IB Physics HLTopic 1 — Motion, Forces & EnergyPaper 1 & 2Work, Energy & Power~10 min read
Calculating Work Done
In physics, “work” has a very precise meaning — and it’s not the same as being tired. You only do work when a force actually moves something. Push against a locked door all day and you’ve done zero work on it, because it hasn’t budged. Slide a box across the floor, though, and you’ve transferred energy to it — that transfer is the work done. Here we’ll turn that idea into two simple equations and learn the one trick that trips most people up: what to do when the force and the motion don’t point the same way.
📘 What you need to know
Work done is the energy transferred when a force moves an object — the two are the same thing
Work is measured in joules (J), where 1 J = 1 N m
When the force is along the motion: W = Fs
When the force is at an angle to the motion: W = Fs cos θ
Only the component of the force parallel to the displacement does work
A force at 90° to the motion does no work at all
The area under a force–displacement graph equals the work done
Work when the force is along the motion
When a constant force acts in the same direction as the movement (parallel to it), the work done is just force times distance:
Work done — force along the motionW = Fs
where W is the work done in joules (J), F is the constant force in newtons (N), and s is the displacement in metres (m). Because a joule is a newton-metre, this equation is really just the definition of the joule in action.
When the force (green) points the same way as the displacement (blue), all of it does work: W = Fs.
WE 1
A worker pushes a trolley 12 m across a flat floor with a constant horizontal force of 45 N, in the direction of motion. Calculate the work done on the trolley.
Step 1 — force is along the motion, so use W = Fs
W = F × s
Step 2 — substituteW = 45 × 12W = 540 JThat 540 J of energy has been transferred from the worker to the trolley.
Here’s the mental test for whether work is being done: did the force move something in its own direction? If yes, work is done and energy is transferred. If the thing didn’t move, or moved sideways to the force, that part of the force did no work — no matter how hard you pushed.
Work when the force is at an angle
Often the force isn’t lined up with the motion. Think of pulling a sled with a rope held at an angle, or a suitcase on wheels. The rope pulls partly forwards and partly upwards, but the case only moves forwards. Only the forward part of the pull does any work.
To find that forward part we resolve the force into components and take the one parallel to the displacement. If the force F acts at an angle θ to the direction of motion, the useful component is F cos θ, and the work done becomes:
Work done — force at an angleW = Fs cos θ
The rope pulls with force F at angle θ. Only the horizontal component, F cos θ (blue), lies along the motion and does work — so W = Fs cos θ.
Notice what happens at the two extremes. When θ = 0 (force straight along the motion), cos 0 = 1, so W = Fs — back to the simple case. When θ = 90° (force at right angles to the motion), cos 90° = 0, so W = 0. A force perpendicular to the motion does no work at all — which is why carrying a heavy bag horizontally does no work against gravity, since gravity points down while you move sideways.
WE 2
A child pulls a sled 15 m along flat ground using a rope held at 30° to the horizontal, pulling with a force of 80 N. Calculate the work done on the sled.
Step 1 — force is at an angle, so use W = Fs cos θ
W = F × s × cos θ
Step 2 — substitute the valuesW = 80 × 15 × cos 30°W = 1200 × 0.866W = 1040 J (3 s.f.)Only the horizontal part of the pull (80 cos 30° ≈ 69 N) does work — the upward part just lifts a little, moving nothing horizontally.
Work done against friction
When you drag something along a rough surface, your force does work on the object, but friction acts backwards and does negative work — it takes energy away, usually turning it into heat and sound. It’s often useful to track each separately.
Push does work ON crate
but →
Friction does negative work
so net =
(F − friction) × distance
WE 3
A crate is pushed 8.0 m across a rough floor with a constant horizontal force of 120 N. A frictional force of 30 N opposes the motion. Find (a) the work done by the push, (b) the work done against friction, and (c) the net work done on the crate.
(a) work done by the pushW = 120 × 8.0 = 960 J(b) work done against frictionW = 30 × 8.0 = 240 J(c) net work = resultant force × distanceW = (120 − 30) × 8.0 = 90 × 8.0Net work = 720 JThe 240 J lost to friction becomes heat — energy is still conserved, just spread into the surroundings.
Work from a force–displacement graph
If a force isn’t constant, you can’t just multiply. Instead, plot the force against displacement and find the area under the graph — that area is the work done. For a constant force the graph is a flat line and the area is a rectangle (height F × width s), which gives back W = Fs. For a changing force, like stretching a spring, the area might be a triangle or a more complex shape.
🛠️ Working out work done
Identify the force doing the work and the displacement of the object.
Check the direction. Is the force along the motion, or at an angle to it?
Along the motion → use W = Fs.
At an angle → use W = Fs cos θ, taking the component parallel to the displacement.
If the force changes, find the area under the force–displacement graph instead.
💡 Top tips
Only the parallel part counts. Always take the component of the force that lies along the displacement — that’s what cos θ gives you.
Perpendicular means zero work. A force at 90° to the motion (like the normal force, or gravity on a horizontal move) does no work.
Watch for spare numbers. Exam questions often hand you masses or speeds you don’t need, just to test whether you know what matters.
cos θ only when angled. If the force is already along the motion, don’t sneak in a cosine — you’d be shrinking a force that shouldn’t shrink.
Quick recap: Work done is energy transferred when a force moves an object, measured in joules. Use W = Fs when the force is along the motion, and W = Fs cos θ when it’s at an angle, taking only the component parallel to the displacement. A force at 90° does no work, and the area under a force–displacement graph gives the work done.
⚠ Common mistakes
Using the full force when it’s at an angle — you must take the F cos θ component along the motion
Adding a cos θ when the force is already parallel to the motion (then θ = 0 and cos 0 = 1)
Thinking a force with no movement does work — no displacement means no work
Forgetting that a perpendicular force does zero work
Getting distracted by unnecessary values like mass when only force and distance are needed
You’ve now got the two work equations under your belt. The natural next question is where all that transferred energy goes — and the answer is usually motion. When a resultant force does work on an object, it speeds it up, and that’s exactly the store we look at next: kinetic energy.
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