Year 11 Physics Module 2: Dynamics 35 min Lesson 6 of 15

Work and Kinetic Energy

Work is a deposit of energy. Kinetic energy is the balance. The work-energy theorem says the deposit always equals the change in balance — exactly.

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Think First

A waiter carries a tray of food horizontally across a restaurant at constant height and constant speed. He exerts an upward force on the tray the entire time. Has he done any work on the tray? Predict your answer and explain your reasoning.

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Formula Reference — This Lesson

$W = Fs\cos\theta$ | $KE = \tfrac{1}{2}mv^2$

Work-Energy Theorem: $W_{net} = \Delta KE = KE_f - KE_i$ | Find v: $v = \sqrt{2KE/m}$ | Note: If $\theta = 90°$, $\cos 90° = 0$, so $W = 0$ (perpendicular force does no work)

W=Fs
Formula Reference — Work, Kinetic Energy and the Work-Energy Theorem

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$W = Fs\cos\theta$

Work Done

W = work (J) | F = force magnitude (N) | s = displacement (m) | θ = angle between force vector and displacement vector

Applies when: a constant force acts on an object over a displacement. θ is measured between the force direction and displacement direction.

Common trap: If force is perpendicular to motion (θ = 90°), cos90° = 0 so W = 0 — no work done regardless of how large F is. A normal force never does work on a horizontally moving object.

$KE = \tfrac{1}{2}mv^2$

Kinetic Energy

KE = kinetic energy (J) | m = mass (kg) | v = speed (m/s)

Applies when: calculating the energy of any moving object

Common trap: KE depends on v² — doubling speed quadruples KE, not doubles it. A car at 60 km/h has 4× the KE of the same car at 30 km/h.

$W_{net} = \Delta KE$

Work-Energy Theorem

W_net = net work done (J) | ΔKE = KE_f − KE_i = change in kinetic energy (J)

Applies when: relating the total work done by all forces to the resulting change in kinetic energy

Common trap: Must use net work — the sum of work done by all forces including friction. Using only applied force gives the wrong ΔKE.

$\Delta U = mg\Delta h$

Change in Gravitational Potential Energy

ΔU = change in gravitational PE (J) | m = mass (kg) | g = 9.8 m/s² | Δh = change in height (m)

Applies when: an object moves vertically in a uniform gravitational field (near Earth's surface)

Common trap: Δh is positive when moving up (ΔU positive — PE increases) and negative when moving down (ΔU negative — PE decreases). Always define which direction is positive height before substituting.

Know

Definition of work in physics (W = Fs cosθ)
Definition of kinetic energy (KE = ½mv²)
The work-energy theorem: W_net = ΔKE
ΔU = mgΔh for gravitational PE

Understand

Why a perpendicular force does zero work
Why KE scales with v² not v
Why net work (not applied force alone) determines ΔKE
How work, KE and PE are connected through energy conservation

Can Do

Calculate work done by a force at any angle θ
Calculate KE and ΔKE for moving objects
Apply W_net = ΔKE to find unknown speed or force
Calculate ΔU = mgΔh for height changes

Misconceptions to Fix

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Wrong: Zero acceleration means an object is stationary.

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Right: Zero acceleration means constant velocity — the object could be moving at constant speed in a straight line.

📚 Core Content

Key Terms

workthe transfer of energy by a force acting through a displacement

Kinetic energy (KE)the energy stored in an object due to its motion

forceperpendicular to motion (θ = 90°), cos90° = 0 so W = 0 — no work done regardless of how large F is

Always define which directionpositive height before substituting

and PEconnected through energy conservation

Zero accelerationan object is stationary

01Work — W = Fs cosθ

Work — W = Fs cosθ

Work is done on an object when a force moves it — but only the component of force in the direction of motion counts.

In physics, work is the transfer of energy by a force acting through a displacement. It is not effort, tiredness, or time — it is specifically a force causing a displacement. Work has the unit of joules (J), the same unit as energy.

When a force F acts at angle θ to the direction of displacement s, only the component F cosθ contributes to work done — the perpendicular component does nothing useful:

Work Done — Force at Angle to Displacement

Horizontal push on a box (force and motion same direction)

θ: 0°

Work done: W = Fs (maximum)

Reason: cos0° = 1 — full force contributes

Rope pull at 30° above horizontal

θ: 30°

Work done: W = Fs cos30°

Reason: Only horizontal component does work

Normal force on a sliding box

θ: 90°

Work done: W = 0

Reason: cos90° = 0 — perpendicular to motion

Friction opposing motion

θ: 180°

Work done: W = −Fs

Reason: cos180° = −1 — opposes motion (negative work)

Real-World Anchor A waiter carrying a tray horizontally across a restaurant exerts an upward force on the tray — but the tray moves horizontally. The angle between force (up) and displacement (horizontal) is 90°, so W = Fs cos90° = 0. The waiter does zero work on the tray in the physics sense, despite the effort involved. This is counterintuitive but follows directly from the definition.

Vector Protocol — work problems

Step 1 — Identify the angle θ between the force direction and the displacement direction (not the surface)

Step 2 — Draw a diagram showing F, s, and θ — do not guess the angle from the problem text

Step 3 — Check sign: if force component opposes motion, work is negative

02Kinetic Energy — KE = ½mv²

Kinetic Energy — KE = ½mv²

Kinetic energy is the energy an object has because it is moving — and it grows with the square of speed, not linearly.

Kinetic energy (KE) is the energy stored in an object due to its motion. A stationary object has zero KE regardless of its mass. A moving object always has KE, and that KE increases rapidly as speed increases — because of the v² relationship.

Object	Mass (kg)	Speed (m/s)	KE (J)	Note
Cyclist	80	5	1 000	Baseline
Cyclist	80	10	4 000	2× speed → 4× KE
Cyclist	80	20	16 000	4× speed → 16× KE
Car	1200	10	60 000	15× mass → 15× KE (same speed)
Car	1200	20	240 000	2× speed → 4× KE

Why speed limits matter A car at 60 km/h has 4× the kinetic energy of the same car at 30 km/h — not 2×. In a collision, all of that KE must be converted to other forms (deformation, heat, sound). Four times the energy means four times the damage potential. This is the physics basis for strict urban speed limits.

03Work-Energy Theorem

The Work-Energy Theorem — W_net = ΔKE

The net work done on an object equals exactly the change in its kinetic energy — no more, no less.

This is not a coincidence — it is a fundamental relationship derived from Newton's Second Law. When a net force acts on an object over a displacement, it both does work AND changes the object's kinetic energy, and the two are always equal:

$$W_{net} = \Delta KE = KE_f - KE_i = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2$$ Positive $W_{net}$ → object speeds up (KE increases)
Negative $W_{net}$ → object slows down (KE decreases)
Zero $W_{net}$ → constant speed (KE unchanged)

Situation	W_net	ΔKE	Effect on motion
Engine force > friction	Positive	Positive (KE increases)	Object accelerates
Brakes applied	Negative	Negative (KE decreases)	Object decelerates
Constant velocity cruise	Zero	Zero (KE unchanged)	No acceleration
Object falling freely	Positive (gravity)	Positive (KE increases)	Speeds up, PE converts to KE

Common Misconceptions

Work and energy are the same thing.

Work is a process — energy transferred by a force acting through a displacement. Energy is a stored quantity an object possesses. Work is what energy does when it moves from one place to another. They share the same unit (joules) but are conceptually distinct.

A heavier object always has more kinetic energy than a lighter one.

KE = ½mv² — it depends on both mass and speed. A 9 g bullet travelling at 900 m/s has KE = ½ × 0.009 × 900² = 3645 J. A 2000 kg stationary truck has KE = 0 J. The bullet has more KE despite being 200 000 times lighter.

If an object doesn't move, no energy is involved.

A compressed spring, a raised book, or a charged battery all store energy without moving. Gravitational potential energy (ΔU = mgΔh) is stored by position, not motion. Zero displacement means zero work done — but energy can still be stored or transferred by other means.

Interactive: Work-Energy Calculator

Interactive: Work and Energy Transfer (3D)

✏️ Worked Examples

Worked Example 1 Type 3 — Energy Transformation

Problem Setup

Problem type: Type 3 — Energy Transformation. Flat surface, no height change. Verify W_net = ΔKE.

Scenario: A 1200 kg car accelerates from rest to 25 m/s. The engine exerts a constant driving force of 4000 N over a distance of 93.75 m on a flat road. Calculate: (a) work done by the engine, (b) the change in KE, (c) verify W_net = ΔKE.

m = 1200 kg | v_i = 0 m/s | v_f = 25 m/s
F = 4000 N (horizontal, θ = 0°)
s = 93.75 m | flat road (no height change)
Assume frictionless for this example

Solution

W = Fs cosθ = 4000 × 93.75 × cos0° = 375 000 J

Force is horizontal, displacement is horizontal — θ = 0°, cos0° = 1. Full force contributes to work.

KE_i = ½mv_i² = ½ × 1200 × 0² = 0 J

Car starts from rest — initial KE is zero.

KE_f = ½mv_f² = ½ × 1200 × 25² = ½ × 1200 × 625 = 375 000 J

Calculate final KE using the final speed. Note v² = 625 — always square first, then multiply.

ΔKE = KE_f − KE_i = 375 000 − 0 = 375 000 J

Change in KE equals the final minus initial values.

W_net = ΔKE → 375 000 J = 375 000 J ✓

Work done by engine equals change in KE — the work-energy theorem is confirmed. This is the verification step.

What would change if...

The same car braked from 25 m/s to rest over 62.5 m. What was the braking force? Is the work done positive or negative — and why? Use W_net = ΔKE to find the answer.

Worked Example 2 Type 3 — Energy Transformation with angle

Problem Setup

Problem type: Type 3 — Energy Transformation. Force at angle, friction present, find final speed.

Scenario: A 15 kg box starts from rest and is pulled 8 m across a floor by a rope at 30° above horizontal. Applied force = 60 N, μ_k = 0.2. Find the final speed.

m = 15 kg | v_i = 0 | s = 8 m
F_applied = 60 N at θ = 30° above horizontal
μ_k = 0.2
Positive direction: forward (direction of motion)

Solution

W_applied = Fs cosθ = 60 × 8 × cos30° = 60 × 8 × 0.866 = 415.7 J

Work by the rope — only horizontal component (F cosθ) does work along the direction of motion.

F_N = mg − F sinθ = (15 × 9.8) − (60 × sin30°) = 147 − 30 = 117 N

Normal force is reduced because the rope pulls upward (F sinθ component), reducing the downward load on the surface.

f_k = μ_k × F_N = 0.2 × 117 = 23.4 N (opposing motion)

Kinetic friction acts backward — it does negative work on the box.

W_friction = −f_k × s = −23.4 × 8 = −187.2 J

Friction opposes motion so θ = 180° and work is negative — energy removed from the box.

W_net = 415.7 + (−187.2) = 228.5 J

Net work = sum of all work done. Applied force adds energy; friction removes energy.

W_net = ΔKE = ½mv_f² − 0 → v_f = √(2 × 228.5 / 15) = √30.47 = 5.52 m/s

Apply work-energy theorem: W_net = ½mv_f². Rearrange for v_f. Starting from rest so KE_i = 0.

What would change if...

The rope were horizontal (θ = 0°) instead of at 30°. Would the final speed be higher or lower? Consider both the change in W_applied and the change in F_N (and therefore friction) — both change when the angle changes.

Visual Break

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Work Done

W = Fs cosθ — θ is angle between F and displacement
θ = 0°: W = Fs (maximum, force parallel to motion)
θ = 90°: W = 0 (force perpendicular, e.g. normal force)
θ = 180°: W = −Fs (force opposes motion, e.g. friction)

Kinetic Energy

KE = ½mv² — depends on v², not v
Doubling speed → 4× KE (not 2×)
Unit: joule (J) = kg m² s⁻²
KE = 0 for any stationary object regardless of mass

Work-Energy Theorem

W_net = ΔKE = KE_f − KE_i
Must use NET work — include all forces
Positive W_net → object speeds up
Negative W_net → object slows down

Gravitational PE

ΔU = mgΔh — height change in uniform field
Moving up: Δh positive → ΔU positive (PE increases)
Moving down: Δh negative → ΔU negative (PE decreases)
Work done by gravity = −ΔU = mg × drop in height

🏃 Activities

Activity 01 — Pattern C

Compare: Horizontal Push vs Angled Pull

Compare work done, net work, and final speed for two scenarios using the same object and same force magnitude.

A 1000 kg car starts from rest and is moved 50 m. In both cases F = 2000 N and μ_k = 0.2.

Scenario A: Horizontal push (θ = 0°)

Scenario B: Rope at 20° above horizontal

Complete the table for both scenarios. Show all working beside the table.
Which scenario produces the higher final speed? Explain why in terms of work and energy.
Why does the angled pull change the normal force — and what effect does this have on friction?

Type your comparison and explanation below. Complete the table in your book.

Complete the table and answer questions in your book.

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Activity 02 — Pattern C

Apply: Cyclist Down a Hill

Apply the work-energy theorem to a novel scenario involving gravitational PE converting to KE.

A 70 kg cyclist freewheels from rest at the top of a 12 m hill to the bottom. The road distance is 40 m.

Part A — No friction: Assume no friction or air resistance. Use ΔU = mgΔh to find the work done by gravity. Then apply W_net = ΔKE to find the speed at the bottom.
Part B — With friction: Now assume a constant friction force of 80 N over the 40 m road distance. Recalculate the final speed.
Part C — Comparison: How much speed does friction cost the cyclist? Express your answer in m/s and as a percentage reduction.
Part D — Extend: If the hill were twice as tall (24 m) but the road distance stayed at 40 m with the same friction force of 80 N, predict qualitatively whether the percentage speed loss would increase or decrease compared to Part C. Explain your reasoning.

Type your working and answers below.

Complete all four parts in your book showing full working.

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✅ Check Your Understanding

Revisit Your Thinking

Earlier you were asked: A waiter carries a tray horizontally at constant height and speed. He exerts an upward force the entire time. Has he done any work on the tray?

The full answer: No — the waiter does zero work on the tray in the physics sense. The force he exerts is upward; the tray's displacement is horizontal. The angle between force and displacement is 90°, so W = Fs cos90° = 0 J. No energy is transferred to the tray by this force. This surprises most people because the waiter clearly expends effort — but physical work requires a force component in the direction of motion. The waiter's upward force has zero component in the horizontal direction.

This is one of the most important conceptual distinctions in the energy topic: everyday effort and physics work are not the same thing.

Look back at your initial prediction. What did you get right? What changed?

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?

Multiple Choice

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

UnderstandBand 3

7. Define work in physics and explain why a normal force does zero work on a box sliding horizontally across a floor, even though the normal force is large. 3 MARKS

Answer in your book

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ApplyBand 5

8. A 20 kg box starts from rest and is pulled 10 m across a flat floor by a rope at 25° above horizontal. The applied force is 100 N and μ_k = 0.25. Apply the Vector Protocol. Calculate the net work done and the final speed of the box. 3 MARKS

Answer in your book — apply Vector Protocol

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EvaluateBand 6

9. A car of mass m brakes from initial speed v_i to rest. Using W_net = ΔKE, derive an expression for the braking distance s in terms of m, v_i, and braking force F_b. Use your expression to show that doubling the initial speed requires four times the braking distance. Explain why this relationship makes road speed limits so critical to safety. 4 MARKS

Answer in your book — derive the expression algebraically

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Comprehensive Answers

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Activity 01 — Comparison Table

Scenario B is faster despite applying less work — because the angled pull lifts some of the box's weight off the surface, reducing the normal force, which reduces friction more than the reduction in applied work. The net result is higher W_net and a higher final speed.

Activity 02 — Cyclist Down a Hill

Part A: W_gravity = mgΔh = 70 × 9.8 × 12 = 8232 J. W_net = 8232 J. v_f = √(2 × 8232/70) = √235.2 = 15.3 m/s

Part B: W_friction = −80 × 40 = −3200 J. W_net = 8232 − 3200 = 5032 J. v_f = √(2 × 5032/70) = √143.8 = 12.0 m/s

Part C: Speed reduction = 15.3 − 12.0 = 3.3 m/s. As a percentage: 3.3/15.3 × 100 = 21.6% reduction.

Part D: With a 24 m hill: W_gravity = 70 × 9.8 × 24 = 16 464 J. W_friction stays at −3200 J (same road, same force). W_net = 13 264 J. The friction work is a smaller fraction of the total work — so the percentage speed loss would decrease. Taller hills reduce the relative impact of friction because gravity does more work.

Multiple Choice

1. C — 2000 J. W = Fs cosθ = 500 × 4 × cos0° = 2000 J. Force is horizontal, motion is horizontal, θ = 0°.

2. D — Normal force acts vertically upward; displacement is horizontal. θ = 90°, cos90° = 0, W = 0.

3. B — 36 J. KE = ½mv² = ½ × 2 × 36 = 36 J. Common error: forgetting to square v — ½ × 2 × 6 = 6 J is wrong.

4. A — 207 J. W_applied = 80 × 5 × cos40° = 306.4 J. W_friction = −20 × 5 = −100 J. W_net = 206.4 J ≈ 207 J. Option B (400 J) ignores friction. Option D (307 J) ignores the cosθ reduction.

5. C — 4×. KE ∝ v². Doubling v → (2v)² = 4v² → 4× KE. This is the v² relationship — KE grows with the square of speed.

6. B. From W_net = ½mv²: v = √(2W/m). For A: v_A = √(2W/2) = √W. For B: v_B = √(2W/8) = √(W/4) = ½√W. So v_A/v_B = √W / (½√W) = 2. Object A is twice as fast. Option D (4×) confuses the mass ratio with the speed ratio — must take the square root.

Short Answer — Model Answers

Q7 (3 marks): Work is defined as the transfer of energy by a force acting through a displacement in the direction of the force — expressed as W = Fs cosθ, where θ is the angle between the force and the displacement. The normal force acts vertically upward on a horizontally sliding box. The displacement of the box is horizontal. The angle between the normal force and the displacement is θ = 90°. Since cos90° = 0, the work done by the normal force is W = Fs × 0 = 0 J, regardless of the magnitude of the force or the distance travelled.

Q8 (3 marks):

Step 1: Positive direction — forward = positive

W_applied = 100 × 10 × cos25° = 100 × 10 × 0.906 = 906 J

F_N = mg − F sin25° = 196 − 100 × 0.423 = 196 − 42.3 = 153.7 N

f_k = 0.25 × 153.7 = 38.4 N | W_friction = −38.4 × 10 = −384 J

W_net = 906 − 384 = 522 J

v_f = sqrt(2 × W_net / m) = sqrt(2 × 522 / 20) = sqrt(52.2) = 7.22 m/s

Q9 (4 marks) — Band 6 derivation:

Apply W_net = ΔKE: −F_b × s = 0 − ½mv_i²

Rearrange: s = mv_i² / (2F_b)

If v_i doubles to 2v_i: s_new = m(2v_i)² / (2F_b) = 4mv_i² / (2F_b) = 4s

Therefore doubling initial speed requires four times the braking distance.

Road safety implication: A car travelling at 60 km/h requires 4× the braking distance of the same car at 30 km/h — not 2× as many drivers assume. At 60 km/h, a car has 4× the KE of a car at 30 km/h. All of this KE must be removed by the brakes over the stopping distance. Since braking force is limited by the friction between tyres and road, the only way to dissipate 4× the energy is over 4× the distance. This is the physics basis for lower urban speed limits — small increases in speed lead to disproportionately large increases in stopping distance and crash energy.

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