# A.3 Work, energy and power

> IB Physics HL · IB Diploma Programme Physics HL
> Source: https://www.owlsprep.com/study/ib-physics-hl-u1-a-3-work-energy-and/

This sub-topic introduces core relationships between work, energy, and power for mechanical systems. You will learn to calculate work done by forces, apply energy conservation, and solve problems involving power and efficiency for IB Physics HL exams.

**Prerequisites:** [Forces and Newton's laws of motion](https://www.owlsprep.com/study/ib-physics-hl-u1-a-2-forces/); [Kinematics in one and two dimensions](https://www.owlsprep.com/study/ib-physics-hl-u1-a-1-kinematics/)

## Learning objectives

- Define work done by a force and calculate it for constant and variable forces
- Apply the work-energy theorem to relate total work to change in kinetic energy
- Calculate power and efficiency for mechanical systems
- Distinguish between conservative and non-conservative forces and apply conservation of mechanical energy
- Solve problems involving energy changes with non-conservative forces

## Work Done by Constant and Variable Forces

**Work done by a constant force** — The scalar product of force and displacement, equal to the energy transferred to or from an object. Calculated as $W = Fs\cos\theta$, where $\theta$ is the angle between the force and displacement vectors.

*Notation:* W

*Example:* A 10 N force pulling 2 m at $30^\circ$ to displacement has $W = 10 \times 2 \times \cos 30^\circ = 17.3 \text{ J}$.

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 $25^\circ$ 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: $W_T = Ts\cos\theta$
2. $$W_T = 40 \times 10 \times \cos 25^\circ \approx 362.4 \text{ J}$$
3. Friction acts opposite displacement, so $\theta = 180^\circ$, $\cos 180^\circ = -1$:
4. $$W_f = 5 \times 10 \times (-1) = -50 \text{ J}$$
5. Normal force and gravity are perpendicular to displacement, so $\cos 90^\circ = 0$, 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.

## Work-Energy Theorem and Conservation of Energy

**Work-Energy Theorem** — The total work done by all forces acting on an object equals the change in the object's kinetic energy: $W_{\text{total}} = \Delta E_k$. Kinetic energy is defined as $E_k = \frac{1}{2}mv^2$.

Conservative forces store energy as potential energy: work done by a conservative force is $W_c = -\Delta E_p$. 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: $E_{k,i} = 0$, $E_{p,i} = mgh$
2. $$E_i = 0 + (2)(9.8)(10) = 196 \text{ J}$$
3. Final state: $E_{p,f} = 0$, $E_{k,f} = \frac{1}{2}mv^2$, so $E_f = \frac{1}{2}mv^2$
4. Conservation of energy: $E_i = E_f$
5. $$196 = \frac{1}{2}(2)v^2 \implies v^2 = 196 \implies v = 14 \text{ m s}^{-1}$$

## Power and Efficiency

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

*Notation:* P

*Example:* A 500 W engine does 500 J of work every second.

No real process converts 100% of input energy to useful output. Efficiency ($\eta$) 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 $2 \text{ m s}^{-1}$ 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.

## Conservative and Non-Conservative Forces

**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: $W_{nc} = \Delta (E_k + E_p)$. 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 $20 \text{ m s}^{-1}$. Calculate work done by friction.

1. Use $W_{nc} = (E_{k,f} + E_{p,f}) - (E_{k,i} + E_{p,i})$
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. $$W_{nc} = (10000 + 0) - (0 + 24500) = -14500 \text{ J}$$

## Common pitfalls

- **Wrong:** Adding magnitudes of work instead of accounting for negative work from opposing forces
  - Why it fails: Work is a scalar quantity that can be negative, so total work is an algebraic sum, not a magnitude sum
  - Correct: Calculate work for each force individually with the correct sign based on angle, then add algebraically
- **Wrong:** Assuming any force acting on a moving object does non-zero work
  - Why it fails: Work is zero if the force is perpendicular to displacement
  - Correct: Always check the angle between force and displacement; perpendicular forces do no work
- **Wrong:** Using $P = Fv$ with the full force magnitude when the force is at an angle to motion
  - Why it fails: The relationship $P = Fv$ only works for the component of force parallel to velocity
  - Correct: Use $P = F_{\parallel}v$, where $F_{\parallel}$ is the parallel component of force
- **Wrong:** Applying conservation of mechanical energy when friction or other non-conservative forces are present
  - Why it fails: Conservation of mechanical energy only holds when no non-conservative forces do work
  - Correct: Only use $E_{initial} = E_{final}$ if the problem states friction/air resistance can be ignored
- **Wrong:** Calculating efficiency as input energy divided by output energy
  - Why it fails: This gives values greater than 100%, which violates the second law of thermodynamics
  - Correct: Efficiency is always useful output energy divided by total input energy, so it is always less than 100%

## Cheatsheet

| Quantity | Formula | Key Notes |
| --- | --- | --- |
| Work (constant F) | W = Fs\cos\theta | $\theta$ = angle between F and s |
| Kinetic Energy | $E_k = \frac{1}{2}mv^2$ | Always positive |
| Work-Energy Theorem | $W_{total} = \Delta E_k$ | Total work = change in KE |
| Conservation of ME | $E_{k1} + E_{p1} = E_{k2} + E_{p2}$ | Only if no non-conservative work |
| Power | $P = \frac{W}{\Delta t} = F_{\parallel}v$ | Unit: watts (1 W = 1 J/s) |
| Efficiency | $\eta = \frac{P_{out}}{P_{in}} \times 100\%$ | $\eta < 100\%$ always |
| Non-conservative work | $W_{nc} = \Delta (E_k + E_p)$ | Calculates energy losses |

## 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.

- [A.4 Rotational mechanics](https://www.owlsprep.com/study/ib-physics-hl-u1-a-4-rotational-mechanics/)
- [A.5 Special relativity: energy and momentum (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u1-a-5-special-relativity-energy/)
- [A.6 Circular motion and gravitation (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u1-a-6-circular-motion-and/)

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