Change force branches at a stopping event
A 2 kg object starts sliding right with speed v0 on a level plane. Gravity is 10 m/s², μk=0.3 and μs=0.5. A constant rightward drive F lies between 0 and 10 N. During rightward slip kinetic friction is 6 N leftward, so acceleration is (F−6)/2 m/s².
Learning outcome: Compare F=0,4,6,10 N at v0=2 m/s. Predict whether a stop exists and whether it lies in the 2 s interval. Retain the position and velocity at the event before evaluating the final state.
Starting recap
Draw recipient-specific interactions, resolve signed components and identify the domain of each force/contact model before substituting.
All lessons are open directly. Your saved experiments remain when you leave and return.
Try
With F=4 N and v0=2 m/s, acceleration is−1 m/s² and the stop occurs exactly at 2 s after 2 m. Immediately after stopping, static friction balances4 N rather than retaining the 6 N sliding value.
Open the experimentPredict
After a stop with drive below static capacity, continuing the old sliding formula would:
Experiment
Compare F=0,4,6,10 N at v0=2 m/s. Predict whether a stop exists and whether it lies in the 2 s interval. Retain the position and velocity at the event before evaluating the final state.
Stop a sliding object without inventing reverse friction
Worked teaching model · synthetic values · A 2 kg object slides right on a level plane with g=10, mu_k=.3 and mu_s=.5. Drive stays between 0 and 10 N. Observe 2 s; after any stop the available static friction can balance the drive. No reversed kinetic branch is imposed.
Reset keeps your prediction, captures and practice history.
Velocity with stop-and-stick event
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: admissible velocity (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Velocity with stop-and-stick event
admissible velocity
| time (s) | rightward velocity (m/s) |
|---|---|
| 0 | 2 |
| 0.05 | 1.9 |
| 0.1 | 1.8 |
| 0.15 | 1.7 |
| 0.2 | 1.6 |
| 0.25 | 1.5 |
| 0.3 | 1.4 |
| 0.35 | 1.3 |
| 0.4 | 1.2 |
| 0.45 | 1.1 |
| 0.5 | 1 |
| 0.55 | 0.9 |
| 0.6 | 0.8 |
| 0.65 | 0.7 |
| 0.7 | 0.6 |
| 0.75 | 0.5 |
| 0.8 | 0.4 |
| 0.85 | 0.3 |
| 0.9 | 0.2 |
| 0.95 | 0.1 |
| 1 | 0 |
| 1.05 | 0 |
| 1.1 | 0 |
| 1.15 | 0 |
| 1.2 | 0 |
| 1.25 | 0 |
| 1.3 | 0 |
| 1.35 | 0 |
| 1.4 | 0 |
| 1.45 | 0 |
| 1.5 | 0 |
| 1.55 | 0 |
| 1.6 | 0 |
| 1.65 | 0 |
| 1.7 | 0 |
| 1.75 | 0 |
| 1.8 | 0 |
| 1.85 | 0 |
| 1.9 | 0 |
| 1.95 | 0 |
| 2 | 0 |
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.
| Quantity | Value | Unit |
|---|---|---|
| Initial sliding acceleration | -2 | m/s² |
| Stop time in continued sliding model | 1 | s |
| Position after 2 s | 1 | m |
| Velocity after 2 s | 0 | m/s |
| Stopped by interval end | 1 | flag |
| Static capacity after a stop | 10 | N |
| Friction magnitude at interval end | 2 | N |
The friction law changes at zero slip; an extrapolated negative velocity is not the next state here.
Before a stop, kinetic friction is 6 N leftward. If F<6 N, the stop time is v0/((6−F)/2). Afterward static friction balances F because F≤10 N, so the object stays at rest. If F=6 N it keeps its initial speed; larger drive accelerates it rightward. This is not a physical stopping-distance or traction guarantee.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
The friction law changes at zero slip; an extrapolated negative velocity is not the next state here.
Before a stop, kinetic friction is 6 N leftward. If F<6 N, the stop time is v0/((6−F)/2). Afterward static friction balances F because F≤10 N, so the object stays at rest. If F=6 N it keeps its initial speed; larger drive accelerates it rightward. This is not a physical stopping-distance or traction guarantee.
Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.
Understand
Apply
Public practice checks a specific claim. It does not establish independent mastery.
Why does a force law need a domain as well as a numerical coefficient?
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.
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.
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.
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.
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.