{0}(2.5-x^2)~dx$
$W = (2.5x-\frac{1}{3}x^3)~\Big\vert^{2.0}_{0}$
$W = (2.5)(2.0)-\frac{1}{3}(2.0)^3- [(2.5)(0)-\frac{1}{3}(0)^3)]$
$W = 5.0-\frac{8}{3}-0$
$W = \frac{7}{3}~J$
The work done on the block by the force is $~~\frac{7}{3}~J$
Since the block started from rest, the kinetic energy of the block will be equal to the work done on the block by the force.
As it passes through $x=2.0~m$, the kinetic energy of the block is $~~\frac{7}{3}~J$.
