Answer
The particle will be turned back before it reaches $~~x = 15~cm$
Work Step by Step
We can find the total energy in the system:
$E = K+U$
$E = \frac{1}{2}m~v_{max}^2+0$
$E = (\frac{1}{2})(2.0~kg)~(0.85~m/s)^2$
$E = 0.7225~J$
At $x = 15~cm$, we can see on the graph that the potential energy would be greater than $1.0~J$
Since $~~1.0~J~~$ is greater than the total energy in the system, the particle can not reach the point $x = 15~cm$
The particle will be turned back before it reaches $~~x = 15~cm$
