Answer
At $x=5.00 \mathrm{m}$ the potential energy is zero, and the kinetic energy is
$$
K=\frac{1}{2} m v^{2}=\frac{1}{2}(2.00 \mathrm{kg})(3.45 \mathrm{m} / \mathrm{s})^{2}=11.9 \mathrm{J} \text { . }
$$
The total energy, therefore, is great enough to reach the point $x=0$ where $U=11.0 \mathrm{J}$ , with a little "left over" $(11.9 \mathrm{J}-11.0 \mathrm{J}=0.9025 \mathrm{J}) .$ This is the kinetic energy at $x=0$ , which means the speed there is
$$
v=\sqrt{2(0.9025 \mathrm{J}) /(2 \mathrm{kg})}=0.950 \mathrm{m} / \mathrm{s}
$$
It has now come to a stop, therefore, so it has not encountered a turning point.
Work Step by Step
At $x=5.00 \mathrm{m}$ the potential energy is zero, and the kinetic energy is
$$
K=\frac{1}{2} m v^{2}=\frac{1}{2}(2.00 \mathrm{kg})(3.45 \mathrm{m} / \mathrm{s})^{2}=11.9 \mathrm{J} \text { . }
$$
The total energy, therefore, is great enough to reach the point $x=0$ where $U=11.0 \mathrm{J}$ , with a little "left over" $(11.9 \mathrm{J}-11.0 \mathrm{J}=0.9025 \mathrm{J}) .$ This is the kinetic energy at $x=0$ , which means the speed there is
$$
v=\sqrt{2(0.9025 \mathrm{J}) /(2 \mathrm{kg})}=0.950 \mathrm{m} / \mathrm{s}
$$
It has now come to a stop, therefore, so it has not encountered a turning point.
