IB Physics SLTopic A.3 — Work, Energy & PowerPaper 1 & 2Calculating Work Done~7 min read
Calculating Work Done
“Work” in physics has a strict meaning: it’s the energy transferred when a force moves something through a distance. No movement means no work, no matter how hard something is pushing.
📘 What you need to know
Work done is measured in joules, and 1 N m = 1 J
When a force acts parallel to the displacement, work done is calculated with W = Fs
When a force acts at an angle to the displacement, only the component parallel to the motion does work: W = Fscosθ
The work done by the resultant force on a system equals the change in that system’s energy
On a force–displacement graph, the area under the line is equal to the work done
What Actually Counts as Work?
Mechanical work is the transfer of energy that happens when an external force causes an object to move over some distance. If you push against a wall until you’re exhausted but the wall doesn’t budge, you’ve transferred zero mechanical work to it — however tired you feel, no displacement means no work done.
Force Parallel to Displacement
Work done equationW = Fs
Where W is work done in joules, F is the constant force applied in newtons, and s is the displacement in metres, measured along the same line as the force.
When the force and the displacement point the same way, work done is simply force multiplied by distance
When the Force Is at an Angle
Often the force isn’t lined up with the direction of motion at all — think of a rope pulling a sled upward and forward while the sled only moves horizontally. In that situation, only the part of the force that actually points along the direction of travel contributes to the work done.
Work done at an angleW = Fscosθ
Here θ is the angle between the force and the direction of motion. When θ = 0°, cosθ = 1 and the equation collapses back to W = Fs — the angled version is really just the general case.
Only the horizontal component, Fcos θ, is aligned with the sled’s motion — the vertical part of the pull does no work at all
Reading Work Done From a Graph
If you plot the force applied against the displacement it causes, the area under that graph is equal to the work done — this works even when the force isn’t constant, since you can simply find the area of whatever shape the graph makes.
🧭 Recipe: Work Done When a Force Acts at an Angle
List the knowns — the applied force, the angle it makes with the direction of motion, the displacement, and any other forces (like friction) acting along the line of motion
Resolve the force — find the component of the applied force that’s parallel to the displacement using Fcos θ
Combine forces along the line of motion — add or subtract any resistive forces acting on the same line to get a resultant force
Multiply by displacement — work done = resultant force × displacement
Quick recap: W = Fs when the force lines up with the motion; W = Fscos θ when it doesn’t. Only the parallel component of a force ever does work.
WE 1
A warehouse worker pushes a crate across a frictionless floor using a constant horizontal force of 45 N. The crate moves 8.0 m in the direction of the push. Calculate the work done on the crate.
Step 1 — Identify the equation
Force is parallel to displacement, so W = Fs
Step 2 — SubstituteW = 45 × 8.0= 360 J
WE 2
A dog pulls a sled using a rope angled at 35° above the horizontal, applying a force of 60 N. The sled moves 15 m horizontally against a constant friction force of 8.0 N. Calculate the work done on the sled in the direction of its motion.
Step 1 — Resolve the pulling force horizontallyF cos θ = 60 × cos(35°) ≈ 49.1 NStep 2 — Find the resultant force along the motionResultant = 49.1 − 8.0 ≈ 41.1 NStep 3 — Multiply by displacementW = 41.1 × 15≈ 617 J
💡 Top tips
Only the component of a force parallel to the displacement ever contributes to work done
If a force acts perpendicular to the motion (θ = 90°), it does zero work, since cos 90° = 0
Work done can come out negative when a force like friction opposes the direction of motion
Remember W = Fs is just the special case of W = Fscos θ where θ = 0°
⚠ Common mistakes
Using the full force in the equation instead of resolving it into the component parallel to displacement
Reaching for sin θ instead of cos θ — cos θ is used whenever θ is measured from the direction of motion
Forgetting to account for resistive forces such as friction when finding the net work done
Mixing up displacement with distance travelled when the path isn’t a straight line
Up next: Kinetic Energy — where we connect the work you’ve just calculated to the energy an object gains from moving.
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