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concepts/mechanics/u3/m3-3-potential-energy.tex

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+\subsection{Conservative Forces and Potential Energy}
+
+This subsection introduces conservative forces through path-independent work and uses that idea to define potential energy differences for an interacting system.
+
+\dfn{Conservative force and potential energy difference}{Let $\vec{F}_c$ denote a force associated with some interaction, let $i$ and $f$ denote initial and final positions, and let $C$ denote a path from $i$ to $f$. The force $\vec{F}_c$ is called \emph{conservative} if the work
+\[
+W_c(i\to f)=\int_C \vec{F}_c\cdot d\vec{r}
+\]
+depends only on the endpoints $i$ and $f$, not on the path $C$.
+
+For a conservative force, the corresponding \emph{potential energy difference} of the interacting system is defined by
+\[
+\Delta U=U_f-U_i=-\int_i^f \vec{F}_c\cdot d\vec{r}.
+\]
+Thus the work done by the conservative force is
+\[
+W_c=-\Delta U.
+\]}
+
+\thm{Equivalent conservative-force relations}{Let $\vec{F}_c$ denote a conservative force and let $d\vec{r}$ denote an infinitesimal displacement. Then the following relations hold:
+\[
+\oint \vec{F}_c\cdot d\vec{r}=0,
+\]
+so the work done by $\vec{F}_c$ around any closed loop is zero.
+
+Equivalently, for any two paths $C_1$ and $C_2$ connecting the same endpoints,
+\[
+\int_{C_1} \vec{F}_c\cdot d\vec{r}=\int_{C_2} \vec{F}_c\cdot d\vec{r}.
+\]
+
+The local potential-energy relation is
+\[
+dU=-\vec{F}_c\cdot d\vec{r}.
+\]
+For one-dimensional motion along the $x$-axis,
+\[
+F_x=-\frac{dU}{dx}.
+\]
+More generally, one may write lightly
+\[
+\vec{F}_c=-\nabla U.
+\]}
+
+\nt{Potential energy is a property of a \emph{system}, not of a single isolated object. For example, gravitational potential energy belongs to the Earth-object system, and spring potential energy belongs to the block-spring system. A conservative force can transfer energy between kinetic and potential forms without making the potential difference depend on the path. By contrast, nonconservative forces such as kinetic friction and air resistance have path-dependent work, so a single-valued potential-energy function for that interaction is not defined in this AP sense.}
+
+\pf{Why zero closed-loop work gives a well-defined $\Delta U$}{Assume that for every closed path,
+\[
+\oint \vec{F}_c\cdot d\vec{r}=0.
+\]
+Take two paths $C_1$ and $C_2$ from the same initial point $i$ to the same final point $f$. Traverse $C_1$ from $i$ to $f$ and then traverse $C_2$ backward from $f$ to $i$. This makes a closed loop, so
+\[
+\int_{C_1} \vec{F}_c\cdot d\vec{r}+\int_{f\to i\text{ on }C_2} \vec{F}_c\cdot d\vec{r}=0.
+\]
+Reversing the limits changes the sign of the second integral, giving
+\[
+\int_{C_1} \vec{F}_c\cdot d\vec{r}=\int_{C_2} \vec{F}_c\cdot d\vec{r}.
+\]
+So the work depends only on the endpoints. Therefore the quantity
+\[
+U_f-U_i=-\int_i^f \vec{F}_c\cdot d\vec{r}
+\]
+is path independent and is a well-defined potential-energy difference.}
+
+\qs{Worked example}{A block of mass $m=0.50\,\mathrm{kg}$ is attached to an ideal horizontal spring of spring constant $k=200\,\mathrm{N/m}$ on a frictionless track. Let $x$ denote the block's displacement from the spring's equilibrium position, with positive $x$ to the right. Initially the block is held at rest at $x_i=+0.15\,\mathrm{m}$ and then released. Find:
+
+\begin{enumerate}
+    \item the change in spring potential energy $\Delta U_s$ as the block moves to $x_f=0$,
+    \item the work done by the spring during that motion, and
+    \item the block's speed $v_f$ when it passes through equilibrium.
+\end{enumerate}}
+
+\sol For an ideal spring, the spring force is
+\[
+\vec{F}_s=-kx\,\hat{\imath}.
+\]
+Since the motion is one-dimensional, the potential-energy relation gives
+\[
+F_x=-\frac{dU_s}{dx}.
+\]
+So
+\[
+-kx=-\frac{dU_s}{dx}
+\qquad \Rightarrow \qquad
+\frac{dU_s}{dx}=kx.
+\]
+Integrate with respect to $x$:
+\[
+U_s(x)=\int kx\,dx=\tfrac12 kx^2+C.
+\]
+Choose the usual reference $U_s=0$ at $x=0$, so $C=0$ and
+\[
+U_s(x)=\tfrac12 kx^2.
+\]
+
+At the initial position,
+\[
+U_{s,i}=\tfrac12 (200)(0.15)^2=100(0.0225)=2.25\,\mathrm{J}.
+\]
+At the final position $x_f=0$,
+\[
+U_{s,f}=\tfrac12 (200)(0)^2=0.
+\]
+Therefore the change in spring potential energy is
+\[
+\Delta U_s=U_{s,f}-U_{s,i}=0-2.25\,\mathrm{J}=-2.25\,\mathrm{J}.
+\]
+
+Because the spring force is conservative,
+\[
+W_s=-\Delta U_s=+2.25\,\mathrm{J}.
+\]
+So the spring does positive work on the block as the spring relaxes toward equilibrium.
+
+The track is frictionless, so the spring is the only force doing work on the block in the horizontal direction. Thus
+\[
+W_{\text{net}}=\Delta K.
+\]
+Since the block starts from rest,
+\[
+K_i=0,
+\qquad
+K_f=W_{\text{net}}=2.25\,\mathrm{J}.
+\]
+Then
+\[
+\tfrac12 mv_f^2=2.25.
+\]
+Substitute $m=0.50\,\mathrm{kg}$:
+\[
+\tfrac12 (0.50)v_f^2=2.25
+\qquad \Rightarrow \qquad
+0.25v_f^2=2.25
+\qquad \Rightarrow \qquad
+v_f^2=9.0.
+\]
+Hence
+\[
+v_f=3.0\,\mathrm{m/s}.
+\]
+
+So the results are
+\[
+\Delta U_s=-2.25\,\mathrm{J},
+\qquad
+W_s=+2.25\,\mathrm{J},
+\qquad
+v_f=3.0\,\mathrm{m/s}.
+\]