Work Calculator

work = force × distance × cos θ, where θ is the angle between the force and the displacement

F

The force doing the work, in newtons, with kilonewtons, pounds-force and kilograms-force in the same field. Only the component along the motion counts, so the same force accomplishes less the further it is aimed away from the direction of travel: at 60 degrees half of this value is doing work and half of it is not

d

How far the object moves while the force acts, in metres, feet, centimetres, inches or kilometres. It is the distance travelled by the point where the force is applied, which is the same thing as the object's displacement for a simple push and not the same thing at all for a lever. Zero is rejected: no movement, no work, however hard you push

θ

The angle between the force and the direction of motion, in degrees, radians or gradians. It runs from 0 to 180, and its cosine carries the whole story: 0 gives the full effect, 60 halves it, 90 removes it entirely, and anything past 90 makes the work negative. A force pulling backwards at 180 does exactly the negative of the same force pulling forwards

W

The work done by that force, in joules as the main row with kilojoules and foot-pounds underneath. It is signed, and the sign is information rather than an error: a negative value means the force was taking energy out of the motion, which is exactly what friction and braking do. The joule is one newton-metre, so 100 N over 5 m is 500 joules

F cos θ

The part of the force lying along the displacement, in newtons — the force field multiplied by the cosine of the angle. It is the honest version of that field, and reading it next to the work is the quickest way to see where the answer came from: 100 N at 60 degrees does the work of 50 N, and 100 N at 90 degrees does the work of nothing at all

Use this page when you need to know what a pull, a push or a drag actually achieves: how much of a rope's tension is doing useful work when you haul a sled at an angle, how much energy a motor has to supply over a given travel, or how much work friction removes when a crate slides to a stop. The cosine is what makes the page worth opening, because the answer is usually smaller than people expect and sometimes zero — a force at right angles to the motion does no work at all, and a force pointing backwards does negative work. That cosine is also the reason the work energy theorem takes the form it does: the work done on an object equals its change in kinetic energy, which is the one place in this group of pages where that statement belongs.

1. The defaults: 100 N over 5 m, straight along the motion

   1. Force 100 N, distance 5 m, angle 0 degrees
   2. The whole force lies along the motion: F cos 0 = 100 × 1 = 100 N, which is the second row of the panel
   3. Work: W = 100 × 5 = 500 J
   4. In kilojoules: 500 ÷ 1000 = 0.5 kJ
   5. In foot-pounds: 500 ÷ 1.3558179483 = 368.7811, which prints as 368.78 ft·lb

   This is the case the formula collapses to when nothing is angled, and it is worth seeing once before the angles start: work is measured in joules, a joule is one newton-metre, and 500 J is a firm push sustained across a small room. Compare it with the calories on a food label later and the scale becomes clear — 500 J is about an eighth of a food calorie, which is why lifting and carrying feel like work to your muscles but register as almost nothing in the energy budget of the object. Your body is spending far more than 500 J to do it, because muscles are not perfectly efficient and holding a load costs energy even when nothing moves.

2. The same push at 60 degrees: half the force does the work

   1. Force 100 N, distance 5 m, angle 60 degrees
   2. Along the motion: F cos 60 = 100 × 0.5 = 50 N — half of the force, and the panel says so
   3. Work: W = 50 × 5 = 250 J
   4. In kilojoules: 250 ÷ 1000 = 0.25 kJ
   5. In foot-pounds: 250 ÷ 1.3558179483 = 184.3905, which prints as 184.39 ft·lb

   The companion row is the point here. Change only the angle and both the effective force and the work are halved, because the cosine appears in both: 100 N becomes 50 N of useful pull, and the work falls with it. A reader who expects the work to depend only on the magnitude of the force will be wrong by a factor of two on this case and by an infinite factor on the next one. That is why the page prints the along-the-motion force at all — it turns a cosine into something you can see on the same screen as the work it produces.

3. Holding a crate level and walking: the work is exactly zero

   1. Force 100 N, distance 5 m, angle 90 degrees — the force is straight up, the motion is straight sideways
   2. Along the motion: F cos 90 = 100 × 0 = 0 N
   3. Work: W = 0 × 5 = 0 J
   4. Kilojoules and foot-pounds are zero too, so all three energy rows read 0
   5. The force is real and the distance is real, and the work is still nothing

   This is the row that makes the definition click, and it is the one people argue with. You carry a heavy crate across a room, your arms ache, and the mechanical work you did on the crate is exactly zero — because the force your arms supply is upward and the crate moves sideways, so the force and the displacement are perpendicular and their product vanishes. The energy your body burned went into heat, into holding the muscles tensed, and into the biochemistry of that tension. It did not go into the crate. This page measures only the second thing.

4. Friction: 180 degrees, and the work comes out negative

   1. Force 100 N, distance 5 m, angle 180 degrees — the force points exactly backwards
   2. Along the motion: F cos 180 = 100 × (−1) = −100 N
   3. Work: W = −100 × 5 = −500 J
   4. In kilojoules: −0.5 kJ; in foot-pounds: −368.78 ft·lb
   5. Every row is the 0-degree case with its sign flipped

   Negative work is not a mistake and not a sign convention you can ignore; it is what friction, drag and braking do all day. A sled sliding to a stop has the friction force pointing backwards along the motion, so the friction does −500 J of work, and the sled loses exactly 500 J of kinetic energy to it. Read the two together as an audit: the push puts energy in, the friction takes energy out, and the difference is what the object keeps. A version of this page that returned the absolute value of the work would be unable to state that, and it would look completely normal.

Only a constant force is covered: if the force changes direction or strength as the object moves, the work becomes an integral and these rows do not apply. The distance is how far the point where the force is applied moves, which for a lever or a rolling wheel is not the distance the object travels. Only one force is tracked, and real objects usually have several acting at once. Holding a crate still burns energy in your muscles, but the mechanical work is zero.

What is the work formula?

W = F × d × cos θ: the force times the distance it acts over times the cosine of the angle between the force and the motion. A 100 N pull over 5 m straight along the direction of travel is 500 J, while the same pull at 60 degrees is 250 J and at 90 degrees is zero. The cosine is not decoration — it selects the part of the force that is actually doing the work, which is why the page also prints that part as its own row.

Why does the angle change the work?

Because only the component of the force along the displacement moves the object in the direction it is going. Split the force into one part along the motion and one part across it: the along part does all the work, and the across part does none, however large it is. At 60 degrees the useful part is half the force; at 90 degrees it is nothing. Pulling a sled by a rope held at 60 degrees really does waste half your effort as far as the sled's motion is concerned.

Can work be negative?

Yes, and the sign is meaningful. When the force points opposite to the motion the cosine is negative, so the work is negative: the force is removing energy from the object rather than adding it. Friction and braking do negative work, which is how a sliding crate loses speed. This page keeps the sign rather than taking an absolute value, so 100 N of friction over 5 m reads −500 J, and that is the figure to add to the push's +500 J to see the net change.

Is carrying a box across a room work?

Not in the sense this page defines. Your arms push upwards on the box while the box moves horizontally, so the force and the displacement are at 90 degrees and the mechanical work is zero. Your muscles still consume energy, and quite a lot of it, but that energy goes into heat and into maintaining the tension rather than into the box. The distinction matters because it is exactly the one that makes the physics definition useful: it measures energy transferred to the object.

How does work relate to kinetic energy?

The work done on an object equals its change in kinetic energy — the work energy theorem. Push a sled with 500 J of net work and it gains 500 J of kinetic energy; let friction do −500 J and it loses that much. That is why the kinetic energy calculator takes a mass and a speed and gives energy, while this page takes a force, a distance and an angle and gives the same quantity from the other direction. The two meet in the middle, and the meeting point is that theorem.

• Work: The Scientific Definition (College Physics 2e, §7.1) — the definition of work as force times displacement times the cosine of the angle between them, and the cases where it comes out zero or negative — OpenStax
• Work (physics) — the sign convention, the work energy theorem, and why a force perpendicular to the displacement does no work — Wikipedia
• NIST Guide for the Use of the International System of Units (SI), Appendix B.8 — Factors for Units Listed Alphabetically: the foot pound-force row (1 ft·lbf = 1.355 818 J) that this page's third row is converted with — NIST