Unit 2 of AP Physics 1: Forces and Newton's Laws.
This unit covers Newton's three laws, free-body diagrams, friction and net force — essential concepts for AP Physics 1. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers Newton's three laws, free-body diagrams, friction and net force — essential concepts for AP Physics 1. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 Newton's First Law
An object remains at rest or in constant velocity unless acted upon by a net external force. Students must understand that constant velocity (including zero) means zero net force, and that inertia is the tendency of an object to resist changes in its state of motion. The AP exam frequently tests whether students can identify situations of equilibrium versus acceleration.
Key Points
- Zero net force → constant velocity (not necessarily zero velocity)
- Inertia is proportional to mass; more mass = more resistance to change in motion
- A moving object with no net force does NOT slow down — friction must be explicitly present to cause deceleration
- Equilibrium: ΣF = 0, object is either at rest or moving at constant velocity
A book slides across a frictionless surface at 3 m/s. What is the net force on the book?
Since the surface is frictionless, no horizontal force acts on the book after it is released. By Newton's First Law, with zero net force the book continues at 3 m/s indefinitely. The answer is 0 N — a common trap is assuming a moving object must have a force keeping it moving.
2 Newton's Second Law
The net force on an object equals its mass times its acceleration: ΣF = ma. Students must apply this in component form (ΣFx = max, ΣFy = may) and recognize that net force and acceleration always point in the same direction. This is the most heavily tested law on the AP exam, appearing in nearly every mechanics free-response question.
Key Points
- ΣF = ma applies to the net (vector sum) of all forces, not individual forces
- Acceleration is in the same direction as net force — if a is upward, ΣF is upward
- Double the net force → double the acceleration; double the mass → half the acceleration
- In systems of connected objects, treat the system as one object to find acceleration: a = ΣF_net / m_total
Two blocks, 3 kg and 5 kg, are connected by a massless string on a frictionless surface. A 16 N force pulls the 5 kg block. Find the acceleration of the system and the tension in the string.
Treating the system as one 8 kg object: a = 16 N / 8 kg = 2 m/s². To find tension, isolate the 3 kg block: T = ma = (3 kg)(2 m/s²) = 6 N. Students who incorrectly apply ΣF = ma to only one block without isolating it will get the wrong tension.
3 Newton's Third Law
For every action force, there is an equal and opposite reaction force acting on a different object. These force pairs are always equal in magnitude, opposite in direction, and act on different objects — they never cancel each other. The AP exam tests whether students can correctly identify third-law pairs and avoid the misconception that they cancel.
Key Points
- Third-law pairs act on different objects; they cannot be added in the same free-body diagram
- The forces in a pair are always the same type (e.g., both normal forces, both gravitational)
- A heavier object and lighter object exert equal-magnitude forces on each other during collision
- Third-law pairs produce different accelerations if the masses differ (a = F/m)
A 60 kg person stands on a 1000 kg elevator accelerating upward at 2 m/s². What force does the person exert on the elevator floor?
First find the normal force the floor exerts on the person: N − mg = ma → N = m(g + a) = 60(9.8 + 2) = 708 N upward. By Newton's Third Law, the person exerts 708 N downward on the floor — equal in magnitude, opposite in direction, acting on the elevator.
4 Free-Body Diagrams
A free-body diagram (FBD) represents all forces acting on a single object as vectors originating from a point or the object's center. Students must correctly identify all forces present (gravity, normal, tension, friction, applied) and omit forces the object exerts on other things. FBDs are required on AP free-response and graded for completeness and correct direction.
Key Points
- Every force arrow must have a label (e.g., F_g, N, T, f) and point in the correct direction
- Draw forces on ONE object only; do not include forces that object exerts on others
- Weight (F_g = mg) always points straight down toward Earth's center
- On an incline, choose axes parallel and perpendicular to the surface to simplify components
Draw and label the FBD for a 5 kg block sliding down a 30° frictionless incline.
Two forces act on the block: weight (mg = 49 N) straight downward, and normal force perpendicular to the incline surface. Decomposing weight: component along incline = mg sin30° = 24.5 N (down the slope, causing acceleration), component perpendicular = mg cos30° = 42.4 N (balanced by N). Since there is no friction, the net force is 24.5 N down the incline.
5 Friction
Friction is a contact force that opposes relative motion or attempted motion between surfaces. Static friction (f_s ≤ μ_s N) acts when surfaces are not sliding; kinetic friction (f_k = μ_k N) acts when they are. Students must know that static friction is variable up to its maximum and that μ_s > μ_k for any surface pair.
Key Points
- Kinetic friction is constant: f_k = μ_k N; it does not depend on speed or contact area
- Static friction adjusts to match applied force until it reaches f_s(max) = μ_s N, then the object moves
- Normal force N is not always equal to mg — on inclines or with vertical applied forces, recalculate N
- Friction force direction always opposes the direction of motion (kinetic) or impending motion (static)
A 10 kg box sits on a surface with μ_s = 0.5 and μ_k = 0.3. A horizontal force of 40 N is applied. Does the box move, and if so, what is its acceleration? (g = 10 m/s²)
Maximum static friction = μ_s × N = 0.5 × (10)(10) = 50 N. Since the applied force (40 N) is less than 50 N, the box does not move and static friction equals exactly 40 N. If the applied force had been 55 N, kinetic friction would be f_k = 0.3 × 100 = 30 N, giving a = (55 − 30)/10 = 2.5 m/s².
6 Net Force
Net force is the vector sum of all forces acting on an object and is the quantity that determines acceleration via ΣF = ma. Students must add forces as vectors using components, recognizing that forces in opposite directions subtract. The AP exam regularly presents multi-force scenarios requiring component decomposition before applying Newton's Second Law.
Key Points
- Net force is a vector: add x-components separately from y-components
- If net force is zero in all directions, the object is in equilibrium (a = 0)
- A nonzero net force always produces acceleration in the direction of that net force
- In 2D problems, solve ΣFx = max and ΣFy = may independently
Three forces act on an object: 10 N east, 6 N west, and 8 N north. Find the magnitude and direction of the net force.
ΣFx = 10 − 6 = 4 N east; ΣFy = 8 N north. The magnitude of the net force is √(4² + 8²) = √80 ≈ 8.9 N. The direction is arctan(8/4) = arctan(2) ≈ 63° north of east. Students must use vector addition, not simply add the magnitudes, which is the most common error on this type of question.
Questions, answered.
What is Forces and Newton's Laws?
Forces and Newton's Laws is Unit 2 of AP Physics 1, covering Newton's three laws, free-body diagrams, friction and net force.
How to study for AP Physics 1 Unit 2?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 30+ review questions across 5 different game modes.