What is the difference between work and power?

Work is energy transferred by a force (W = Fs cos θ), measured in joules. Power is how fast that happens (P = W ÷ Δt), measured in watts. Lifting a box onto a shelf takes the same work whether you do it slowly or quickly, but doing it quickly takes more power.

Is work a vector or a scalar?

A scalar. It has no direction, even though it is calculated from two vectors. It can be negative, which means the force is taking energy away from the object, as friction does.

Does the path matter for gravitational potential energy?

No. ΔU = mgΔh depends only on the change in vertical height, because gravity is a conservative force. A steep path and a gentle path between the same two heights give the same ΔU. The path does matter for the work done against friction.

Which of these forces is non-conservative?

the electrostatic force between two charges

Year 11 Physics · Fundamentals of mechanics · Work and energy · 12 min read

Work, energy and power

Energy gives you a second way into mechanics problems. Instead of following forces and accelerations moment by moment, you compare the energy at the start with the energy at the end. When a question asks for a speed at a different height, energy is usually the shortest route.

This guide covers work, kinetic energy, gravitational potential energy, conservative and non-conservative forces, conservation of mechanical energy and power. Then the mistakes that cost marks, a quick quiz and a full worked example of a cyclist coasting down a hill.

What you need to know

Compare conservative forces with non-conservative forces.
For total mechanical energy to be conserved, state the conditions.
For accelerated straight-line motion, use W = F_∥s = Fs cos θ to link the work done on an object to its change in kinetic energy.
Within a uniform gravitational field, solve problems where gravitational potential energy changes, using ΔU = mgΔh.
Use KE = ½mv² to solve kinetic energy problems.
Solve problems in which mechanical energy is conserved, using W = ΔU and W = ΔKE.
For an object in a uniform gravitational field, analyse its energy changes and the work done, including how its gravitational potential energy varies with height and which forces do work.
Use P = W ÷ Δt = ΔE ÷ Δt to solve power problems (this dot point sits in the electricity part of the syllabus, but it applies to any energy transfer).

The equations you will use:

W = F_∥s = Fs cos θθ is the angle between the force and the displacement. Work is in joules (J).

KE = ½mv²Kinetic energy, in J. It depends on the square of the speed.

ΔU = mgΔhChange in gravitational potential energy near Earth's surface (a uniform field).

W = ΔKEThe work–energy theorem: the net work done on an object equals its change in kinetic energy.

KE_i + U_i = KE_f + U_fConservation of mechanical energy, when only conservative forces (such as gravity) do work.

P = W ÷ Δt = ΔE ÷ ΔtPower: the rate of doing work or transferring energy, in watts (1 W = 1 J s⁻¹).

Dot points: PY11-FM-36, PY11-FM-37, PY11-FM-38, PY11-FM-39, PY11-FM-40, PY11-FM-41, PY11-FM-42, PY11-EM-27 (2025 Physics syllabus, Year 11), in our own words.

Mini lesson

Work: a force acting through a distance

A force does work on an object when the object moves and the force has a part along the motion. W = Fs cos θ, where θ is the angle between the force and the displacement. Only the part of the force along the displacement, F cos θ, does any work.

Work is a scalar measured in joules, but it can be negative. A force with a part along the motion does positive work (it gives the object energy). A force at right angles to the motion does no work: the normal force on a box sliding across a floor, for example. A force against the motion, like friction, does negative work (it takes energy away).

Only the part of F along the displacement does work: W = (F cos θ) × s. The vertical part, F sin θ, is at right angles to the motion, so it does none.

Kinetic energy and the work–energy theorem

A moving object has kinetic energy, KE = ½mv². Because the speed is squared, doubling the speed makes the kinetic energy four times as big. That is why a car's braking distance roughly quadruples when its speed doubles: the brakes have to do four times the work.

The work–energy theorem links forces to energy: the net work done on an object (the work done by all the forces together) equals its change in kinetic energy, W = ΔKE. Positive net work speeds it up; negative net work slows it down.

Gravitational potential energy

Lifting an object at a steady speed takes a force mg through a height Δh, so the work done on it is mgΔh. That energy is stored as gravitational potential energy: ΔU = mgΔh, where Δh is the change in vertical height.

This formula works near Earth's surface, where the gravitational field is uniform (g = 9.8 m s⁻² everywhere). Only the change in height matters, never the path. Raise a 2.0 kg bag 1.5 m onto a shelf by any route and its U rises by 2.0 × 9.8 × 1.5 = 29 J. When it falls, gravity does positive work and U decreases by the same amount.

Conservative and non-conservative forces

A conservative force does work that depends only on where the object starts and finishes, not on the path between. Gravity is conservative: going down any slope of height h, it does work mgh, and over a round trip back to the start it does zero work. The energy it takes away is stored as potential energy and can be fully recovered.

A non-conservative force, such as friction, air resistance or drag, does work that depends on the path: a longer path means more (negative) work. That energy is not stored; it is transferred out of the mechanical energy, mostly as thermal energy and sound, and can't be recovered.

Gravity does the same work, mgh, on both paths from A to B: it is conservative. Friction does more negative work on the longer path: it is non-conservative.

Conservation of mechanical energy

Mechanical energy is KE + U. It stays constant when only conservative forces do work: no friction or air resistance doing work, and any other force (such as a normal force or a smooth track) acting at right angles to the motion so that it does no work. Then KE_i + U_i = KE_f + U_f.

For an object starting from rest and dropping a height h, mgh = ½mv², so v = √(2gh). The mass cancels: a heavy and a light object reach the same speed. When friction does act, mechanical energy decreases, and the amount lost equals the work done against friction.

With no friction or air resistance, U turns into KE but the total stays the same. Halfway down in height, the energy is half U and half KE.

Power

Power is how fast work is done or energy is transferred: P = W ÷ Δt = ΔE ÷ Δt, in watts (1 W = 1 J s⁻¹). Two students who climb the same stairs do the same work against gravity; the one who climbs faster has the greater power.

Example: a 70 kg student climbs stairs 4.0 m high in 5.0 s. The work done against gravity is mgΔh = 70 × 9.8 × 4.0 = 2744 J, so the power is 2744 ÷ 5.0 ≈ 550 W.

Common mistakes

Calculating the work done by an angled pull as F × s.Only the part along the displacement does work: W = Fs cos θ, with θ the angle between the force and the displacement.
Saying the normal force or the weight does work on a box sliding along a level floor.Both act at right angles to the motion, so cos 90° = 0 and they do no work. Only forces with a part along the motion do work.
Thinking that doubling the speed doubles the kinetic energy.KE = ½mv², so doubling v makes KE four times as big. Forgetting to square v is the most common KE slip.
Putting the distance along a slope into mgΔh.Δh is the change in vertical height. A 150 m road that drops 12 m has Δh = 12 m.
Using KE_i + U_i = KE_f + U_f when friction or air resistance is doing work.Mechanical energy is conserved only when non-conservative forces do no work. Otherwise the mechanical energy lost equals the work done against friction.
Mixing up power and energy: 'a more powerful motor does more work'.Power is work per second. A more powerful motor does the same work faster, or more work in the same time. Energy is in J; power is in W.

Worked example

A cyclist and her bike have a combined mass of 65 kg. She starts from rest at the top of a hill and coasts, without pedalling, down 150 m of road to the bottom, which is 12 m lower than the start. At the bottom her speed is 13.0 m s⁻¹.

(a) Calculate the speed she would have at the bottom if no mechanical energy were lost. (b) Calculate the mechanical energy lost to resistive forces. (c) Calculate the average resistive force acting along the road.

(6 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. With no losses, mgh = ½mv² gives v = √(2gh) ≈ √(2 × 10 × 12) ≈ 15 m s⁻¹. She only reaches 13 m s⁻¹, and (13 ÷ 15)² ≈ 0.75, so about a quarter of her roughly 65 × 10 × 12 ≈ 7800 J of potential energy is lost, around 2 kJ. Spread over 150 m, that is an average force of roughly 13 to 15 N.

K · Knowns

I sketch the hill with the values on it. The key point is that the height for mgΔh is the 12 m vertical drop, not the 150 m of road; the 150 m is the distance the resistive force acts over.

m = 65 kg, u = 0, v = 13.0 m s⁻¹ at the bottom
Δh = 12 m drop (vertical), s = 150 m along the road
g = 9.8 m s⁻²

Knowns on the sketch: 65 kg, from rest, a 12 m vertical drop over 150 m of road, 13.0 m s⁻¹ at the bottom. Unknown: the average resistive force, which acts back up the road.

N · Not known

Three answers: an ideal speed, an energy loss, and an average force.

(a) v_ideal = ? m s⁻¹ (b) E_lost = ? J (c) F = ? N

O · Organise

With no losses, all the potential energy she loses becomes kinetic energy, and the mass cancels. The real loss is the potential energy lost minus the kinetic energy gained. The resistive force acts against her motion, so its work is −Fs (θ = 180°), and that work is the mechanical energy lost. The normal force does no work, because it is at right angles to the road.

(a) mgΔh = ½mv² → v = √(2gΔh)
(b) E_lost = mgΔh − ½mv²
(c) Fs = E_lost → F = E_lost ÷ s

W · Work it out

I keep full values and round only the final answers.

(a) v = √(2 × 9.8 × 12) = √235.2 = 15.34 m s⁻¹
U lost = 65 × 9.8 × 12 = 7644 J
KE gained = ½ × 65 × 13.0² = 5492.5 J
(b) E_lost = 7644 − 5492.5 = 2151.5 J
(c) F = 2151.5 ÷ 150 = 14.34 N

U · Units

√(m s⁻² × m) gives m s⁻¹; energy in joules; J ÷ m = N. Two or three significant figures to match the data.

(a) 15.3 m s⁻¹ (b) 2.15 × 10³ J ≈ 2.2 kJ (c) 14 N

D · Direction

Speed and energy are scalars, so (a) and (b) need no direction. The resistive force is a vector: it acts back up the road, against her motion, which is why its work is negative. As a sense check, her real speed, 13.0 m s⁻¹, is below the ideal 15.3 m s⁻¹, as it must be if energy is lost.

Average resistive force ≈ 14 N, acting up the road (opposite to her motion)

(a) 15.3 m s⁻¹. (b) About 2.2 kJ (2150 J). (c) About 14 N, acting back up the road.

Check against the prediction. I predicted about 15 m s⁻¹, a loss of around 2 kJ and a force of 13 to 15 N. 15.3 m s⁻¹, 2.15 kJ and 14 N all match. About 28% of the potential energy was lost, close to my quarter.

Exam advice

What markers look for

Before choosing equations, ask what is conserved. When a speed at a different height is asked for, state that mechanical energy is conserved (and why) and show it mathematically, rather than reaching for forces first.NESA marking feedback, 2025, summarised
If a question asks about energy, answer with energy: name each transformation (for example, gravitational potential energy to kinetic energy, and kinetic energy to thermal energy through friction). Explanations that only talked about forces missed the point.NESA marking feedback, 2024, summarised
Finish every number with a unit, and the right one: joules for work and energy, watts for power. Reporting answers in the correct SI units was one of the areas students needed to improve.NESA marking feedback, 2023, summarised
Show the equation, the substitution and the answer. Correct working earns marks even when the final number slips; a bare wrong answer earns none.General advice from NESA marking feedback, 2021 to 2025, summarised

Summaries in our own words. Read NESA's own feedback with the exam packs: 2023 · 2024 · 2025.

Where this leads in Year 12

Energy runs through Year 12. For gravity beyond Earth's surface, ΔU = mgΔh is replaced by U = −GMm ÷ r, and conservation of mechanical energy gives escape velocity and the energy of orbiting satellites. A charged particle accelerated by an electric field gains kinetic energy equal to the work done on it. Electromagnetic induction and transformers are energy conservation again, with power P = VI.

FAQ

What is the difference between work and power?: Work is energy transferred by a force (W = Fs cos θ), measured in joules. Power is how fast that happens (P = W ÷ Δt), measured in watts. Lifting a box onto a shelf takes the same work whether you do it slowly or quickly, but doing it quickly takes more power.
Is work a vector or a scalar?: A scalar. It has no direction, even though it is calculated from two vectors. It can be negative, which means the force is taking energy away from the object, as friction does.
When is mechanical energy conserved?: When only conservative forces, such as gravity, do work. In practice: no friction or air resistance doing work, and any other forces (like a normal force) acting at right angles to the motion. Then KE + U stays constant.
Does the path matter for gravitational potential energy?: No. ΔU = mgΔh depends only on the change in vertical height, because gravity is a conservative force. A steep path and a gentle path between the same two heights give the same ΔU. The path does matter for the work done against friction.

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