Mistake Master
Newton's second law
Topic 2.4 named the no-net-force case: the velocity stays put. Newton's second law names what happens when the net force isn't zero. The center of mass accelerates, in the direction of the net force, with size $|\vec F_\text{net}|/m$. That's the whole law. The rest is reading it carefully.
§1
What Newton's second law actually says.
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When the net force on a system isn't zero, the system's center of mass accelerates.
Topic 2.4 named the default state. Zero net force means constant velocity, no exceptions. This topic asks the next question. Suppose the net force is nonzero. What happens?
The answer is Newton's second law. The system's center of mass accelerates, in the direction of the net force, with size proportional to the net force and inversely proportional to the mass. Both pieces matter, the size and the direction, and the next two sections take them in turn.
The figure shows a block under a single horizontal force. Pay attention to the two arrows above it. They look similar but they are different physical things. The blue arrow is the velocity, the pink arrow is the acceleration, and Newton's second law is the rule that ties one to the force.
§2
The equation, and what proportional means.
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In symbols:
$$\vec a = \dfrac{\vec F_\text{net}}{m}$$
Read the equation as a recipe. Add up the external forces on the system as vectors. The result is $\vec F_\text{net}$. Divide by the system's mass. The result is $\vec a$, with both a size and a direction.
The "proportional and inversely proportional" language is just the equation in words. If you double the net force at fixed mass, $\vec a$ doubles. If you double the mass at fixed force, $\vec a$ halves. If you flip the force around, $\vec a$ flips with it.
The thing to be careful about, the source of the most-tested misconception in this topic, is what scales with $\vec F_\text{net}$. It is not velocity. The velocity is the running total: it changes whenever the object accelerates. What scales with $\vec F_\text{net}$ is the acceleration. A constant force gives a constant acceleration, which is to say, a velocity that climbs without bound. Push for a second and you've added a little to $v$. Push for ten seconds and you've added ten times that little. The push doesn't tap out.
$\vec F_\text{net}$ scales $\vec a$, not $\vec v$. A steady force doesn't hold a steady speed. It builds speed at a steady rate.
§3
Direction: $\vec a$ points where $\vec F_\text{net}$ points.
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$\vec a$ points where $\vec F_\text{net}$ points. That is the directional half of the law. It says nothing about where the velocity points.
Take a ball you toss straight up. While the ball rises, $\vec v$ points up. While it falls, $\vec v$ points down. Gravity, the only force on it, points down the entire time. So $\vec a$ points down the entire time, including on the way up.
This trips students because rising and falling feel like such different motions that the underlying physics seems like it should flip too. It doesn't. The same downward $g$ acts on the ball at every instant, including at the top of the arc when $\vec v$ is exactly zero. The acceleration is still $g$ downward right then. (Topic 1.4 made the same point about velocity and acceleration being independent vectors; if it's still feeling slippery, that's the lesson to revisit.)
Read the problem for what the velocity is doing (speeding up, slowing, reversing, zero), not which way it's pointing. The doing tells you about $\vec a$, and therefore about $\vec F_\text{net}$. The pointing tells you about $\vec v$ alone.
§4
Worked example: block on a frictionless incline.
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Setup. A block of mass $m$ rests on a frictionless ramp inclined at angle $\theta$ from horizontal. The block is released from rest. Find the size of the acceleration along the slope.
The trick is the choice of axes. Pick $+x'$ along the slope (positive going down-slope, the way the block will move) and $+y'$ perpendicular to the slope (positive pointing out of the slope). With those axes, the normal force lies entirely along $+y'$, and gravity has both an $x'$ and a $y'$ component.
The decomposition uses the angle $\theta$ in the standard way. The component of $\vec F_g$ pointing into the slope (along $-y'$) has size $mg\cos\theta$, and the component pointing down the slope (along $+x'$) has size $mg\sin\theta$. (If you haven't done this kind of decomposition before, sketch it: drop a perpendicular from the tip of $\vec F_g$ onto the slope direction. The two right triangles you get out have the angle $\theta$ in matching positions, and the trig comes out as stated.)
Now apply $\vec a = \vec F_\text{net}/m$ along each axis.
The block doesn't lift off the slope or sink into it, so $a_{y'} = 0$. The forces along $y'$ are $+F_N$ (out of slope) and $-mg\cos\theta$ (into slope). Newton's second law in $y'$:
$$F_N - mg\cos\theta = 0 \quad\Longrightarrow\quad F_N = mg\cos\theta.$$
Nothing balances the $mg\sin\theta$ component along the slope. The only force along $x'$ is $+mg\sin\theta$. Newton's second law in $x'$:
$$mg\sin\theta = ma_{x'} \quad\Longrightarrow\quad a_{x'} = g\sin\theta.$$
If $\theta = 0$ the slope is flat. $\sin 0 = 0$, so $a_{x'} = 0$. The block doesn't accelerate on a level surface. If $\theta = 90°$ the slope is vertical. $\sin 90° = 1$, so $a_{x'} = g$. Free fall, with the slope pretending to be there. Both make sense.
Notice what happened to the mass. The acceleration $g\sin\theta$ doesn't depend on $m$. A bowling ball and a marble released side-by-side on the same ramp slide with the same acceleration. (This is a general feature of gravity: more mass means more gravitational pull and equally more inertia, and the two cancel in $\vec a$.)
§5
Worked example: an elevator that accelerates upward.
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Setup. Mei (mass $m$) stands on a bathroom scale inside an elevator. The elevator is accelerating upward at rate $a$. The scale reads the size of the contact force pressing up on Mei. Find that reading.
Two real forces act on Mei. Gravity, $F_g = mg$, pulls down. The scale (which is just the part of the elevator floor in contact with Mei) pushes up with normal force $F_N$. There's no friction here, no tension on Mei, no other contact forces. (Mei is also in contact with the air, but air drag at standing rest inside an elevator is negligible.)
Pick $+y$ upward. Then gravity contributes $-mg$ and the normal contributes $+F_N$. Apply Newton's second law along $y$, with the elevator's acceleration $a$ also pointing up:
$$\sum F_y = F_N - mg = ma$$
$$F_N = m(g + a)$$
Set $a = 0$ and recover $F_N = mg$, the elevator-at-rest answer from Topic 2.4. Set $a = -g$ (the elevator falling freely) and recover $F_N = 0$. Both make sense.
The scale reads $m(g+a)$, which is more than $mg$. While the elevator accelerates upward, Mei feels heavier than usual.
The colloquial name for $F_N$ in this kind of problem is apparent weight. It's a useful phrase as long as you remember what it means. The scale reading is the size of the contact force on Mei, which equals $mg$ only when the acceleration is zero. With the elevator accelerating up, the scale reads more. With it accelerating down (or decelerating while moving up, which is the same thing), the scale reads less. In free fall, the scale reads zero. None of that changes Mei's actual mass. It only changes the contact force.
Don't write $F_N = mg$ as a rule. Derive $F_N$ from Newton's second law for the specific situation, every time. On level ground at rest you get $F_N = mg$. On a ramp you get $F_N = mg\cos\theta$. In an accelerating elevator you get $F_N = m(g+a)$. Same procedure, different answer.
§6
Worked example: Atwood machine.
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Setup. Two blocks hang from a light, inextensible cord that runs over an ideal (frictionless, massless) pulley. The blocks have masses $m_1$ and $m_2$, with $m_2 > m_1$ (so block 2 is the heavier one). The system is released from rest. Find the size of the acceleration of each block and the tension in the cord.
The new piece in this problem is the constraint between the two blocks. The cord is inextensible, so as block 2 moves down by some amount, block 1 moves up by the same amount. That means the two blocks have the same magnitude of velocity (and therefore the same magnitude of acceleration) at every instant. Call this common magnitude $a$.
Sign convention. Let $+x$ be the direction of motion of the system, so block 1 going up and block 2 going down both count as $+x$. With this choice, both blocks have $a_x = +a$, and the whole system's motion is described by a single number.
The cord pulls up on each block with tension $T$. The same $T$ on both sides: that is what ideal pulley means (massless, frictionless, just redirects the cord without adding or losing tension).
Now apply Newton's second law to each block separately.
Forces on block 1: $T$ up, $m_1 g$ down. The $+x$ direction for block 1 is upward, so $T$ contributes $+T$ and gravity contributes $-m_1 g$:
$$T - m_1 g = m_1 a$$
Forces on block 2: $T$ up, $m_2 g$ down. The $+x$ direction for block 2 is downward, so gravity contributes $+m_2 g$ and tension contributes $-T$:
$$m_2 g - T = m_2 a$$
Two equations, two unknowns ($a$ and $T$). Add them to eliminate $T$:
$$(m_2 - m_1) g = (m_1 + m_2) a \quad\Longrightarrow\quad a = \dfrac{m_2 - m_1}{m_1 + m_2}\, g$$
Substitute back into the block-1 equation to get $T$:
$$T = m_1(g + a) = \dfrac{2 m_1 m_2}{m_1 + m_2}\, g$$
Three limits, all of which should match physical intuition. If $m_1 = m_2$ (equal masses), $a = 0$ and $T = mg$: the system sits in equilibrium and each block hangs at its own weight. If $m_1 \to 0$ (one side massless), $a \to g$ and $T \to 0$: block 2 is in free fall and the cord goes slack. If $m_2 \gg m_1$ (one side enormous), $a \to g$ and $T \to 2 m_1 g$: block 1 gets yanked upward at $g$ and the cord tension is twice block 1's weight. All three pass.
The pattern generalizes. In any connected-bodies problem, the cord constraint gives you a single $a$ that all the bodies share, and Newton's second law on each body separately gives one equation per body. Always free-body each body alone, never the combined system: the internal tension cancels in a system FBD and you lose the information needed to find $T$.
§7
Three mistakes that cost real points.
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Three failure modes show up over and over on Newton's-second-law problems. Each one has a family of distractors built around it on the AP exam.
"If the force is steady, the velocity is steady."
This is the impetus instinct in its purest form. A constant push feels like it should hold a constant speed. Take the push away and the object slows; press harder and it speeds up; press at a steady level and it cruises. Real-world experience reinforces the intuition because friction and drag almost always conspire to make a steady push produce a steady terminal speed. The intuition is wrong, but it's earned.
Fix. The thing that scales with force is acceleration, not velocity. $\vec a = \vec F_\text{net}/m$. A constant net force gives a constant acceleration, which means the velocity grows linearly with time. If you see a problem where a steady applied force is paired with a steady velocity, friction or drag is balancing the push, so the net force is zero and the first law is what's running the show.
"Net force points the way the object is moving."
A natural shortcut. See motion, infer force in the direction of motion. It works for the simple case where an object accelerates from rest, because there $\vec v$ and $\vec a$ end up pointing the same way. It fails the moment the object slows, reverses, or moves through an extreme like the top of a toss.
Fix. $\vec F_\text{net}$ points where $\vec a$ points, which is not the same as where $\vec v$ points. If the object is slowing, $\vec a$ opposes $\vec v$, so $\vec F_\text{net}$ opposes $\vec v$. At the top of a toss, $\vec v$ is momentarily zero but $\vec a$ (and therefore $\vec F_\text{net}$) is still $g$ downward. Read the problem for what the velocity is doing, not which way it's pointing.
"Normal force always equals $mg$."
True for an object at rest on level ground. False everywhere else, in three predictable cases: an inclined surface, an accelerating elevator (or any vertical acceleration), and a situation where another applied force has a vertical component. The shortcut $F_N = mg$ gets the right answer often enough that students stop deriving it, and then the AP exam finds the cases where it's wrong.
Fix. Never quote $F_N = mg$ as a rule. Write Newton's second law along the vertical (or perpendicular) axis for the specific setup, then solve for $F_N$. On level ground at rest, $\sum F_y = 0$ gives $F_N = mg$. On a ramp, $\sum F_{y'} = 0$ along the slope-perpendicular axis gives $F_N = mg\cos\theta$. In an accelerating elevator, $\sum F_y = ma$ gives $F_N = m(g+a)$. Same procedure each time.