Year 11 Physics Module 2: Dynamics 35 min Lesson 4 of 15

Newton's Laws and Friction

The packages kept moving. The van stopped. Two different objects, one moment — and Newton's Laws explain exactly why.

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Think First

A delivery driver brakes suddenly at 60 km/h. The van stops. The packages in the back fly forward and slam into the cabin. What force caused the packages to move forward — and why didn't the same thing happen to the driver?

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📐

Formula Reference — This Lesson

$F_{net} = ma$  |  $f = \mu F_N$
Fnet = net force (N)  |  m = mass (kg)  |  a = acceleration (m/s²) f = friction force (N)  |  μ = coefficient of friction  |  FN = normal force (N)
Find a: $a = F_{net}/m$   |   Find m: $m = F_{net}/a$   |   Normal force (flat): $F_N = mg$   |   Note: $\mu_s > \mu_k$ always

F=ma
Formula Reference — Newton's Laws and Friction

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$F_{net} = ma$
Newton's Second Law
Fnet = net force (N) | m = mass (kg) | a = acceleration (m/s²). Direction of a is always the same as direction of Fnet.
Applies when: a single object experiences a constant net force
Common trap: Fnet is the net force — not the applied force alone. Always subtract friction and other opposing forces before substituting into F = ma.
$f = \mu F_N$
Friction Force
f = friction force (N) | μ = coefficient of friction (dimensionless) | FN = normal force (N)
Applies when: an object is on a surface — use μs for static friction (not yet sliding), μk for kinetic friction (sliding)
Common trap: μs > μk always. It takes more force to start an object moving than to keep it moving. Never use μk when the object is stationary.
$F_N = mg$
Normal Force (flat surface)
FN = normal force (N) | m = mass (kg) | g = 9.8 m/s²
Applies when: surface is horizontal and no vertical forces other than weight and normal act on the object
Common trap: On a slope, FN = mg cosθ — not mg. On a flat surface with a force applied at an angle downward, FN increases above mg.
$F_{net} = 0 \Rightarrow a = 0$
Newton's First Law (equilibrium condition)
When net force is zero, acceleration is zero — object is at rest or moving at constant velocity.
Applies when: object is stationary or moving at constant velocity in a straight line
Common trap: Constant velocity does not mean no forces — it means forces are balanced. A car cruising at 100 km/h has driving force and friction both acting, cancelling each other.

Know

  • Newton's First Law — formal statement
  • Newton's Second Law — Fnet = ma
  • f = μFN and what μ represents
  • Difference between static and kinetic friction

Understand

  • Why packages fly forward when a van brakes
  • Why F in F = ma is always the net force
  • Why μs > μk and what this means practically
  • Why constant velocity ≠ no forces

Can Do

  • Draw a FBD for a braking or accelerating object
  • Calculate Fnet including friction
  • Find acceleration using Fnet = ma
  • Distinguish Newton 1 from Newton 2 scenarios

Choose how you work — type your answers below or write in your book.

Misconceptions to Fix

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Wrong: Friction always opposes motion and is always unwanted.

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Right: Friction can be static and enable motion (walking, driving); it is essential for many everyday activities.

📚 Core Content

01Newton's First Law — Inertia Revisited

Newton's First Law — Inertia Revisited

An object does not change what it is doing unless something forces it to.

In L01 we met Newton's First Law conceptually. Now we apply it to the delivery van scenario. When the driver brakes, a friction force from the road acts on the van — decelerating it. But the packages? Nothing horizontal acts on them. No seatbelt, no harness. So by Newton's First Law, they continue at the van's original velocity — forward — while the van slows beneath them.

The driver does not fly forward because the seatbelt exerts a backward force on them — providing the net force needed to decelerate with the van. The packages have no such force. They are not "thrown" forward — they simply continue moving forward while the van stops around them.

Van

Horizontal force?: Yes — friction from road (backward)
What happens: Decelerates and stops
Law: Newton 2: Fnet = ma

Packages

Horizontal force?: No horizontal force
What happens: Continue at original velocity
Law: Newton 1: a = 0

Driver (belted)

Horizontal force?: Yes — seatbelt force (backward)
What happens: Decelerates with the van
Law: Newton 2: Fnet = ma
Real-World Anchor Crumple zones in cars work by increasing the time over which deceleration occurs — reducing the force required to decelerate the occupant to rest. This is Newton's Second Law applied to road safety engineering.
02Newton's Second Law

Newton's Second Law — Fnet = ma

The harder you push, the faster it accelerates — and the heavier it is, the less it responds to the same push.

Formal statement: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The direction of acceleration is always the same as the direction of the net force.

Critical point F in Fnet = ma is always the net force — the vector sum of all forces. For the braking van, the only horizontal force is friction (backward). So Fnet = −f, not the braking force applied by the driver's foot. Forgetting to account for all forces is the most common error in Module 2 calculations.
ScenarioMass (kg)Fnet (N)Acceleration (m/s²)
Small car braking900−5400−6.0
Delivery van braking1800−5400−3.0
Delivery van, harder braking1800−10800−6.0

Notice: same force on double the mass gives half the acceleration. Same mass with double the force gives double the acceleration. Fnet = ma captures both relationships simultaneously.

Free Body Diagram — Braking Van

The Excalidraw diagram below shows all forces on the braking van. Note that weight and normal force cancel vertically — the only unbalanced force is friction, acting backward.

Free Body Diagram — Braking Delivery Van
Vector Protocol — apply before every FBD
Step 1 — Positive direction defined: forward = positive
Step 2 — FBD drawn: W down, N up, f backward (negative)
Step 3 — Net force equation: Fnet = −f = ma, solve for a
03Friction — Static vs Kinetic

Friction — Static vs Kinetic

It takes more force to get something moving than to keep it moving — because static friction is stronger than kinetic friction.

Friction arises from microscopic interactions between surfaces. There are two types relevant to this module:

Static friction (fs)

When it acts: Object is stationary — opposing tendency to move
Formula: fs ≤ μs × FN (up to a maximum)
Key feature: Adjusts to match applied force up to maximum — then object slides

Kinetic friction (fk)

When it acts: Object is already sliding
Formula: fk = μk × FN (constant)
Key feature: Constant value — does not depend on speed

For any given surface pair, μs > μk always. The maximum static friction is always greater than kinetic friction. This is why a parked van on a slope stays put — but once it starts rolling, it is harder to stop again.

Van on a slope A delivery van parked on a 10° slope stays stationary because static friction is sufficient to oppose the weight component along the slope (mg sinθ). If the handbrake is released, the van begins to roll — now kinetic friction acts, which is smaller than static friction was, so the van accelerates down the slope.

Common Misconceptions

Friction always opposes motion.
Friction opposes relative sliding motion — not absolute motion. The friction between a car tyre and the road actually propels the car forward. The tyre pushes backward on the road; the road pushes forward on the tyre. Without friction, a car cannot accelerate at all.
Heavier objects experience more friction, so they decelerate faster.
Heavier objects do have more friction (f = μmg), but they also have more mass resisting deceleration. In a = Fnet/m = μmg/m = μg, the mass cancels. On the same surface, all objects decelerate at the same rate regardless of mass.
A smooth surface has zero friction.
No real surface is perfectly frictionless. "Smooth" in physics problems means frictionless as an approximation. In reality, μ approaches zero but never equals it — even ice has μ ≈ 0.03.
Interactive: Newton's Laws Demonstrator
Interactive: Friction Forces (3D)
Navigation: ↗ Drag to rotate  ·  Scroll to zoom  ·  Double-click to reset. Watch how static friction adjusts to match the applied force — up to its maximum value μₛN — before kinetic friction takes over.

✏️ Worked Examples

Worked Example 1 Type 2 — Kinetic: Single Object

Problem Setup

Problem type: Type 2 — Kinetic Single Object. One object decelerating under a constant net force (friction only, horizontal surface).

Scenario: A 1200 kg delivery van decelerates from 60 km/h to rest. The braking friction force is 8000 N. Find the deceleration and the coefficient of kinetic friction.

  • m = 1200 kg
  • vi = 60 km/h = 16.67 m/s, vf = 0
  • f = 8000 N (friction, opposing motion)
  • Positive direction: forward (direction of initial motion)

Solution

1
Positive direction: forward = positive. FBD: W down, N up, f = −8000 N (backward)
Vector Protocol Steps 1 and 2 first — always define direction and draw FBD before writing equations.
2
Fnet = −f = −8000 N
Only horizontal force is friction, acting backward (negative). Net vertical force = 0 since N = W.
3
a = Fnet / m = −8000 / 1200 = −6.67 m/s²
Newton's Second Law: Fnet = ma, rearranged to a = Fnet/m. Negative confirms deceleration (opposing positive direction).
4
FN = mg = 1200 × 9.8 = 11760 N
Normal force on a flat surface equals weight. Needed to find μ.
5
μk = f / FN = 8000 / 11760 = 0.68
Rearrange f = μk × FN to find μk. This is the coefficient of kinetic friction between the tyres and road surface.

What would change if...

The van were fully loaded to 2000 kg with the same braking force of 8000 N. Would the deceleration change? Would μk change? Calculate the new deceleration and explain the difference in stopping distance.

Worked Example 2 Type 2 — Kinetic: Single Object

Problem Setup

Problem type: Type 2 — Kinetic Single Object. Horizontal push with kinetic friction opposing motion.

Scenario: A 15 kg box is pushed across a warehouse floor with a horizontal applied force of 80 N. The coefficient of kinetic friction is μk = 0.35. Find the acceleration of the box.

  • m = 15 kg
  • Fapplied = 80 N (forward, horizontal)
  • μk = 0.35
  • Positive direction: forward (direction of applied force)

Solution

1
FN = mg = 15 × 9.8 = 147 N
Normal force on flat surface = weight. Need this before calculating friction.
2
fk = μk × FN = 0.35 × 147 = 51.45 N (backward)
Kinetic friction because box is already sliding. Acts backward (opposing motion) — assign negative sign.
3
Fnet = Fapplied − fk = 80 − 51.45 = 28.55 N
Net force = all forces combined. Applied force forward (+80 N), friction backward (−51.45 N).
4
a = Fnet / m = 28.55 / 15 = 1.90 m/s²
Newton's Second Law. Positive acceleration confirms the box accelerates forward — consistent with Fnet being forward.

What would change if...

At what applied force would the box move at constant velocity? Set up the equation and solve. What does this tell you about the relationship between applied force and friction at constant velocity?

Visual Break

Object in question Is the object accelerating? Velocity is changing in magnitude or direction No Newton 1 Fnet = 0 Yes Newton 2 Fnet = ma Is friction present? Surface is rough or drag acts Yes No Fnet = Fapp directly Subtract friction first f = μFn, then Fnet = Fapp − f a = Fnet ÷ m

Copy into your books

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Newton's First Law

  • Object stays at rest or constant velocity unless acted on by a net force
  • Inertia = resistance to change in motion (proportional to mass)
  • Packages fly forward in a braking van — no horizontal force acts on them
  • Constant velocity ≠ no forces — forces are balanced (Fnet = 0)

Newton's Second Law

  • Fnet = ma — net force, not applied force alone
  • Direction of acceleration = direction of net force
  • Always subtract friction before applying F = ma
  • Larger mass → smaller acceleration for same net force

Friction

  • f = μFN — friction depends on surface type (μ) and normal force
  • Static friction (μs): resists motion up to a maximum
  • Kinetic friction (μk): constant once sliding begins
  • μs > μk always — harder to start than to maintain motion

Key Vocabulary

  • Net force: vector sum of all forces on one object
  • Coefficient of friction (μ): dimensionless measure of surface interaction
  • Static friction: friction on a stationary object
  • Kinetic friction: friction on a sliding object

🏃 Activities

Activity 01 — Pattern A

Free Body Diagrams and Law Classification

Draw FBDs for four scenarios using the Vector Protocol. Classify each as Newton 1 or Newton 2. Identify friction type.

  1. A delivery van cruising at constant 80 km/h on a flat road (driving force = friction).
  2. The same van braking hard to a stop (friction only, no driving force).
  3. A 20 kg box being pushed across a floor at constant velocity (applied force = friction).
  4. The same box being pushed with a larger force, accelerating across the floor.

For each: (a) draw and label the FBD, (b) apply the Vector Protocol checklist, (c) state which Newton's Law applies and why, (d) identify whether friction is static or kinetic.

Type notes here — draw your FBDs in your book.

Complete all four FBDs in your book using the Vector Protocol.

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Activity 02 — Pattern A

Newton's Law Scenario Classification

For each scenario below, classify which law applies, state the direction of friction (if present), and determine whether the object is in equilibrium.

#ScenarioNewton 1 or 2?Friction directionEquilibrium?
1Skateboarder coasting at constant speed on flat ground
2Rocket accelerating upward from launch pad
3Book sliding off a tilted table, accelerating down the slope
4Car travelling at constant speed around a bend
5A package at rest on the floor of a van moving at constant velocity
6A cyclist pedalling harder to accelerate from 20 km/h to 30 km/h

Type any working or explanations here.

Complete the table in your book with justifications for each answer.

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✅ Check Your Understanding

Revisit Your Thinking

Earlier you were asked: What force caused the packages to fly forward when the van braked — and why didn't the same thing happen to the driver?

The full answer: no force caused the packages to fly forward. That is the key insight. The packages simply continued moving forward at the van's original velocity because no horizontal force acted on them — Newton's First Law. The van decelerated around them.

The driver did not fly forward because the seatbelt exerted a backward (negative) force on them, providing the net force needed to decelerate with the van — Newton's Second Law. The seatbelt is the difference between following Newton's First Law (packages) and being subject to Newton's Second Law (belted driver).

Look back at your initial response. What did you get right? What are you now thinking differently about?

Annotate your initial response in your book with what you now understand differently.

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Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

MC

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

UnderstandBand 3

7. Explain why a heavier vehicle takes longer to stop than a lighter vehicle travelling at the same speed, even when both vehicles brake with the same coefficient of kinetic friction. Use Newton's Second Law in your answer. 3 MARKS

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ApplyBand 5

8. A 20 kg box is pushed with a force of 120 N directed at 25° below the horizontal across a flat floor. The coefficient of kinetic friction is μk = 0.3. Apply the Vector Protocol and calculate the acceleration of the box. 3 MARKS

Answer in your book — apply all Vector Protocol steps
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EvaluateBand 6

9. A 1500 kg car accelerates from rest to 100 km/h in 8.2 s on a flat road. The engine driving force is 6500 N. Calculate: (a) the acceleration, (b) the friction force opposing motion, (c) the coefficient of kinetic friction, (d) explain what would happen to the acceleration if the road became wet, reducing μk from 0.30 to 0.15. 4 MARKS

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Comprehensive Answers

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Activity 02 — Scenario Classification

  1. Newton 1 — friction acts backward (opposing tendency to slide), equilibrium (Fnet = 0)
  2. Newton 2 — air resistance and gravity act (for atmospheric flight), not in equilibrium
  3. Newton 2 — friction acts up the slope (opposing sliding direction), not in equilibrium
  4. Newton 2 — centripetal acceleration toward centre of bend, not in equilibrium (direction changing)
  5. Newton 1 — package moves with van at constant velocity, static friction from van floor keeps it in place, equilibrium
  6. Newton 2 — friction from road acts backward, net forward force accelerates cyclist, not in equilibrium

Multiple Choice

1. C — Constant velocity means Fnet = 0. The driving force must exactly equal friction. There are forces present — they are balanced. Option A is wrong: friction and driving force both act.

2. B — 39.2 N. FN = mg = 10 × 9.8 = 98 N. fk = μk × FN = 0.4 × 98 = 39.2 N. Option A (4 N) multiplies μ by m only, forgetting g.

3. D — a = Fnet/m. If Fnet doubles and m is constant, a doubles. Direct proportionality.

4. A — 246 N. The applied force at 30° below horizontal has a downward component = 100 sin30° = 50 N, which adds to weight. FN = mg + F sin30° = 20×9.8 + 50 = 196 + 50 = 246 N. Option B ignores the applied force component.

5. C — 40 m. a = Fnet/m = −5000/1000 = −5 m/s². Using v² = u² + 2as: 0 = 400 + 2(−5)s → s = 400/10 = 40 m.

6. B — Net force on system = weight of B (driving) minus friction on A (opposing). Friction on A = μk × FN(A) = 0.2 × 5 × 9.8. Total mass = 5 + 3. Option A ignores friction entirely. Option C uses B's mass for friction on A.

Short Answer — Model Answers

Q7 (3 marks): A heavier vehicle has a greater friction force during braking (f = μmg — larger m means larger f). However, it also has greater mass resisting deceleration. Applying Newton's Second Law: a = Fnet/m = μmg/m = μg. The mass cancels, so the deceleration is the same for both vehicles on the same surface. However, since both start at the same speed, the stopping distance depends on deceleration alone — which is identical. The heavier vehicle does NOT take longer to stop if the friction coefficient is the same. If the question implies the same braking force (not coefficient), then the heavier vehicle has smaller deceleration (a = F/m, larger m gives smaller a) and takes longer to stop.

Q8 (3 marks) — Vector Protocol:

Step 1: Positive direction — forward = positive, upward = positive
Step 2: FBD — W = 196 N down, Fapplied = 120 N at 25° below horizontal, FN upward, fk backward. Note: applied force has a downward component, increasing FN above mg.
FN = mg + F sin25° = 196 + 120 × 0.423 = 196 + 50.7 = 246.7 N
fk = μk × FN = 0.3 × 246.7 = 74.0 N (backward)
Fnet = F cos25° − fk = 120 × 0.906 − 74.0 = 108.7 − 74.0 = 34.7 N (forward)
a = Fnet / m = 34.7 / 20 = 1.74 m/s²

Q9 (4 marks):

(a) Convert: 100 km/h = 27.78 m/s. a = Δv/Δt = 27.78/8.2 = 3.39 m/s²
(b) Fnet = ma = 1500 × 3.39 = 5083 N. Ffriction = F_drive − Fnet = 6500 − 5083 = 1417 N
(c) μk = f / FN = f / mg = 1417 / (1500 × 9.8) = 1417 / 14700 = 0.096
(d) On wet road: f_new = μk(wet) × mg = 0.15 × 14700 = 2205 N. Fnet(new) = 6500 − 2205 = 4295 N. a_new = 4295/1500 = 2.86 m/s². The reduced friction means less resistance, so net force increases and acceleration increases. The car accelerates faster on a wet road for the same driving force — this is why wet-road handling requires more careful throttle control, as acceleration responses become more sensitive.

Consolidation Game

Newton's Laws and Friction

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