# Work, energy and power

> IB Physics SL · IB Physics 2025 SL
> Source: https://www.owlsprep.com/study/ib-physics-sl-u1-work-energy-and-power/

This sub-topic introduces core relationships between force, energy, motion and power, foundational for all mechanics topics in IB Physics SL. You will learn to calculate work done, energy transformations, and power output of physical systems.

**Prerequisites:** [Newton's laws of motion](https://www.owlsprep.com/study/ib-physics-sl-u1-newtons-laws-of-motion/); [Linear kinematics](https://www.owlsprep.com/study/ib-physics-sl-u1-kinematics-linear-motion/)

## Learning objectives

- Define work done by a constant force and calculate it for different force-displacement angles
- Distinguish between kinetic and gravitational potential energy
- Apply the work-energy theorem to solve mechanics problems
- Calculate power and efficiency in physical systems

## Work Done by a Constant Force

**Work** — Work is the energy transferred when a force moves an object through a displacement. For a constant force, it equals the product of the parallel component of force and displacement.

*Notation:* W

$$W = F s \cos\theta$$

$\theta$ is the angle between the force vector and the displacement vector. If $\theta = 90^\circ$, $\cos\theta = 0$ so work done is zero, even if a force is applied.

> **info**
>
> A common example of zero work is a person carrying a heavy box walking horizontally: the vertical supporting force is perpendicular to displacement, so it does no work on the box.

**Worked example:** A 50 N force pulls a box across a horizontal floor at 30° to the horizontal. The box moves 4 m horizontally. Calculate the work done by the applied force.

1. Identify given values: $F = 50$ N, $s = 4$ m, $\theta = 30^\circ$
2. Substitute into the work formula:
3. $$W = 50 \times 4 \times \cos(30^\circ)$$
4. Calculate the final result:
5. $$W = 200 \times \frac{\sqrt{3}}{2} \approx 173 \text{ J}$$

## Kinetic Energy and the Work-Energy Theorem

**Kinetic Energy** — Kinetic energy is the energy an object has due to its motion. It is proportional to the object's mass and the square of its speed.

*Notation:* E_k

*Example:* A 1000 kg car moving at 10 m/s has 50,000 J of kinetic energy

$$E_k = \frac{1}{2} m v^2$$

The **work-energy theorem** connects net work done on an object to its change in kinetic energy:

$$W_{net} = \Delta E_k = E_{k,\text{final}} - E_{k,\text{initial}}$$

**Worked example:** A 0.5 kg ball moving at 4 m/s is slowed to rest by constant friction over 2 m. Calculate the magnitude of the frictional force.

1. Calculate initial kinetic energy:
2. $$E_{k,i} = \frac{1}{2} m v_i^2 = 0.5 \times 0.5 \times (4)^2 = 4 \text{ J}$$
3. Final kinetic energy is 0, so $\Delta E_k = 0 - 4 = -4$ J. The negative sign matches the negative work done by friction.
4. Apply the work-energy theorem: net work equals work done by friction
5. $$-f s = \Delta E_k \implies -f (2) = -4 \implies f = 2 \text{ N}$$

## Gravitational Potential Energy

**Change in Gravitational Potential Energy** — Gravitational potential energy is energy stored due to an object's position in a gravitational field. Near Earth's surface, we only calculate changes in potential energy, not absolute values.

*Notation:* $\Delta E_p$

$$\Delta E_p = m g \Delta h$$

> **info**
>
> $\Delta h$ is the change in vertical height, not total distance moved. We can set zero potential energy anywhere, because only changes matter for calculations.

**Worked example:** A 2 kg brick is lifted from the ground to a shelf 1.5 m high. Calculate the change in gravitational potential energy. Take $g = 9.8 \text{ m s}^{-2}$.

1. Identify values: $m = 2$ kg, $\Delta h = 1.5$ m, $g = 9.8$ m s⁻²
2. Substitute into the potential energy formula:
3. $$\Delta E_p = 2 \times 9.8 \times 1.5 = 29.4 \text{ J}$$
4. The positive change means energy is stored as gravitational potential energy in the brick-Earth system.

## Power and Efficiency

**Power** — Power is the rate at which work is done or energy is transferred. When a force is parallel to velocity, power can be written as the product of force and speed.

*Notation:* P

$$P = \frac{W}{\Delta t} = F v$$

Efficiency describes how much of the total input energy is converted to useful output energy, because all real processes waste some energy as heat.

$$\eta = \frac{P_{\text{useful out}}}{P_{\text{total in}}} \times 100\%$$

**Worked example:** A 1000 kg car accelerates from rest to 20 m/s in 10 s. The engine has an efficiency of 25%. Calculate the total power input from the fuel.

1. Calculate useful kinetic energy output:
2. $$E_{\text{useful}} = \frac{1}{2} m v^2 = 0.5 \times 1000 \times (20)^2 = 200 000 \text{ J}$$
3. Calculate useful power output:
4. $$P_{\text{out}} = \frac{E_{\text{useful}}}{t} = \frac{200 000}{10} = 20 000 \text{ W} = 20 \text{ kW}$$
5. Rearrange the efficiency formula to solve for total input power:
6. $$P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{20 000}{0.25} = 80 000 \text{ W} = 80 \text{ kW}$$

## Common pitfalls

- **Wrong:** Ignoring the angle $\theta$ and just calculating $W = Fs$ for angled forces
  - Why it fails: Only the component of force parallel to displacement contributes to work; perpendicular components do no work
  - Correct: Always multiply by $\cos\theta$ when force is at an angle to displacement, and remember work is zero when force is perpendicular
- **Wrong:** Using work done by a single force instead of net work in the work-energy theorem
  - Why it fails: The work-energy theorem relates the total work from all forces acting on an object to the change in kinetic energy
  - Correct: Sum the work done by all individual forces to get net work before equating to $\Delta E_k$
- **Wrong:** Using absolute height instead of change in height for gravitational potential energy
  - Why it fails: Zero potential energy is an arbitrary reference point, only changes in height are physically meaningful
  - Correct: Always use the difference between final and initial vertical height to calculate $\Delta E_p$
- **Wrong:** Using percentage efficiency directly in calculations (e.g. 25 instead of 0.25)
  - Why it fails: Efficiency is a ratio between 0 and 1, percentage is just for reporting final results
  - Correct: Divide the percentage efficiency by 100 before substituting into the efficiency formula

## 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 non-negative |
| Work-Energy Theorem | $W_{net} = \Delta E_k$ | Net work = change in KE |
| Gravitational PE change | $\Delta E_p = mg\Delta h$ | $\Delta h$ = vertical change |
| Power | $P = \frac{W}{\Delta t} = Fv$ | $Fv$ for parallel F and v |
| Efficiency | $\eta = \frac{P_{out}}{P_{in}} \times 100\%$ | $0 < \eta < 100\%$ for all real systems |

## What's next

Work, energy and power is a foundational concept that extends to nearly every topic in IB Physics SL. Understanding energy transformations is key for thermal physics, circular motion, and electricity, where energy transfer is a core theme. Mastery of this sub-topic will also make solving complex mechanics problems much simpler, as energy methods often avoid needing to calculate acceleration. Next, you will build on this knowledge to explore conservation of mechanical energy, then extend to momentum for systems of multiple objects.

- [Circular motion and gravitation](https://www.owlsprep.com/study/ib-physics-sl-u1-circular-motion-and-gravitation/)
- [The particulate nature of matter](https://www.owlsprep.com/study/ib-physics-sl-u2-overview/)
- [Temperature and thermal energy](https://www.owlsprep.com/study/ib-physics-sl-u2-temperature-and-thermal-energy/)

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