A $2.00 \mathrm{~kg}$ particle moves along an $x$ axis in one-dimensional motion while a conservative force along that axis acts on it. The potential energy $U(x)$ associated with the force is plotted in Fig. 8-10a. That is, if the particle were placed at any position between $x=0$ and $x=7.00 \mathrm{~m}$, it would have the plotted value of $U$. At $x=6.5 \mathrm{~m}$, the particle has velocity $\vec{v}_0=(-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}$
From Figure, determine the particle's speed at $x_1=4.5 \mathrm{~m}$.

(1) The particle's kinetic energy is given by Eq. 7-1 $\left(K=\frac{1}{2} m v^2\right)$. (2) Because only a conservative force acts on the particle, the mechanical energy $E_{\mathrm{mec}}(=K+U)$ is conserved as the particle moves. (3) Therefore, on a plot of $U(x)$ such as Fig. 8-10a, the kinetic energy is equal to the difference between $E_{\mathrm{mec}}$ and $U$.
Calculations: At $x=6.5 \mathrm{~m}$, the particle has kinetic energy
$$
\begin{aligned}
K_0 & =\frac{1}{2} m v_0^2=\frac{1}{2}(2.00 \mathrm{~kg})(4.00 \mathrm{~m} / \mathrm{s})^2 \\
& =16.0 \mathrm{~J} .
\end{aligned}
$$
Because the potential energy there is $U=0$, the mechanical energy is
$$
E_{\text {mec }}=K_0+U_0=16.0 \mathrm{~J}+0=16.0 \mathrm{~J} .
$$
This value for $E_{\mathrm{mec}}$ is plotted as a horizontal line in Fig. 8-10a. From that figure we see that at $x=4.5 \mathrm{~m}$, the potential energy is $U_1=7.0 \mathrm{~J}$. The kinetic energy $K_1$ is the difference between $E_{\text {mec }}$ and $U_1$ :
$$
K_1=E_{\text {mec }}-U_1=16.0 \mathrm{~J}-7.0 \mathrm{~J}=9.0 \mathrm{~J} .
$$
Because $K_1=\frac{1}{2} m v_1^2$, we find
$$
v_1=3.0 \mathrm{~m} / \mathrm{s}
$$
