When a force pushes an object through a distance, it transfers energy, it does work. And when that work speeds an object up, it shows up exactly as a gain in kinetic energy. This is the work–energy theorem, one of physics' most useful shortcuts: instead of tracking forces moment by moment, you can jump straight from "force times distance" to "change in motion".
You hold a heavy box perfectly still in your arms for two minutes. Your muscles ache, and you feel like you are "working hard".
Before reading: in the physics sense, are you doing any work on the box while holding it still? Write your prediction and one sentence explaining why.
Force times distance, in the direction of motion
Work in physics has a precise meaning: a force only does work when it moves something through a distance, pushing without moving anything does no work at all.
Work is the energy transferred when a force acts over a distance. On the NESA formula sheet:
$$W = Fs \qquad (W = Fs\cos\theta \text{ if } F \text{ is at angle } \theta \text{ to the motion})$$
Work is measured in joules (J): one joule is one newton acting through one metre. If the object does not move ($s = 0$), no work is done, which is why holding a box still does no physical work on it, even though your muscles tire.
Work is energy transferred by a force acting through a distance: $W = Fs$ (or $W = Fs\cos\theta$ when the force is at an angle to the motion), measured in joules. No displacement means no work, no matter how much force is applied.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? True or false: a person holding a heavy bag still does no work on the bag in the physics sense.
Net work done equals the change in kinetic energy
We just defined work as $W = Fs$. That raises a question: where does that energy go when a force speeds an object up? This card answers it → into kinetic energy; the net work equals the change in $KE$.
When the net force does work on a moving object, that work doesn't vanish, it appears exactly as a change in the object's energy of motion.
The work–energy theorem states that the net work done on an object equals its change in kinetic energy:
$$W = Fs = \Delta KE = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2$$
Here $u$ is the initial speed and $v$ the final speed. Positive net work speeds the object up; negative net work (like friction or braking) slows it down. This lets you skip the forces-and-acceleration steps and go straight from work to speed.
A $1500\ \text{kg}$ car travelling at $20\ \text{m s}^{-1}$ brakes to a stop. The braking force is $6000\ \text{N}$. Use the work–energy theorem to find the stopping distance.
Work–energy theorem: net work equals change in kinetic energy, $W = Fs = \Delta KE = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2$. For braking, the braking work removes all the kinetic energy. Here $KE = \tfrac{1}{2}(1500)(20)^2 = 3.0 \times 10^5\ \text{J}$ and $s = KE/F = 50\ \text{m}$.
Add the highlighted theorem and method to your notes before the check below.
Quick check: A $2.0\ \text{kg}$ object speeds up from rest to $6.0\ \text{m s}^{-1}$. The net work done on it is…
Pattern A, Calculate then reason
A constant horizontal force of $40\ \text{N}$ acts on a $5.0\ \text{kg}$ trolley (initially at rest) over a distance of $10\ \text{m}$ on a frictionless surface. In your book: (1) calculate the work done by the force; (2) use the work–energy theorem to find the trolley's final speed; (3) explain in one sentence what fraction of the original speed you would expect if only a quarter of that work were done.
Odd one out: three of these forces do work that does NOT depend on the path taken. Which one does?
Dot point: Compare conservative forces and non-conservative forces
Energy you can get back, and energy lost as heat
We just used work to change kinetic energy. That raises a question: is the energy a force transfers always recoverable? This card answers it → it depends on whether the force is conservative (like gravity) or non-conservative (like friction).
Some forces store energy you can get back later, while others quietly turn it into heat that is gone for good.
A conservative force, such as gravity, does work that depends only on the start and end positions, not on the path taken. Energy moved by a conservative force is stored (as potential energy) and can be fully recovered: lift a ball and let it fall, and it returns the energy as motion. A non-conservative force, such as friction or air resistance, dissipates energy as heat and sound. Its work depends on the path (a longer path means more friction loss), and the energy is not recoverable. This distinction is what makes mechanical energy conserved in some situations (Lesson 26) but not others.
A conservative force (e.g. gravity) does path-independent work and stores recoverable energy as potential energy. A non-conservative force (e.g. friction, air resistance) does path-dependent work and dissipates energy as heat, which cannot be recovered. This determines when mechanical energy is conserved.
Pause, copy the highlighted comparison into your book before the check below.
Last check: Decide if the statement is true or false.
Friction is a conservative force because the energy it removes can be fully recovered.
Match each situation with the work done by the stated force:
Dot point: Relate the work done on an object to the change in its kinetic energy during accelerated rectilinear motion using W = Fs = ΔKE
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A $1200\ \text{kg}$ car accelerates from $8.0\ \text{m s}^{-1}$ to $20\ \text{m s}^{-1}$ along a straight road. Using the work–energy theorem, calculate the net work done on the car.
1 mark: states $W = \Delta KE = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2$ · 1 mark: $\tfrac{1}{2}(1200)(20)^2 = 240\,000\ \text{J}$ · 1 mark: $\tfrac{1}{2}(1200)(8.0)^2 = 38\,400\ \text{J}$ · 1 mark: $W = 240\,000 - 38\,400 \approx 2.0 \times 10^5\ \text{J}$
ApplyBand 5(4 marks) 2. A $1500\ \text{kg}$ car accelerates from $8.0\ \text{m s}^{-1}$ to $20\ \text{m s}^{-1}$ over $120\ \text{m}$. Use the work-energy theorem to determine the net force acting on it.
1 mark: $KE_i = 48\ \text{kJ}$ · 1 mark: $KE_f = 300\ \text{kJ}$ · 1 mark: $\Delta KE = 252\ \text{kJ} = W = Fs$ · 1 mark: $F = 2100\ \text{N}$
Dot point: Relate the work done on an object to the change in its kinetic energy during accelerated rectilinear motion using W = Fs = ΔKE
UnderstandBand 5(3 marks) 3. Compare conservative and non-conservative forces, giving one example of each, and explain how the distinction is tested.
1 mark: conservative, work done is path-independent and recoverable · 1 mark: non-conservative, work depends on the path and is dissipated · 1 mark: the test is whether the work around a closed loop is zero, with valid examples
Dot point: Compare conservative forces and non-conservative forces
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): By the work–energy theorem, the net work equals the change in kinetic energy: $W = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2$. The final kinetic energy is $\tfrac{1}{2}(1200)(20)^2 = 240\,000\ \text{J}$ and the initial kinetic energy is $\tfrac{1}{2}(1200)(8.0)^2 = 38\,400\ \text{J}$. Therefore $W = 240\,000 - 38\,400 = 201\,600\ \text{J} \approx 2.0 \times 10^5\ \text{J}$. This positive net work is what accelerated the car.
Q2 (4 marks): The work-energy theorem states that the net work done equals the change in kinetic energy, $W = Fs = \Delta KE$. The initial and final kinetic energies are $$KE_i = \tfrac{1}{2}(1500)(8.0)^2 = 48\,000\ \text{J}$$ $$KE_f = \tfrac{1}{2}(1500)(20)^2 = 300\,000\ \text{J}$$ so $$\Delta KE = 300\,000 - 48\,000 = 252\,000\ \text{J}$$ Since the net force acts along the $120\ \text{m}$ displacement, $$F = \frac{W}{s} = \frac{252\,000}{120} = 2100\ \text{N}$$ in the direction of motion. Note the theorem gives the NET force, so the engine must supply more than this to overcome resistive forces as well.
Q3 (3 marks): A conservative force does work that depends only on the start and end positions, not on the route taken, and the energy it transfers is stored and fully recoverable. Gravity is the standard example: raising a book and lowering it back returns exactly the energy spent. A non-conservative force does work that depends on the path taken and dissipates energy irreversibly, usually as heat. Kinetic friction is the standard example: a longer or more winding route means more sliding and more energy lost, and none of it returns when the object comes back. The test is the work done around a closed loop that returns to the starting point. For a conservative force that total is exactly zero; for a non-conservative force it is negative, because energy has left the system.
You can now calculate work, apply the work–energy theorem $W = \Delta KE$, and classify forces as conservative or non-conservative. Next lesson adds gravitational potential energy to the energy toolkit.
↩ Back to focus-area hubAt the start you predicted whether holding a box still does work. The answer: no work is done in the physics sense, because $W = Fs$ and there is no displacement ($s = 0$), your muscles tire for biological reasons, not because work is done on the box. If you said work was done, revisit Card 1.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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