IB PHYSICS • SPACE, TIME AND MOTION

Apply Work, Energy & Power — Apply A.3 Work, energy and power in problem-solving and explanations

Master how forces transfer energy and how quickly machines deliver it in real-world scenarios.

Historical Context & Motivation

Long before physicists had formal equations, engineers and inventors wrestled with a practical question: how much effort does it take to move something, and how fast can a machine do it? The concepts of work, energy, and power grew out of centuries of experimentation with levers, waterwheels, and steam engines. Understanding this history helps you see why these ideas are so central to physics—and why they remain essential for solving real problems today.

1687
Newton's Principia
Isaac Newton publishes his laws of motion, laying the foundation for defining force and displacement—the two ingredients of work.
1807
Thomas Young Coins 'Energy'
English polymath Thomas Young introduces the word 'energy' in a physics context, linking it to the motion of objects and the capacity to do work.
1829
Coriolis Defines 'Work'
French mathematician Gaspard-Gustave de Coriolis formally defines mechanical work as force multiplied by displacement, giving engineers a precise quantity to calculate.
1845
Joule's Paddle-Wheel Experiment
James Prescott Joule demonstrates the mechanical equivalent of heat, proving that energy can transform between forms but is never lost—establishing conservation of energy.
1882
Watt as the Unit of Power
The unit of power—the watt—is formally adopted in honor of James Watt, who quantified the rate at which steam engines could deliver work.

These milestones reveal a recurring theme: physicists kept refining the idea that forces acting over distances transfer something measurable—energy—and that the rate of that transfer matters just as much as the total amount. In the IB Physics A.3 topic, you will use these ideas to solve problems about objects speeding up, slowing down, rising, falling, and everything in between.

Core Principles & Definitions

Before diving into calculations, you need a solid grip on the three big ideas and how they connect. Each one builds on the previous concept, forming a chain from force to energy to the rate at which energy is transferred.

1

Work (W)

Work is done when a force causes a displacement. Only the component of force parallel to the displacement counts. Measured in joules (J).
2

Kinetic Energy (Eₖ)

The energy an object has because it is moving. Doubling the speed quadruples the kinetic energy, since it depends on speed squared.
3

Gravitational Potential Energy (Eₚ)

Energy stored by an object's position in a gravitational field. It increases linearly with height above a chosen reference level.
4

Conservation of Energy

Energy cannot be created or destroyed—only transferred or transformed. The total energy of an isolated system remains constant.
5

Power (P)

The rate of doing work or the rate of energy transfer. A powerful engine does the same work in less time. Measured in watts (W).
KEY TAKEAWAY
Think of energy like money in a bank account. Work is like making a deposit or withdrawal—it changes the balance. Power is how quickly you make that transaction. Two people can deposit the same amount, but the one who does it faster has more power.

Visual Explanation — Work and Energy Transfers

The cyan arrow is the applied force F at angle θ to the displacement. The green dashed arrow shows the parallel component F cos θ that actually does work. The pink component F sin θ is perpendicular and does no work.

This diagram captures the most important detail of the work equation: only the component of force along the displacement does work. If you push a suitcase across the floor at an angle, the horizontal part of your push moves the suitcase, while the vertical part simply presses it into the ground. When the force is perpendicular to the displacement (θ = 90°), cos 90° = 0 and no work is done at all. This is why the normal force and gravity do no work on a box sliding on a level surface—they act vertically while the box moves horizontally.

Mathematical Framework

The equations in this section are the toolkit you will use for almost every problem in A.3. Make sure you understand what each variable represents and the conditions under which each equation applies.

WORK DONE BY A CONSTANT FORCE
W = F d cos θ
W = work (J), F = magnitude of the applied force (N), d = displacement (m), θ = angle between the force and displacement vectors. When force and displacement are in the same direction, θ = 0° and cos θ = 1, so W = Fd.
KINETIC ENERGY
Eₖ = ½ m v²
Eₖ = kinetic energy (J), m = mass (kg), v = speed (m s⁻¹). Because velocity is squared, doubling speed gives four times the kinetic energy.
GRAVITATIONAL POTENTIAL ENERGY
Eₚ = m g Δh
Eₚ = gravitational potential energy (J), m = mass (kg), g = gravitational field strength (≈ 9.8 m s⁻² near Earth's surface), Δh = change in height (m) relative to a chosen reference.
WORK–ENERGY THEOREM
W_net = ΔEₖ = ½ m v² − ½ m u²
The net work done on an object equals the change in its kinetic energy. Here u is the initial speed and v is the final speed. If net work is positive, the object speeds up; if negative, it slows down.
POWER
P = W / t = F v cos θ
P = power (W), W = work (J), t = time (s). The alternative form P = Fv cos θ is useful when an object moves at constant velocity against a resistive force.
💡 IB Exam Tip
The IB data booklet provides these equations. Your job is to recognize which equation to use and to keep track of signs. Work done against friction is negative (energy leaves kinetic energy and enters thermal energy). Always define your reference level for Eₚ before starting a problem.

Energy Transfers — Bar Chart Model

One of the most powerful tools for solving energy problems is the energy bar chart. It lets you visualize how energy is distributed at different moments during a process, making it much harder to forget a term or mix up a sign. The diagram below shows a ball being thrown upward: at the bottom it has maximum kinetic energy, and at the top that energy has been fully converted to gravitational potential energy.

At the bottom, all energy is kinetic (cyan bar). Midway up, energy is split equally between kinetic and gravitational potential (amber bar). At the peak, all energy is potential. The dashed line shows that the total mechanical energy remains constant when no friction acts.

When friction or air resistance is present, the total bar height decreases from one snapshot to the next because some mechanical energy has been transferred to thermal energy in the surroundings. In that case, you would add a third bar labeled Ethermal that grows as the other two shrink. Drawing these bar charts before writing equations is a strategy recommended by IB examiners, because it forces you to account for every energy store involved.

📊 Strategy: Using Bar Charts in Exams
Sketch a quick bar chart for the initial and final states of your system. Label every energy type present. If total bar height changes, identify where the missing energy went. Then translate your chart into the equation: Eₖ,initial + Eₚ,initial = Eₖ,final + Eₚ,final + Wfriction.

Worked Example — Roller Coaster with Friction

A 600 kg roller coaster car starts from rest at the top of a 35 m hill. It reaches the bottom of the hill with a speed of 22 m s⁻¹. Determine the work done by friction and the average friction force if the track length from top to bottom is 80 m.

Roller Coaster Energy Problem
1
Step 1 — Identify Given ValuesMass m = 600 kg, initial height h = 35 m, initial speed u = 0 m s⁻¹, final speed v = 22 m s⁻¹, final height = 0 m (reference level at bottom), track length d = 80 m, g = 9.8 m s⁻².
2
Step 2 — Calculate Initial EnergyAt the top the car is at rest, so Eₖ,initial = 0. The gravitational potential energy is Eₚ,initial = mgh = 600 × 9.8 × 35.
Eₚ,initial = 205 800 J
3
Step 3 — Calculate Final EnergyAt the bottom, the height is zero, so Eₚ,final = 0. The kinetic energy is Eₖ,final = ½ mv² = ½ × 600 × 22².
Eₖ,final = ½ × 600 × 484 = 145 200 J
4
Step 4 — Find Work Done by FrictionBy conservation of energy: Eₖ,initial + Eₚ,initial + W_friction = Eₖ,final + Eₚ,final. Rearranging: W_friction = (Eₖ,final + Eₚ,final) − (Eₖ,initial + Eₚ,initial) = (145 200 + 0) − (0 + 205 800).
W_friction = −60 600 J (negative because friction removes energy from the system)
5
Step 5 — Find the Average Friction ForceSince friction acts opposite to displacement, θ = 180° and cos 180° = −1. Using W = Fd cos θ: −60 600 = F × 80 × (−1), so F = 60 600 / 80.
F_friction ≈ 758 N
CHECKING YOUR ANSWER
Always verify that friction does negative work on the system. If you get a positive value for work done by friction, revisit your signs. Also, the final kinetic energy must be less than the initial potential energy when friction is present—if it's larger, something has gone wrong.

Strengths and Limitations of the Energy Approach

You have two main tools for solving mechanics problems: Newton's second law (forces and acceleration) and the energy approach. Each has its strengths and weaknesses, and recognizing when to use which can save you significant time on exams.

Comparing the energy approach with Newton's second law for solving mechanics problems
FeatureEnergy ApproachNewton's Second Law
Best forFinding speeds, heights, or distances when the path doesn't matterFinding forces, accelerations, or analysing motion at a specific instant
Path dependenceWorks regardless of path shape (for conservative forces); only start and end states matterRequires knowledge of the path to resolve forces along it
Vector vs. scalarScalar quantities—no need to break into components along x and yVector equations—must resolve forces into components
Time informationDoes not directly give time; must combine with kinematics if time is neededCan find time through kinematic equations once acceleration is known
Friction handlingFriction appears as negative work or energy lost to thermal energyFriction is treated as a force opposing motion in the free-body diagram
WHEN TO CHOOSE ENERGY
Use the energy approach whenever a problem asks "how fast" or "how high" without caring about time or the exact path. Think of it like checking your bank balance—you don't need to know every individual transaction to see how much you have left, you just compare the initial and final balances. If the problem asks "how long" or "what is the acceleration," reach for Newton's laws instead.

Connection to Advanced Theory

The work–energy ideas you learn in A.3 are not just useful for toy problems—they form the backbone of more advanced physics. The table below shows how the same concepts extend into higher-level topics you may encounter in HL Physics, university physics, or engineering courses.

How A.3 concepts extend into advanced physics
A.3 ConceptAdvanced Extension
W = Fd cos θ (constant force)W = ∫ F · ds — work is the integral of force over a path, allowing variable forces to be handled
Eₚ = mgΔh (near Earth's surface)Eₚ = −GMm/r — the full gravitational potential energy formula for any distance from a planet, used in orbital mechanics
Eₖ = ½mv²Relativistic kinetic energy: Eₖ = (γ − 1)mc², needed when objects approach the speed of light
Conservation of mechanical energyFirst law of thermodynamics (ΔU = Q − W) — energy conservation including heat transfer and internal energy
Power P = W/tInstantaneous power P = dW/dt — a calculus-based formulation that handles time-varying work

You don't need to memorize these advanced formulas now, but it's worth knowing that every equation in A.3 is a simplified version of a deeper, more general principle. When you master the simplified forms, you are building the intuition that makes the advanced versions feel natural later. In the IB HL course, you will already encounter the gravitational potential formula Ep = −GMm/r in the astrophysics and gravitation topics.

Practice Problems

PROBLEM 1CONCEPTUAL
A car travels at constant speed around a circular track on level ground. The engine exerts a forward force that exactly balances friction. Is the net work done on the car positive, negative, or zero? Explain your reasoning.
PROBLEM 2BASIC CALCULATION
A student pushes a 12 kg box across a frictionless floor with a horizontal force of 40 N over a distance of 6.0 m. The box starts from rest. What is the final speed of the box?
PROBLEM 3INTERMEDIATE
A 0.50 kg ball is thrown vertically upward with an initial speed of 14 m s⁻¹. It reaches a maximum height of 8.5 m. Determine the average force of air resistance acting on the ball during its ascent.
PROBLEM 4APPLIED
An electric motor lifts a 200 kg elevator cabin at a constant speed of 3.0 m s⁻¹. If the motor operates at an efficiency of 85 %, what is the electrical power input to the motor?
PROBLEM 5CRITICAL THINKING
A skier (mass 65 kg) starts from rest at the top of a 120 m slope inclined at 30° to the horizontal. A constant friction force of 50 N acts on the skier. At the bottom, the slope levels out onto a flat section where the same friction force continues. Using energy methods, find the skier's speed at the bottom of the slope and how far the skier travels along the flat before stopping.

Lesson Summary

Work is the transfer of energy when a force acts over a displacement, calculated as W = Fd cos θ. Only the component of force parallel to the displacement contributes. Kinetic energy (½mv²) depends on speed squared, while gravitational potential energy (mgΔh) depends linearly on height. The work–energy theorem states that the net work on an object equals its change in kinetic energy, providing a powerful shortcut for many mechanics problems.

Conservation of energy means the total energy of an isolated system is constant—it only changes form. When friction is present, mechanical energy decreases as thermal energy increases. Power (P = W/t or P = Fv) measures the rate of energy transfer and is crucial for understanding machines and engines. Use energy bar charts to visualize transfers, choose the energy approach when the path doesn't matter and time isn't needed, and always check that friction does negative work.

Varsity Tutors • IB Physics • Apply Work, Energy & Power — Apply A.3 Work, energy and power in problem-solving and explanations