## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$v = 6.3~m/s$
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 $x = 6$: $K_2+U_2 = K_4+U_4$ $\frac{1}{2}mv^2= 0+U_4 - U_2$ $v^2= \frac{(2)(U_4 - U_2)}{m}$ $v= \sqrt{\frac{(2)(U_4 - U_2)}{m}}$ $v= \sqrt{\frac{(2)(8~J -4~J)}{0.200~kg}}$ $v = 6.3~m/s$