Energy, work and power

CIE A-Level Mathematics· Unit 3: Mechanics· 7 min read

1. Work Done by a Constant Force★★☆☆☆⏱ 15 min

📘 Definition

Work done by a constant force

W = F s \cos\theta

The energy transferred when a force moves an object through a displacement. \theta is the angle between the force and displacement vectors.

Example:

Work done by a 10 N force pulling an object 2 m at 30° to the horizontal is $10 \times 2 \times cos 3 0^{\circ} \approx 17.3$ J.

Work done can be positive, negative or zero. If the force acts in the same direction as displacement ( $θ = 0^{\circ}$ ), $cos θ = 1$ so $W = F s$ , meaning positive work (energy is added to the object). If the force opposes motion ( $θ = 18 0^{\circ}$ ), work is negative, meaning energy is removed from the object. If the force is perpendicular to displacement ( $θ = 9 0^{\circ}$ ), work done is zero.

📐 Worked Example

A 5 kg block is pulled 4 m up a rough slope inclined at 20° to the horizontal by a rope parallel to the slope. Friction has magnitude 8 N, and the block moves at constant speed. Calculate the total work done by all forces acting on the block.

1
List all forces parallel to displacement (along the slope): weight component is down the slope:
2
$mg \sin\theta = 5 \times 9.8 \times \sin 20^\circ \approx 16.8 \text{ N}$
3
Tension up the slope balances the sum of friction and weight component for constant speed:
4
$T = 16.8 + 8 = 24.8 \text{ N}$
5
Calculate work done by each force: tension is parallel to displacement:
6
$W_T = 24.8 \times 4 = 99.2 \text{ J}$
7
Friction and weight component oppose displacement, so their work is negative:
8
$W_f + W_g = (-8 \times 4) + (-16.8 \times 4) = -99.2 \text{ J}$
9
Total work done is the sum of all individual work values:
10
$W_{\text{total}} = 99.2 - 99.2 = 0 \text{ J}$
11
This matches the work-energy principle: no change in kinetic energy means total work done is zero.

Exam tip:

Always check the angle between the force and displacement; many candidates incorrectly use $sin θ$ instead of $cos θ$ .

2. Kinetic and Gravitational Potential Energy★★☆☆☆⏱ 15 min

📘 Definition

Kinetic Energy (KE)

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

The energy an object possesses due to its motion, measured in joules (J)

Example:

A 2 kg object moving at 3 m s⁻¹ has $K E = 0.5 \times 2 \times 3^{2} = 9$ J.

📘 Definition

Gravitational Potential Energy (GPE)

\Delta PE = mg\Delta h

The change in energy of an object when it changes height by $Δ h$ in a uniform gravitational field

Example:

Lifting a 10 kg mass by 2 m increases GPE by $10 \times 9.8 \times 2 = 196$ J.

The work-energy principle states that the total work done by all forces acting on an object equals the change in its kinetic energy: $W_{total} = Δ K E$ . This principle holds whether or not there are resistive forces, making it very flexible for problem solving.

📐 Worked Example

A car of mass 1200 kg is moving at 15 m s⁻¹. The brakes apply a constant braking force of 4500 N. Find the distance the car travels before stopping.

1
Calculate initial kinetic energy of the car:
2
$KE_{\text{initial}} = \frac{1}{2} \times 1200 \times 15^2 = 135000 \text{ J}$
3
Final KE is 0 when the car stops, so change in KE is:
4
$\Delta KE = 0 - 135000 = -135000 \text{ J}$
5
Work done by braking force $F$ over distance $d$ is $W = - F d$ . By work-energy principle: $W = Δ K E$
6
$-4500d = -135000$
7
Solve for stopping distance $d$ :
8
$d = \frac{135000}{4500} = 30 \text{ m}$

3. Conservation of Mechanical Energy★★★☆☆⏱ 20 min

When no non-conservative forces (like friction or air resistance) do work on a system, the total mechanical energy (sum of kinetic and potential energy) is conserved. This means we can relate energy at the start of motion to energy at the end without needing to calculate acceleration or time.

📘 Definition

Conservation of Mechanical Energy

In the absence of non-conservative forces, total mechanical energy remains constant: energy is only converted between KE and PE, not created or destroyed.

📐 Worked Example

A ball of mass 0.5 kg is dropped from rest from a height of 10 m above the ground. Ignoring air resistance, find its speed when it hits the ground.

1
Set GPE = 0 at ground level. Initial energy: ball is at rest so $K E_{initial} = 0$
2
$PE_{\text{initial}} = mgh = 0.5 \times 9.8 \times 10 = 49 \text{ J}$
3
Final energy at ground level: $P E_{final} = 0$ , so all energy is kinetic:
4
$KE_{\text{final}} = \frac{1}{2}mv^2$
5
By conservation of energy, total initial energy equals total final energy:
6
$0 + 49 = \frac{1}{2}(0.5)v^2 + 0$
7
Solve for speed $v$ :
8
$v^2 = 196 \implies v = 14 \text{ m s}^{-1}$

4. Power and Efficiency★★★☆☆⏱ 15 min

📘 Definition

Power

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

The rate of doing work, measured in watts (W), where 1 W = 1 J s⁻¹. For a constant force $F$ in the direction of motion at speed $v$ , power equals the product of force and speed.

📘 Definition

Efficiency

\text{Efficiency} = \frac{\text{useful power output}}{\text{power input}} \times 100\%

A measure of how much input energy is converted to useful output energy. Efficiency is always less than 100% for real systems due to energy losses like heat from friction.

📐 Worked Example

A car engine has a useful power output of 60 kW and moves at constant speed 20 m s⁻¹ along a horizontal road. Calculate the total resistive force acting on the car.

1
Convert power from kilowatts to watts for consistency of units:
2
$P = 60 \text{ kW} = 60000 \text{ W}$
3
At constant speed, driving force equals total resistive force by Newton's first law: $F_{d r i v in g} = F_{r es i s t i v e}$
4
Use the instantaneous power formula $P = F v$ :
5
$F = \frac{P}{v} = \frac{60000}{20} = 3000 \text{ N}$
6
Therefore total resistive force acting on the car is 3000 N.

5. Common Pitfalls

Wrong move:

Using $W = F s$ for a force at an angle to displacement, ignoring the $cos θ$ term

Why:

Work done depends on the component of force parallel to displacement, not the full force magnitude

Correct move:

Always use $W = F s cos θ$ where $θ$ is the angle between the force and displacement vectors

Wrong move:

Taking $Δ P E$ as negative when an object gains height

Why:

Gravitational potential energy increases when height increases, so the change should be positive

Correct move:

Use $Δ P E = m g (h_{f ina l} - h_{ini t ia l})$ , which is positive when the object moves upwards

Wrong move:

Applying conservation of mechanical energy without accounting for work done against friction

Why:

Conservation of total mechanical energy only holds when there are no non-conservative forces doing work

Correct move:

If friction is present, use: $Initial total energy = Final total energy + Work done against friction$

Wrong move:

Forgetting to convert power from kilowatts to watts before calculation

Why:

Power must be in watts to get force in newtons, leading to answers 1000x too large or small

Correct move:

Always multiply kilowatts by 1000 to get watts before substituting into formulas

Wrong move:

Calculating efficiency as input power divided by output power

Why:

This gives a value greater than 100% for most real systems, which is impossible

Correct move:

Use: $Efficiency = \frac{Useful output power}{Total input power} \times 100%$

6. Quick Reference Cheatsheet

Quantity	Formula	Key Notes
Work done (constant force)	$W = F s cos θ$	$θ$ = angle between $F$ and $s$
Kinetic Energy	$K E = \frac{1}{2} m v^{2}$	Energy due to motion
Change in GPE	$Δ P E = m g Δ h$	Positive when height increases
Work-Energy Principle	$W_{t o t a l} = Δ K E$	Works for all forces
Conservation of Energy (no resistance)	$K E_{1} + P E_{1} = K E_{2} + P E_{2}$	Total mechanical energy constant
Average Power	$P = \frac{W}{t}$	Units: watts (W) = J/s
Instantaneous Power	$P = F v$	Force parallel to velocity
Efficiency	$\frac{P _{o u t}}{P _{in}} \times 100%$	Always < 100% for real systems

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 32
Energy conservation on a slope
2022 · 31
Power and resistive force calculation
2021 · 33
Work done against friction problem

Going deeper

syllabusOfficial CIE 9709 SyllabusFull syllabus for Mechanics Unit 3

What's Next

Understanding energy, work and power is foundational for all advanced Mechanics topics in A-level Mathematics, including circular motion, momentum, and rigid body equilibrium. Conservation of energy is a particularly powerful problem-solving tool that often simplifies calculations compared to using Newton's laws directly, especially for motion along curved paths where force varies with position. Mastery of this topic also prepares you to tackle combined Mechanics questions that draw on multiple core concepts, which are common in CIE A-level exams.