content(warnings): Add W1-W2, X1, N1, F1 — mechanics misconceptions + cross-refs + notes

concepts/mechanics/u2/m2-1-newton-laws.tex

@@ -32,6 +32,8 @@ This is the main working law for AP mechanics.
 These two forces act on different bodies, so they do not cancel on a single free-body diagram.
 \end{enumerate}}
 
+\wc{Net force causes acceleration, not motion}{A body can move at constant velocity with \emph{zero} net force (Newton's law I). Net force determines \emph{how} motion changes, not \emph{if} motion exists. An object with zero net force does not stop --- it continues at whatever velocity it already had.}
+
 \pf{Why the vector law becomes component equations}{Choose Cartesian axes with unit vectors $\hat{\imath}$ and $\hat{\jmath}$. Let the acceleration be
 \[
 \vec{a}=a_x\hat{\imath}+a_y\hat{\jmath},
@@ -56,6 +58,8 @@ In practice, one first draws only the actual forces on the free-body diagram, th
 
 \cor{Zero net force and equilibrium}{If $\vec{F}_{\mathrm{ext}}=\vec{0}$, then Newton II gives $\vec{a}=\vec{0}$. This does \emph{not} mean the velocity must be zero. A body can have zero net force while moving with a nonzero constant velocity. A special case is equilibrium: if a body is initially at rest and $\vec{F}_{\mathrm{ext}}=\vec{0}$, then it remains at rest.}
 
+See also the kinematics in Unit 1 for the constant-velocity case, and Unit 3 for the work-energy connection to zero net force.
+
 \qs{Worked example}{A block of mass $m=5.0\,\mathrm{kg}$ rests on a frictionless incline that makes an angle $\theta=30^\circ$ with the horizontal. Let $g=9.8\,\mathrm{m/s^2}$ denote the magnitude of the gravitational field near Earth. Choose the system to be the block.
 
 Find the acceleration of the block and the magnitude of the normal force exerted by the incline on the block.}
@@ -114,6 +118,9 @@ Therefore,
 \[
 N=mg\cos\theta.
 \]
+
+\wc{Normal force is not always equal to weight}{On a horizontal surface with only gravity and normal force, $N=mg$. But on an incline $N=mg\cos\theta$, and if another force pushes on the object, $N$ adjusts accordingly. The normal force is \emph{whatever value is required} to prevent the object from penetrating the surface. Never assume $N=mg$.}
+
 Substitute the stated values:
 \[
 N=(5.0\,\mathrm{kg})(9.8\,\mathrm{m/s^2})\cos 30^\circ.