A force trace reaches 50 N. That observation tells you its peak, but it does not tell you the change in the object's momentum. A brief peak and a sustained force can have very different effects. You also need to know whether the trace shows one applied force or the total external force.
For physiotherapy learners reading movement data, the useful task is to connect the plotted quantity to the claim being made. This article uses an original cart example to distinguish peak force, average force, applied impulse and net impulse. It does not prescribe an exercise or assign an injury threshold to a force value.
Identify the system, direction and interval
Impulse is the signed area under a force-time curve. The impulse of the net external force equals the system's change in momentum. For constant mass in one direction, Jnet = m(vfinal − vinitial). Average force is impulse divided by the specified elapsed time; it is not generally the largest force recorded. OpenStax: impulse and collisions.
Those statements leave practical choices to make. Which object is the system? Which direction is positive? Where does the interval start and end? Are all external forces included? A correct calculation attached to the wrong force trace answers the wrong question.
For the following example, the system is a cart with constant mass 40 kg. Positive means rightward. Its initial horizontal velocity is +1.0 m/s, and the interval lasts 1.0 s. A hand pushes horizontally to the right; a stipulated resistance acts at −10 N throughout the interval. These are the only horizontal external forces. Vertical support and weight cancel. The cart continues travelling right throughout, so the assumed resistance direction does not need to change.
Worked example: integrate the supplied force record
Assume the hand force is exactly constant within each interval in the table. The abrupt changes are an idealisation for teaching; these are not sparse samples requiring an interpolation rule.
| Interval | Duration | Hand force | Hand impulse |
|---|---|---|---|
| 0 to 0.2 s | 0.2 s | +50 N | +10 N·s |
| 0.2 to 0.6 s | 0.4 s | +30 N | +12 N·s |
| 0.6 to 1.0 s | 0.4 s | +20 N | +8 N·s |
| Total | 1.0 s | — | +30 N·s |
Each row's area is force multiplied by duration. The hand contributes +30 N·s. The resistance contributes (−10 N) × (1.0 s) = −10 N·s. Adding the impulses gives Jnet = +20 N·s.
Now connect net impulse to momentum:
Δv = Jnet/m = 20/40 = +0.5 m/s.
The final velocity is therefore +1.5 m/s. An answer of +0.5 m/s reports the change rather than the final velocity. An answer of +1.75 m/s results from treating the hand's +30 N·s as the net impulse and omitting the resistance.
The units also provide a useful check. A newton-second is a kilogram-metre per second, the unit of momentum. Dividing by kilograms gives metres per second. Multiplying force by seconds does not give energy.
Three numbers from the same record
The peak hand force is 50 N. The average hand force over the full second is 30 N. The average net force is 20 N. All three numbers are correct, but they describe different things.
A simple arithmetic mean of the three listed hand-force levels would be (50 + 30 + 20)/3, or approximately 33.3 N. That is wrong for the time average because the intervals have unequal durations. The 50 N level lasts half as long as each of the other two levels. Weighting each value by its duration gives the correct 30 N.
This distinction becomes especially useful when a report compresses a trace into a single summary. Read the label as a complete phrase: “average net horizontal force over the full one-second interval.” If the word “net,” direction or interval disappears, a later reader may reasonably reconstruct a different quantity.
Counterexample: the same peak, a different outcome
Reset the cart to the same initial conditions. Keep the same −10 N resistance for one second. This time, the hand supplies +50 N for only the first 0.2 s and zero force for the remaining 0.8 s.
The peak hand force is still 50 N. However, hand impulse is now +10 N·s, resistance impulse remains −10 N·s, and net impulse is zero. The final velocity is therefore +1.0 m/s, matching the initial velocity.
Zero net impulse over the complete interval does not mean the cart was stationary or never accelerated. During the first 0.2 s, its net force is +40 N, so its velocity increases by 0.2 m/s. During the remaining 0.8 s, the −10 N force decreases velocity by 0.2 m/s. It keeps moving right throughout. The endpoint momentum changes cancel while the intermediate motion remains different.
The comparison disproves a tempting shortcut: identical peaks do not establish identical momentum changes. It also shows why specifying the integration interval matters. If you stop the analysis at 0.2 s, you obtain a positive net impulse rather than zero.
Same impulse does not require the same peak either
Consider a third hand-force profile under the original conditions. It rises linearly from zero to 100 N and falls linearly back to zero within the first 0.6 s, making a triangle. Hand force is zero during the remaining 0.4 s. The triangle's area is ½ × 0.6 × 100 = 30 N·s.
Its hand impulse matches the first record, even though its peak is twice as large. With the same resistance impulse, its net impulse is again +20 N·s and final velocity again +1.5 m/s. These endpoint agreements do not establish equal acceleration histories or equal forces at every instant.
The triangle calculation is justified because linear sides were explicitly supplied. Connecting a few measured points with straight lines would introduce an additional approximation. An unexplained sampling gap cannot be converted into an exact area by drawing a convenient shape through it.
Transfer exercise: distinguish braking force from net braking force
A different cart has mass 30 kg. Over 0.6 s, its rightward velocity decreases from +0.8 to +0.2 m/s. The only horizontal forces are a hand force and constant resistance of −5 N.
Calculate the net impulse, the average net force and the average hand force. Can you recover the peak hand force from this information?
Reasoned feedback: net impulse is 30 × (0.2 − 0.8) = −18 N·s. Average net force is −18/0.6 = −30 N. Resistance contributes −5 × 0.6 = −3 N·s, leaving hand impulse of −15 N·s and average hand force of −25 N.
The negative signs indicate direction, not a negative mass or an invalid answer. The cart still moves right at the end; a leftward net force can reduce rightward momentum without reversing motion. Peak hand force is not determined because many different force histories could share this impulse and duration.
Use the graph's axes to choose the question
A force-time area addresses impulse. A force-displacement calculation addresses work. Swapping the horizontal axis changes the quantity, even if the vertical axis still says “force.” The companion work and power article develops that different calculation.
For your next practice trace, write four items before calculating: system, positive direction, time interval and included external forces. Then state whether your answer is a peak, an average, an impulse, a velocity change or a final velocity. This small habit makes the interpretation checkable rather than merely numerical.
Sources
- OpenStax, University Physics Volume 1: impulse and collisions. Relevant definitions and impulse–momentum relations inspected. All cart parameters, force histories, counterexamples and exercise calculations are original constructions, not textbook examples or clinical observations.
Source checked 2 November 2026. This article teaches a defined mechanics task; it makes no claim about individual tissue loading or clinical safety from a force peak.