Example 1
A crate moved across a hall by a constant 200 N over 10 m. The product is 2,000 J. This is the reference example on the page.
W = 2,000 J
Work Calculator
Enter the force in newtons and the distance in meters. The calculator finds the work of a constant force along the displacement from W = F·s and also reports joules, watt-hours, and the work over half the distance and over twice the distance. For 200 N and 10 m the result is 2,000 J. The model does not include an angle cos α or power P = W/t.
Result
Enter a force and a distance. Work, watt-hours, and the W(s) chart will appear here.
Force from mass is on the F = m·a calculator, average power is on the P = W/t calculator, and energy from speed is on the Ek = ½mv² calculator.
How it works
F200 N
s10 m
W2,000 J
Mechanical work of a constant force along the displacement is the product W = F·s. The unit is the joule: 1 J = 1 N·m. For 200 N and 10 m: 200 · 10 = 2,000 J. The same 200 N over 5 m gives 1,000 J, and 50 N over 4 m gives 200 J. At zero distance, for example when a load is held in place, the work is 0 J, because s = 0.
The calculator uses force F in newtons and distance s in meters. A comma and a period are the same separator. A negative distance is rejected. The force may be negative: the component then points opposite the positive direction of the distance, and the work comes out negative. Zero force or zero distance gives 0 J.
What the result panel shows. Besides the work in a chosen unit (J, kJ, MJ), the calculator reports the same value in joules and watt-hours, the entered force, the entered distance, and the work over half the distance and over twice the distance. For 200 N and 10 m that is 2,000 J, or 2 kJ, or 0.5556 Wh. Half the distance gives 1,000 J, and twice the distance gives 4,000 J. The chart draws the straight line W(s) from a quarter of the entered distance to twice that distance, when the force is not zero and the distance is greater than zero. At zero distance and a nonzero force, the chart uses a reference span from 1 m to 10 m.
Limits of the model. The formula is a constant force parallel to the displacement. The calculator does not evaluate cos θ, so the angle has to be handled first and the component F cos θ entered as F. There is no time: average power P = W/t is on the P = W/t page. Energy from speed is on the Ek = ½mv² page. Spring energy ½kx² is on the Hooke’s law page. When lifting at constant speed, the work is often close to m g h if the entered force is close to the weight, but this form does not take mass or height. That energy is on the potential-energy page.
The unit switch in the header changes force to lbf, distance to feet, and work to ft·lb, and it converts numbers already entered in the fields. Values in the text stay in SI units as the reference.
The examples on the page use the same product. 200 N over 10 m gives 2,000 J. The Fill this example button enters those numbers into the fields and refreshes the panel and the chart.
Formula
W = F · s
[W] = N · m = J
Model: work of a constant force along the displacement (Coriolis’s 1829 definition of work, so that the change in ½mv² equals the work). The textbook form with an angle, W = F s cos θ, is the scalar product. This calculator multiplies the two entered numbers and does not evaluate the cosine. Order of calculation: the product F·s in SI units, then the same value in kJ and in Wh, plus the work at s/2 and at 2s.
Theory, derivation, and domain
Definition and the joule in SI
Mechanical work of a constant force parallel to the displacement is W = F·s. It is the energy transferred by that force along that distance, not a measure of fatigue. The SI unit is the joule: 1 J = 1 N·m = 1 kg·m²/s². The name commemorates James Prescott Joule. The unit is defined in the SI Brochure (BIPM, 9th edition). In the United States, NIST SP 330 describes the same SI units.
For 200 N and 10 m the order is short. Step 1: force F = 200 N. Step 2: distance s = 10 m. Step 3: the product W = 200 · 10 = 2,000 J. Those 2,000 J are 2 kJ and 0.5556 Wh, because 1 Wh = 3,600 J. The same force over 5 m gives 1,000 J, and over 20 m it gives 4,000 J, because distance enters the formula linearly. 50 N over 4 m gives 200 J.
Angle and the scalar product
When the force is not parallel to the displacement, textbooks write W = F s cos θ, or W = F⃗ · s⃗. The cosine extracts the component of the force along the distance. This calculator has no angle field and does not evaluate cos θ. The parallel component is entered as F. At 200 N and 60°, that component is 200 · ½ = 100 N, and the work over 10 m is 1,000 J. Entering 200 N instead of 100 N shows 2,000 J, the product of the numbers in the fields.
Coriolis and the work–energy theorem
Gaspard-Gustave Coriolis, in Du calcul de l’effet des machines (1829), defined work so that it equals the change in ½mv². For a particle of constant mass, the work–energy theorem says the net work equals the change in kinetic energy: W = ΔEk. This page computes W from force and distance. Energy from mass and speed is on the Ek = ½mv² page. Both report joules, from different fields. Doubling the distance doubles the work. Doubling the speed on the kinetic-energy page quadruples Ek, because speed is squared there. That is a different formula.
Power, a spring, and a lift
Average power is P = W/t. 2,000 J in 10 s is 200 W, but time is not a field on this form. On a spring the force grows with the extension, so the product of a constant F and x is not the elastic energy; the matching formula is ½kx². When lifting at constant speed, the force is often close to the weight mg and the work is often close to m g h. This form does not take m, g, or h. It also does not find friction: if the force field already holds a friction value, the product is the work of that force, not a separate drag model.
Outside the formula
The formula does not cover a force that changes along the path. That case needs the integral ∫ F ds, not one product. It also does not cover a loop whose net displacement is zero while the path length is positive: the field s takes the displacement along the chosen force, not the length of an arbitrary curve. The calculator does not apply a relativistic correction and has no power mode.
How to use
- Enter the force in newtons, for example 200. After the header is switched to US, the field takes lbf. Convert mass to newtons on the F = m·a page first.
- Enter the distance in meters, for example 10. For an angle other than zero, enter the component F cos θ as F, not the full force.
- Click Calculate or wait for the live result. For 200 N and 10 m the calculator shows 2 kJ, which is 2,000 J.
- Check the panel: joules, watt-hours, force, distance, and the work over half the distance and over twice the distance. The chart is the straight line W(s).
- Clearing the form removes the entered numbers and hides the panel. The examples enter ready pairs of force and distance, including 200 N and 10 m, 200 N and 5 m, and 50 N and 4 m.
Symbol glossary
W is the work of a constant force along the displacement, F is the force, and s is the distance. For 200 N and 10 m the result is 2,000 J.
- W
- Work [J], W = F·s. 1 J = 1 N·m. The panel also shows kJ and Wh.
- F
- Force [N] along the positive direction of s. A negative value flips the sign of the work. It is not a mass.
- s
- Distance [m], the displacement along F. Zero gives W = 0. A negative value is rejected.
- J
- Joule, 1 J = 1 N·m. 2,000 J = 2 kJ = 0.5556 Wh, because 1 Wh = 3,600 J.
- θ
- Angle between the force and the distance. W = F s cos θ is the textbook form. This calculator does not evaluate the cosine.
- P
- Power [W], P = W/t. This form has no time. 2,000 J in 10 s is 200 W on the power page.
Worked examples
Example 2
The same 200 N force over half the distance, 5 m. The work falls linearly to 1,000 J, not with the square of the distance.
W = 1,000 J
Example 3
A smaller force: 50 N over 4 m. The product is 200 J.
W = 200 J
Example 4
A full cart through a supermarket: about 80 N over 40 m of aisle. The work is 3,200 J.
W = 3,200 J
Example 5
A short push on a car: 600 N over 3 m. The work is 1,800 J.
W = 1,800 J
Example 6
A wheelbarrow of soil: 350 N over 25 m of garden. The work is 8,750 J.
W = 8,750 J
Example 7
A vacuum pulled across the floor: 40 N over 15 m. The work is 600 J.
W = 600 J
Example 8
A dresser lifted onto a van: 500 N through 0.8 m. The work is 400 J. A short distance gives a small product.
W = 400 J
Example 9
A winch on a trailer ramp: 1,200 N over 2 m. The work is 2,400 J.
W = 2,400 J
Example 10
Drag while riding into a headwind: 30 N held over 5,000 m. The work is 150,000 J, or 150 kJ.
W = 150,000 J
Example 11
A sofa moved across the floor: 450 N over 6 m. The work is 2,700 J.
W = 2,700 J
Example 12
A sled pulled along a path: 70 N over 80 m. The work is 5,600 J.
W = 5,600 J
Common questions
What does this calculator compute, and from which formula?
The calculator finds the work of a constant force along the displacement from W = F·s. It multiplies the entered force by the entered distance. For 200 N and 10 m the result is 2,000 J.
Which units should I enter for force and distance?
Enter force in newtons and distance in meters. The result is in joules. After the header is switched to US units, force is in lbf, distance is in feet, and work is in ft·lb. A comma and a period are the same decimal separator. Convert mass to newtons on the F = m·a page first: 80 kg is not 80 N.
Why does time not change the work?
Time is not in the formula W = F·s, so the same 2,000 J comes out slowly or quickly. The rate is power, P = W/t. 2,000 J in 10 s is 200 W, and the power page computes that value, not this form.
What if the force is at an angle to the distance?
The calculator does not evaluate cos θ. Enter the component parallel to the distance in the force field. At 200 N and 60°, that component is 100 N, and the work over 10 m is 1,000 J. Entering the full 200 N would give 2,000 J, which is the product of the numbers in the fields, not the parallel component.
How is this work different from kinetic energy?
Work on this page comes from force and distance. Kinetic energy Ek = ½mv² comes from mass and speed. The unit is the same and the fields are different. The work–energy theorem links the two: the net work equals the change in Ek. This calculator does not find a speed.
What do zero and a minus sign mean?
A negative distance is rejected. A negative force means the component points opposite the positive direction of the distance, so the work is negative. Zero force or zero distance gives 0 J. Both fields need a number before the panel appears. An empty field is not a zero.
Does W = F·s apply to a spring?
No. On a spring the force is not constant, and the elastic energy is ½kx². This product stays with a constant force along the distance. The Hooke’s law page computes the spring energy.
What do the panel and the chart show?
The panel reports the work, the same value in joules and watt-hours, the entered force, the entered distance, and the work over half the distance and over twice the distance. For 200 N and 10 m that is 2,000 J, or 2 kJ, or 0.5556 Wh. Half the distance gives 1,000 J, and twice the distance gives 4,000 J. The chart draws the straight line W(s) when the force is not zero. The Fill this example button enters the example numbers and refreshes the panel and the chart. The fields start empty.
Does the result include friction or a changing force?
No. The result is the product of one force and one distance. Friction can be that force if its value is entered, but the calculator does not find friction on its own. A force that changes along the path needs an integral, not this product.
Where is potential energy for a lift?
On the potential-energy page. When a body is lifted at constant speed, the force is often close to the weight and the work is often close to m g h. This form does not take mass or height, so it does not check that equality.
How do I calculate work with W = F·s?
Multiply force in newtons by distance in meters. For 200 N and 10 m: 200·10 = 2,000 J.
What is W for 200 N over 10 m?
2,000 J, or 2 kJ. With the same force over 5 m the result is 1,000 J.
Must I multiply by cos α when the force is at an angle?
Not on this page. The form has only F and s. The full relation W = F·s·cos α needs a separate angle; here the force is taken along the path.
How is W = F·s different from power P = W/t?
Work is energy transferred with displacement. Power is the rate of that work. This calculator does not divide the result by time.
Does the calculator use the work–energy theorem ΔEk?
No. The result is W = F·s. The kinetic-energy change ½m(v² − v₀²) is not a mode on this page.
Sources
The definition, the unit, and the work–energy theorem come from physics and metrology sources. The page does not invent publication identifiers.
- Wikipedia — Work (physics)
- Feynman Lectures on Physics, Vol. I, Ch. 13 — Work and Potential Energy
- BIPM — The International System of Units (SI), 9th edition
- NIST SP 330 — The International System of Units
- Coriolis, G.-G., Du calcul de l’effet des machines, 1829.
- Young, H. D.; Freedman, R. A., University Physics, chapter on work.
Page updated November 3, 2026.