Lesson 5: Newton's Laws of MotionAP

Every unit so far described motion; CED Unit 2 (Dynamics) explains what causes it to change: forces. A force is an interaction between two objects — a push or pull with an agent, a receiver, a magnitude and a direction. Newton's three laws are the complete operating system for classical mechanics, and nearly every AP question in Units 2–3 is one of them in disguise.

Newton's first law and inertia

An object with no net force on it keeps its velocity — rest stays at rest, motion stays straight-line and steady. This is not "objects naturally stop"; it is the statement that changes in velocity need a net force. Inertia is the property behind it, measured by mass: the more mass, the harder the velocity is to change. The first law also defines inertial reference frames — the frames in which the law holds (the ground, usually; an accelerating car, no).

Newton's second law

The quantitative heart: F_net = ma, applied component-wise: ΣF_x = ma_x, ΣF_y = ma_y. Note what it says and does not say. It does not say force causes velocity; it says the vector sum of forces causes acceleration, in the same direction as the net force. A 10 N push balanced by 10 N of friction produces no acceleration at all. The unit of force, the newton, is defined by this law: 1 N = 1 kg·m/s². Mass (kg) is intrinsic; weight is a force — gravity's pull, F_g = mg, measured in newtons. A 60 kg astronaut has mass 60 kg everywhere, but weighs about 590 N on Earth and 95 N on the Moon.

Newton's third law

Forces come in pairs: if A pushes B, B pushes A with an equal force in the opposite direction. The pair acts on different objects, so the two forces never cancel each other in a single free-body diagram. The pair members are the same type of interaction (both gravitational, both contact, both electric). The classic trap: a book resting on a table — gravity (Earth pulls book down) and the normal force (table pushes book up) are not a third-law pair, because they act on the same object and are different interaction types. Gravity's real partner is the book pulling the Earth upward.

Solving dynamics problems: the system approach

The CED expects you to choose a system deliberately. Treat connected objects as one system and internal forces (like the tension between them) vanish from the analysis; analyse one object and they appear. The standard routine:

• Sketch the situation and choose axes (tilt them along an incline).
• Draw a free-body diagram for each object: only forces acting on it, drawn at its centre.
• Write ΣF = ma in components for each object.
• Solve the simultaneous equations; check signs against the diagram.

Worked example 1 — net force with friction

A 5.0 kg crate is pushed across a level floor with a horizontal force of 30 N. Kinetic friction opposes the motion with 10 N. Find the acceleration. Given: m = 5.0 kg, push = +30 N, friction = −10 N, vertical forces cancel (N = mg, no vertical acceleration). Equation: ΣF_x = ma_x.

ΣF_x = 30 − 10 = 20 N
a = ΣF_x / m = 20 / 5.0 = 4.0 m/s² in the direction of the push

Check: units N/kg = m/s² ✓. If friction equalled 30 N the acceleration would be zero even though the crate might still be sliding — force balance means constant velocity, not rest.

Worked example 2 — the elevator and apparent weight

A 60 kg person stands on a scale in an elevator accelerating upward at 2.0 m/s² (use g = 10 m/s²). What does the scale read? Given: forces on the person: gravity −mg = −600 N (down) and normal force N (up). Equation: ΣF_y = ma_y with a_y = +2.0.

N − 600 = 60 × 2.0 → N = 600 + 120 = 720 N

The scale reads the normal force, 720 N — you feel "heavier". Accelerating downward at 2.0 m/s² gives N = 600 − 120 = 480 N. In free fall (a = −10), N = 0: apparent weightlessness is a normal force of zero, not an absence of gravity — an idea Lesson 8 needs for orbiting astronauts. Check: for the stationary elevator N = 600 N = mg, recovering the familiar case between the two accelerating ones.

Real-world application: why seatbelts and headrests exist

In a crash, a car's velocity changes by ~15 m/s in about 0.1 s, requiring a ≈ 150 m/s² of anything rigidly attached. The seatbelt's job is to make the passenger decelerate with the car over the crumple time rather than continue at 15 m/s (first law) until something else supplies the force. Headrests exist because of the third law/first law combination: the seat pushes your torso forward, and without a headrest your head's inertia leaves it behind, injuring the neck.

Experimental design note

The canonical Unit 2 investigation measures acceleration while varying net force on a fixed mass (cart pulled by hanging weights), then while varying mass at fixed force. Plot a against F_net: a straight line through the origin with slope 1/m confirms the second law. Declare your system honestly: if the hanging mass is outside the system, the tension is an external force and is slightly less than the hanging weight — a subtlety the College Board rewards.

Common mistakes (from College Board scoring commentary)

• Naming gravity and the normal force as a Newton's-third-law pair.
• Writing "F = ma" for a single force rather than the net force.
• Believing a moving object must have a net force in its direction of motion.
• Drawing forces the object exerts on others in its own free-body diagram.

AP Exam Experience

MCQ 1. A 4.0 kg object experiences forces of 6.0 N east and 8.0 N north. The magnitude of its acceleration is:

• (A) 1.5 m/s²
• (B) 2.0 m/s²
• (C) 2.5 m/s²
• (D) 3.5 m/s²

Answer: (C). The forces are perpendicular: F_net = √(36 + 64) = 10 N, so a = 10/4.0 = 2.5 m/s². Adding the forces as scalars (14 N) is the distractor path to (D).

MCQ 2. A book rests on a table. The Newton's-third-law partner of the gravitational force on the book is:

• (A) The normal force from the table on the book
• (B) The force of the book pushing down on the table
• (C) The gravitational force of the book pulling up on the Earth
• (D) The friction force from the table on the book

Answer: (C). Third-law pairs are the same interaction with the two objects swapped: Earth pulls book ↔ book pulls Earth. (A) acts on the same object and is a different interaction type; (B) pairs with the table's normal force on the book.

MCQ 3. An object moves at constant velocity in a straight line. Which must be true?

• (A) No forces act on it
• (B) The net force on it is zero
• (C) All forces on it are individually zero
• (D) It has no acceleration because it has no forces

Answer: (B). Constant velocity means a = 0, and the second law then requires only that the vector sum of forces is zero; balanced forces are perfectly possible and common.

FRQ. A 2.0 kg block on a horizontal frictionless surface is pulled by a horizontal force of 12 N while a second force of unknown magnitude acts at 60° above the horizontal in the opposite horizontal sense, so the block accelerates at 2.0 m/s² in the direction of the 12 N force. (The opposing force also has an upward component; the surface is strong enough that the block stays on it.) (a) Draw a free-body diagram of the block, labelling all forces. (b) Determine the magnitude of the unknown force. (c) Determine the normal force from the surface (g = 10 m/s²). (d) State how the acceleration would change if the angle of the unknown force were decreased toward horizontal, keeping its magnitude fixed, and justify your answer. Scoring guideline. (a) 1 point: weight down, normal force up, 12 N pull one way, unknown force up-and-opposite; no extra forces. (b) 2 points: ΣF_x = 12 − F cos60° = ma = 4.0, so F cos60° = 8.0 and F = 16 N (1 point setup, 1 point value). (c) 1 point: ΣF_y = 0: N + F sin60° − mg = 0 → N = 20 − 16 × 0.87 ≈ 6.1 N. (d) 1 point: acceleration decreases, because a smaller angle increases the opposing horizontal component F cosθ, reducing the net force in the pull direction.

Worked example 3 — contact forces between blocks

Two blocks, 2.0 kg (A) and 3.0 kg (B), sit in contact on a frictionless floor. A horizontal 20 N force pushes A into B. Find (a) the acceleration of the pair, (b) the force A exerts on B. Method: first treat A + B as one system, then isolate B.

(a) a = F_net/(m_A + m_B) = 20/5.0 = 4.0 m/s²
(b) For B alone, the only horizontal force is A's push: F_AB = m_B a = 3.0 × 4.0 = 12 N

Check: for A alone: 20 − F_BA = 2.0 × 4.0 → F_BA = 12 N — equal and opposite to F_AB, the third law satisfied inside the arithmetic. The contact force (12 N) is not the applied 20 N; internal forces redistribute according to mass. If the push had been applied to B instead, the contact force would be m_A a = 8.0 N — same system acceleration, different internal force: a favourite exam discrimination.

Choosing the system: the meta-skill

Notice what happened: the two-step rhythm "system first for acceleration, isolate for internal forces" solves Atwood machines, connected blocks, and elevators carrying passengers with identical logic. The College Board's system concept (CED: the system schema) is really this: internal forces cancel in pairs, external forces do not. When a problem asks for an interaction between parts, you must cut the system there; when it asks only for the motion of the whole, keep the system whole and save the algebra.