Determine the sign of work from force and displacement
Work is an energy transfer associated with a force acting through displacement. For a constant force and straight displacement, W=Fd cosθ, where θ is the angle between the force and displacement vectors. The product has units N m=J. Time does not appear in this expression; how rapidly work occurs is a separate power question.
Learning outcome: Hold the force at 12 N and compare angles 0°,60°,90°,120° and 180°. Predict signed work, then final kinetic energy and speed. Explain why a backward component reduces energy even while the object still moves forward.
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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The model’s 2 kg object begins at 8 m/s, so Ki=64 J. A 12 N force acting at 60° over 3 m does W=12×3×0.5=18 J. Final K=82 J and speed√82≈9.06 m/s. At 120°, work is−18 J instead; the force magnitude and distance are unchanged.
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At which angle is the work zero? Does that require the force itself to vanish?
Experiment
Hold the force at 12 N and compare angles 0°,60°,90°,120° and 180°. Predict signed work, then final kinetic energy and speed. Explain why a backward component reduces energy even while the object still moves forward.
Determine signed work from a force component
Worked teaching model · synthetic values · A 2 kg point object moves 3 m along a straight ideal guide, starting at +8 m/s. Selected force has constant angle to displacement; guide forces are perpendicular and do zero work. No other tangential force. Supported settings leave positive kinetic energy throughout.
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| Quantity | Value | Unit |
|---|---|---|
| Force component along displacement | 6 | N |
| Work over 3 m | 18 | J |
| Initial kinetic energy | 64 | J |
| Final kinetic energy | 82 | J |
| Final speed | 9.05539 | m/s |
Accessible plot coordinates
| position (m) | force component (N) |
|---|---|
| 0 | 6 |
| 3 | 6 |
Only the force component along displacement contributes to this work.
W=Fd cosθ and Wnet=ΔK. A force can be nonzero while its work is zero if it is perpendicular to displacement. The work sign concerns energy transfer, not the magnitude of the force.
Only the force component along displacement contributes to this work.
W=Fd cosθ and Wnet=ΔK. A force can be nonzero while its work is zero if it is perpendicular to displacement. The work sign concerns energy transfer, not the magnitude of the force.
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Why does carrying a supported object horizontally at constant height involve zero work by its vertical support force in this ideal point-object model, even though a person may use metabolic energy?
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