A load can return to its starting position after a demanding task. Its total change in height is zero. Does that mean no work occurred, no energy was used, or the task was easy? Those are three different claims.
For Canadian physiotherapy students studying movement science, the useful skill is to name the object, force and time interval before interpreting a number. This lesson uses an original classroom model. It does not calculate a person's muscle forces, prescribe an exercise or establish a safe loading limit.
Give force, work and power different jobs
Force describes an interaction and is measured in newtons. Mechanical work describes energy transferred by a force acting through displacement. For a constant force, work = force × displacement × cos θ, where θ is the angle between them. A joule is one newton-metre in this work calculation. When the force changes along the path, its contributions must be added over that path. OpenStax: Work.
Average mechanical power divides the work by the time taken and is measured in watts: 1 W = 1 J/s. It describes a rate, not a second name for force. OpenStax: Power.
Before calculating, complete the phrase “work done by this force, on this object, during this interval.” “The person did 24 joules” leaves all three choices unclear.
Set up a lift, hold and return
Original fictional example. A small rigid load has a weight of 60 N. A hand raises it vertically by 0.40 m in 2 seconds, holds it stationary for 5 seconds, then lowers it to its starting height in 3 seconds.
Use these supplied assumptions:
- The load is the object being analysed; the person's body is outside that boundary.
- Each moving phase starts and finishes at rest. The load translates without rotating or deforming.
- Gravity and the hand's contact force are the only forces acting on the load. Ignore air resistance and other contacts.
- The gravitational force is constant at 60 N downward. The hand's force is upward, but its magnitude can change while starting and stopping the movement.
The last distinction prevents a hidden contradiction. A hand force fixed at exactly 60 N upward would balance gravity at every instant and could not accelerate this load from rest. We can calculate the total work without pretending that the force stayed constant.
Find the hand's work without inventing its force history
Take upward displacement as positive. During the lift, gravity acts opposite to the 0.40 m displacement, so its work is:
Work by gravity = −60 × 0.40 = −24 J.
The work–energy theorem relates net work from all forces to the change in the load's kinetic energy. Both endpoints are at rest, so that change is zero. OpenStax: Work–Energy Theorem.
For this two-force model:
Work by hand + work by gravity = 0, giving work by hand = +24 J.
The net work is zero, while the hand does positive work. There is no contradiction: two different quantities are being reported. During acceleration and deceleration the load can temporarily gain and lose kinetic energy; equal endpoint values do not mean it remained stationary throughout.
Nor does the 24 J result reveal the peak hand force. Many force and speed histories can produce the same displacement, duration and endpoint conditions. A request for peak force would require more information.
Follow the signs through the whole task
During the stationary hold, the hand supplies 60 N upward under the stated assumptions, but its point of contact does not move. Its external mechanical work on the load is zero.
During lowering, gravity and displacement point downward. Gravity therefore does +24 J. Because the load again starts and ends at rest, the hand's work is −24 J. Negative work here describes the direction of mechanical energy transfer associated with that force. It is not a negative effort score or a judgement about the movement.
| Phase | Work by hand on load | Work by gravity on load | Net work on load | Average power of hand on load |
|---|---|---|---|---|
| Raise 0.40 m in 2 s | +24 J | −24 J | 0 J | +12 W |
| Hold still for 5 s | 0 J | 0 J | 0 J | 0 W |
| Lower 0.40 m in 3 s | −24 J | +24 J | 0 J | −8 W |
Across the complete ten-second cycle, the signed hand work sums to zero. The upward phase still contains 24 J of positive work, and the downward phase contains 24 J of negative work in magnitude. Reporting only the signed total hides those phases.
Adding the two magnitudes gives 48 J, but call it the sum of the magnitudes of the phase work. It is neither the net work nor a measurement of metabolic energy expenditure. Likewise, the signed average hand power over all ten seconds is zero, although the phase averages are not.
A motionless load does not make muscle activity free
Zero work on the external load does not establish zero biological energy use. In a primary study, Elder and colleagues measured increased pulmonary oxygen consumption during electrically stimulated isometric knee-extension exercise in seven healthy young adults. That particular experiment demonstrates an oxygen cost during a task without joint movement; it does not supply an energy-conversion factor for our fictional hand-held load. Elder et al., 2006, abstract.
For this classroom task, no oxygen-consumption measurements, muscle activation data or metabolic model have been supplied. Converting the 48 J sum into calories “burned” would therefore add an unsupported assumption. Doubling the hold duration also cannot be interpreted as doubling metabolic expenditure from the information given.
The external-load model does not resolve what happens inside each muscle or tendon. Controlled lowering is often discussed alongside eccentric muscle action, but the hand's negative external work alone does not identify the behaviour of every contributing muscle. Do not turn an object-level energy result into a complete account of the person.
Change the time, then change the question
Suppose the same 60 N load rises through the same 0.40 m in 1 second, still starting and finishing at rest with the same two-force boundary. The hand's work remains +24 J, but its average power becomes 24 J ÷ 1 s = +24 W.
The average power doubles because the time halves. That does not establish that peak force doubled, that the task used twice the metabolic energy, or that it was appropriate for a particular learner or patient.
Average power also does not describe every instant. A constant power assumption has not been supplied, so the +24 W average cannot be treated as the peak. OpenStax: average and instantaneous power.
Now change a different fact: the load finishes the lift moving upward, with 3 J more kinetic energy than at the start. Gravity still does −24 J, but net work must now be +3 J. The hand must therefore do 27 J. The old answer depended on equal endpoint kinetic energies, not merely on knowing the change in height.
Transfer exercise: an assistant changes the boundary
Original fictional exercise. A load weighing 100 N rises vertically by 0.30 m in 1.5 seconds, starting and finishing at rest. Two hands share the lift. Their combined upward contact force and gravity are the only external forces; the load does not rotate or deform.
A student writes: “The combined hand work is 30 J, the average combined power is 20 W, and each hand did 15 J.” Which parts are supported?
Reasoned feedback. Gravity does −100 × 0.30 = −30 J. Zero change in kinetic energy means the combined hand work is +30 J. Dividing by 1.5 seconds gives +20 W average combined power.
The equal split is unsupported. The prompt gives no sharing rule or individual force histories. One hand could contribute more work than the other. An additional measurement or assumption is needed to divide the combined result.
If the second hand is then removed in a different trial, do not interpret a difference in the first person's contribution as a change in their underlying capacity without accounting for the changed assistance. The calculation describes a task under specified conditions.
When reading an exercise report, keep four entries together: the object analysed, whose force is included, the movement interval and whether the result is signed work, a phase magnitude or average power. That small record helps prevent a valid mechanical calculation from becoming an unsupported physiological conclusion.
Sources and further reading
- OpenStax University Physics Volume 1: Work — constant and variable force, displacement and gravitational work.
- OpenStax University Physics Volume 1: Work–Energy Theorem — net work and change in kinetic energy.
- OpenStax University Physics Volume 1: Power — average versus instantaneous mechanical power.
- Elder et al., 2006: Oxygen cost of dynamic or isometric exercise relative to recruited muscle mass — primary human experiment; PubMed abstract read, not a conversion factor for the fictional task.