Work, Energy & Power

This lesson covers work, energy, and power at A-Level. These concepts provide an alternative approach to solving mechanics problems — instead of using forces and acceleration directly, we use energy methods. This approach is particularly powerful for problems where the force varies or the motion is complex.

Work Done

The work done by a constant force $F$ moving an object through displacement $d$ in the direction of the force is:

$W = Fd$

If the force acts at angle $\theta$ to the direction of motion:

$W = Fd\cos\theta$

Scenario	Work done
Force in direction of motion	$W = Fd$ (positive)
Force at angle $\theta$ to motion	$W = Fd\cos\theta$
Force perpendicular to motion	$W = 0$ (no work done)
Force opposing motion (friction)	$W = -Fd$ (negative — energy dissipated)

The unit of work is the joule (J): 1 J = 1 Nm.

Exam Tip: The normal reaction and weight do no work when an object moves horizontally, because they are perpendicular to the direction of motion. Only forces (or components of forces) in the direction of motion do work.

Kinetic Energy

The kinetic energy of an object of mass $m$ moving at speed $v$ is:

$KE = \frac{1}{2}mv^2$

The Work-Energy Theorem

The net work done on an object equals the change in its kinetic energy:

$W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Gravitational Potential Energy

The gravitational potential energy of an object of mass $m$ at height $h$ above a reference level is:

$GPE = mgh$

When an object moves up by height $h$ , it gains $mgh$ of gravitational potential energy. When it moves down, it loses GPE.

Conservation of Energy

In the absence of non-conservative forces (like friction), the total mechanical energy is conserved:

$KE_1 + GPE_1 = KE_2 + GPE_2$

$\frac{1}{2}mu^2 + mgh_1 = \frac{1}{2}mv^2 + mgh_2$

When friction or other resistive forces are present:

$KE_1 + GPE_1 = KE_2 + GPE_2 + W_{\text{friction}}$

where $W_{\text{friction}} = F_{\text{friction}} \times d$ is the energy dissipated by friction.

Example: A 2 kg ball is dropped from a height of 5 m. Find its speed just before hitting the ground (ignore air resistance).

$mgh = \frac{1}{2}mv^2$ $2 \times 9.8 \times 5 = \frac{1}{2} \times 2 \times v^2$ $v^2 = 98$ $v = 9.9 \text{ m/s}$

Power

Power is the rate of doing work (or the rate of energy transfer):

$P = \frac{W}{t}$

For a vehicle moving at constant velocity $v$ with a driving force $F$ :

$P = Fv$

The unit of power is the watt (W): 1 W = 1 J/s.

Maximum Speed

At maximum speed, the driving force equals the total resistance (acceleration = 0):

$F_{\text{drive}} = F_{\text{resistance}}$

$P = F_{\text{resistance}} \times v_{\max}$

$v_{\max} = \frac{P}{F_{\text{resistance}}}$

Exam Tip: The equation $P = Fv$ is extremely useful. Remember that at maximum speed, acceleration is zero, so the driving force equals the total resistance. Use $P = Fv$ to find the maximum speed.

Work Done Against Gravity on an Inclined Plane

When an object moves a distance $d$ up a slope inclined at $\theta$ :

Height gained: $h = d\sin\theta$
Work done against gravity: $W = mgh = mgd\sin\theta$

Summary

Work done: $W = Fd\cos\theta$ . Only the component of force in the direction of motion does work.
Kinetic energy: $KE = \frac{1}{2}mv^2$ . Work-energy theorem: $W_{\text{net}} = \Delta KE$ .
GPE: $GPE = mgh$ . Conservation of energy: $KE + GPE = \text{constant}$ (no friction).
Power: $P = Fv$ . At maximum speed, driving force = resistance.
Energy dissipated by friction: $W_{\text{friction}} = F_{\text{friction}} \times d$ .

Exam Tip: Energy methods are powerful alternatives to $F = ma$ and SUVAT. If a question asks about speed at a certain point and there are height changes or friction, consider using conservation of energy rather than equations of motion.

A-Level Deep Dive: Work, Energy and Power

Spec mapping

Beyond Mathematics 7357 spec — Further Maths / Physics overlap. Work, energy and power are not part of the AQA A-Level Mathematics 7357 core specification. They are typically reserved for AQA Further Mathematics 7367 (Mechanics options) and AQA A-Level Physics 7408 — section 3.4 Mechanics and materials. However, candidates whose centres deliver Mechanics 2 modules, or who are co-studying Physics, encounter these ideas regularly, and the synoptic links to A-Level Mathematics 7357 (kinematics, forces, integration) are deep enough that a rigorous mathematical treatment is worthwhile. This deep dive covers $W = Fd\cos\theta$ , kinetic energy $\tfrac{1}{2}mv^2$ , gravitational potential energy $mgh$ , power $P = Fv = dW/dt$ , and the principle of conservation of mechanical energy.

Worked example with full mark scheme

Question (8 marks):

A block of mass $5\,\text{kg}$ slides down a rough plane inclined at $30°$ to the horizontal. The coefficient of friction between the block and the plane is $\mu = 0.2$ . The block starts from rest and slides a distance of $4\,\text{m}$ down the slope. Take $g = 9.8\,\text{m s}^{-2}$ .

(a) Find the work done by gravity on the block. (2)

(b) Find the work done against friction. (3)

Solution with mark scheme:

(a) The component of gravity along the slope is $mg\sin\theta$ , and the displacement along the slope is $d = 4\,\text{m}$ . Work done by gravity:

$W_g = mg\sin\theta \cdot d = 5 \cdot 9.8 \cdot \sin 30° \cdot 4 = 5 \cdot 9.8 \cdot 0.5 \cdot 4 = 98\,\text{J}$

M1 — identifying $mg\sin\theta$ as the force-component along the displacement direction. A1 — $98\,\text{J}$ .

(b) The normal reaction is $R = mg\cos\theta = 5 \cdot 9.8 \cdot \cos 30° = 49 \cdot \tfrac{\sqrt{3}}{2} \approx 42.44\,\text{N}$ .

The friction force is $F_f = \mu R = 0.2 \cdot 42.44 \approx 8.487\,\text{N}$ .

Friction acts opposite to motion, so work done against friction:

$W_f = F_f \cdot d = 8.487 \cdot 4 \approx 33.95\,\text{J}$

M1 — computing $R = mg\cos\theta$ . M1 — $F_f = \mu R$ . A1 — $W_f \approx 33.9\,\text{J}$ (accept $33.94$ – $34.0$ ).

$W_{\text{net}} = \Delta\text{KE} = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2$

Since $u = 0$ , we have $W_g - W_f = \tfrac{1}{2}mv^2$ , so:

$98 - 33.95 = \tfrac{1}{2} \cdot 5 \cdot v^2 \implies 64.05 = 2.5 v^2 \implies v^2 = 25.62 \implies v \approx 5.06\,\text{m s}^{-1}$

M1 — applying the work-energy theorem with the net work (gravity minus friction). M1 — substituting and rearranging for $v$ . A1 — $v \approx 5.06\,\text{m s}^{-1}$ (accept $5.0$ – $5.1$ ).

Total: 8 marks. The crux is that the work-energy theorem replaces a Newton's-second-law plus suvat calculation with a single energy balance — quicker, less error-prone, and decoupled from the kinematic detail.

Specimen question modelled on Further Maths / Physics format

Question (6 marks): A car of mass $1200\,\text{kg}$ travels along a straight horizontal road. The engine delivers a constant power of $30\,\text{kW}$ . Resistive forces total $400\,\text{N}$ .

(a) Find the maximum speed of the car. (3)

(b) Find the acceleration of the car at the instant when its speed is $15\,\text{m s}^{-1}$ . (3)

Mark scheme decomposition:

(a) At maximum speed, acceleration is zero, so the driving force equals the resistance: $F = 400\,\text{N}$ . Using $P = Fv$ :

$v_{\max} = \dfrac{P}{F} = \dfrac{30000}{400} = 75\,\text{m s}^{-1}$

M1 — recognising "maximum speed $\Leftrightarrow$ zero acceleration $\Leftrightarrow$ driving force = resistance".
M1 — applying $P = Fv$ .
A1 — $75\,\text{m s}^{-1}$ .

(b) At $v = 15\,\text{m s}^{-1}$ , the driving force is $F = P/v = 30000/15 = 2000\,\text{N}$ . Net force: $2000 - 400 = 1600\,\text{N}$ . Acceleration:

$a = \dfrac{F_{\text{net}}}{m} = \dfrac{1600}{1200} = \dfrac{4}{3} \approx 1.33\,\text{m s}^{-2}$

M1 — applying $P = Fv$ to find driving force at the given speed.
M1 — applying Newton's second law with the net force.
A1 — $\tfrac{4}{3}\,\text{m s}^{-2}$ .

Total: 6 marks. This is the canonical "variable driving force at constant power" problem — the driving force decreases with speed because power is fixed, which is why acceleration tapers as the car speeds up.

Synoptic links

Connects to:

Kinematics (7357 section J): the work-energy theorem replaces $v^2 = u^2 + 2as$ for problems with non-constant force. Where suvat fails (variable acceleration), energy methods succeed.
Forces and Newton's laws (7357 section K): work is the line integral of force along displacement. Resolving forces correctly along the direction of motion is the prerequisite — every $\cos\theta$ in $W = Fd\cos\theta$ comes from the same resolution skills used in inclined-plane statics.
Integration (7357 section H): for variable force $F(x)$ , work is $W = \int_a^b F(x)\,dx$ . Power as $P = dW/dt$ links work and time via differentiation; instantaneous power $P = Fv$ falls out of the chain rule $dW/dt = (dW/dx)(dx/dt)$ .
A-Level Physics (7408 section 3.4): identical formulae, but Physics emphasises efficiency $\eta = W_{\text{useful}}/W_{\text{input}}$ and energy-resource transformations. Physics extends to electrical work $W = QV$ and thermal energy $Q = mc\Delta T$ .
Engineering / mechanical engineering degree: energy methods underpin structural analysis (strain energy $U = \tfrac{1}{2}k x^2$ ), thermodynamics (first law $\Delta U = Q - W$ ), and fluid mechanics (Bernoulli's equation is conservation of energy per unit volume).

Mark-scheme literacy

For energy/power questions in Further Maths and Physics, AO marks distribute roughly:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Stating $W = Fd\cos\theta$ , $\text{KE} = \tfrac{1}{2}mv^2$ , $P = Fv$ ; substituting correctly
AO2 (reasoning / interpretation)	25–35%	Identifying which forces do work; choosing energy method over Newton's laws; recognising "max speed" implies zero acceleration
AO3 (problem-solving / modelling)	10–20%	Applying conservation of energy in non-trivial geometry; deciding when energy is dissipated as heat

Examiner-rewarded phrasing: "by the work-energy theorem, net work = change in KE"; "since the surface is rough, mechanical energy is not conserved — the deficit equals the work done against friction"; "at maximum speed, $a = 0$ , so driving force = total resistance". Phrases that lose marks: writing "energy is conserved" without qualifying mechanical energy on a rough surface; using $P = Fv$ at maximum speed without noting why $F$ equals resistance; computing $mg\cos\theta$ as the component along the slope (it is the normal component).

Grade-band model answers

3-mark question

Question: A force of $20\,\text{N}$ acts at $60°$ above the horizontal on a box, dragging it $5\,\text{m}$ horizontally across the floor. Find the work done by the force.

Grade C response (~140 words):

Work done = force times distance times $\cos\theta$ :

$W = Fd\cos\theta = 20 \cdot 5 \cdot \cos 60° = 100 \cdot 0.5 = 50\,\text{J}$ .

So the work done is $50\,\text{J}$ .

Examiner commentary: Full marks (3/3). The candidate correctly identifies that only the component of force along the displacement direction contributes to work. The $\cos 60° = 0.5$ evaluation is clean. A common error this candidate avoided: using $\sin 60°$ (which would give the vertical component, which does no work as the box moves horizontally).

Grade A response (~190 words):*

The work done by a force $\mathbf{F}$ acting through a displacement $\mathbf{d}$ is $W = \mathbf{F} \cdot \mathbf{d} = Fd\cos\theta$ , where $\theta$ is the angle between the force and the displacement.

Work, Energy & Power

Work, Energy & Power

Work Done

Kinetic Energy

The Work-Energy Theorem

Gravitational Potential Energy

Conservation of Energy

Power

Maximum Speed

Work Done Against Gravity on an Inclined Plane

Summary

A-Level Deep Dive: Work, Energy and Power

Spec mapping

Worked example with full mark scheme

Specimen question modelled on Further Maths / Physics format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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