Physics/Lessons/Mechanics/Work, Energy & Power
Physics · Mechanics · AS Level

Work, Energy & Power

Push a stalled car and you're doing work. Stand and push against a brick wall until your arms shake, and — according to physics, however unfair it feels — you've done no work at all. This lesson pins down exactly what "work" means in physics, and follows it through to energy and power, the two ideas built directly on top of it.

10 min read

Defining work done

Work is the amount of energy transferred when a force moves an object. In physics, work is done whenever a force causes a displacement — and the precise amount depends not just on how big the force is and how far the object moves, but on the angle between them. The general definition is:

$W = Fs\cos\theta$

where $F$ is the force, $s$ is the displacement, and $\theta$ is the angle between the direction of the force and the direction of the displacement. Work is measured in joules (J), and despite being built from a force and a displacement — both vectors — work itself is a scalar: it has a size but no direction.

DerivationW = Fs cos θ, from resolving the force
1
Suppose a force $F$ acts on an object at an angle $\theta$ to the direction the object actually moves in, and it moves a distance $s$.
2
Only the component of the force acting along the direction of motion contributes to moving the object that way. Resolving $F$ into components gives $F\cos\theta$ along the direction of motion, and $F\sin\theta$ perpendicular to it.
The perpendicular component doesn't push the object along its path at all — it acts sideways to the motion.
3
Work done is this along-motion component multiplied by the distance moved: $W = (F\cos\theta)\times s = Fs\cos\theta$.
4
Check the two limiting cases. When $\theta = 0$ (force and displacement in the same direction), $\cos\theta = 1$, giving the familiar $W = Fs$. When $\theta = 90°$ (force perpendicular to displacement), $\cos\theta = 0$, giving $W = 0$.
A force acting perpendicular to an object's motion — like the normal reaction force on something sliding along a level surface — does no work on it at all, however large that force is.
▪ derived

This is counter-intuitive at first. Carry a heavy suitcase at constant height across a level floor and it feels like work, but the force you exert (upward, supporting its weight) is perpendicular to the direction you're moving (horizontal), so by this definition you're doing zero work on the suitcase. That effort goes into muscle tension and heat, none of which counts as mechanical work on the case.

Work from a force-displacement graph

When a force is constant, $W = Fs$ is all you need. But plenty of real forces change as an object moves — a spring pulling harder the further it's stretched. For these, plotting force against displacement and reading the area under the graph gives the work done, whatever shape the curve takes: breaking the motion into tiny steps where the force is essentially constant, the work in each step is a thin sliver of area, and the total is the area under the whole curve.

A force-displacement graph with force on the vertical axis and displacement on the horizontal axis. The line is not straight — it curves, rising from the origin, reaching a peak, then falling, representing a varying force such as stretching a spring or an irregular pushing force. The area under the curve between two marked displacement values is shaded and labelled 'work done = area under graph'.
For a varying force, the work done is still the area under the force-displacement graph, even though W = Fs no longer applies directly.

Power: how quickly work gets done

Two cranes might lift the identical crate to the identical height, doing exactly the same work, but if one takes five seconds and the other fifty, they're not equivalent machines. Power captures that difference: the rate at which work is done, or equivalently the rate at which energy is transferred.

$P = \dfrac{\Delta W}{\Delta t}$

measured in watts (W), where one watt is one joule of work done, or one joule of energy transferred, per second.

DerivationP = Fv, for a constant force
1
Start from the definition $P = \dfrac{\Delta W}{\Delta t}$.
2
For a constant force acting in the direction of motion, work done is $W = Fs$, so $\Delta W = F\,\Delta s$.
F stays fixed, so only the displacement s is changing here.
3
Substitute: $P = \dfrac{F\,\Delta s}{\Delta t} = F\left(\dfrac{\Delta s}{\Delta t}\right)$.
4
Recognise $\dfrac{\Delta s}{\Delta t}$ as velocity $v$, giving $P = Fv$.
This is hugely useful in practice — for a vehicle travelling at constant velocity against a resistive force, the driving force needed is just P divided by v, straight from this relationship.
▪ derived

Efficiency

No real machine converts all the power fed into it into useful output — some is always lost, typically as heat through friction or resistance. Efficiency measures how much of the input power ends up doing something useful:

$\text{efficiency} = \dfrac{\text{useful output power}}{\text{input power}}$

often multiplied by 100 for a percentage. Efficiency can never exceed 100%, which would mean getting more useful energy out than you put in, violating conservation of energy.

Conservation of energy

The principle of conservation of energy: energy cannot be created or destroyed, only transferred from one form to another or one place to another. The total energy of an isolated system never changes, even as it shuffles between stores — kinetic, gravitational potential, elastic, thermal. Two of these stores matter enormously for mechanics at this level.

Gravitational potential energy

Lift an object of mass $m$ through a height $\Delta h$, moving it at a slow, constant velocity so nothing is being wasted accelerating it. The change in its gravitational potential energy equals the work you do against gravity to lift it.

DerivationΔEp = mgΔh, from W = Fs
1
To lift the object at constant velocity, the upward force you apply must equal its weight: $F = mg$.
Constant velocity means zero acceleration, so by Newton's first law the resultant force is zero — the applied force exactly balances weight.
2
The force acts straight upward, in the same direction as the displacement $\Delta h$, so $\theta = 0$ and $W = Fs$ applies directly: $W = mg\,\Delta h$.
3
All of this work goes into increasing the object's gravitational potential energy, since its kinetic energy hasn't changed (constant velocity, zero acceleration).
This is exactly what "potential energy" means — energy stored as a result of position, released again as kinetic energy if the object is allowed to fall.
4
So $\Delta E_p = mg\,\Delta h$.
▪ derived

Kinetic energy

Kinetic energy is the energy an object has because it's moving. Its formula can be derived from the work-energy theorem — the idea that the work done by a resultant force on an object equals the change in that object's kinetic energy.

DerivationEk = ½mv², via the work-energy theorem
1
Consider an object of mass $m$, starting from rest ($u=0$), accelerated by a constant resultant force $F$ over a distance $s$, reaching final velocity $v$.
2
The work done by the resultant force is $W = Fs$, and by Newton's second law, $F = ma$, so $W = mas$.
3
Use the equation of motion $v^2 = u^2 + 2as$ with $u=0$: $v^2 = 2as$, which rearranges to $s = \dfrac{v^2}{2a}$.
This lets us eliminate s and a separately, replacing them with the single ratio v²/2a.
4
Substitute into $W = mas$: $W = ma\times\dfrac{v^2}{2a} = \dfrac{1}{2}mv^2$.
The a cancels cleanly, leaving an expression purely in terms of mass and final velocity.
5
Since the object started from rest, all of this work has gone into giving it kinetic energy: $E_k = \tfrac{1}{2}mv^2$.
▪ derived

Energy conservation in falling, rising, and resisted motion

Put gravitational potential energy and kinetic energy together and you get one of the most useful problem-solving ideas in mechanics. For an object falling freely with no resistive forces, mechanical energy — the sum of kinetic and gravitational potential energy — stays exactly constant: whatever $E_p$ is lost reappears exactly as $E_k$ gained. A ball thrown straight up converts kinetic energy into potential energy on the way up and back again on the way down, with the total unchanged at every instant.

Once a resistive force like drag or friction enters, mechanical energy on its own is no longer conserved — some is continuously transferred into heat as the resistive force does negative work. A skydiver at terminal velocity loses gravitational potential energy steadily but gains no kinetic energy, since speed is constant — all that lost $E_p$ converts to heat in the surrounding air. Total energy is still exactly conserved; only the mechanical portion shrinks.

Worked exampleSpeed at the bottom of a frictionless incline
1
A block of mass $2.0$ kg is released from rest at the top of a frictionless slope of length $5.0$ m, inclined at $20°$ to the horizontal. Find its speed at the bottom.
This is the same inclined-plane geometry as resolving weight along a slope — but here it's energy conservation doing the work, not a force calculation.
A block sits near the top of a frictionless incline of length L equals 5 metres, inclined at 20 degrees to the horizontal. An angle arc at the base of the slope marks the 20 degree angle between the incline and the horizontal ground. The vertical height of the incline is labelled h.
The block slides down the frictionless slope of length $L = 5$ m at $20°$ to the horizontal.
2
The height dropped is $h = L\sin\theta$, using the length of the slope $L$ as the hypotenuse: $h = 5.0\sin20° = 5.0 \times 0.342 = 1.71$ m.
3
With no friction, mechanical energy is conserved: loss in gravitational PE = gain in KE, so $mgh = \frac12mv^2$.
4
The mass $m$ cancels from both sides: $gh = \frac12v^2$, so $v = \sqrt{2gh}$.
Notice the final speed doesn't depend on the mass at all — only on how far it's fallen vertically, exactly as with any object in free fall.
5
$v = \sqrt{2 \times 9.81 \times 1.71} = \sqrt{33.5} = 5.8\,\text{m}\,\text{s}^{-1}$ (to 2 s.f.).
▪ answer
Worked examplePower output and efficiency of a lifting motor
1
An electric motor lifts a $50$ kg load a height of $12$ m in $20$ s at constant velocity. The motor draws $400$ W of input power from the mains. Find the motor's useful power output and its efficiency.
2
Useful work done is the gain in gravitational PE: $W = mgh = 50 \times 9.81 \times 12 = 5886$ J.
3
Useful output power: $P = \dfrac{W}{t} = \dfrac{5886}{20} = 294\,\text{W}$ (to 3 s.f.).
4
Efficiency: $\dfrac{\text{useful output power}}{\text{input power}} = \dfrac{294}{400} = 0.735$.
5
As a percentage, the motor is $73.5\%$ efficient — the remaining $26.5\%$ of the input power is lost, mostly as heat in the motor's windings and bearings.
▪ answer

Key takeaways

  • Work done is $W = Fs\cos\theta$; only the component of a force along the direction of motion contributes, so a force perpendicular to displacement does zero work.
  • Power is the rate of doing work, $P = \Delta W/\Delta t$, and for a constant force in the direction of motion this becomes the useful form $P = Fv$.
  • The area under a force-displacement graph gives work done even when the force varies, extending W = Fs to non-constant forces.
  • Efficiency is useful output power divided by input power, and can never exceed 100% without violating conservation of energy.
  • Conservation of energy: energy is never created or destroyed, only transferred between stores or from place to place.
  • Gravitational potential energy change is $\Delta E_p = mg\Delta h$, derived from the work done lifting an object at constant velocity against its weight.
  • Kinetic energy is $E_k = \tfrac{1}{2}mv^2$, derived from the work-energy theorem using F = ma and v² = u² + 2as.
  • Mechanical energy (kinetic + potential) is conserved only when no resistive forces act; otherwise the "missing" energy is transferred to heat, while total energy remains exactly conserved.
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