SCIENCE · INVESTIGATE AND EXPLAIN

Add work across changes of force and direction

A force can change momentum without transferring kinetic energy if it is perpendicular to motion. For a constant force over a straight displacement, work is the component along that displacement times its length: W=Fd cosθ. Resolve force and displacement in one frame before multiplying. The result is a signed scalar, not another vector arrow.

Learning outcome: Compare a positive first segment with a resisting second, then reverse both force signs. Keep separate applied contributions, total work and intermediate kinetic energy. Explain how a perpendicular guide force can matter to momentum without adding energy.

Starting recap

Choose the system, interval and reference frame before using a conservation statement. Keep vector momentum, scalar energy and transferred power distinct.

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At F1=−2 N and F2=−2 N, K falls from 10 to 6 to 4 J; final speed is 2 m/s. Both negative work segments are accessible, despite the speed decreasing throughout.

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Predict

Net work on a constant-mass point determines its change in:

Your prediction and reason

Experiment

Compare a positive first segment with a resisting second, then reverse both force signs. Keep separate applied contributions, total work and intermediate kinetic energy. Explain how a perpendicular guide force can matter to momentum without adding energy.

Add signed work, not force magnitudes

Worked teaching model · synthetic values · A 2 kg point starts with K=10 J and travels first 2 m east then 1 m north on an ideal frictionless guide. Applied tangential forces are separately specified on each segment. Guide reactions do no work; ideal direction-change limit is not a constructed sharp track.

First segment tangential force: 4 N Second segment tangential force: -2 N

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Energy along the two-segment path

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: piecewise work-energy account (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Energy along the two-segment path

piecewise work-energy account

traveled path length (m)	kinetic energy (J)
0	10
2	18
3	16

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
First applied work	8	J
Second applied work	-2	J
Guide-reaction work	0	J
Total work	6	J
Kinetic energy after first segment	18	J
Final kinetic energy	16	J
Final speed	4	m/s

Work is a signed scalar sum of force components along each displacement, not the sum of force magnitudes.

The first force acts over 2 m and the second over 1 m, so Wnet=2F1+F2. All supported states retain positive kinetic energy along both segments. The ideal guide changes direction without transferring energy; it may transfer momentum, so kinetic-energy accounting does not imply total point momentum is conserved.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Work is a signed scalar sum of force components along each displacement, not the sum of force magnitudes.

Observation to attach to a capture

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

Understand

Define the energy or momentum system

Build the algebraic account

The original packet follows a 2 kg point with initial kinetic energy 10 J along two segments: 2 m east and 1 m north. The tangential applied force on the first is F1 and on the second is F2. Their works are2F1 and F2 joules. An ideal frictionless guide supplies the direction-changing reaction without work; the sharp-corner limit is a mathematical simplification, not a proposed constructed track.

Check an original packet

Only total work equals the change of kinetic energy. Here all other force work is zero by stipulation, so Kmid=10+2F1 and Kfinal=10+2F1+F2. For F1=4 N and F2=−2 N, the first stage adds8 J and the second removes2 J. Final K is 16 J, giving speed 4 m/s because m=2 kg. A selected positive contribution alone would overstate the endpoint energy.

Test the domain and evidence

All supported control settings keep K positive along both segments; that check is necessary before treating the endpoint as reachable. If an added resisting force stopped the point earlier, its work would have to be integrated only along the actual accessible path. The guide can transfer momentum while doing no work, so the energy account is not a claim of conserved point momentum.

Worked original packet

At F1=−2 N and F2=−2 N, K falls from 10 to 6 to 4 J; final speed is 2 m/s. Both negative work segments are accessible, despite the speed decreasing throughout.

Bounded investigation

Changed-case numerical practice

Try the changed case before feedback

A 3 kg point starts with kinetic energy 6 J. Applied works are+12 J and−4 J and all other work is zero. Find final kinetic energy in joules.

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and limits

Original content checked 30 October 2026 against OpenStax College Physics 2e. All packets and diagrams are synthetic teaching evidence. No physical launches, collisions, spring releases, explosive processes or equipment-safety judgments are requested. Retained self-reports and public numerical feedback do not certify supervised laboratory practice or independent mastery.

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What extra information would be needed if an applied force varied continuously along a segment?

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