From 1ee990373b9b2fd0c478aa08527dc1997f5253dc Mon Sep 17 00:00:00 2001 From: Krishna Ayyalasomayajula Date: Mon, 4 May 2026 23:36:54 -0500 Subject: [PATCH] =?UTF-8?q?content(warnings):=20Add=20W6-W7,=20X2-X3,=20N4?= =?UTF-8?q?=20=E2=80=94=20energy=20misconceptions=20+=20cross-refs?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- concepts/mechanics/u2/m2-7-drag-terminal.tex | 2 ++ concepts/mechanics/u3/m3-1-work.tex | 2 ++ concepts/mechanics/u3/m3-2-work-energy.tex | 2 ++ concepts/mechanics/u3/m3-3-potential-energy.tex | 4 ++++ 4 files changed, 10 insertions(+) diff --git a/concepts/mechanics/u2/m2-7-drag-terminal.tex b/concepts/mechanics/u2/m2-7-drag-terminal.tex index 3aabed1..3ac15a5 100644 --- a/concepts/mechanics/u2/m2-7-drag-terminal.tex +++ b/concepts/mechanics/u2/m2-7-drag-terminal.tex @@ -36,6 +36,8 @@ In particular, if the object is released from rest, then 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: \[ m\frac{dv}{dt}=mg-bv. diff --git a/concepts/mechanics/u3/m3-1-work.tex b/concepts/mechanics/u3/m3-1-work.tex index ddd862b..be61ab3 100644 --- a/concepts/mechanics/u3/m3-1-work.tex +++ b/concepts/mechanics/u3/m3-1-work.tex @@ -19,6 +19,8 @@ Therefore the \emph{total work} done by the force as the particle moves along th W=\int_C \vec{F}\cdot d\vec{r}=\int_C F_{\parallel}\,ds. \]} +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 \[ W=\int_C \vec{F}\cdot d\vec{r}. diff --git a/concepts/mechanics/u3/m3-2-work-energy.tex b/concepts/mechanics/u3/m3-2-work-energy.tex index 4a87047..5edd7f4 100644 --- a/concepts/mechanics/u3/m3-2-work-energy.tex +++ b/concepts/mechanics/u3/m3-2-work-energy.tex @@ -35,6 +35,8 @@ W_{\text{net}}=\Delta K=K_f-K_i=\tfrac12 mv_f^2-\tfrac12 mv_i^2. \] Thus the net work done on a body equals the change in its kinetic energy.} +\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.} \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, diff --git a/concepts/mechanics/u3/m3-3-potential-energy.tex b/concepts/mechanics/u3/m3-3-potential-energy.tex index ce71ca4..9de0e95 100644 --- a/concepts/mechanics/u3/m3-3-potential-energy.tex +++ b/concepts/mechanics/u3/m3-3-potential-energy.tex @@ -17,6 +17,8 @@ Thus the work done by the conservative force is 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: \[ \oint \vec{F}_c\cdot d\vec{r}=0, @@ -146,3 +148,5 @@ W_s=+2.25\,\mathrm{J}, \qquad v_f=3.0\,\mathrm{m/s}. \] + +The same spring restoring force $-kx$ governs the simple harmonic motion studied in Unit 7 (Sections 7.1--7.4).