How this instrument works
Work is an accounting entry for energy moved by force. Gaspard-Gustave Coriolis named this quantity in 1826, while analysing how efficiently machines turn effort into motion, and defined it exactly as this sheet does — force applied, times distance travelled, times the cosine of the angle between them. That cosine carries all the subtlety. Force pointed sideways to travel contributes nothing; only its component lying along your path shifts any energy.
Set the angle to 90 degrees and your answer collapses to zero, however hard you push. No algebraic quirk causes that. It explains why an orbiting moon burns no fuel to keep circling, why floor reaction never adds energy to a sliding crate, and why carrying luggage down a level corridor does nothing to that luggage, whatever your arms report. Past 90 degrees the cosine turns negative and so does this result: friction, drag, and brake pads drain energy rather than supply it.
One joule is one newton acting through one metre — small enough that raising a medium apple from waist to shoulder spends roughly one. James Prescott Joule earned that naming right with ten years of paddle-wheel experiments during the 1840s, churning water to measure how much stirring yields how much warmth, which is why heat and mechanical effort now share one unit. This formula assumes force holds steady in size and keeps a fixed angle to straight-line travel; loosen any of that and honesty demands an integral along your route, as with stretching springs or rope wrapped around a pulley.
- Enter Force applied — newtons, kilonewtons, or pounds-force from its unit menu.
- Enter Distance moved along your path: millimetres through kilometres, or feet.
- Set Angle between force and motion in degrees or radians; leave it at 0 for straight-ahead pushing.
- Read Work done in joules, or flip its output unit to kilojoules, calories, or watt-hours.
- Try 90 degrees to watch everything vanish, then 180 to see negatives appear.
Worked example — 10 N pushing over 5 metres
A trolley on a level floor, shoved dead ahead with a steady Force applied of 10 N across Distance moved of 5 m. Angle between force and motion sits at 0, so cos θ = 1 and no projection is lost: Work done = 10 × 5 × 1 = 50 J exactly. Fifty joules is modest — near enough what hoisting a five-kilogram shopping bag from floor to countertop demands.
Now tilt that same shove. Swing it 60 degrees off your direction of travel and cos 60° = 0.5, halving your tally to 25 J. Swing it fully 90 degrees and every joule disappears, because a force square across your path drives nothing along it. Your arm registers an identical 10 N in all three cases, which is precisely why effort makes such a poor guide to what physics counts.
Questions
Why is a newton-metre of work called a joule, but a newton-metre of torque never is?
Both amount to one newton times one metre dimensionally; geometry separates them. Work takes whichever force component lies along displacement — a dot product yielding one scalar you can add to any other energy in your ledger. Torque takes whichever component lies across its lever arm — a cross product yielding one vector along its rotation axis. Since neither can ever be added to its counterpart, SI keeps them visibly apart by convention: energy in joules, torque in newton-metres. NIST SP 811 states this rule outright, and mixing them causes genuine confusion in gearbox and fastener specifications.
Which angle am I supposed to enter?
Whichever one sits between your force and where your object genuinely moves — not the ramp's slope, and not a rope's tilt from horizontal, unless travel happens to run horizontally. Push a mower whose handle angles 40 degrees below horizontal across a flat lawn, and 40 degrees is correct. Drag a sledge up that same incline and travel now runs along that slope, so what you want is whatever separates rope from slope, often much smaller. Getting this wrong is easily the commonest slip on homework problems.
Can this result come out negative, and what does that mean?
Yes, for any angle beyond 90 degrees, and it signals energy leaving your object rather than entering. Friction, air resistance, and brake pads all act opposite to travel, giving cos θ = −1 and a negative tally. Lowering a crate at steady speed tells that same story: your upward hold and its downward journey sit 180 degrees apart, so you subtract energy from that crate while gravity adds it. Sum every contribution with its sign and you get whatever net figure the work–energy theorem needs.
Does holding something heavy still count if nothing moves?
No. Mechanical work requires displacement, and d = 0 zeroes this product regardless of how large your force grows. Standing under a loaded barbell exhausts you because muscle fibres cycle through microscopic contractions and burn chemical fuel as heat, yet not one joule reaches that bar. Physiological effort and physical work are separate ledgers, and only this second one is what your instrument computes.
How does this connect to kinetic energy?
Through the work–energy theorem: net work on any body equals change in its kinetic energy, ΣW = ½mv² − ½mv₀². Those 50 J from our example above, applied to a 1 kg cart starting at rest on a frictionless floor, leave it travelling at 10 m/s. Such equivalence is why both quantities live in joules, and why stopping distance is really one big work problem — pads and tyres must remove every joule your vehicle carries before it halts.
How do I turn this into power?
Divide by elapsed time: one joule per second is one watt. Spread our example's 50 J over 2 seconds and you have 25 W; compress it into half a second and it becomes 100 W. Work asks how much energy shifted, power asks how briskly. Given long enough, one feeble motor matches something strong in total joules — rate, not total, decides whether your machine suits its job.