Work Calculator – Work, Force and Distance (W = F × d)

See the force at an angle

The arrow is your force, the box is what moves. Tilt the angle and watch what happens: the arrow swings up, the share of the force that actually works — cos (θ) — shrinks, and the work falls with it. The diagram and the steps redraw live with your numbers.

Worked examples: everyday pushes, pulls and lifts

Four familiar jobs, worked with the same formula. Type any of them into the calculator to check.

Push, pull and lift calculators nearby

The tools this one is paired with most — conversions and heavier math.

How it works

One formula, three directions — solve for whichever piece is missing.

The math behind the default example

Two things worth understanding about work

Why a perpendicular push does nothing at all — and where the energy actually goes.

Work in more units: kJ, calories, BTU, kilograms and long distances

The main calculator answers in joules, kilojoules and foot-pounds, because those are the units its formula is built on and the units most textbooks use. Two more energy units turn up constantly outside textbooks, and both are worth knowing by their exact conversion rather than by feel: the calorie, used for food and heat, and the British thermal unit (BTU), used for heating and cooling equipment. The Advanced panel above converts any energy into all of them at once, and the table below gives the conversions for round energies so a food label, a heater rating or a textbook figure can be read in joules without a second tool.

The factors are: 1 kJ = 1,000 J exactly; 1 cal (thermochemical) = 4.184 J exactly, by definition; 1 kcal = 4,184 J, which is why the “calories” on a food label are really kilocalories — a 500 kcal meal is 500,000 small calories, and the label shortens the unit, not the energy; and 1 BTU (International Table) = 1,055.06 J, defined as the energy needed to raise one pound of water by one degree Fahrenheit. The calorie trap is the one that actually bites: a snack labelled 200 calories holds 200 kcal, which is 836,800 J, enough energy, in principle, to lift a 70 kg person about 1,219 metres straight up. The body is nowhere near that efficient at turning food into lifting, which is why the number feels absurd — but the arithmetic is exact, and it shows why joules feel small and food calories feel large: they are measuring the same thing on scales a thousand apart.

Mass and weight need the same care, because everyday language treats them as one thing and physics does not. Mass, in kilograms, is how much matter an object has; weight, in newtons, is the force gravity exerts on that mass, W = m × g, with g = 9.80665 m/s² on Earth. A 10 kg mass therefore weighs 98.07 N on Earth (22.05 lbf), and a 70 kg person weighs 686.47 N (154.32 lbf). The Advanced panel does this conversion, and it matters for work because lifting is work against weight: lifting that 70 kg person 3 m takes 686.47 × 3 = 2,059.4 J, which is the table row above. On the Moon, where g is 1.62 m/s², the same 10 kg mass weighs only 16.2 N, and lifting it 3 m would take just 48.6 J — same mass, same distance, far less work, because the force changed. Whenever a problem gives a mass and asks for work, the first step is always this conversion; entering the mass itself as the force is the most common homework error on this page.

Long distances get the same treatment. The main calculator works in metres and feet, which suit rooms, workshops and sports fields, but a road, a railway or a hiking trail is measured in kilometres or miles, and converting first avoids a string of zeros. The Advanced panel converts any kilometre or mile distance into metres and feet and the other long unit: 5 km is 5,000 m, 16,404.2 ft and 3.11 miles, while 5 miles is 8,046.72 m, 26,400 ft and 8.05 km. The factors are exact: 1 mile = 1,609.344 m by definition, and 1 ft = 0.3048 m, so a mile is exactly 5,280 ft. If a journey is given in miles and a force in pounds-force, convert both before multiplying — the main calculator’s own unit selects do the short conversions, and the Advanced panel does the long ones, so the formula itself never has to change.

Friction, slopes and net work: when forces fight each other

The main calculator handles one force at a time, which is the right place to start, but real pushes are rarely alone. A crate pushed up a ramp faces two opponents at once: gravity, pulling it back down the slope, and friction, gripping it against the surface. The Advanced slope panel above works out all three forces and the work done against each one, and this section explains the method with the panel’s own default numbers so every figure can be checked by hand.

Take a 20 kg crate pushed 5 m up a 30° slope, with a friction coefficient μ = 0.3. The weight is 20 × 9.80665 = 196.13 N, straight down. The slope only feels part of that weight pressing into it: the normal force, perpendicular to the surface, is weight × cos (θ) = 196.13 × cos (30°) = 169.86 N. Friction is μ times the normal force, 0.3 × 169.86 = 50.96 N, pointing back down the slope. The work against gravity is weight × sin (θ) × distance = 196.13 × 0.5 × 5 = 490.33 J — the same as lifting the crate through its vertical rise, which is 5 × sin (30°) = 2.5 m, so 196.13 × 2.5 = 490.33 J by either route. The work against friction is 50.96 × 5 = 254.78 J. The push itself must supply both, so the applied work is 490.33 + 254.78 = 745.12 J. The table below shows how the split changes with the slope angle, holding the crate, the friction and the distance fixed.

Two patterns in that table deserve attention. First, the friction work falls as the slope steepens, because the normal force shrinks — a steeper slope presses the crate less firmly into the surface, so friction has less grip, even though the total work rises. Second, the gravity work rises with sin (θ), so it is zero on the flat and dominates on steep slopes. At 45° the push is doing more than three times the work it did on the flat for the same 5 m of travel, and almost all of the increase is gravity, not friction. This is why ramps are built shallow: a shallow ramp trades a longer distance for a smaller force, and the work against gravity for a given height is the same whatever the ramp angle — weight times vertical rise, always — while the friction work grows with the longer path. The cheapest ramp is therefore a balance: shallow enough that the force stays manageable, steep enough that the extra friction distance does not eat the saving.

Net work ties the whole picture together, and it is simpler than it looks. Net work is the sum of the work done by every force, with signs: the push does positive work, gravity and friction do negative work, because they oppose the movement. Pushed at a steady speed, the crate gains no kinetic energy, so by the work-energy theorem the net work must be zero — and indeed it is: +745.12 J from the push, −490.33 J from gravity, −254.78 J from friction, totalling 0.01 J of rounding, which is zero. If the push were stronger and the crate sped up, the net work would be positive and exactly equal to the gain in kinetic energy, ½mv²; if the push weakened and the crate slowed, the net work would be negative. That is the general rule the main calculator’s single-force answers sit inside: every force contributes its own signed work, and only the total decides whether the object speeds up, slows down or carries on unchanged.

What a kilowatt-hour really runs: appliance equivalents

Joules are the right unit for pushes and lifts, but household energy is sold and metered in kilowatt-hours, and the two scales are so far apart that neither feels intuitive in the other’s world. One kilowatt-hour is 3,600,000 J — a watt is one joule per second, so a kilowatt for one hour is 1,000 × 3,600 joules. The main calculator’s default answer, 500 J, is 0.000139 kWh: a rounding error on a power bill, and a reminder that human-scale pushes are tiny beside household energy. The table below turns the relationship around: for common appliances, how long each one takes to use a single kilowatt-hour, using typical nameplate wattages from the U.S. Department of Energy’s Energy Saver factsheet “Estimating Appliance and Home Electronic Energy Use” (published via Virginia Tech Extension). The wattage in the middle column is the representative value used for the time; the range the factsheet gives is in the notes.

The refrigerator row needs a warning, because it is the one most often misread. The 725 W figure is the nameplate rating — what the compressor draws while it is actually running — but a refrigerator cycles on and off, and the factsheet notes that the running time is roughly a third of the plugged-in time. So the 1 hour 23 minutes above is running time; in wall-clock terms the same kilowatt-hour takes about 4 hours 8 minutes of being plugged in. The same distinction applies, less dramatically, to anything with a thermostat: a heater that cycles, an oven holding temperature, or a coffee maker on a warming plate all draw their rated watts only part of the time. Nameplate watts answer “how fast does it use energy while on”; only the duty cycle answers “how much does it use in a day”. For the other rows the distinction matters less, because microwaves, toasters, hair dryers and vacuums are used in short bursts at full power, so their running time and their switched-on time are nearly the same.

Read the other way, the table gives a feel for power itself, which is work divided by time. A microwave at 1,000 W does the main calculator’s default 500 J of work in half a second — 500 ÷ 1,000 = 0.5 s — while a laptop at 50 W would take 10 seconds to deliver the same 500 J. Neither machine is “doing work” in the pushing sense, but the energy bookkeeping is identical: watts times seconds equals joules, always. That is also why the power card on the main calculator matters more than it first appears: the same 500 J done in 20 seconds is 25 W, barely a laptop charger, while the same 500 J done in 0.5 seconds is 1,000 W, a microwave. The work is unchanged; only the rate changed, and the rate is what sizes motors, heaters and power supplies.

Work in lifts, sports and machines

Once the formula is familiar, work starts showing up everywhere, and most of the examples need nothing beyond the main calculator plus the mass-to-weight conversion. Lifting is the cleanest case: the force is the weight, the distance is the height, and the angle is 0°, because the lift points the way the load moves. A 10 kg crate lifted 1.5 m, the example already on this page, takes 98.1 × 1.5 = 147.15 J. Scale it up and the numbers stay honest: a 70 kg person climbing 3 m of stairs does 686.47 × 3 = 2,059.4 J of work against gravity, which is the energy row in the units table above — about half a food calorie, which is why climbing stairs feels like effort but barely dents a day’s food intake. A crane lifting a 500 kg load 10 m does 500 × 9.80665 × 10 = 49,033.25 J, about 49.03 kJ; a 2,000 W crane motor doing that lift at full power would take 49,033.25 ÷ 2,000 = 24.52 seconds, which is the power card’s logic applied to a machine instead of a person.

Sports give the same formula with the angle doing the interesting work. A rower pulling with 100 N over a 20 m stretch of water at 0° does 100 × 20 = 2,000 J per stroke cycle in the ideal case; a shot-putter’s throw is messier, because the force varies through the throw and the honest calculation adds the work up in small pieces, but the textbook estimate still starts from average force times distance. The zero-work cases are the ones coaches argue about: holding a heavy weight still overhead does no work in the physics sense, because the distance is zero, however much the muscles burn — the burning is internal chemistry, not mechanical work on the weight. Carrying a bag horizontally is the same story from the other side: the upward force is at 90° to the movement, so cos (90°) = 0 and that force does no work, even though the carrier is plainly tired. Both cases are already on this page’s guide, and both are worth repeating, because they are the two places where everyday language and physics language most often part company.

Machines exist to reshape the force-distance trade without changing the work. A ramp, a lever and a pulley all let a smaller force do a job over a longer distance, or a larger force over a shorter one, and in the frictionless ideal the work is identical either way — which is why the slope table above shows the gravity work depending only on the height gained, never on the ramp’s angle. Real machines add friction, and friction’s work is always negative, always heat, and always the reason a machine delivers less useful work than it takes in. That is the practical meaning of efficiency: useful work out divided by work in, with the gap lost to friction. So when a job looks too big for the force available, the answer is never to wish the work smaller — the work is fixed by the load and the distance — but to spread it over a longer distance, a longer time, or both, and to keep the friction bill as small as the design allows.

Work and power questions, answered

More physics and unit tools

Converters and math tools that sit next to this one — plus the other “work” calculator, for hours.