Work-Energy Principle

The work–energy principle states that the net work done on a particle equals the change in its kinetic energy. It is the space-integral counterpart of the impulse–momentum principle (the time-integral): where impulse links force to a change in momentum, work links force to a change in kinetic energy. Its great strength is that it relates speeds to distances without ever needing the time — ideal whenever a question gives you "how far" rather than "how long".

1 Where this sits in AQA 7367

Work, energy and power sit in the Mechanics option, Paper 3 (7367/3M), and recur whenever a collision or motion question asks about energy. Paper 3 is weighted AO1 40% / AO2 25% / AO3 35%, and energy methods are prized in AO3 because they often crack a problem that direct force–acceleration analysis would make messy (variable forces, slopes, friction over a distance). It builds on $\vec F = m\vec a$ , kinematics, and definite integration from A-Level Maths. Energy questions frequently form the second half of a longer problem whose first half is a collision: a typical structure is "find the velocity after the collision (momentum + restitution), then find how far the body slides before stopping (work–energy)". Recognising which principle governs which stage — and switching cleanly between them — is a hallmark of a strong Further-Mechanics candidate.

2 Core theory

Work done by a force

For a constant force $F$ acting through a displacement $d$ at angle $\theta$ to the displacement,

$W = Fd\cos\theta.$

$\theta = 0$ (force along motion): $W = Fd$ (positive work).
$\theta = 90^\circ$ (force perpendicular): $W = 0$ — e.g. the normal reaction, and the tension in circular motion, do no work.
$\theta = 180^\circ$ (force opposing): $W = -Fd$ (negative work, e.g. friction or air resistance).

The $\cos\theta$ factor is simply the projection of the force onto the direction of motion: only the component of $\vec F$ along the displacement transfers energy. A force at right angles changes direction but not speed, so it does no work — which is exactly why a satellite in a circular orbit, with gravity always perpendicular to its motion, keeps a constant speed.

For a variable force along the line of motion,

$W = \int_{x_1}^{x_2} F(x)\,\mathrm{d}x \quad(\text{the area under the force–displacement graph}).$

This is the displacement-space analogue of the impulse integral $\int F\,\mathrm{d}t$ : the constant-force result $W = Fd$ is just the special case where $F$ is constant over the interval.

The work–energy principle

Starting from $F = m\dfrac{\mathrm{d}v}{\mathrm{d}t}$ and using the chain rule $\dfrac{\mathrm{d}v}{\mathrm{d}t} = v\dfrac{\mathrm{d}v}{\mathrm{d}x}$ ,

$\int F\,\mathrm{d}x = \int m v\,\frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = \int_u^v m v\,\mathrm{d}v = \tfrac12 mv^2 - \tfrac12 mu^2,$

$\boxed{\;W_{\text{net}} = \tfrac12 mv^2 - \tfrac12 mu^2 = \Delta(\text{KE})\;}.$

Net work done equals change in kinetic energy. If several forces act, $W_{\text{net}}$ is the sum of the works done by each.

Energy forms and conservation

Kinetic energy: $KE = \tfrac12 mv^2$ .
Gravitational potential energy: raising mass $m$ by height $h$ does work $mgh$ against gravity, stored as $GPE = mgh$ . The zero of GPE is arbitrary — choose a convenient reference level (often the lowest point of the motion) and measure $h$ from there; only differences in GPE are physical.
For conservative forces only (gravity, elastic), total mechanical energy is constant: $\tfrac12 mv^2 + mgh = \text{constant}.$ When a non-conservative force such as friction also acts, the shortfall $\Delta(KE + GPE)$ equals the (negative) work done by that force — energy is not destroyed, merely transferred to heat. This "energy audit" view — total energy in = total energy out + energy dissipated — is often the clearest way to lay out a multi-force problem.

Power

Power is the rate of doing work; for a force along the motion,

$P = \frac{\mathrm{d}W}{\mathrm{d}t} = Fv,$

measured in watts $(1\ \mathrm{W} = 1\ \mathrm{J\,s^{-1}})$ . For a vehicle, the driving force is $F = P/v$ — note this rises as $v$ falls, which is why cars can pull harder uphill at low speed.

The relation $P = Fv$ follows from $P = \dfrac{\mathrm{d}W}{\mathrm{d}t}$ and $\mathrm{d}W = F\,\mathrm{d}x$ : $P = F\dfrac{\mathrm{d}x}{\mathrm{d}t} = Fv$ . For vehicle problems the standard three-line setup is: (1) find the driving force $F = P/v$ at the speed in question; (2) write Newton's second law along the motion, $F - (\text{resistances}) = ma$ ; (3) read off the acceleration, or set $a=0$ for maximum speed. At maximum speed the acceleration is zero, so the driving force exactly balances the total resistance — this is the single most common power exam scenario.

When to use work–energy versus $F = ma$

A reliable rule of thumb: if a problem links speed and distance and does not mention time, reach for the work–energy principle; if it links speed and time (or asks for an acceleration at an instant), use Newton's second law. Energy methods shine on slopes, variable forces, and multi-stage journeys where tracking the time would be painful, because they care only about the endpoints, not the path's time history.

3 Worked examples with mark scheme

Worked Example 1 — variable force

A force $F = 3x^2\ \mathrm{N}$ acts on a particle along its line of motion from $x = 0$ to $x = 4\ \mathrm{m}$ . Find the work done.

$W = \int_0^4 3x^2\,\mathrm{d}x = \big[x^3\big]_0^4 = 64\ \mathrm{J}.$

Mark scheme: M1 integrate $F$ with respect to $x$ ; A1 $64\ \mathrm{J}$ . The work is the area under the force–displacement curve.

Worked Example 2 — power on an incline

A car of mass $1000\ \mathrm{kg}$ drives up a slope where $\sin\alpha = \tfrac{1}{20}$ at a steady $25\ \mathrm{m\,s^{-1}}$ against a resistance of $400\ \mathrm{N}$ . Find the engine power. (Take $g = 9.8\ \mathrm{m\,s^{-2}}$ .)

At constant speed there is no acceleration, so the driving force balances resistance and the weight component down the slope:

\begin{aligned} F &= 400 + mg\sin\alpha = 400 + 1000(9.8)\big(\tfrac{1}{20}\big) = 400 + 490 = 890\ \mathrm{N}\\ P &= Fv = 890 \times 25 = 22250\ \mathrm{W} = 22.25\ \mathrm{kW}. \end{aligned}

Mark scheme: M1 driving force = resistance + weight component (no acceleration); A1 $F = 890\ \mathrm{N}$ ; M1 $P = Fv$ ; A1 $22.25\ \mathrm{kW}$ . (4 marks.)

Worked Example 3 — work–energy with friction on a slope

A block of mass $2\ \mathrm{kg}$ slides from rest $3\ \mathrm{m}$ down a rough plane inclined at $30^\circ$ ; the coefficient of friction is $0.3$ . Find its speed at the bottom. (Take $g = 9.8\ \mathrm{m\,s^{-2}}$ .)

Normal reaction: $R = mg\cos 30^\circ = 2(9.8)\dfrac{\sqrt3}{2} = 9.8\sqrt3 \approx 16.97\ \mathrm{N}$ .

\begin{aligned} \text{work by gravity} &= mg\,d\sin 30^\circ = 2(9.8)(3)(0.5) = 29.4\ \mathrm{J}\\ \text{work by friction} &= -\mu R\,d = -0.3(9.8\sqrt3)(3) \approx -15.27\ \mathrm{J}\\ W_{\text{net}} &= 29.4 - 15.27 = 14.13\ \mathrm{J}\\ \tfrac12(2)v^2 &= 14.13 \;\Rightarrow\; v = \sqrt{14.13} \approx 3.76\ \mathrm{m\,s^{-1}}. \end{aligned}

Mark scheme: M1 $R = mg\cos\theta$ ; M1 work by gravity $mgd\sin\theta$ ; M1 work by friction $-\mu R d$ ; M1 apply $W_{\text{net}} = \Delta KE$ ; A1 $3.76\ \mathrm{m\,s^{-1}}$ . (5 marks.) Note the normal reaction does no work — it is perpendicular to the motion.

Worked Example 4 — smooth horizontal push

A box of mass $5\ \mathrm{kg}$ , initially at rest, is pushed $4\ \mathrm{m}$ along a smooth horizontal floor by a constant horizontal force of $20\ \mathrm{N}$ . Find its final speed.

The only force doing work is the applied force (the floor is smooth, and gravity and the normal reaction are perpendicular to the motion):

$W = Fd = 20 \times 4 = 80\ \mathrm{J}; \qquad \tfrac12(5)v^2 - 0 = 80 \Rightarrow v^2 = 32 \Rightarrow v = 4\sqrt2 \approx 5.66\ \mathrm{m\,s^{-1}}.$

Mark scheme: M1 $W = Fd$ ; A1 $80\ \mathrm{J}$ ; M1 $W = \Delta KE$ ; A1 $4\sqrt2\ \mathrm{m\,s^{-1}}$ . This is the cleanest possible work–energy problem — one force, one distance, no time anywhere in sight.

Worked Example 5 — maximum height by energy

A ball of mass $0.5\ \mathrm{kg}$ is thrown vertically upward at $10\ \mathrm{m\,s^{-1}}$ . Ignoring air resistance, find the maximum height. (Take $g = 9.8\ \mathrm{m\,s^{-2}}$ .)

At the highest point $v = 0$ . By conservation of energy (only gravity acts), the kinetic energy converts entirely to gravitational potential energy:

$\tfrac12 m u^2 = mgh \;\Rightarrow\; \tfrac12(0.5)(10^2) = (0.5)(9.8)h \;\Rightarrow\; 25 = 4.9h \;\Rightarrow\; h \approx 5.10\ \mathrm{m}.$

Mark scheme: M1 equate KE to GPE (or apply $W_{\text{net}} = \Delta KE$ with work $-mgh$ ); A1 $5.10\ \mathrm{m}$ . Note the mass cancels — the height depends only on the launch speed and $g$ , so a heavy and a light ball thrown at the same speed reach the same height.

Worked Example 6 — power on a level road

A car of mass $1200\ \mathrm{kg}$ travels at a constant $30\ \mathrm{m\,s^{-1}}$ on a level road against a resistance of $600\ \mathrm{N}$ . Find the power of the engine.

At constant speed the acceleration is zero, so the driving force equals the resistance: $F = 600\ \mathrm{N}$ . Then

$P = Fv = 600 \times 30 = 18000\ \mathrm{W} = 18\ \mathrm{kW}.$

Mark scheme: M1 state $a = 0\Rightarrow F = \text{resistance}$ ; M1 $P = Fv$ ; A1 $18\ \mathrm{kW}$ . The mass is a red herring at constant speed — it would only matter if the car were accelerating.

Worked Example 7 — work–energy over a full journey

A particle of mass $2\ \mathrm{kg}$ is projected up a rough slope ( $\theta = 30^\circ$ , $\mu = 0.25$ ) at $6\ \mathrm{m\,s^{-1}}$ . Find how far up the slope it travels before momentarily coming to rest. (Take $g = 9.8\ \mathrm{m\,s^{-2}}$ .)

Going up, both gravity (component down the slope) and friction (down the slope, opposing motion) do negative work. With $R = mg\cos\theta$ :

\begin{aligned} \text{weight component} &= mg\sin 30^\circ = 2(9.8)(0.5) = 9.8\ \mathrm{N}\\ \text{friction} &= \mu mg\cos 30^\circ = 0.25(2)(9.8)(0.866) = 4.24\ \mathrm{N}\\ \text{total opposing} &= 9.8 + 4.24 = 14.04\ \mathrm{N}. \end{aligned}

Work–energy, coming to rest: $(14.04)d = \tfrac12(2)(6^2) = 36\ \mathrm{J}$ , so $d = \dfrac{36}{14.04} \approx 2.56\ \mathrm{m}$ .

Mark scheme: M1 $R = mg\cos\theta$ ; M1 both opposing forces; M1 apply $W_{\text{net}} = \Delta KE$ with final KE zero; A1 $\approx 2.56\ \mathrm{m}$ . The work–energy route avoids ever computing the (constant) deceleration or the time — a clear win over $F = ma$ plus kinematics.

4 Specimen-style exam question

(Specimen-style.) A car of mass $1200\ \mathrm{kg}$ has a maximum engine power of $36\ \mathrm{kW}$ . It moves on a level road against a constant resistance of $600\ \mathrm{N}$ . (a) Find the maximum speed. (b) Find the acceleration when travelling at $20\ \mathrm{m\,s^{-1}}$ .

(a) At maximum speed the acceleration is zero, so driving force = resistance: $F = 600\ \mathrm{N}$ . Then

$P = Fv \Rightarrow 36000 = 600\,v_{\max} \Rightarrow v_{\max} = 60\ \mathrm{m\,s^{-1}}.$

(b) At $v = 20\ \mathrm{m\,s^{-1}}$ the driving force is $F = \dfrac{P}{v} = \dfrac{36000}{20} = 1800\ \mathrm{N}$ . By Newton's second law,

$F - 600 = ma \Rightarrow 1800 - 600 = 1200a \Rightarrow a = 1\ \mathrm{m\,s^{-2}}.$

Work-Energy Principle

Work-Energy Principle

1 Where this sits in AQA 7367

2 Core theory

Work done by a force

The work–energy principle

Energy forms and conservation

Power

When to use work–energy versus $F = ma$

3 Worked examples with mark scheme

Worked Example 1 — variable force

Worked Example 2 — power on an incline

Worked Example 3 — work–energy with friction on a slope

Worked Example 4 — smooth horizontal push

Worked Example 5 — maximum height by energy

Worked Example 6 — power on a level road

Worked Example 7 — work–energy over a full journey

4 Specimen-style exam question

5 Synoptic links

More in Mathematics

Work-Energy Principle

Work-Energy Principle

1 Where this sits in AQA 7367

2 Core theory

Work done by a force

The work–energy principle

Energy forms and conservation

Power

When to use work–energy versus F=maF = maF=ma

3 Worked examples with mark scheme

Worked Example 1 — variable force

Worked Example 2 — power on an incline

Worked Example 3 — work–energy with friction on a slope

Worked Example 4 — smooth horizontal push

Worked Example 5 — maximum height by energy

Worked Example 6 — power on a level road

Worked Example 7 — work–energy over a full journey

4 Specimen-style exam question

5 Synoptic links

More in Mathematics

When to use work–energy versus $F = ma$