Use a force-position graph to calculate work
When force changes with position, multiply force by displacement only after accounting for that change. Partition the path into small intervals and add force component times interval length. On a force-versus-position graph, this is signed area. Straight segments allow exact triangle and trapezoid geometry without calculus.
Learning outcome: Compare endpoint pairs(8,−4),(4,−4) and(−8,−4) N. Sketch the line and signed areas first. Use both geometric splitting and the trapezoid expression. Check that the final kinetic energy equals 50 J plus the calculated work.
Starting recap
Review signed force components, graph areas, speed squared and system boundaries. Distinguish an interaction force from an energy store or transfer.
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Try
A force changes linearly from 8 N at x=0 to−4 N at x=4 m. Trapezoidal area is(8−4)×4/2=8 J. It crosses zero at x=8/3 m: positive triangular area is 32/3 J and negative-area magnitude 8/3 J. Their signed sum is 8 J, so final kinetic energy is 58 J.
Open the experimentPredict
If the endpoint forces are equal and opposite, can the force be nonzero along most of the path yet do zero net work?
Experiment
Compare endpoint pairs(8,−4),(4,−4) and(−8,−4) N. Sketch the line and signed areas first. Use both geometric splitting and the trapezoid expression. Check that the final kinetic energy equals 50 J plus the calculated work.
Read work from a changing force graph
Worked teaching model · synthetic values · Net horizontal force varies linearly between endpoints over 0–4 m. A 2 kg object starts with 50 J kinetic energy and moves positively. Work is exact trapezoidal area; supported settings retain positive kinetic energy, so the full path is reachable.
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Axes rescale with settings; compare numerical values, not screen angles or apparent lengths. Solid/points show the selected model relationship or observations; the dashed comparison is explained below.
| Quantity | Value | Unit |
|---|---|---|
| Exact signed work | 8 | J |
| Initial kinetic energy | 50 | J |
| Final kinetic energy | 58 | J |
| Final speed | 7.61577 | m/s |
| Average net force over position interval | 2 | N |
Accessible plot coordinates
| position (m) | net force (N) |
|---|---|
| 0 | 8 |
| 4 | -4 |
The signed trapezoid area is work; the slope of this force-position graph is not work.
W=(Fstart+Fend)Δx/2 for a linear graph. Opposite-signed regions cancel in net work; total positive area plus negative-area magnitude would not be the work–energy result.
The signed trapezoid area is work; the slope of this force-position graph is not work.
W=(Fstart+Fend)Δx/2 for a linear graph. Opposite-signed regions cancel in net work; total positive area plus negative-area magnitude would not be the work–energy result.
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Public practice checks a specific claim. It does not establish independent mastery.
What extra information would be needed to calculate work from a force-versus-time record? Why is multiplying its area by distance not correct?
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