Evaluate power on the same trajectory whose work you claim
A unit-mass point starts from rest at x(0) = 0 with net force F=k t. Integrating Newton's law gives v=k t²/2 and x=k t³/6. The force depends on time, so work can be evaluated as integral F(t)v(t)dt along this particular trajectory rather than pretending the endpoint force acts through the entire displacement.
Learning outcome: Derive velocity first, integrate Fv and compare with kinetic energy. Retain a counterexample to the endpoint-force-times-distance shortcut.
Starting recap
Use vector dot products for work, gradients for conservative force, definite integrals for accumulation and explicit energy/frame boundaries. Check accessible states before assigning trajectories.
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With k = 1 and T = 2, endpoint velocity2 m/s, work2 J, average power1 W and endpoint power4 W. Doubling k at fixed T quadruples work.
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For k = 2 and T = 2, average net power is:
Experiment
Derive velocity first, integrate Fv and compare with kinetic energy. Retain a counterexample to the endpoint-force-times-distance shortcut.
Integrate power along the force-driven trajectory
Worked teaching model · synthetic values · Unit-mass point starts from rest at x0=0 under net force F=k t. Work is integral Fv dt using the same trajectory, no other forces. Net power and energy are not device input efficiency or operating specifications.
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Power evaluated on the force-driven trajectory
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: F(t) times v(t) (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Power evaluated on the force-driven trajectory
F(t) times v(t)
| time (s) | net power (W) |
|---|---|
| 0 | 0 |
| 0.04 | 0.000128 |
| 0.08 | 0.001024 |
| 0.12 | 0.003456 |
| 0.16 | 0.008192 |
| 0.2 | 0.016 |
| 0.24 | 0.027648 |
| 0.28 | 0.043904 |
| 0.32 | 0.065536 |
| 0.36 | 0.093312 |
| 0.4 | 0.128 |
| 0.44 | 0.170368 |
| 0.48 | 0.221184 |
| 0.52 | 0.281216 |
| 0.56 | 0.351232 |
| 0.6 | 0.432 |
| 0.64 | 0.524288 |
| 0.68 | 0.628864 |
| 0.72 | 0.746496 |
| 0.76 | 0.877952 |
| 0.8 | 1.024 |
| 0.84 | 1.185408 |
| 0.88 | 1.362944 |
| 0.92 | 1.557376 |
| 0.96 | 1.769472 |
| 1 | 2 |
| 1.04 | 2.249728 |
| 1.08 | 2.519424 |
| 1.12 | 2.809856 |
| 1.16 | 3.121792 |
| 1.2 | 3.456 |
| 1.24 | 3.813248 |
| 1.28 | 4.194304 |
| 1.32 | 4.599936 |
| 1.36 | 5.030912 |
| 1.4 | 5.488 |
| 1.44 | 5.971968 |
| 1.48 | 6.483584 |
| 1.52 | 7.023616 |
| 1.56 | 7.592832 |
| 1.6 | 8.192 |
| 1.64 | 8.821888 |
| 1.68 | 9.483264 |
| 1.72 | 10.1769 |
| 1.76 | 10.90355 |
| 1.8 | 11.664 |
| 1.84 | 12.45901 |
| 1.88 | 13.28934 |
| 1.92 | 14.15578 |
| 1.96 | 15.05907 |
| 2 | 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.
| Quantity | Value | Unit |
|---|---|---|
| Endpoint velocity | 4 | m/s |
| Endpoint position | 2.666667 | m |
| Integrated net work | 8 | J |
| Kinetic energy increase | 8 | J |
| Average net power | 4 | W |
| Endpoint net power | 16 | W |
| Endpoint divided by average power | 4 | ratio |
Integrating net power along the same trajectory reproduces the kinetic-energy change.
For mass1 from rest, v=k t²/2 and x=k t³/6. Fv=k²t³/2 integrates to k²T⁴/8, equal to vfinal²/2. Endpoint power is four times interval-average power in this particular cubic-power ramp, not a universal ratio and not an efficiency claim.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
Integrating net power along the same trajectory reproduces the kinetic-energy change.
For mass1 from rest, v=k t²/2 and x=k t³/6. Fv=k²t³/2 integrates to k²T⁴/8, equal to vfinal²/2. Endpoint power is four times interval-average power in this particular cubic-power ramp, not a universal ratio and not an efficiency claim.
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Which initial condition made the power a pure cubic in time?
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