Answer
The maximum kinetic energy of the block is $~~2.64~J$
Work Step by Step
$F(x) = (2.5-x^2)\hat{i}~N$
Initially, the force is directed in the +x direction. The block will accelerate in the +x direction until $F(x) = 0$. Then the force is directed in the -x direction so the block will start to decelerate. The block reaches its maximum kinetic energy when $F(x) = 0$
We can find $x$ when $F(x) = 0$:
$2.5-x^2 = 0$
$x^2 = 2.5$
$x = \sqrt{2.5}$
$x = 1.58~m$
We can find the work done on the block by the force from $x=0$ to $x=1.58~m$:
$W = \int^{1.58}_{0}(2.5-x^2)~dx$
$W = (2.5x-\frac{1}{3}x^3)~\Big\vert^{1.58}_{0}$
$W = (2.5)(1.58)-\frac{1}{3}(1.58)^3- [(2.5)(0)-\frac{1}{3}(0)^3)]$
$W = 3.95-1.31-0$
$W = 2.64~J$
The work done on the block by the force is $~~2.64~J$
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.
The maximum kinetic energy of the block is $~~2.64~J$.
