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SCIENCE · INVESTIGATE AND EXPLAIN

Integrate mechanical loss once within a declared system

A2 kg point in a stationary medium experiences linear resistance with coefficient1 kg/s and no drive. From initial velocity v0, the solution is v=v0 exp(−t/2). Its kinetic energy is v0² exp(−t) joules because m/2=1 in this packet.

Learning outcome: Integrate drag power, compare with endpoint kinetic change and write both point-only and point-plus-medium balances without double counting.

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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Try

For v(0) = 2 and T = 1, remaining kinetic energy4/e≈1.471518 J and internal gain4(1−1/e)≈2.528482 J. Their sum remains4 J.

Open the experiment

Predict

Including the medium in the stipulated closed boundary means:

Experiment

Integrate drag power, compare with endpoint kinetic change and write both point-only and point-plus-medium balances without double counting.

Integrate dissipation into a larger energy account

Worked teaching model · synthetic values · Horizontal point m2 kg with drag coefficient1 kg/s starts at selected positive velocity in a stationary medium, no other forces. v=v0 exp(−t/2). All mechanical loss becomes internal/thermal energy of the selected point-plus-medium boundary by stipulation; no measured heating distribution.

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Mechanical and internal conversion within a chosen boundary

energy (J)time (s)02160

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: mechanical kinetic energy (solid).2: internal/thermal gain (dashed). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Mechanical and internal conversion within a chosen boundary

mechanical kinetic energy

time (s)energy (J)
016
0.0415.37263
0.0814.76986
0.1214.19073
0.1613.6343
0.213.09969
0.2412.58605
0.2812.09254
0.3211.61838
0.3611.16282
0.410.72512
0.4410.30458
0.489.900534
0.529.512329
0.569.139345
0.68.780986
0.648.436679
0.688.105872
0.727.788036
0.767.482663
0.87.189263
0.846.907368
0.886.636527
0.926.376305
0.966.126286
15.886071
1.045.655275
1.085.433528
1.125.220477
1.165.015779
1.24.819107
1.244.630147
1.284.448597
1.324.274165
1.364.106572
1.43.945551
1.443.790844
1.483.642203
1.523.49939
1.563.362177
1.63.230344
1.643.103681
1.682.981984
1.722.865058
1.762.752718
1.82.644782
1.842.541079
1.882.441442
1.922.345711
1.962.253735
22.165365

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.

internal/thermal gain

time (s)energy (J)
00
0.040.627369
0.081.230138
0.121.809273
0.162.365699
0.22.900308
0.243.413954
0.283.90746
0.324.381615
0.364.837179
0.45.274879
0.445.695417
0.486.099466
0.526.487671
0.566.860655
0.67.219014
0.647.563321
0.687.894128
0.728.211964
0.768.517337
0.88.810737
0.849.092632
0.889.363473
0.929.623695
0.969.873714
110.11393
1.0410.34473
1.0810.56647
1.1210.77952
1.1610.98422
1.211.18089
1.2411.36985
1.2811.5514
1.3211.72584
1.3611.89343
1.412.05445
1.4412.20916
1.4812.3578
1.5212.50061
1.5612.63782
1.612.76966
1.6412.89632
1.6813.01802
1.7213.13494
1.7613.24728
1.813.35522
1.8413.45892
1.8813.55856
1.9213.65429
1.9613.74627
213.83464

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
QuantityValueUnit
Final velocity1.471518m/s
Initial mechanical energy16J
Final mechanical energy2.165365J
Integrated drag work-13.83464J
Point-plus-medium internal energy gain13.83464J
Mechanical plus internal total16J

The point loses mechanical energy while a larger stipulated boundary retains the corresponding internal energy.

Drag power is−bdrag v²=−v0² exp(−t) W with the supplied m2,bdrag1. Integrating gives work−v0²(1−exp(−T)), equal to Kfinal−Kinitial. Count that loss once: point-only external work and point-plus-medium thermal conversion are two descriptions with different boundaries, not two independent losses.

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

The point loses mechanical energy while a larger stipulated boundary retains the corresponding internal energy.

Drag power is−bdrag v²=−v0² exp(−t) W with the supplied m2,bdrag1. Integrating gives work−v0²(1−exp(−T)), equal to Kfinal−Kinitial. Count that loss once: point-only external work and point-plus-medium thermal conversion are two descriptions with different boundaries, not two independent losses.

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

Understand

Integrate mechanical loss once within a declared system

A2 kg point in a stationary medium experiences linear resistance with coefficient1 kg/s and no drive. From initial velocity v0, the solution is v=v0 exp(−t/2). Its kinetic energy is v0² exp(−t) joules because m/2=1 in this packet.

Drag power on the point is Fv=−v²=−v0² exp(−t) watts. Integrating gives work−v0²(1−exp(−T)), equal to the kinetic-energy change. The sign follows from force opposing velocity, not from replacing velocity with an unsigned quantity prematurely.

For v(0) = 4 m/s and T = 2 s, initial kinetic energy16 J falls to16e⁻²≈2.165365 J. Integrated drag work is approximately−13.834635 J. In the larger stipulated point-plus-medium boundary, internal/thermal energy gains13.834635 J, closing the initial16 J account.

Do not subtract point-only drag work and then subtract the same thermal gain again in the larger system. They are alternative boundary descriptions of one conversion. The packet does not determine where a real medium heats, how much heat escapes, or a measurable temperature rise; those require additional material and boundary information.

Changed-case check

Predict before reading the comparison. For v(0) = 2 and T = 1, remaining kinetic energy4/e≈1.471518 J and internal gain4(1−1/e)≈2.528482 J. Their sum remains4 J.

Retained investigation

Integrate drag power, compare with endpoint kinetic change and write both point-only and point-plus-medium balances without double counting. What evidence is missing before converting the internal gain into a predicted temperature rise?

Source and boundary

Primary conceptual reference checked30 October2026: OpenStax University Physics Volume1. All functions, numerical packets and teaching prose are original. Bounded supplied models are not measured experiments, apparatus instructions, independent practical skill or institutional credit. Numerical algorithm properties are derived for the disclosed model; do not treat them as universal solver guarantees.

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Including the medium in the stipulated closed boundary means:

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What evidence is missing before converting the internal gain into a predicted temperature rise?

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