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SCIENCE · INVESTIGATE AND EXPLAIN

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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Try

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.

Open the experiment

Predict

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.

Reset keeps your prediction, captures and practice history.

0360position (m)force component (N)

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.

Calculated evidence ledger — synthetic teaching model
QuantityValueUnit
Force component along displacement6N
Work over 3 m18J
Initial kinetic energy64J
Final kinetic energy82J
Final speed9.05539m/s
Accessible plot coordinates
Primary plotted points
position (m)force component (N)
06
36

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.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

1. Orient: 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.

2. Explain and derive

A force component parallel to displacement does positive work, an opposite component does negative work, and a perpendicular force does zero work. A nonzero force does not guarantee nonzero work. The normal force of an ideal fixed guide can redirect motion without doing work when it remains perpendicular to each displacement.

The net work on a point object equals its change in kinetic energy: Wnet=Kf−Ki. Include the work of every relevant force, not only the force you happen to measure. For constant net force along motion, combining F=ma with v²−v₀²=2 aΔx yields FΔx=½m(v²−v₀²), showing why the work–energy relation matches the motion equations.

State the system before interpreting a transfer. For an object alone, gravitational work changes its kinetic energy. For object plus Earth, the same gravitational interaction can instead be represented by a change in internal potential energy. Do not count both the gravitational work and the corresponding potential-energy change as independent input terms in one ledger.

3. Work an example

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.

4. Predict and investigate

Predict first: At which angle is the work zero? Does that require the force itself to vanish?

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. Use the lesson controls and retain the settings with your evidence.

5. Transfer and explain

Try a new practice case

A constant 8 N force opposes an object’s 3 m displacement. Find the work done by that force.

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

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?

Retain a first attempt, signed ledger, units, system boundary, assumptions and a changed-case prediction. Public worked checks are practice, not independent mastery evidence. Synthetic outputs are not physical measurements.

Application-point work and a deforming system

A point-object work–energy equation is not a complete internal-energy account for a deforming system. Work uses displacement of the force’s application point, which need not equal the center-of-mass displacement. Two equal outward pulls can stretch a system while its center stays fixed; zero center displacement does not mean zero external work.

Original predict/check: slowly stretch an ideal spring of stiffness 40 N/m symmetrically so each end moves outward 0.10 m. Total extension is 0.20 m and each holding force rises linearly from zero to 8 N. Each endpoint contributes triangular work (0+8)×0.10/2=0.40 J, giving 0.80 J total, equal to 40×0.20²/2. The center stays fixed and final bulk kinetic energy is negligible. Halving both endpoint displacements gives one quarter of that work, 0.20 J. This explicitly qualifies the framework’s configuration-change wording: general external work can change internal potential energy, not necessarily only kinetic energy. The actual point-object theorem remains valid within its stated model.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 3 topic mapping: 3.2. Checked 30 October 2026; this is a content cross-reference, not AP authorization.

Apply

A 10 N force acts at 60° to a straight displacement of 4 m. What work does that force do?

Public practice checks a specific claim. It does not establish independent mastery.

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?

Continue

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