#### Answer

$v = 6.3~m/s$

#### Work Step by Step

From $x=2$ to $x = 4$, the particle's kinetic energy is transformed into potential energy.
To find the maximum speed at $x = 2$, we can assume that the kinetic energy at $x = 4$ is zero.
We can find the maximum speed the particle could have at $x = 2$ and not reach $x = 6$:
$K_2+U_2 = K_4+U_4$
$\frac{1}{2}mv^2= 0+U_4 - U_2$
$v^2= \frac{(2)(U_4 - U_2)}{m}$
$v= \sqrt{\frac{(2)(U_4 - U_2)}{m}}$
$v= \sqrt{\frac{(2)(8~J -4~J)}{0.200~kg}}$
$v = 6.3~m/s$