IB Physics Topic 1.3: Work, energy and power | Notes & Flashcards | Aimnova

IB Physics — Work, energy and power

Topic 1.3 of IB Physics covers Work, energy and power, which is part of Unit 1: Space, time and motion. Students explore key concepts including Work done & force-distance graphs, Kinetic energy & the work-energy principle, Gravitational PE & conservation of energy, and more. A strong understanding of work, energy and power is essential for IB Physics exams and builds the foundation for connected topics across the syllabus.

Key concepts in Work, energy and power

Key Idea: This topic is the energy toolkit: forces do work, which is stored as kinetic, gravitational or elastic energy, transferred at a rate set by power, and never destroyed — only shared out and partly wasted as heat. It is examined on both papers: quick definition and 'double the speed' MCQs on Paper 1A, and multi-step calculations (force–distance areas, slide-to-rest, falling-body speeds, spring collisions, power against drag, efficiency) on Paper 2.

📐 The six given equations

Every equation below is given in the data booklet (Theme A.3) — tap any one for the booklet entry. Knowing which to reach for is the skill, so the table after them sorts that out.

W = Fs\cos\theta

Work done by a force. When the force is along the motion θ = 0 and cos 0 = 1, so W = Fs.

$W$: work done = energy transferred (J)
$F$: force applied (N)
$s$: distance moved (m)
$\theta$: angle between the force and the direction of motion (°)

E_k = \tfrac{1}{2}mv^{2} = \frac{p^{2}}{2m}

Kinetic energy — ½mv² when you know the speed, p²/2m when you know the momentum.

$E_k$: kinetic energy — the energy of motion (J)
$m$: mass (kg)
$v$: speed (m s⁻¹)
$p$: momentum, p = mv (kg m s⁻¹)

\Delta E_p = mg\,\Delta h

Change in gravitational PE — equal to the kinetic energy gained in a free fall.

$\Delta E_p$: change in gravitational PE (J)
$m$: mass (kg)
$g$: gravitational field strength (≈ 9.8 N kg⁻¹ on Earth)
$\Delta h$: change in height (m)

E_H = \tfrac{1}{2}k\,\Delta x^{2}

Elastic potential energy stored in a stretched or squashed spring. Convert any cm to m before squaring Δx.

$E_H$: elastic potential energy stored in the spring (J)
$k$: spring constant — the stiffness (N m⁻¹)
$\Delta x$: extension or compression from the natural length (m)

P = \frac{\Delta W}{\Delta t} = Fv

Power. Use ΔW/Δt with energy and time; use Fv when a force moves at a steady speed (e.g. cruising against drag).

$P$: power — energy transferred per second (W = J s⁻¹)
$\Delta W$: work done / energy transferred (J)
$\Delta t$: time taken (s)
$F$: force (N) — at constant speed, equal to the resistive force
$v$: speed in the direction of the force (m s⁻¹)

\eta = \frac{\text{useful out}}{\text{total in}}

Efficiency — the useful fraction of the energy (or power) supplied. Always between 0 and 1; × 100 for a %.

$\eta$: efficiency — the useful fraction (0 to 1; × 100 for a %), no unit
$\text{useful out}$: useful work or power that comes out (J or W)
$\text{total in}$: total work or power put in (J or W)

🧭 Which one when?

🔁 Conservation, transfers & waste

When something slides to a stop, friction takes away all its kinetic energy: friction force × distance = Eₖ, so distance = Eₖ ÷ friction force.

✏️ IB-style worked examples

IB-style question — work from a graph, then a speed

A 3.0 kg trolley starts from rest on a smooth track. A constant net force of 12 N acts on it as it moves 5.0 m, shown on a force–distance graph. (a) State what the area under the graph represents. (b) Find the trolley's final speed.

Solution:

(a) The area under a force–distance graph is the work done by the force:
$\text{area} = W = Fs = 12 \times 5.0 = 60\ \text{J}$
(b) From rest, all this work becomes kinetic energy — use the given formula:
$E_k = \tfrac{1}{2}mv^{2}$
Set Eₖ = 60 J, put in m = 3.0 and solve for v:
$60 = \tfrac{1}{2}\times 3.0 \times v^{2} \;\Rightarrow\; v^{2} = 40$
Take the square root — keep the unit:
$v = \sqrt{40} = 6.3\ \text{m s}^{-1}$

Final answer:

(a) the work done on the trolley (= 60 J); (b) v = 6.3 m s⁻¹. The area gives energy in joules — you still need ½mv² to get the speed.

IB-style question — falling-body speed (mass cancels)

A 0.60 kg ball falls from rest through a height of 2.0 m. Air resistance is negligible and g = 9.8 N kg⁻¹. Find its speed just before it lands.

Solution:

Energy is conserved, so set the PE lost equal to the KE gained:
$mg\,\Delta h = \tfrac{1}{2}mv^{2}$
The mass cancels — put in g = 9.8 and Δh = 2.0:
$9.8 \times 2.0 = \tfrac{1}{2}v^{2} \;\Rightarrow\; v^{2} = 39.2$
Take the square root — keep the unit:
$v = \sqrt{39.2} = 6.3\ \text{m s}^{-1}$

Final answer:

v = 6.3 m s⁻¹ — and it would be the same for any mass, because m cancels out.

IB-style question — power against drag, then efficiency

A van cruises at a steady 20 m s⁻¹ against a total resistive force of 600 N. (a) Find the useful power the van delivers to overcome drag. (b) The engine is supplied with 18 kW of power. Find its efficiency.

Solution:

(a) At constant speed the driving force equals the resistive force, so use the given P = Fv:
$P = Fv = 600 \times 20 = 12\,000\ \text{W} = 12\ \text{kW}$
(b) Efficiency is the useful power out over the total power in — use the given formula:
$\eta = \frac{\text{useful out}}{\text{total in}} = \frac{12}{18}$
Work it out, then × 100 for a percentage:
$\eta = 0.67 = 67\%$

Final answer:

(a) 12 kW; (b) η ≈ 0.67 (67%). The other ~6 kW is wasted, mostly as thermal energy (heat).

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

Pick the equation from what you're GIVEN: force × distance → W = Fs cos θ; mass & speed → Eₖ = ½mv²; height → ΔEₚ = mgΔh; spring stretch → EH = ½kΔx²; energy & time (or steady-speed force) → P; useful vs total → η.
Area under a force–distance graph = the work done — a rectangle for a flat line, a triangle for a sloping spring line. The area is energy in joules, not a speed.
Always square the variable that's squared: v in ½mv² and Δx in ½kΔx². Double it → four times the energy (the classic 'stopping distance' point).
Conservation of energy: set PE lost = KE gained → mgΔh = ½mv²; the mass cancels, so don't carry it through.
Slide-to-rest against friction: friction force × distance = Eₖ, so distance = Eₖ ÷ friction force.
At constant speed use P = Fv with F as the resistive force; convert kW ↔ W and cm ↔ m before substituting.
Efficiency is a fraction ≤ 1 — if you get more than 100%, you've divided total by useful instead of useful by total.

IB Physics — Work, energy and power

Key concepts in Work, energy and power

Key Idea: This topic is the energy toolkit: forces do work, which is stored as kinetic, gravitational or elastic energy, transferred at a rate set by power, and never destroyed — only shared out and partly wasted as heat. It is examined on both papers: quick definition and 'double the speed' MCQs on Paper 1A, and multi-step calculations (force–distance areas, slide-to-rest, falling-body speeds, spring collisions, power against drag, efficiency) on Paper 2.

📐 The six given equations

Every equation below is given in the data booklet (Theme A.3) — tap any one for the booklet entry. Knowing which to reach for is the skill, so the table after them sorts that out.

W = Fs\cos\theta

Work done by a force. When the force is along the motion θ = 0 and cos 0 = 1, so W = Fs.

$W$: work done = energy transferred (J)
$F$: force applied (N)
$s$: distance moved (m)
$\theta$: angle between the force and the direction of motion (°)

E_k = \tfrac{1}{2}mv^{2} = \frac{p^{2}}{2m}

Kinetic energy — ½mv² when you know the speed, p²/2m when you know the momentum.

$E_k$: kinetic energy — the energy of motion (J)
$m$: mass (kg)
$v$: speed (m s⁻¹)
$p$: momentum, p = mv (kg m s⁻¹)

\Delta E_p = mg\,\Delta h

Change in gravitational PE — equal to the kinetic energy gained in a free fall.

$\Delta E_p$: change in gravitational PE (J)
$m$: mass (kg)
$g$: gravitational field strength (≈ 9.8 N kg⁻¹ on Earth)
$\Delta h$: change in height (m)

E_H = \tfrac{1}{2}k\,\Delta x^{2}

Elastic potential energy stored in a stretched or squashed spring. Convert any cm to m before squaring Δx.

$E_H$: elastic potential energy stored in the spring (J)
$k$: spring constant — the stiffness (N m⁻¹)
$\Delta x$: extension or compression from the natural length (m)

P = \frac{\Delta W}{\Delta t} = Fv

Power. Use ΔW/Δt with energy and time; use Fv when a force moves at a steady speed (e.g. cruising against drag).

$P$: power — energy transferred per second (W = J s⁻¹)
$\Delta W$: work done / energy transferred (J)
$\Delta t$: time taken (s)
$F$: force (N) — at constant speed, equal to the resistive force
$v$: speed in the direction of the force (m s⁻¹)

\eta = \frac{\text{useful out}}{\text{total in}}

Efficiency — the useful fraction of the energy (or power) supplied. Always between 0 and 1; × 100 for a %.

$\eta$: efficiency — the useful fraction (0 to 1; × 100 for a %), no unit
$\text{useful out}$: useful work or power that comes out (J or W)
$\text{total in}$: total work or power put in (J or W)

🧭 Which one when?

🔁 Conservation, transfers & waste

When something slides to a stop, friction takes away all its kinetic energy: friction force × distance = Eₖ, so distance = Eₖ ÷ friction force.

✏️ IB-style worked examples

IB-style question — work from a graph, then a speed

Solution:

(a) The area under a force–distance graph is the work done by the force:
$\text{area} = W = Fs = 12 \times 5.0 = 60\ \text{J}$
(b) From rest, all this work becomes kinetic energy — use the given formula:
$E_k = \tfrac{1}{2}mv^{2}$
Set Eₖ = 60 J, put in m = 3.0 and solve for v:
$60 = \tfrac{1}{2}\times 3.0 \times v^{2} \;\Rightarrow\; v^{2} = 40$
Take the square root — keep the unit:
$v = \sqrt{40} = 6.3\ \text{m s}^{-1}$

Final answer:

(a) the work done on the trolley (= 60 J); (b) v = 6.3 m s⁻¹. The area gives energy in joules — you still need ½mv² to get the speed.

IB-style question — falling-body speed (mass cancels)

A 0.60 kg ball falls from rest through a height of 2.0 m. Air resistance is negligible and g = 9.8 N kg⁻¹. Find its speed just before it lands.

Solution:

Energy is conserved, so set the PE lost equal to the KE gained:
$mg\,\Delta h = \tfrac{1}{2}mv^{2}$
The mass cancels — put in g = 9.8 and Δh = 2.0:
$9.8 \times 2.0 = \tfrac{1}{2}v^{2} \;\Rightarrow\; v^{2} = 39.2$
Take the square root — keep the unit:
$v = \sqrt{39.2} = 6.3\ \text{m s}^{-1}$

Final answer:

v = 6.3 m s⁻¹ — and it would be the same for any mass, because m cancels out.

IB-style question — power against drag, then efficiency

Solution:

(a) At constant speed the driving force equals the resistive force, so use the given P = Fv:
$P = Fv = 600 \times 20 = 12\,000\ \text{W} = 12\ \text{kW}$
(b) Efficiency is the useful power out over the total power in — use the given formula:
$\eta = \frac{\text{useful out}}{\text{total in}} = \frac{12}{18}$
Work it out, then × 100 for a percentage:
$\eta = 0.67 = 67\%$

Final answer:

(a) 12 kW; (b) η ≈ 0.67 (67%). The other ~6 kW is wasted, mostly as thermal energy (heat).

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

Pick the equation from what you're GIVEN: force × distance → W = Fs cos θ; mass & speed → Eₖ = ½mv²; height → ΔEₚ = mgΔh; spring stretch → EH = ½kΔx²; energy & time (or steady-speed force) → P; useful vs total → η.
Area under a force–distance graph = the work done — a rectangle for a flat line, a triangle for a sloping spring line. The area is energy in joules, not a speed.
Always square the variable that's squared: v in ½mv² and Δx in ½kΔx². Double it → four times the energy (the classic 'stopping distance' point).
Conservation of energy: set PE lost = KE gained → mgΔh = ½mv²; the mass cancels, so don't carry it through.
Slide-to-rest against friction: friction force × distance = Eₖ, so distance = Eₖ ÷ friction force.
At constant speed use P = Fv with F as the resistive force; convert kW ↔ W and cm ↔ m before substituting.
Efficiency is a fraction ≤ 1 — if you get more than 100%, you've divided total by useful instead of useful by total.

Key concepts in Work, energy and power

📐 The six given equations

🧭 Which one when?

🔁 Conservation, transfers & waste

✏️ IB-style worked examples

IB-style question — work from a graph, then a speed

IB-style question — falling-body speed (mass cancels)

IB-style question — power against drag, then efficiency

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 1.3

Study resources — 1.3 Work, energy and power

Work done & force-distance graphs

Kinetic energy & the work-energy principle

Gravitational PE & conservation of energy

Elastic potential energy

Power & efficiency

Energy in collisions & systems (Sankey/energy transfers)

Ready to study Work, energy and power?

Ready to practice?

Key concepts in Work, energy and power

📐 The six given equations

🧭 Which one when?

🔁 Conservation, transfers & waste

✏️ IB-style worked examples

IB-style question — work from a graph, then a speed

IB-style question — falling-body speed (mass cancels)

IB-style question — power against drag, then efficiency

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 1.3

Study resources — 1.3 Work, energy and power

Work done & force-distance graphs

Kinetic energy & the work-energy principle

Gravitational PE & conservation of energy

Elastic potential energy

Power & efficiency

Energy in collisions & systems (Sankey/energy transfers)

Ready to study Work, energy and power?

Ready to practice?