SCIENCE · INVESTIGATE AND EXPLAIN

Integrate dissipation only along an accessible branch

Combine position-dependent friction with kinetic-energy feasibility and an enlarged thermal inventory.

Learning outcome: Compare K 0=4 and 8 at c=1. Independently locate the first stop, retain the target flag, and reconcile friction work with thermal energy at the physical endpoint. Explain why a new contact branch is needed after rest.

Starting recap

Derive the integral or derivative, preserve the system boundary, and reject physically inaccessible continuations.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

For K 0=8,c=1, the target 2 m is reached with loss 6 J and K=2 J; stop is sqrt(20)−2≈2.47214 m. With K 0=4,c=1, stop=sqrt(12)−2≈1.46410 m, before the target.

Open the experiment

Predict

What should happen if the algebraic right-sliding K becomes negative before the target?

Your prediction and reason

Experiment

Compare K 0=4 and 8 at c=1. Independently locate the first stop, retain the target flag, and reconcile friction work with thermal energy at the physical endpoint. Explain why a new contact branch is needed after rest.

Stop a dissipative integral at its physical boundary

Worked teaching model · synthetic values · Mass 2 kg initially rightward; resistance magnitude 2+cx N while sliding right. Target x=2 m; if it stops earlier, do not continue the right-sliding law into negative K or a fabricated reversal.

Initial kinetic energy: 8 J Friction slope c: 1 N/m

Reset keeps your prediction, captures and practice history.

Allowed kinetic energy branch

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Allowed kinetic energy branch: numerical coordinates

position (m)	energy (J)
0	8
0.05	7.89875
0.1	7.795
0.15	7.68875
0.2	7.58
0.25	7.46875
0.3	7.355
0.35	7.23875
0.4	7.12
0.45	6.99875
0.5	6.875
0.55	6.74875
0.6	6.62
0.65	6.48875
0.7	6.355
0.75	6.21875
0.8	6.08
0.85	5.93875
0.9	5.795
0.95	5.64875
1	5.5
1.05	5.34875
1.1	5.195
1.15	5.03875
1.2	4.88
1.25	4.71875
1.3	4.555
1.35	4.38875
1.4	4.22
1.45	4.04875
1.5	3.875
1.55	3.69875
1.6	3.52
1.65	3.33875
1.7	3.155
1.75	2.96875
1.8	2.78
1.85	2.58875
1.9	2.395
1.95	2.19875
2	2

Calculated values under the declared model
Quantity	Value	Unit
Stopping position	2.47214	m
Evaluated physical endpoint	2	m
Target reached	1	flag
Friction work to endpoint	-6	J
Final kinetic energy	2	J
Thermal increase in enlarged system	6	J

Do not turn a negative algebraic kinetic energy into an imaginary speed or a prediction of reversal.

Solve K0−2x−cx²/2=0. The rationalized root avoids division by c near zero and equals K0/2 at c=0. Mechanical energy decreases for the object while the declared object/surface thermal inventory increases; this is not destruction of total energy.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Do not turn a negative algebraic kinetic energy into an imaginary speed or a prediction of reversal.

Observation to attach to a capture

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

Understand

Declare the contact law and direction

A 2 kg object starts moving right on a horizontal surface. The stipulated sliding resistance has magnitude 2+cx N, directed left while rightward sliding continues. This is a spatially varying phenomenological contact model, not a claim that a friction coefficient universally grows with distance. No drive is supplied.

Derive the energy branch

Integrating the opposing force from zero to x gives work −2x−cx²/2. Thus K(x)=K 0−2x−cx²/2. This also follows from m v dv/dx=F while velocity is nonzero. The expression describes only the right-sliding branch; extending it past zero K would violate its physical domain.

Find the first stop

Solve K=0 for the nonnegative root. For c>0, xstop=(sqrt(4+2cK 0)−2)/c, equivalently 2K 0/(2+sqrt(4+2cK 0)), which avoids cancellation near zero slope. At c=0 the limit is K 0/2. The workbench evaluates at the smaller of the target 2 m and stopping distance.

Do not fabricate a reversal

Once the object stops, the assumed rightward-sliding resistance law no longer supplies a valid acceleration rule. With no driving force, there is no reason to invent a leftward launch from a negative K or continued leftward friction. A different static-contact model would govern any later imposed force. The displayed flag distinguishes a reached target from a stopped-short path.

Move the system boundary without losing energy

For an object-only mechanical account, friction transfers energy out. In an enlarged object-and-surface inventory, the assigned lost mechanical energy becomes a thermal increase under the packet assumptions. Total energy is not destroyed. Real partition between object and surface, deformation, sound and temperature would require further measurements; one energy ledger does not determine them.

Worked original case

For K 0=8,c=1, the target 2 m is reached with loss 6 J and K=2 J; stop is sqrt(20)−2≈2.47214 m. With K 0=4,c=1, stop=sqrt(12)−2≈1.46410 m, before the target.

Investigation

Public numerical transfer practice

Try a new practice case

For K 0=6 J and constant resistance 2 N, what is the stopping distance in metres?

Numerical result in the requested units

This public, unsaved check is not a secure assessment. Explain the derivative, integral or boundary in the retained course evidence workspace.

Sources and evidence boundary

Original teaching, values and calculations, checked 30 October 2026 against the official calculus-based Mechanics framework and OpenStax work and kinetic energy. The numerical packets are supplied or generated, not measured apparatus records. No physical launch, machinery, loading or field intervention is requested. Public feedback and virtual captures do not establish independent mastery or supervised laboratory participation.

Apply

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

Why does changing the system boundary change the transfer label but not eliminate the energy balance?

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Integrate gravitational work with a defensible reference Next: Integrate power without confusing one interaction with net work

Save a science backup or import one

Science work is stored locally in this browser, separately from mathematics and AI. It does not sync automatically. Use this science-only backup; this checkout's older all-subject export does not include science.

A legacy file is available only when the complete snapshot fits the 36 MB limit. For larger snapshots, prepare a complete file set below; never discard notebooks to make a smaller file.

Choose a science backup (maximum 36 MB)

Matching imported notebook drafts replace their current drafts. Omitted lessons and known first attempts remain. Export your current work first if you want to retain both drafts.

Complete file sets and course backups

A file set can hold more work than one science file. Keep its manifest and every part together. Each file is limited to 36 MB. This format needs this version of LearnAI or a compatible newer version; older versions can use only the legacy single-file backup above.

Export scope

The set includes every saved science course and global continuation at this moment. Later edits need a new export. A conflicting tab exports its own visible work, which may differ from saved work in another tab.

Choose the manifest and all parts together

0 file(s) selected. File order and filenames do not determine identity. Missing, changed or conflicting parts block the entire restore; no good part is applied by itself.

Science storage: loading. No account or remote computation is required.