Topic 2.5: Newton's Second Law, Part 1 — Single-Object Systems

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Before We Start: Yesterday's Recap

Yesterday, equilibrium meant $F_N = mg$ for someone standing still. What do you think has to change about that equation if the person is accelerating instead of standing still?

Today's New Concepts

AP CED Alignment: Unit 2, Topic 2.5 (Part 1 of 2)

Newton's Second Law

The net force on an object equals its mass times its acceleration
Symbol: $F_{net} = ma$, or $a = \dfrac{F_{net}}{m}$
Units: force in newtons (N), mass in kilograms (kg), acceleration in meters per second squared (m/s²)
Equilibrium (Topic 2.4) is the special case where $a = 0$ — this is the same law, just the general case where $a$ can be anything

Worked Example: Standard process for any Newton's-second-law problem: draw the FBD, write $\Sigma F$ symbolically along the axis of interest with correct signs, set it equal to $ma$, solve symbolically for the unknown, THEN substitute numbers.

Rocket with constant thrust and shrinking mass: the same F_thrust arrow produces a bigger acceleration arrow once fuel has burned off, since a = F/m

Elevator and Scale Problems

Choosing up as positive: $F_N - mg = ma \implies F_N = m(g+a)$
If accelerating upward ($a>0$): $F_N > mg$ (feels heavier)
If accelerating downward ($a<0$): $F_N < mg$ (feels lighter)
The scale reading tells you the direction of ACCELERATION, not necessarily the direction of motion — the object could be moving either way

Worked Example: A person's scale reads more than their normal weight — this means the elevator (and person) are accelerating upward, but they could be moving upward and speeding up, OR moving downward and slowing down.

FBD comparing a person on a scale accelerating up (F_N greater than F_g, feels heavier) versus accelerating down (F_N less than F_g, feels lighter), up-positive convention

Single-Object Incline Problems

Same rotated-axis strategy as Topic 2.2, now combined with Newton's second law: write $F_{net}$ along the ramp-parallel axis equal to $ma$
On a frictionless incline, if gravity's parallel component is the only force along the ramp, $a = g\sin\theta$ — independent of mass, since mass cancels

Worked Example: A block pulled up a ramp by a rope at an angle, with friction opposing: $F_p\cos\theta_2 - F_f - mg\sin\theta_1 = ma$, solved symbolically for $a$ before any numbers are substituted.

FBD of a block on an incline with an applied force at an angle to the ramp and friction, rotated x'/y' axes, gravity decomposed into parallel and perpendicular components

Keep This Sheet!

These sheets build into your semester study guide. Keep them in order in a binder or folder — you'll want to flip back through them before quizzes, unit tests, and when AP exam review starts in the spring.