Work–Energy Theorem Calculator (Final Speed and Stopping Distance)
Work–Energy Theorem Calculator
Net work equals the change in kinetic energy, so give me a mass, a starting speed, the net force and the distance and you get the final speed — plus the stopping distance, the acceleration, and the cross-check against v² = u² + 2as that must agree with it.
Net work to final speed, and the stopping distance
A 1200 kg car at 20 m/s, net force −5000 N (braking), over 30 m
W_net = ΔKE, and the three things it does not need
- W_net
- the net work: every force’s work added with its own sign. Not the work of the force the question happens to name. For a constant net force along the path it is simply Fnet s, which is what this page takes.
- ΔKE
- the change in kinetic energy, ½mv² − ½mu². Note that it is a difference of squares and not the square of a difference: a car slowing from 30 to 20 m/s loses 2.5 times as much energy as one slowing from 20 to 10, although both lose 10 m/s.
- the square
- the reason stopping distances are quoted the way they are. Kinetic energy goes as v², so doubling the speed quadruples the energy and quadruples the distance a given braking force needs. This is the single most useful consequence of the theorem outside an exam hall.
- what it does not need
- time, the path, and the direction. The theorem is a statement about start and end states only, which is exactly why it is easier than the kinematic equations for a problem where the force is known and the duration is not. It is also why it works unchanged on a curved path, and why it survives a force that varies as long as you can find the area under the force–distance graph.
- the check
- for a constant force the theorem and v² = u² + 2as are the same algebra — substitute a = F/m and the one becomes the other. So the two routes must give identical answers, which is what the check row on this page is for. If they ever differed, one of the two would be wrong; they cannot both be right and different.
Worked example
A 1200 kg car at 20 m/s, net force −5000 N (braking), over 30 m
KINETIC ENERGY AT THE START. KE = ½mu² = 0.5 × 1200 × 20² = 240 000 J, which is 240 kJ. Worth a sanity check against the unit: a joule is a newton metre, so 240 kJ is the work a 5000 N brake would do over 48 m, and that is the stopping distance before we start.
THE NET WORK OVER THE 30 m. W = Fnet s = −5000 × 30 = −150 000 J. Negative, because the force opposes the motion. There is no cosine here: the net force has been resolved along the path already, which is what “net force along the path” means on the input above.
THE THEOREM, APPLIED. ΔKE = Wnet, so the final kinetic energy is 240 000 − 150 000 = 90 000 J. The car has lost 62.5 per cent of its energy and has not yet lost two thirds of its speed — the asymmetry is the square.
BACK OUT THE SPEED. v = √(2 KE / m) = √(2 × 90 000 / 1200) = √150 = 12.247 m/s, which is 44.09 km/h. So two thirds of the way through the stopping distance, a car that started at 72 km/h is still doing 44 km/h. That is the number worth remembering from this page.
THE CHECK, AND THE STOPPING DISTANCE. The acceleration is F/m = −5000/1200 = −4.1667 m/s², and v² = u² + 2as = 400 + 2(−4.1667)(30) = 150, so v = 12.247 m/s — identical, as it must be. The full stopping distance is ½mu²/f = 240 000/5000 = 48.0 m. Double the starting speed to 40 m/s and it becomes 192 m: four times, not twice.
Why stopping distance goes as the square of the speed
| Starting speed | In km/h | Kinetic energy of a 1200 kg car | Distance to stop at 5000 N | Speed still left after 30 m |
|---|---|---|---|---|
| 10 m/s | 36 | 60 kJ | 12.0 m | stopped at 12.0 m |
| 14 m/s | 50.4 | 117.6 kJ | 23.5 m | stopped at 23.5 m |
| 20 m/s | 72 | 240 kJ | 48.0 m | 12.247 m/s (44.1 km/h) |
| 25 m/s | 90 | 375 kJ | 75.0 m | 19.365 m/s (69.7 km/h) |
| 30 m/s | 108 | 540 kJ | 108.0 m | 25.495 m/s (91.8 km/h) |
| 40 m/s | 144 | 960 kJ | 192.0 m | 36.056 m/s (129.8 km/h) |
One statement, no time in it, and the square that catches everybody
The work–energy theorem says the net work done on a body equals the change in its kinetic energy, and that is the entire content. Wnet = ½mv² − ½mu². What makes it powerful is not what it contains but what it leaves out: there is no time in it, no path, and no direction. A body dragged 30 m in a straight line and a body taken 30 m along a curve under the same net force along its path arrive with the same speed. That is why the theorem is usually the shorter route than the kinematic equations — a question that gives you a force and a distance and asks for a speed has no time in it either, and the theorem is the tool with the same shape as the question.
The square is where the marks go. Kinetic energy depends on v², so the energy a body carries is not proportional to its speed. Doubling the speed quadruples the energy, and therefore quadruples the distance a given braking force needs. A car at 40 m/s does not need twice the stopping distance of a car at 20 m/s; it needs four times. The same square is behind the fact that a 10 m/s change of speed is not a fixed amount of energy: slowing from 30 to 20 m/s sheds 2.5 times as much energy as slowing from 20 to 10, because the difference of the squares is 500 against 300 per kilogram, not 100 against 100.
Net, net, net. The theorem takes the resultant of every force along the path, and almost every wrong answer on this topic comes from feeding it the work of one named force instead. If a question mentions an engine force and a drag force, the net force is the difference; if it mentions a tension and a friction, the net is the difference; if the body moves at a constant speed, the net force is zero and the net work is zero however much work the engine is doing. That last case is worth sitting with, because it feels wrong and is not: a car cruising on a level road at a steady 25 m/s has an engine doing tens of kilowatts of work, and a net work of exactly nothing, because the drag is removing it as fast as the engine supplies it.
What the theorem cannot do, and the degenerate case this page handles explicitly. If you ask for a distance longer than the stopping distance, the arithmetic wants a negative kinetic energy. There is no such thing, so this page reports a final speed of zero and tells you where the body actually stopped. That is not a rounding of the model; it is the point at which a constant retarding force stops being a description of anything. After the body has stopped, friction does not keep taking energy out, and whether it stays still or slides back depends on facts the theorem does not know. The other limit is the constant-force assumption itself: brakes fade, drag rises with the square of speed and a spring gets harder the further you pull it, so for a varying force the net work is the area under the force–distance graph and not F s. The elastic potential energy page is the standard worked case of that, where the area under a straight line gives the ½kx².
And the cross-check that is worth doing every time. For a constant force, the theorem and v² = u² + 2as are algebraically the same statement: put a = F/m into the kinematic equation, multiply through by ½m, and you have the theorem. So the two must agree exactly, and this page computes both. If they ever disagreed on your own paper, the fault is arithmetic rather than physics — usually a mass in grams, a speed in km/h, or a sign on the force.
Frequently asked questions
When should I use the work–energy theorem instead of v² = u² + 2as?
When the question gives you forces and distances and does not mention time, which is most of them. For a constant force the two are the same algebra and give the same answer, so it is a matter of which is less work. The theorem wins outright in three situations the kinematic equations cannot handle: a force that varies with position, where the work is the area under the force–distance graph; a curved or awkward path, since the theorem does not care about the route; and a problem with several forces, where adding works is easier than resolving a resultant acceleration. Use v² = u² + 2as when the question is about uniform acceleration and the time matters.
Does the theorem work if the force is not constant?
Yes — the theorem itself never assumed a constant force. What changes is how you get the net work: instead of F s you need the area under the force–distance graph, which for a straight-line graph is a triangle and for anything else is an integral. This page assumes the force is constant over the distance, because that is what school questions specify, so a varying force needs the area computed first and the answer typed into the net-force box as an average. The classic non-constant case is a spring, where the force rises linearly with extension and the area under the line gives ½kx².
Why is my final speed zero when the car should have kept moving?
Because the distance you typed is longer than the distance the body can travel against that net force. The net work exceeds the kinetic energy available, the arithmetic asks for a negative kinetic energy, and the honest answer is that the body stopped on the way. The stopping-distance row tells you where. The theorem cannot tell you what happens afterwards — a braked car stays put, a block on a slope might slide back, a compressed spring pushes the body the other way — because those depend on forces the theorem was never given.
Does the mass cancel out?
Not in general, and the place it does is worth knowing. If you are given the net force, the mass matters: the same net work gives a light body much more speed than a heavy one. But if the retarding force is friction on a level surface, friction is μmg, so the mass appears on both sides and the stopping distance u²/(2μg) has no mass in it at all. That is why a loaded lorry and an empty one stop in roughly the same distance on the same surface, and it is a result that surprises people every time. It fails the moment the force is not proportional to the weight — air resistance, a fixed braking force, a rope.
Is kinetic energy a vector? My answer came out negative.
Kinetic energy is a scalar and is never negative: it is ½mv², and the square makes the direction of motion irrelevant. A body moving at −20 m/s has exactly the same kinetic energy as one moving at +20. What can be negative is the change in kinetic energy, and the net work that caused it. So if a final kinetic energy comes out negative, nothing has gone wrong with the physics of energy — you have asked for more energy to be removed than the body had, which is the stopping case above. And the final speed this page prints is a magnitude: it does not tell you which way the body is going, which is one of the things the theorem gives up in exchange for having no vectors in it.
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References
- PHYSICS (042), Class XI-XII, Central Board of Secondary Education senior secondary curriculum document for the session 2025-26 (cbseacademic.nic.in, read 11 November 2026). Cited for what the syllabus asks for, never as any kind of endorsement. Unit IV, Chapter 5, names the work-energy theorem explicitly alongside “work done by a constant force and a variable force” and “kinetic energy”, which is the pairing this page implements: the constant-force case in the calculator and the variable-force case described as the area under a force–distance graph.
- Cambridge International AS & A Level Physics 9702, syllabus for use in 2025, 2026 and 2027, Version 1 (cambridgeinternational.org, read 11 November 2026). Quoted by outcome number so that the difference from the Indian syllabus is checkable rather than asserted. The theorem is not a separately numbered outcome in 9702 at all. The nearest is 5.1.2, “recall and apply the principle of conservation of energy”, with the kinetic-energy formula under 5.1.4. That is a real difference between the two syllabuses rather than an omission in either: a Cambridge candidate is expected to reach the same result through energy conservation, and the phrase “work-energy theorem” may not appear on the paper.
- The International System of Units (SI), BIPM, 9th edition 2019. The copy served at the time of writing is V4.01 dated June 2026 and the derived-unit table in it was read in the French text, which is why the wording is paraphrased here rather than quoted: the newton is kg m s−2; the joule is énergie, travail, quantité de chaleur — energy, work and quantity of heat, one unit for all three — equal to N m and to kg m2 s−2; the watt is J/s and kg m2 s−3. The brochure also puts the radian in the table with the expression “1”: a plane angle in the SI is a quantity of unit one, which is the formal reason a trigonometric function takes a radian and not a degree.