diff --git a/concepts/mechanics/u2/m2-7-drag-terminal.tex b/concepts/mechanics/u2/m2-7-drag-terminal.tex
index 1e6e37e..3aabed1 100644
--- a/concepts/mechanics/u2/m2-7-drag-terminal.tex
+++ b/concepts/mechanics/u2/m2-7-drag-terminal.tex
@@ -10,6 +10,10 @@ The negative sign means that the drag force always points opposite the velocity.
 
 If an object falls vertically through the fluid and eventually moves with constant velocity, that steady velocity is called the \emph{terminal velocity}. At terminal velocity, the net force is zero, so the acceleration is zero.}
 
+\wc{Terminal velocity is not zero speed}{Terminal velocity is a \emph{nonzero constant speed}: the object is still moving, but its acceleration is zero because drag balances the driving force. The object approaches $v_T$ asymptotically as $t\to\infty$; it never stops.}
+
+\wc{Drag force is not kinetic friction}{Drag $F_D=bv$ or $F_D=\tfrac12 C\rho A v^2$ is velocity-dependent and acts in fluids. Kinetic friction $f_k=\mu_k N$ is approximately velocity-independent and acts at solid-solid contacts. They are physically distinct interactions.}
+
 \thm{Vertical-fall ODE and terminal-speed result}{Choose the vertical axis positive downward. Let $v(t)$ denote the downward velocity of an object of mass $m$ at time $t$, let $g$ denote the gravitational field strength, and let $b>0$ denote the linear-drag coefficient. Then the vertical equation of motion is
 \[
 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 5979134..ddd862b 100644
--- a/concepts/mechanics/u3/m3-1-work.tex
+++ b/concepts/mechanics/u3/m3-1-work.tex
@@ -39,6 +39,8 @@ W=F\,\Delta s.
 
 \nt{Work is positive when the force has a component in the same direction as the displacement, negative when that component is opposite the displacement, and zero when the force is perpendicular to the displacement. For a general force, the value of $W=\int_C \vec{F}\cdot d\vec{r}$ can depend on the path $C$, not just on the endpoints. In AP problems, this often appears when the force changes with position or when different paths make the parallel component $F_{\parallel}$ different. Typical zero-work cases include a normal force on motion along a surface or a centripetal force in uniform circular motion, because those forces are perpendicular to the instantaneous displacement.}
 
+\wc{Holding something still does no work}{If you hold a heavy object stationary, you expend biological energy but do \emph{zero mechanical work} because $\Delta\vec{r}=\vec{0}$. Work requires both a force \emph{and} a displacement along that force: $W=\int\vec{F}\cdot d\vec{r}$.}
+
 \pf{Why the line integral gives total work}{Break the path into many small displacement vectors $\Delta \vec{r}_1,\Delta \vec{r}_2,\dots,\Delta \vec{r}_n$. Over each small segment, the work is approximately
 \[
 \Delta W_k\approx \vec{F}_k\cdot \Delta \vec{r}_k.
diff --git a/concepts/mechanics/u6/m6-4-rolling.tex b/concepts/mechanics/u6/m6-4-rolling.tex
index 8d194e3..ce51953 100644
--- a/concepts/mechanics/u6/m6-4-rolling.tex
+++ b/concepts/mechanics/u6/m6-4-rolling.tex
@@ -41,6 +41,8 @@ Thus the wheel's center moves at $2.4\,\mathrm{m/s}$, while the top point moves
 4. For a body rolling without slipping down an incline of angle $\beta$, choose positive down the incline. Let $a_{\mathrm{cm}}$ denote the center-of-mass acceleration magnitude along the incline. If $f_s$ denotes the magnitude of the static friction force, then $Mg\sin\beta-f_s=Ma_{\mathrm{cm}}$, $f_sR=I_{\mathrm{cm}}\alpha$, $a_{\mathrm{cm}}=R\alpha$. Therefore, $a_{\mathrm{cm}}=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)}$, $f_s=\frac{I_{\mathrm{cm}}}{R^2}a_{\mathrm{cm}}$.
 For an object accelerating down the incline, the static friction force on the object points up the incline.}
 
+\nt{When several objects roll without slipping down the same incline from the same height, the one with the smallest $I_{\mathrm{cm}}/(MR^2)$ ratio reaches the bottom first. Race order: solid sphere ($\tfrac25=0.40$) $>$ solid cylinder ($\tfrac12=0.50$) $>$ hollow cylinder ($\simeq1$) $>$ thin hoop ($1.0$). Mass and radius do not appear in the comparison --- only the shape-dependent coefficient matters.}
+
 \qs{Worked example}{A solid cylinder of mass $M=2.0\,\mathrm{kg}$ and radius $R=0.20\,\mathrm{m}$ is released from rest on an incline that makes an angle $\beta=30^\circ$ with the horizontal. The cylinder rolls without slipping through a vertical drop $h=0.75\,\mathrm{m}$. Let $g=9.8\,\mathrm{m/s^2}$.
 
 Find:
diff --git a/concepts/mechanics/u7/m7-3-shm-energy.tex b/concepts/mechanics/u7/m7-3-shm-energy.tex
index de8fbb2..8aa6ca8 100644
--- a/concepts/mechanics/u7/m7-3-shm-energy.tex
+++ b/concepts/mechanics/u7/m7-3-shm-energy.tex
@@ -151,3 +151,5 @@ v=1.60\,\mathrm{m/s},
 \qquad
 v_{\max}=2.0\,\mathrm{m/s}\text{ at }x=0.
 \]
+
+\nt{In simple harmonic motion, total mechanical energy $E = K + U$ is constant. At maximum displacement ($x=\pm A$), $K=0$ and $U=E$ (all energy is potential). At equilibrium ($x=0$), $U=0$ and $K=E$ (all energy is kinetic). At $x=\pm A/\sqrt{2}$, $K=U=E/2$. The energy exchange oscillates at twice the frequency of the motion itself.}