A.3 Work, energy and power

IB Physics HL· Theme A: Space, time and motion, Topic A.3· 45 min read

1. Work Done by Constant and Variable Forces★★☆☆☆⏱ 15 min

📘 Definition

Work done by a constant force

W

The scalar product of force and displacement, equal to the energy transferred to or from an object. Calculated as $W = F s cos θ$ , where $θ$ is the angle between the force and displacement vectors.

Example:

A 10 N force pulling 2 m at $3 0^{\circ}$ to displacement has $W = 10 \times 2 \times cos 3 0^{\circ} = 17.3 J$ .

Work can be positive or negative: positive work adds energy to the object, while negative work removes energy (e.g., work done by friction). For variable forces, work done equals the area under a force vs. displacement graph.

📐 Worked Example

A student pulls a 5 kg sled 10 m along horizontal ice with a rope at $2 5^{\circ}$ to the horizontal. Tension is 40 N, and friction force is 5 N. Calculate the total work done on the sled.

1
Calculate work done by tension: $W_{T} = T s cos θ$
2
$W_T = 40 \times 10 \times \cos 25^\circ \approx 362.4 \text{ J}$
3
Friction acts opposite displacement, so $θ = 18 0^{\circ}$ , $cos 18 0^{\circ} = - 1$ :
4
$W_f = 5 \times 10 \times (-1) = -50 \text{ J}$
5
Normal force and gravity are perpendicular to displacement, so $cos 9 0^{\circ} = 0$ , their work is 0.
6
Sum all work for total work:
7
$W_{\text{total}} = 362.4 - 50 + 0 + 0 = 312.4 \text{ J} \approx 310 \text{ J (2 s.f.)}$

Exam tip:

Always check the angle between force and displacement. Perpendicular forces always do zero work, even if the force is large.

2. Work-Energy Theorem and Conservation of Energy★★☆☆☆⏱ 15 min

📘 Definition

Work-Energy Theorem

The total work done by all forces acting on an object equals the change in the object's kinetic energy: $W_{total} = Δ E_{k}$ . Kinetic energy is defined as $E_{k} = \frac{1}{2} m v^{2}$ .

Conservative forces store energy as potential energy: work done by a conservative force is $W_{c} = - Δ E_{p}$ . The law of conservation of mechanical energy states that if only conservative forces do work, total mechanical energy $E = E_{k} + E_{p}$ is constant.

📐 Worked Example

A 2 kg ball is dropped from rest from a height of 10 m. Ignoring air resistance, calculate its speed just before impact using conservation of energy.

1
Take ground as zero potential energy. Initial state: $E_{k, i} = 0$ , $E_{p, i} = m g h$
2
$E_i = 0 + (2)(9.8)(10) = 196 \text{ J}$
3
Final state: $E_{p, f} = 0$ , $E_{k, f} = \frac{1}{2} m v^{2}$ , so $E_{f} = \frac{1}{2} m v^{2}$
4
Conservation of energy: $E_{i} = E_{f}$
5
$196 = \frac{1}{2}(2)v^2 \implies v^2 = 196 \implies v = 14 \text{ m s}^{-1}$

3. Power and Efficiency★★★☆☆⏱ 15 min

📘 Definition

Power

P

The rate of doing work or transferring energy. SI unit is the watt ( $1 W = 1 J s^{- 1}$ ). For a force moving at constant speed, $P = F_{∥} v$ , where $F_{∥}$ is the component of force parallel to velocity.

Example:

A 500 W engine does 500 J of work every second.

No real process converts 100% of input energy to useful output. Efficiency ( $η$ ) measures the proportion of useful output energy:

\eta = \frac{\text{useful power output}}{\text{total power input}} = \frac{\text{useful work output}}{\text{total energy input}} \times 100\%

📐 Worked Example

A 2000 kg elevator accelerates upwards from rest to $2 m s^{- 1}$ in 4 s, reaching a height of 4 m. Calculate the minimum power output of the motor.

1
Total energy gained = change in potential energy + change in kinetic energy
2
$\Delta E_p = mgh = (2000)(9.8)(4) = 78400 \text{ J}$
3
$\Delta E_k = \frac{1}{2}mv^2 - 0 = \frac{1}{2}(2000)(2^2) = 4000 \text{ J}$
4
Total minimum energy input (ignoring losses):
5
$\Delta E_{\text{total}} = 78400 + 4000 = 82400 \text{ J}$
6
Power = energy / time:
7
$P = \frac{82400}{4} = 20600 \text{ W} = 20.6 \text{ kW}$

Exam tip:

Efficiency is always output divided by input. If you get an efficiency over 100%, you swapped the values and need to correct it.

4. Conservative and Non-Conservative Forces★★★☆☆⏱ 10 min

📘 Definition

Conservative vs Non-Conservative Forces

Conservative forces have work done independent of path, and zero work over a closed path. They store energy as potential energy. Non-conservative forces have work done that depends on path, and dissipate energy as heat.

When non-conservative forces do work, the work they do equals the change in total mechanical energy: $W_{n c} = Δ (E_{k} + E_{p})$ . This lets us calculate energy losses in real systems.

📐 Worked Example

A 50 kg skier starts from rest at the top of a 50 m high slope. Their speed at the bottom is $20 m s^{- 1}$ . Calculate work done by friction.

1
Use $W_{n c} = (E_{k, f} + E_{p, f}) - (E_{k, i} + E_{p, i})$
2
Take bottom of slope as $E_{p} = 0$ :
3
$E_{p,i} = mgh = (50)(9.8)(50) = 24500 \text{ J}, \quad E_{k,i} = 0$
4
$E_{k,f} = \frac{1}{2}(50)(20^2) = 10000 \text{ J}, \quad E_{p,f} = 0$
5
$W_{nc} = (10000 + 0) - (0 + 24500) = -14500 \text{ J}$

5. Common Pitfalls

Wrong move:

Adding magnitudes of work instead of accounting for negative work from opposing forces

Why:

Work is a scalar quantity that can be negative, so total work is an algebraic sum, not a magnitude sum

Correct move:

Calculate work for each force individually with the correct sign based on angle, then add algebraically

Wrong move:

Assuming any force acting on a moving object does non-zero work

Why:

Work is zero if the force is perpendicular to displacement

Correct move:

Always check the angle between force and displacement; perpendicular forces do no work

Wrong move:

Using $P = F v$ with the full force magnitude when the force is at an angle to motion

Why:

The relationship $P = F v$ only works for the component of force parallel to velocity

Correct move:

Use $P = F_{∥} v$ , where $F_{∥}$ is the parallel component of force

Wrong move:

Applying conservation of mechanical energy when friction or other non-conservative forces are present

Why:

Conservation of mechanical energy only holds when no non-conservative forces do work

Correct move:

Only use $E_{ini t ia l} = E_{f ina l}$ if the problem states friction/air resistance can be ignored

Wrong move:

Calculating efficiency as input energy divided by output energy

Why:

This gives values greater than 100%, which violates the second law of thermodynamics

Correct move:

Efficiency is always useful output energy divided by total input energy, so it is always less than 100%

6. Quick Reference Cheatsheet

Quantity	Formula	Key Notes
Work (constant F)	W = Fs\cos\theta	$θ$ = angle between F and s
Kinetic Energy	$E_{k} = \frac{1}{2} m v^{2}$	Always positive
Work-Energy Theorem	$W_{t o t a l} = Δ E_{k}$	Total work = change in KE
Conservation of ME	$E_{k 1} + E_{p 1} = E_{k 2} + E_{p 2}$	Only if no non-conservative work
Power	$P = \frac{W}{Δ t} = F_{∥} v$	Unit: watts (1 W = 1 J/s)
Efficiency	$η = \frac{P _{o u t}}{P _{in}} \times 100%$	$η < 100%$ always
Non-conservative work	$W_{n c} = Δ (E_{k} + E_{p})$	Calculates energy losses

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.

2025 · 1
Work done by variable force
2024 · 2
Power and efficiency calculation
2023 · 1
Work-energy theorem application
2022 · 2
Conservative force energy problem

What's Next

Work, energy and power form the foundation of all mechanics topics in IB Physics, and energy concepts are extended across every area of the syllabus. Energy methods allow you to solve complex problems that would be very difficult to tackle using only Newton's laws, especially for systems with variable motion and multiple interacting objects. Mastering these basics will make it much easier to understand energy concepts in thermal physics, electricity, and modern physics later in your course. These concepts are also heavily tested in both Paper 1 and Paper 2 exams, so regular practice is key.