Library Work, Energy & Power
Physics (IAL)

Work, Energy & Power

Revise Work, Energy & Power for Physics (IAL) — revision notes and instant AI marking. Free to start.

📖 Revision notes · preview
Edexcel IAL Physics · Revision Guide

Work, Energy & Power

The Big Idea Whenever a force moves something, energy gets transferred from one place (or one form) to another — and no matter how it moves around, the total amount of energy never changes.
Summary — What This Chapter Covers
  • Work is done whenever a force causes something to move through a distance — it's how energy actually gets "delivered" from a force to an object.
  • If the force isn't perfectly aligned with the motion, you only count the part of the force that is aligned with the motion (that's where the cos θ / sin θ comes from).
  • Kinetic energy is the energy something has purely because it's moving — the faster it goes, the more it has, and it goes up with the square of speed, not speed itself.
  • Gravitational potential energy is "stored height" — energy an object has because it's been lifted up against gravity.
  • The Principle of Conservation of Energy says energy is never created or destroyed — only transferred between forms (like GPE ↔ KE) or converted to less useful forms (like heat or sound).
  • Power is simply how fast energy is being transferred — the same amount of work done quickly means high power; done slowly means low power.
  • Efficiency measures how much of the energy you put into a system actually comes out as something useful, versus how much is "wasted" (usually as heat).
1. Work
What is "work" in physics?

In everyday English, "work" can mean almost anything — studying, sitting in a meeting, even just feeling tired. In physics it means something very specific:

Work is the amount of energy transferred when an external force causes an object to move over a certain distance.

Here's the key insight that trips people up: if there's no movement, no work is done — no matter how hard you push. Imagine leaning your whole body weight against a brick wall for ten minutes. You'll be exhausted, sweating, and your muscles will genuinely be using energy — but in the physics sense, zero work is done on the wall, because the wall didn't move anywhere. Work only "counts" once there's displacement.

Think of it like this Work is like a "delivery receipt" for energy. A force can want to give energy to an object all day long, but the receipt only gets signed — the transfer only actually happens — once the object has moved. No movement, no delivery, no work done.
Key Formula — Force Parallel to Motion
ΔW = F Δs
  • ΔW = work done, in joules (J)
  • F = force applied, in the same direction as the motion (N)
  • Δs = distance moved (m)

If the force pushes in the direction the object is already moving, the object gains energy. If the force pushes against the motion (like friction), the object loses energy — work is being done against it, converting its kinetic energy into other forms like heat and sound.


When the force is at an angle

Most of the time in real life, forces aren't perfectly parallel to the direction of travel. Think about pulling a sledge with a rope over your shoulder — the rope pulls up and forward, but the sledge only moves forward along the ground.

In these cases, you can't use the whole force — only the component of the force that's actually pointing in the direction of motion does any useful work. The rest of the force (the part pointing "sideways" to the motion) does no work at all, because there's no displacement in that direction.

Key Formula — Force at an Angle θ
W = Fs cos θ   or   W = Fs sin θ
  • Use cos θ when θ is measured from the horizontal (the direction of motion)
  • Use sin θ when θ is measured from the vertical
  • Always pick out the component that's parallel to the displacement — that's the only part that does work
Worked Example
Problem: A barrel of weight 2.5 × 10³ N sits on a frictionless slope inclined at 40° to the horizontal. A force parallel to the slope pushes it 6.0 m up the slope at constant speed. Find the work done.
Step 1: Since the barrel moves at constant speed, the applied force must balance the component of weight pulling it back down the slope.
Step 2: Resolve the weight into components. The component of weight along the slope (parallel to the motion) is W sin 40°, because 40° here is measured from the horizontal — but the slope itself runs at that angle, so the "along-slope" component uses sin.
Step 3: F = 2.5 × 10³ × sin 40° = 1607 N
Step 4: ΔW = F Δs = 1607 × 6.0 = 9642 J
Answer: ΔW ≈ 9.6 × 10³ J (2 s.f.)
Common Mistake Students often grab whichever force is written in the question and multiply it straight by the distance — without checking whether that force is actually parallel to the motion. Always ask yourself: "is this force pointing the same way the object is moving?" If not, resolve it first.
Practice Question 1
A shopping trolley is pushed with a force of 45 N applied at 30° below the horizontal (pushing down and forward). The trolley moves 8.0 m along a flat, frictionless floor. Calculate the work done on the trolley.
Practice Question 2
A 15 N force acts horizontally on a crate, but the crate doesn't move at all because a wall is blocking it. How much work is done on the crate? Explain your answer.
2. Kinetic Energy
The energy of motion

Kinetic energy (Ek) is the energy an object possesses purely because it's moving. A parked car has zero kinetic energy; the same car doing 100 km/h has a lot. The faster something moves, the more kinetic energy it carries.

Key Formula
Ek = ½ m v²
  • Ek = kinetic energy (J)
  • m = mass (kg)
  • v = velocity / speed (m s⁻¹)
This is the #1 exam trap in this whole chapter Notice that only the velocity is squared — not the mass, and not the ½. Students very commonly write Ek = (½mv)² by accident. Also remember: because v is squared, doubling an object's speed doesn't double its kinetic energy — it quadruples it. A car crash at 60 mph isn't twice as dangerous as one at 30 mph — it's four times as dangerous, energy-wise.

Where does this formula come from? It's not just handed down from nowhere — it comes directly from the work-energy relationship you just learned. If a constant force F accelerates a mass m from rest over a distance d, the work done on it is W = Fd. Using F = ma (Newton's Second Law) and the suvat equation v² = u² + 2as (with u = 0, s = d), you can substitute through and show that the work done equals exactly ½mv². Since all of that work has gone into speeding the object up, that ½mv² is now "stored" in the object as kinetic energy — that's the derivation.

Worked Example
Problem: A body travelling at 12 m s⁻¹ has kinetic energy 1650 J. If its speed increases to 45 m s⁻¹, estimate its new kinetic energy.
Step 1: The mass doesn't change, so find it from the initial data: m = 2Ek / v² = (2 × 1650) / 12² = 23 kg
Step 2: Plug the mass and new speed into Ek = ½mv²: Ek = ½ × 23 × 45² = 23 000 J
Answer: New Ek ≈ 23 000 J (2 s.f.) — notice the speed only went up by roughly 3.75×, but the kinetic energy went up by roughly 14×, because of that squared relationship.
Practice Question
A 900 kg car accelerates from 10 m s⁻¹ to 30 m s⁻¹. Calculate the increase in its kinetic energy.
3. Gravitational Potential Energy
Stored "height" energy

Gravitational potential energy (Ep or GPE) is energy stored in an object because of its position within a gravitational field — essentially, how high up it is. Lift something up and you're doing work against gravity; that work gets "banked" as GPE, ready to be released (usually converted into kinetic energy) the moment the object falls.

Think of it like this GPE is like winding up a spring or drawing back a bowstring — you put effort in to get something into a "loaded" position, and that effort doesn't disappear. It just waits there, stored, until gravity is allowed to act and release it as motion.
Key Formula
ΔEgrav = mg Δh
  • ΔEgrav = change in gravitational potential energy (J)
  • m = mass (kg)
  • g = gravitational field strength (9.81 N kg⁻¹ near Earth's surface)
  • Δh = change in height (m)

This equation is only valid for a uniform gravitational field — i.e. situations reasonably close to the Earth's surface, where g doesn't meaningfully change with height. (You'll meet a different, more general GPE formula for large-scale orbital situations elsewhere in the course.) By convention, ground level is usually taken as the "zero" of potential energy, and everything is measured as a change relative to that.

Just like with kinetic energy, this formula is derived from the work-done concept: lifting a mass m through height h requires overcoming its weight (mg) over that distance, so the work done is W = F × d = mg × Δh — and since all that work has gone into raising the object, that same amount is now stored as GPE.

Worked Example
Problem: A man of mass 74 kg climbs five flights of stairs, each 3.7 m high. Find his approximate gain in GPE.
Step 1: Total height, Δh = 5 × 3.7 = 18.5 m
Step 2: ΔGPE = mgΔh = 74 × 9.81 × 18.5
Answer: ΔGPE ≈ 13 000 J (2 s.f.)
Practice Question
A crane lifts a 500 kg steel beam vertically by 12 m. Calculate the gain in gravitational potential energy.
4. The Principle of Conservation of Energy
Energy is never created or destroyed

This is one of the most fundamental laws in all of physics: in a closed system, the total energy in is always equal to the total energy out. Energy doesn't vanish and it doesn't appear from nowhere — it just changes form, or moves from one place to another.

In this chapter, that principle is most often applied to swapping between kinetic energy and gravitational potential energy. Classic examples:

  • A pendulum swinging — GPE at the top of each swing converts fully to KE at the bottom, and back again.
  • An object in free fall — GPE lost equals KE gained (ignoring air resistance).
  • Skiing or skydiving — gravity does the work of speeding you up as your height (and GPE) decreases.

In an "ideal" calculation, we usually assume all the GPE lost converts to KE gained — but in reality, some energy always leaks away as heat (from friction/air resistance) or sound. Exam questions will often tell you what percentage is "lost" so you can factor that into a more realistic calculation.

Core Idea (not a single formula — a strategy!)
Loss in GPE = Gain in KE (ideal case)
  • This lets you find a final velocity from a height drop, or a height from a known final speed.
  • If energy is "lost" to other forms, multiply one side by the fraction that's actually transferred (e.g. × 0.85 if 15% is lost).
Worked Example
Problem: A skier starts from rest and descends 750 m along a slope at 25° to the horizontal. If 15% of the initial GPE is not transferred to KE, find the skier's final speed.
Step 1: Vertical height dropped, h = 750 sin 25°
Step 2: Since 15% is lost, only 85% of GPE becomes KE: Ek = 0.85 Egrav
Step 3: ½mv² = 0.85 × mgh → the masses cancel → v² = 0.85 × 2gh
Step 4: v = √(0.85 × 2 × 9.81 × 750 sin 25°) = 72.7
Answer: Final speed ≈ 73 m s⁻¹
Notice something neat? The mass cancelled out completely in that example! This is a classic feature of GPE ↔ KE conservation problems — it's why a heavy person and a light person released from rest at the top of the same frictionless slide will reach the bottom at exactly the same speed (in the idealised, no-friction case).
Practice Question
A ball of mass 0.2 kg is dropped from a height of 20 m. Assuming no air resistance, calculate its speed just before hitting the ground.
5. Power
How fast energy is transferred

Power is the rate at which energy is transferred (or equivalently, the rate at which work is done). It doesn't tell you how much total energy was transferred — it tells you how quickly that transfer happened.

Two engines could do exactly the same amount of work (say, lifting an identical crate to the same height) — but if one does it in 2 seconds and the other takes 20 seconds, the first engine is ten times more powerful, even though the total work done is identical.

Key Formula
P = E / t = W / t
  • P = power (Watts, W)
  • E or W = energy transferred / work done (J)
  • t = time taken (s)
  • 1 Watt = 1 joule per second (1 W = 1 J s⁻¹)
Worked Example
Problem: A car engine exerts a force of 500 N over a distance of 1.0 km in 200 s. Find the average power developed.
Step 1: Work done, W = F × d = 500 × 1.0 × 10³ = 5 × 10⁵ J
Step 2: Power = W / t = (5 × 10⁵) / 200 = 2500 W
Answer: P = 2500 W = 2.5 kW
Watch your prefixes Power values in exam questions are often huge, so you'll frequently see kW (×10³), MW (×10⁶), and even GW (×10⁹). A quick sanity check on your powers of ten can save you from an easy silly mistake.
Practice Question
A weightlifter raises a 120 kg barbell 2.0 m off the ground in 1.5 s. Calculate the average power developed.
6. Efficiency
How much energy is actually useful?

Efficiency measures how well a system converts the energy you put in into the energy you actually want out. No real machine is 100% efficient — some energy always escapes as heat, sound, or other "wasted" forms.

What counts as "useful" versus "wasted" depends entirely on the system: in a lightbulb, light is useful and heat is wasted; in a heater, heat is useful and any sound produced is wasted. It's the same physics, just a different goal.

Key Formulas
Efficiency = (Useful energy output / Total energy input) × 100%
Efficiency = (Useful power output / Total power input) × 100%
  • Efficiency has no units — it's a ratio, expressed as a decimal (0 to 1) or a percentage (0% to 100%)
  • Power itself is just P = E / t, so you can switch between the energy and power versions freely
The Golden Rule for Efficiency Questions Decide before you start exactly where the energy is being "lost" from the system, and apply the efficiency percentage to that specific stage — not just tacked onto your final answer at the end. If a pump converts kinetic energy into GPE and you're given the pump's efficiency, the losses happen during that conversion, so multiply the KE side (or the GPE side, whichever direction the energy is flowing) by the efficiency fraction at that exact point in your working — don't just calculate the ideal answer and multiply by the efficiency afterwards as an afterthought, since that can easily lead to the wrong equation setup.
Worked Example 1
Problem: An electric motor with 35% efficiency lifts a 7.2 kg load 5 m in 3 s. Find the power input to the motor.
Step 1: Power output = useful energy transferred ÷ time. Useful energy here is GPE gained: mgh = 7.2 × 9.81 × 5 = 353.16 J
Step 2: Power output = 353.16 / 3 = 117.72 W
Step 3: Efficiency = (Power out / Power in) × 100 → Power in = (Power out × 100) / Efficiency
Step 4: Power in = (117.72 × 100) / 35 = 336 W
Answer: Power input ≈ 336 W
Worked Example 2 — spotting where the loss happens
Problem: A hydraulic ram pump has a 700 kg column of water moving at 3.5 m s⁻¹ when a valve shuts. This kinetic energy is used to lift a small quantity of water by 12 m, and the pump is 20% efficient. Find the mass of water lifted.
Step 1: The pump converts KE → GPE, and its efficiency describes that exact conversion. So: 0.2 × ½mv² = mgΔh (careful — the "m" on the left is the 700 kg water column; the "m" on the right is the unknown mass being lifted!)
Step 2: 0.2 × ½ × 700 × 3.5² = m × 9.81 × 12
Step 3: 857.5 = 117.72 m → m = 7.284
Answer: Mass lifted ≈ 7.3 kg (2 s.f.)
Practice Question
A pulley system is used to raise a 40 kg load by 6 m in 8 s. The motor driving it has a power input of 400 W. Calculate the efficiency of the system.
What to Memorise
Term / FormulaMeaning
Work
ΔW = FΔs
Energy transferred when a force moves an object over a distance, force parallel to motion.
Work at an angle
W = Fs cos θ / Fs sin θ
Only the component of force parallel to the displacement does work. cos θ if θ from horizontal, sin θ if θ from vertical.
Kinetic Energy
Ek = ½mv²
Energy due to motion. Only v is squared — not m, not the ½. Doubling speed quadruples Ek.
Gravitational PE
ΔEgrav = mgΔh
Energy stored due to height, in a uniform gravitational field (g = 9.81 N kg⁻¹ near Earth).
Conservation of Energy Total energy in a closed system is constant. GPE lost = KE gained (in the ideal, no-loss case).
Power
P = E/t = W/t
Rate of energy transfer / rate of doing work. 1 W = 1 J s⁻¹.
Efficiency
(Useful out / Total in) × 100%
Fraction of energy or power that's usefully transferred. No units. Always < 100% in reality.
Concepts Checklist
Exam Tips & Common Mistakes
⚠ Trap: Wrong force component The most common mistake in "work" questions is using a force that isn't actually parallel to the direction of motion. Always resolve first, and double-check whether you need cos θ or sin θ depending on how the angle is defined.
⚠ Trap: Squaring the wrong thing In Ek = ½mv², only the velocity is squared. Writing (½mv)² is a very easy mistake to make under exam pressure — get in the habit of writing the formula out fully before substituting numbers.
✓ Good habit: Watch for cancelling masses In conservation-of-energy problems (GPE ↔ KE), the mass often cancels out completely. If you're asked for a final speed or a drop height and the mass isn't given in the question, that's usually your clue that it's about to cancel — don't panic if you can't find a value for it!
✓ Good habit: Locate the loss before calculating For efficiency questions, work out which stage of the energy transfer the efficiency percentage applies to (before you write any equations). Multiplying your final "ideal" answer by the efficiency at the very end, instead of applying it at the correct point in the chain, is a common source of lost marks — even when your final number looks plausible.
⚠ Trap: Losing/gaining a negative sign Energy is a scalar quantity — it has no direction. If a question asks for a "loss of kinetic energy," give a positive magnitude, not a negative number.
⚠ Trap: Mixing up power prefixes Power answers are often large. Remember: kW = ×10³, MW = ×10⁶, GW = ×10⁹. A slip here can turn a correct method into a wrong final answer.
✓ Exam pattern to expect Multi-step questions typically chain these ideas together: work done → converted to KE → converted to GPE → with an efficiency factor applied somewhere along the way. Always sketch out the energy pathway (what converts into what) before diving into the algebra.
Work, Energy & Power — Revision Guide · Built for offline study · Click any "Show Answer" button to check your work.
🔓 Read the full Work, Energy & Power note — free You're seeing the preview · free account, no card needed
What's inside
📖 Revision notes 🎯 Learn mode ✦ AI flashcards ✓ Instant AI marking 🧊 3D explorers 🧪 Experiments & simulations 📈 Progress tracking

Read the full Work, Energy & Power notes free

That's the preview — create a free account to read the rest, plus flashcards and practice questions with instant AI marking. No credit card.

Unlock the full notes free →