Answer
$x_{max} = 10.75~m$
Work Step by Step
According to part (l), the kinetic energy at $x = 5.0~m$ is $11~J$
When the particle reaches $x = 8.00~m$, a force of $-4.00~N$ acts on it.
We can find $x$, the distance past $8.00~m$ that the particle reaches:
$K+W = 0$
$11~J+(-4.00~N)(x) = 0$
$x = \frac{11~J}{4.00~N}$
$x = 2.75~m$
Then: $~~x_{max} = 10.75~m$
