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content(warnings): Add W6-W7, X2-X3, N4 — energy misconceptions + cross-refs

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Krishna Ayyalasomayajula
2026-05-04 23:36:54 -05:00
parent ddabfdf79f
commit 1ee990373b
4 changed files with 10 additions and 0 deletions
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concepts/mechanics/u2/m2-7-drag-terminal.tex
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@@ -36,6 +36,8 @@ In particular, if the object is released from rest, then
v(t)=v_T\left(1-e^{-bt/m}\right). v(t)=v_T\left(1-e^{-bt/m}\right).
\]} \]}
\nt{The linear drag model $F_D=bv$ is valid for low-Reynolds-number flow (small objects, slow speeds, viscous fluids). For most everyday situations (baseballs, skydivers, cars at highway speed), drag is approximately quadratic: $F_D=\tfrac12 C\rho A v^2$, where $C$ is the drag coefficient, $\rho$ is the fluid density, and $A$ is the cross-sectional area. The AP exam typically uses the linear model for analytical tractability, but recognizes the quadratic model qualitatively.}
\pf{Short derivation of the velocity function}{Start with Newton's second law for vertical fall in the downward-positive direction: \pf{Short derivation of the velocity function}{Start with Newton's second law for vertical fall in the downward-positive direction:
\[ \[
m\frac{dv}{dt}=mg-bv. m\frac{dv}{dt}=mg-bv.

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concepts/mechanics/u3/m3-1-work.tex
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@@ -19,6 +19,8 @@ Therefore the \emph{total work} done by the force as the particle moves along th
W=\int_C \vec{F}\cdot d\vec{r}=\int_C F_{\parallel}\,ds. W=\int_C \vec{F}\cdot d\vec{r}=\int_C F_{\parallel}\,ds.
\]} \]}
This definition of work is the foundation for the work-energy theorem (Section 3.2) and the potential-energy formalism (Section 3.3).
\thm{Line-integral form of work and the constant-force special case}{Let a particle move from an initial point to a final point along a path $C$ under a force $\vec{F}$. Then the work done by that force is \thm{Line-integral form of work and the constant-force special case}{Let a particle move from an initial point to a final point along a path $C$ under a force $\vec{F}$. Then the work done by that force is
\[ \[
W=\int_C \vec{F}\cdot d\vec{r}. W=\int_C \vec{F}\cdot d\vec{r}.

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concepts/mechanics/u3/m3-2-work-energy.tex
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@@ -35,6 +35,8 @@ W_{\text{net}}=\Delta K=K_f-K_i=\tfrac12 mv_f^2-\tfrac12 mv_i^2.
\] \]
Thus the net work done on a body equals the change in its kinetic energy.} Thus the net work done on a body equals the change in its kinetic energy.}
\wc{Net work is the algebraic sum}{Net work $W_{\mathrm{net}}=W_1+W_2+\cdots$ includes the \emph{signs} of individual works. If one force does $+120\,\mathrm{J}$ and another does $-24\,\mathrm{J}$, the net work is $+96\,\mathrm{J}$, \emph{not} $144\,\mathrm{J}$.}
\nt{The work-energy theorem is often more efficient than combining Newton's second law with kinematics when the problem asks for a speed after a known displacement or after a known amount of work. It avoids solving for time and often avoids solving for acceleration explicitly. In AP mechanics, \emph{net work} means the algebraic sum of the work done by all forces on the chosen system. A force parallel to the displacement does positive work, a force opposite the displacement does negative work, and a force perpendicular to the displacement does zero work.} \nt{The work-energy theorem is often more efficient than combining Newton's second law with kinematics when the problem asks for a speed after a known displacement or after a known amount of work. It avoids solving for time and often avoids solving for acceleration explicitly. In AP mechanics, \emph{net work} means the algebraic sum of the work done by all forces on the chosen system. A force parallel to the displacement does positive work, a force opposite the displacement does negative work, and a force perpendicular to the displacement does zero work.}
\pf{Short derivation from Newton II}{Let $m$ denote the constant mass of the body, let $\vec{v}$ denote its velocity, and let $d\vec{r}$ denote its infinitesimal displacement. Start with Newton's second law, \pf{Short derivation from Newton II}{Let $m$ denote the constant mass of the body, let $\vec{v}$ denote its velocity, and let $d\vec{r}$ denote its infinitesimal displacement. Start with Newton's second law,

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concepts/mechanics/u3/m3-3-potential-energy.tex
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@@ -17,6 +17,8 @@ Thus the work done by the conservative force is
W_c=-\Delta U. W_c=-\Delta U.
\]} \]}
\wc{Potential energy belongs to a system}{Gravitational $U=mgh$ requires both the object \emph{and} the Earth. Spring $U=\tfrac12 kx^2$ requires both the block \emph{and} the spring. An isolated single object cannot have potential energy --- it is stored in the \emph{interaction}.}
\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: \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:
\[ \[
\oint \vec{F}_c\cdot d\vec{r}=0, \oint \vec{F}_c\cdot d\vec{r}=0,
@@ -146,3 +148,5 @@ W_s=+2.25\,\mathrm{J},
\qquad \qquad
v_f=3.0\,\mathrm{m/s}. v_f=3.0\,\mathrm{m/s}.
\] \]
The same spring restoring force $-kx$ governs the simple harmonic motion studied in Unit 7 (Sections 7.1--7.4).