Answer
$v = 7.3~m/s$
Work Step by Step
At $x = 7.0~m$, the ball has a mechanical energy of $75~J$
On the graph, we can see that $U = 60~J$ at $x = 13~m$
By conservation of energy, since the ball has a mechanical energy of $75~J$, the ball will be able to reach the point $x = 13~m $ where $U = 60~J$.
We can find the total kinetic energy $K$ at $x = 13~m$:
$E = K+U = 75~J$
$K = 75~J-U$
$K = 75~J-60~J$
$K = 15~J$
The rotational inertia of a solid ball is $I = \frac{2}{5}MR^2$
We can find an expression for the total kinetic energy:
$K = \frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$
$K = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2$
$K = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$
$K = \frac{7}{10}Mv^2$
We can find the ball's speed at $x = 13~m$:
$K = \frac{7}{10}Mv^2 = 15~J$
$v^2 = \frac{(10)(15~J)}{7M}$
$v = \sqrt{\frac{(10)(15~J)}{7M}}$
$v = \sqrt{\frac{(10)(15~J)}{(7)(0.400~kg)}}$
$v = 7.3~m/s$
