Classical Mechanics

The Work-Energy Theorem, Step by Step

P PhysicsFundamentals Editorial Team · September 19, 2026 · 22 min read
Definition

The work energy theorem states that the net work done on a body equals the change in its kinetic energy. Add up the work every force did, each with its own sign, and the total equals ½mv22 – ½mv12. It turns a force and a distance into a speed, with no time in it.

Shove a supermarket trolley across a car park and let go. It rolls on, slows, stops — and you can work out exactly where it stops without knowing how long any of it took, and without ever writing down an acceleration.

That is what this one equation buys you. Count the energy that went in, subtract the energy that friction took back out, and the speed drops out of the arithmetic. It is also why a car braking from 60 m/s needs four times the road it needs from 30 m/s, and not twice.

What Is the Work Energy Theorem?

The work-energy theorem says that the total work done on a body by all the forces acting on it equals the change in its kinetic energy. Energy in, speed up. Energy out, slow down.

One word in that sentence does most of the damage when it is skipped: net. Every force acting gets its own term, with its own sign, and it is their sum that the change in kinetic energy matches — not the work done by the force you happen to be thinking about.

The mechanism is worth a moment. Work done is energy transferred by a force pushing something through a distance, and kinetic energy is the store that energy lands in. Our guide to work done also carries the one-line derivation from a = F/m and v2 = u2 + 2ad, which is where it belongs; this page takes the finished result and puts it to work.

Where the theorem earns its keep is the middle of a problem. You are handed a force and a distance, you are asked for a speed, and nobody has told you how long anything took — that is exactly the gap it fills.

One convention runs through every number below. Take g = 9.81 m/s2, the standard value rounded to three figures and the figure the simulator on this page uses. Real local gravity runs from roughly 9.78 to 9.83 m/s2, so three significant figures is the ceiling on any answer here.

The Work Energy Theorem Formula

Written out in full, the theorem equates one sum of work terms to one difference of kinetic energies.

Wnet = ½mv22 – ½mv12

When the net force is constant and points along the direction of travel, the left-hand side has a shape you can evaluate in one multiplication.

Wnet = Fnet · d
Symbol What it is SI unit Worth knowing
Wnettotal work done by every force actingJnegative when the body is being slowed; the unit is the joule, and the seven SI base units define the twenty-two derived units with special names
Fnetvector sum of the forces, resolved along the motionNadd the forces before you multiply; anything acting sideways contributes nothing
ddistance over which that net force actsmthe length of the skid or the depth of the dent, not the whole journey
mmass of the bodykgcancels when the forces themselves scale with the mass, and not otherwise
v1, v2speed at the start and at the endm/sspeeds, not velocities; the equation is blind to direction

Rearranged, that single relation answers five different questions, and grinding through the algebra by hand every time is thankless. The work-energy theorem calculator solves it for whichever of the five you nominate — either speed, the net force, the distance or the mass — and prints the kinetic energy at both ends beside the answer.

The simulator below is the same equation with a block, a floor and something to rub against. Five sliders set the mass, your push, the friction coefficient, how far you push and how fast the block was already going; the readouts split the work into the push’s share and friction’s share and show the change in kinetic energy they add up to. It also runs as a standalone work energy theorem simulator.

Work-Energy Theorem Lab

One habit of that lab is worth knowing before it trips you up. It prints joules and newtons to one decimal place and speeds to two, so the numbers you see are rounded versions of quantities that are exact underneath — the opening state has a net work of 122.28 J and shows 122.3 J, and a final speed of 5.529014 m/s shown as 5.53 m/s.

The rounding has a consequence that looks like a bug and is not. Because each readout is rounded on its own, the two work boxes do not always add up on screen: a 30 N push over 20.0 m on a 1.0 kg block shows 600.0 J and -49.1 J against a net of 551.0 J, and 600.0 minus 49.1 is 550.9.

The same applies to the kinetic-energy pair. Subtracting the two printed energies is not a reliable check on the net work — one state shows 2.3 J going to 2.1 J beside a net of -0.1 J. The quantities balance perfectly; it is the printed strings that need not.

Not every ratio is spoiled, though, so do not over-learn the lesson. Doubling the push distance doubles the net work exactly, and on screen 1.0 m to 2.0 m really does run 20.4 J to 40.8 J. Try 3.0 m to 6.0 m and it reads 61.1 J then 122.3 J, where doubling the printed figure would have given 122.2 J.

Four Steps That Turn Forces into a Speed

Almost every mistake on this topic is made before the arithmetic starts, in the step where the forces are collected. Work through these four in order and the sums look after themselves.

  1. List every force acting along the motion. The push or the drive, friction, drag, the component of the weight on a slope. Forces perpendicular to the motion — the normal force on a level floor, the tension in a pendulum string — do no work at all, so they do not appear.
  2. Work out each force’s own work, with its sign. A force pushing the body along does positive work, force times distance. A force opposing the motion does negative work, the same product with a minus sign in front.
  3. Add them. The sum is the net work. If it comes out negative, the body is slowing; if it comes out zero, the speed at the end matches the speed at the start, however violent the journey between them was.
  4. Set that total equal to the change in kinetic energy. Wnet = ½mv22 – ½mv12, then rearrange for whatever you were asked for.

Watch what that looks like along a single push. An 8.00 kg crate shoved with 40.0 N across a floor with a friction coefficient of 0.25 collects energy from the push at a steady 40.0 J per metre and loses it to friction at 19.62 J per metre, and the gap between those two lines is the kinetic energy it actually has.

A line graph of energy in joules against distance pushed in metres, for an 8.00 kilogram crate pushed with 40.0 newtons over 6.00 metres against a friction coefficient of 0.25, starting from rest. Three straight lines share one energy scale. The gold line is the work done by the push and climbs from 0 to plus 240.00 joules. The dark line is the kinetic energy and climbs from 0 to plus 122.28 joules. The wine line is the work done by friction and falls from 0 to minus 117.72 joules. At every distance the gold value plus the wine value equals the dark value. The vertical axis is energy in joules, ticked from minus 120 to plus 240; the horizontal axis is distance pushed in metres, ticked 0, 1.5, 3.0, 4.5 and 6.0.
Figure 1. The same 8.00 kg crate, 40.0 N push and friction coefficient of 0.25, tracked over a 6.00 m push from rest. The gold and wine lines add to the dark one at every distance, which is the theorem in picture form. These are exact values, not readings taken off the simulator, which rounds to one decimal place.

Read the far right of that figure and you have problem 1 below already solved: 240.00 J in, 117.72 J out, 122.28 J left as motion. The call-out makes the same point a quarter of the way along, where 60.00 minus 29.43 leaves 30.57 J.

When to Use It Instead of SUVAT

Reach for the theorem when you have a force and a distance but no time, and reach for kinematics when the clock is part of the question. Neither is more accurate than the other.

That is worth stating plainly, because students often suspect the energy route of being a shortcut that costs something. It is not. On a straight path under a constant net force the theorem and the SUVAT equations are the same algebra wearing different clothes, one multiplication by ½m apart, and across a grid of test cases the two answers differ only in the fifteenth decimal place.

The difference is scope. SUVAT needs one steady acceleration to exist; the theorem does not care whether the force held still, provided you can find the total work it did.

  • Use the theorem when the question gives you a force and a distance and asks for a speed, or gives you speeds and asks how far or how hard.
  • Use the theorem when the force varies along the way, because the work is still the area under its force-distance graph even though no single acceleration exists.
  • Use kinematics the moment a time, a direction or a position part-way through is wanted. Energy has none of those.
  • Use an energy balance with heights when the problem is a drop or a climb rather than a push; the conservation of energy calculator carries the mgh terms this equation has no room for.

Why Stopping Distance Goes as the Square of the Speed

Kinetic energy carries the speed squared and the braking force does not, so doubling the speed leaves four times as much energy for the same force to remove over four times the distance.

Set the final speed to zero and rearrange: d = ½mv12 divided by the braking force. Every doubling of v1 quadruples the numerator while the denominator sits still. Here is a 1200 kg car under a constant 5000 N, worked out at six speeds.

Speed before braking Kinetic energy to remove Braking distance at 5000 N
10 m/s60 000 J12 m
20 m/s240 000 J48 m
30 m/s540 000 J108 m
40 m/s960 000 J192 m
50 m/s1 500 000 J300 m
60 m/s2 160 000 J432 m

Every doubling in that column is an exact quadrupling: 12 to 48, 48 to 192, 108 to 432. Not four to a good approximation, not four as a rule of thumb — exactly four times, because the only thing that moved between the two rows was a squared term.

A curve of stopping distance in metres against the speed before braking in metres per second, for a 1200 kilogram car under a constant 5000 newton braking force. The curve is a parabola rising from the origin. Six marked points read 10 metres per second to 12 metres, 20 to 48 metres, 30 to 108 metres, 40 to 192 metres, 50 to 300 metres and 60 to 432 metres. Dashed guide lines join the 30 and 60 metres per second points to the vertical axis, and a note beside them says that doubling the speed multiplies the distance by exactly four, 108 metres to 432 metres. The vertical axis is stopping distance in metres, ticked 0, 100, 200, 300 and 400; the horizontal axis is speed before braking in metres per second, ticked every 10 up to 60.
Figure 2. The same six values drawn as a curve. It is a parabola, exactly: the distance is a constant multiplied by the square of the speed. The dashed pair marks the 30 m/s and 60 m/s cases, 108 m and 432 m, whose ratio is exactly 4.

Two things ordinary speech runs together here, and they should not be. The 108 m is the distance the brakes need once they are gripping; a driver’s real stopping distance also includes everything covered while the news is still travelling from eye to foot, and this theorem says nothing at all about that reaction distance.

Nor is 108 m a figure any particular car on any particular road would deliver. It is what a 1200 kg body under a steady 5000 N does, and real braking is neither perfectly steady nor delivered by a single number.

Does the mass matter? Both answers are right

It depends entirely on what is setting the braking force, and the two ordinary cases answer in opposite directions. Take them one at a time.

At a fixed force, the mass matters exactly. Hold the braking force at 5000 N and let the mass change: from 30.0 m/s a 1000 kg car needs 90 m and a 2000 kg one needs 180 m. Twice the mass, twice the energy, twice the distance.

When grip is the limit, it cancels completely. Braking on a road is usually limited by friction, and that friction force is itself proportional to the mass, so the mass divides out of both sides. Taking a coefficient of 0.70, every vehicle braking from 30.0 m/s needs the same 65.5 m, whatever it weighs.

So “heavier vehicles always take longer to stop” is true of a winch with a fixed pull and false of the ordinary road case, and the opposite slogan fails in the opposite situation. The question to settle first is always what sets the force.

What the Work Energy Theorem Cannot Tell You

The theorem returns a speed, an energy, a force or a distance, and nothing else. Four of its blind spots cause most of the trouble.

It never gives you a time. There is no t in the equation and none can be recovered from it. The simulator does animate the push, but that replay is built from separate kinematics and it stretches or compresses the clock to fit a watchable few seconds — it is not a measurement of how long anything took.

It never gives you a direction. Both speeds enter squared, so the equation cannot tell a body travelling north from one travelling south, and it has nothing to say about where the body was part-way through.

Only the energy of moving along counts. A rolling ball, a spinning flywheel and a crumpling wing all hold energy that ½mv2 cannot see. The model behind this page is a rigid body that slides without turning, which is why the simulator’s block cannot tip or spin.

The work it needs is the total, not a convenient part of it. If the force changes as it acts, the product of a force and a distance is the wrong quantity and you need the area under the curve instead — problem 5 shows how badly that goes if it is ignored.

The lab itself is a model at its own scale, and worth reading as one. Its push is horizontal, so the normal force is exactly mg and the friction force exactly μmg; angle the push and every number changes. It also uses one friction coefficient for both sticking and sliding, so it does not reproduce the extra shove a stationary object often needs to get going.

Where You Meet the Work Energy Theorem

The pattern to look for is a known force acting over a measurable distance, with a speed at one end of it.

Skid marks on a road, the length that the work energy theorem turns back into a speed
A skid is a distance you can measure afterwards. That is half of what the equation needs.
  • Reading a skid backwards. The mark on the road is a measured distance and the friction force can be estimated, so the equation runs in reverse to give the speed that produced it. Any coefficient used in such an estimate is a figure chosen for the calculation, never a measured property of one tyre on one road.
  • Anything that hits. A hammer head, a punch, a piling rig: the energy arriving is modest, the distance it is stopped in is tiny, and the force is whatever the division demands. Problem 6 puts a number on how violent that gets.
  • Crumple zones. The energy is fixed by the mass and the speed before impact, so the only variable a designer controls is how far the structure folds. Longer fold, smaller force — the same division read as a design brief.
  • Gravel arrester beds. A runaway lorry buries its wheels in loose stone, which supplies a large backwards force over a short distance and empties the kinetic-energy store without a collision.
  • A bow, a catapult, a spring launcher. Here the force fades as the string returns, so the work is the area under the force-distance curve rather than a single product — which is precisely the case problem 5 works through.
  • The trolley-and-track practical. Pull a trolley with a known force over a measured distance, time it through a gate, and compare the measured speed with the one this equation predicts. The shortfall is the energy the axles and the track quietly took.

Notice what those six have in common. Not one of them needs a stopwatch, and not one of them needs the force to have been acting for a known length of time.

Common Misconceptions About the Work Energy Theorem

Four beliefs do most of the damage, and the first is far and away the most expensive.

“W = ΔKE means the work I did”

It means the work everything did. Push a crate 6.00 m with 40.0 N and you have done 240.00 J of work on it, but if friction removed 117.72 J over the same distance, the crate’s kinetic energy rose by 122.28 J and not by 240.00 J. Put a single force’s work on the left-hand side and the answer comes out too fast, every time.

“Negative work is just bookkeeping”

Negative work is as physical as positive work — it is energy leaving the body. Friction, drag and brakes all transfer energy out into heat and sound, and the minus sign is what stops you adding that energy to the motion instead of subtracting it.

“The theorem can tell you how long the stop took”

It cannot, and this trips up exam candidates who have just got the right distance. Energy connects forces to distances and speeds; time enters only through a kinematic equation or an average speed, worked out separately afterwards.

“It applies to anything that moves”

It applies cleanly to a rigid body sliding along. Let the body roll, spin or deform and part of the work goes into stores that ½mv2 does not count, so the predicted speed comes out too high for a rolling ball and the absorbed force too low for a panel that folds. Know which model you are in before you trust the number.

Worked Problems

Problem 1
An 8.00 kg crate is pushed from rest across a level floor with a steady horizontal force of 40.0 N for 6.00 m. Take the coefficient of kinetic friction between crate and floor as 0.25. How fast is it moving at the end of the push?
Show Solution
Solution: Step 1: Two forces act along the motion — your push, doing positive work, and friction, doing negative work. The normal force and the weight are perpendicular to the motion and do none. Step 2: The friction force on a level floor. f = μmg = 0.25 × 8.00 kg × 9.81 m/s2 = 19.6200 N. Step 3: The two work terms over the 6.00 m. Wpush = 40.0 N × 6.00 m = 240.000 J, and Wfric = -19.6200 N × 6.00 m = -117.720 J. Step 4: Add them for the net work. Wnet = 240.000 – 117.720 = 122.280 J. The same figure comes from the net force: 40.0 – 19.6200 = 20.3800 N, and 20.3800 × 6.00 = 122.280 J. Step 5: The crate started from rest, so all of that net work is kinetic energy: ½mv22 = 122.280 J. Then v22 = 2 × 122.280 / 8.00 = 244.560 / 8.00 = 30.5700 m2/s2. Step 6: Take the root of the full figure. v2 = sqrt(30.5700) = 5.529014 m/s. Answer: 5.53 m/s. The simulator above opens on exactly this case and prints 122.3 J and 5.53 m/s, because it rounds 122.280 to one decimal place.
Problem 2
A 1200 kg car travelling at 30.0 m/s brakes with a constant net force of 5000 N opposing its motion. How fast is it still going after 40.0 m of braking?
Show Solution
Solution: Step 1: Start with the kinetic energy it has. 30.02 = 900.00 m2/s2, so KE1 = ½mv12 = 0.5 × 1200 kg × 900.00 m2/s2 = 540000 J. Step 2: The braking force opposes the motion, so its work is negative. Wnet = -5000 N × 40.0 m = -200000 J. Step 3: Apply the theorem as a running balance. KE2 = KE1 + Wnet = 540000 – 200000 = 340000 J. Step 4: Turn that energy back into a speed. v22 = 2 × 340000 / 1200 = 680000 / 1200 = 566.666667 m2/s2, so v2 = sqrt(566.666667) = 23.804761 m/s. Answer: 23.8 m/s. The car has shed 200000 J, which is 37.0 % of its kinetic energy, while its speed has dropped by only 20.7 % — energy falls away faster than speed does, because of the squared term.
Problem 3
The same 1200 kg car under the same 5000 N of braking. How far does it need to stop completely from 30.0 m/s, and how far from 60.0 m/s?
Show Solution
Solution: Step 1: A complete stop means every joule of kinetic energy has to be removed, so the braking force multiplied by the distance equals the starting kinetic energy: d = ½mv12 / 5000. Step 2: From 30.0 m/s the store is the 540000 J of problem 2. d = 540000 / 5000 = 108.000000 m. Step 3: From 60.0 m/s, 60.02 = 3600.00 m2/s2, so the store is 0.5 × 1200 × 3600.00 = 2160000 J and d = 2160000 / 5000 = 432.000000 m. Step 4: Compare them. 432.000000 / 108.000000 = 4.000000. Answer: 108 m and 432 m — a factor of exactly four, not an approximation that happens to land close. Both are whole numbers and both come out of the same division. Remember that a driver also covers a reaction distance before any of this begins.
Problem 4
A car brakes to a stop from 27.0 m/s on a level road where the tyres are the limiting factor. Take the coefficient of friction as 0.70. How far does it take, and how much does the answer depend on the mass?
Show Solution
Solution: Step 1: Friction is the only horizontal force, and on the level it is f = μmg. All the kinetic energy must go into its negative work: ½mv2 = μmgd. Step 2: Every term carries one factor of m, so divide it out. d = v2 / (2μg), with no mass anywhere in it. Step 3: Build the denominator. 2μg = 2 × 0.70 × 9.81 = 13.734000 m/s2. Step 4: Build the numerator. v2 = 27.02 = 729.0000 m2/s2. Step 5: Divide. d = 729.0000 / 13.734000 = 53.079948 m. Step 6: Round early and you lose the answer. Taking 13.734 as 13.7 gives 729.0000 / 13.7 = 53.211679 m, which prints as 53.2 m rather than 53.1 m. Answer: 53.1 m, and the same for a small hatchback and a loaded van. Take μ = 0.70 as a figure chosen for this problem, not a measured property of tyres or tarmac.
Problem 5
A spring-loaded bolt pushes a 2.50 kg slider from rest over 0.400 m. The force starts at 300 N and falls steadily to zero across that distance. How fast is the slider moving when the bolt lets go?
Show Solution
Solution: Step 1: The force is not constant, so multiplying a force by a distance is the wrong move. The work done is the area under the force-distance graph, and a force falling linearly to zero draws a triangle. Step 2: Take that area. W = ½ × 300 N × 0.400 m = 60.0000 J. Step 3: The slider starts from rest, so all of it becomes kinetic energy. v2 = 2 × 60.0000 / 2.50 = 120.0000 / 2.50 = 48.000000 m2/s2. Step 4: Take the root. v = sqrt(48.000000) = 6.928203 m/s. Step 5: See what the shortcut costs. Use the peak 300 N for the whole 0.400 m and you get W = 300 × 0.400 = 120.0000 J, then v = sqrt(2 × 120.0000 / 2.50) = sqrt(96.000000) = 9.797959 m/s — too fast by 2.869756 m/s. Answer: 6.93 m/s. This is the one problem on the page that no constant-force tool can do end to end: turn the triangle into joules yourself, and the work and power calculator will handle a steady force over a distance once you have one.
Problem 6
A 4.50 kg hammer head strikes a nail at 12.0 m/s and drives it 12.0 mm into a beam before coming to rest. What average force does the nail experience?
Show Solution
Solution: Step 1: The energy arriving with the head. 12.02 = 144.00 m2/s2, so KE = ½mv2 = 0.5 × 4.50 kg × 144.00 m2/s2 = 324.000 J. Step 2: The head ends at rest, so the net work over the penetration must remove all of it: F × d = -324.000 J. Step 3: Convert the distance and divide. d = 12.0 mm = 0.0120 m, so F = -324.000 / 0.0120 = -27000.0 N. Step 4: Put that in proportion. The head’s own weight is mg = 4.50 × 9.81 = 44.1450 N, and 27000.0 / 44.1450 = 611.620795. Answer: 27000 N, opposing the motion — 611.6 times the weight of the head. A hammer is not violent because it carries much energy; it is violent because it gets rid of what it has in twelve millimetres.
Problem 7
A 15.0 kg box is released sliding up a ramp inclined at 20.0 degrees, moving at 6.00 m/s. Take the coefficient of kinetic friction as 0.18. How far up the slope does it get before stopping?
Show Solution
Solution: Step 1: Going up, two forces do negative work: the component of the weight along the slope, mg sinθ, and friction, μmg cosθ. Nothing does positive work once the box has been released. Step 2: The two trigonometric values, kept in full. sin 20.0° = 0.34202014 and cos 20.0° = 0.93969262. Step 3: The resisting forces. mg sinθ = 15.0 × 9.81 × 0.34202014 = 50.32826 N, and μmg cosθ = 0.18 × 15.0 × 9.81 × 0.93969262 = 24.88964 N. Together they total 50.32826 + 24.88964 = 75.21790 N. Step 4: The kinetic energy to be removed. 6.002 = 36.0000 m2/s2, so KE1 = 0.5 × 15.0 × 36.0000 = 270.000 J. Step 5: Distance is energy divided by force. d = 270.000 / 75.21790 = 3.589571 m. Step 6: Divide the whole balance by m and the mass disappears: d = v12 / (2g(sinθ + μcosθ)). Here g(sinθ + μcosθ) = 9.81 × (0.34202014 + 0.18 × 0.93969262) = 5.014527 m/s2, so 2 × 5.014527 = 10.029054 and 36.0000 / 10.029054 = 3.589571 m — the same answer without the 15.0 kg anywhere in it. Step 7: Round the trigonometry first and it falls apart. Using sin 20.0° = 0.3 and cos 20.0° = 0.9 gives 9.81 × (0.3 + 0.18 × 0.9) = 4.532220 m/s2, then 2 × 4.532220 = 9.064440 and 36.0000 / 9.064440 = 3.971564 m, wrong in the second figure. Answer: 3.59 m up the slope. More on resolving forces on a slope in our guide to inclined plane physics.
Problem 8
A 50.0 kg trolley is pushed from rest with a steady 250 N over 4.00 m of level floor and then released. Take the coefficient of kinetic friction as 0.30. How fast is it going when you let go, and how far does it travel altogether?
Show Solution
Solution: Step 1: Friction acts for the whole journey, the push only for the first part, so treat them as two phases. f = μmg = 0.30 × 50.0 × 9.81 = 147.150 N. Step 2: Phase one, with your hands on it. The net force is 250 – 147.150 = 102.850 N, so Wnet = 102.850 × 4.00 = 411.400 J. Step 3: From rest, that is the kinetic energy at the moment you let go. v2 = 2 × 411.400 / 50.0 = 822.800 / 50.0 = 16.456000 m2/s2, so v = sqrt(16.456000) = 4.056600 m/s. Step 4: Phase two, coasting. Friction alone now removes those 411.400 J, so the extra distance is 411.400 / 147.150 = 2.795787 m. Step 5: Add the two phases. 4.00 + 2.795787 = 6.795787 m. Step 6: Check the whole journey in one line. The push did 250 × 4.00 = 1000.00 J, friction did -147.150 × 6.795787 = -1000.00 J, and the two cancel exactly — which they must, because the trolley began at rest and ended at rest. Answer: 4.06 m/s when you let go, then 2.80 m of coasting, 6.80 m in total.
A signed work ledger for a 50.0 kilogram trolley pushed with 250 newtons over 4.00 metres on a floor with friction coefficient 0.30 and then released. A gold bar above the zero line is the work done by the push, plus 1000.00 joules. A stacked bar below the zero line is the work done by friction, minus 1000.00 joules in total, split into a wine segment of minus 588.60 joules during the 4.00 metre push and a navy segment of minus 411.40 joules during the 2.7958 metre coast, labelled in the drawing as acting over all 6.7958 metres. A third column marks the net work at exactly zero joules, which is why the trolley starts at rest and ends at rest. The vertical axis is work done in joules, ticked minus 1000, minus 500, 0, 500 and 1000.
Figure 3. Problem 8 as a ledger. The push puts in 1000.00 J over 4.00 m; friction takes out 588.60 J across that same push and another 411.40 J across the 2.7958 m of coasting. The two columns are equal and opposite, so the net work is zero and the trolley ends as it began, at rest.

Frequently Asked Questions

What is the work energy theorem formula?
The work energy theorem formula is W net = ½mv2² – ½mv1², where W net is the total work done by every force acting on the body. When the net force is constant and lies along the direction of travel, that total is simply the net force multiplied by the distance, so the equation turns a force and a distance straight into a change of speed.
Does the work energy theorem still work when there is friction?
Yes, provided friction’s work enters the sum with a minus sign. Friction opposes the motion, so it removes energy: its term is the friction force multiplied by the distance slid, taken as negative. Add that to the work done by the push and the total is the change in kinetic energy, whether the total comes out positive, negative or zero.
Why does doubling your speed quadruple the stopping distance?
Because kinetic energy carries the speed squared while the braking force does not. Twice the speed means four times the energy, and the same force needs four times the distance to remove it. A 1200 kg car under a constant 5000 N stops in 108 m from 30.0 m/s and 432 m from 60.0 m/s, an exact factor of four rather than a rough one.
Do heavier vehicles always take longer to stop?
Only when the braking force is fixed. Under a constant 5000 N from 30.0 m/s, 1000 kg needs 90 m and 2000 kg needs 180 m, exactly double. Road braking is usually limited by grip instead, and the friction force grows in proportion to the mass, so the two effects cancel and the distance stops depending on the mass at all.
Can the work energy theorem tell you how long the braking took?
No. It relates a force and a distance to two speeds, and time never appears in it. For a duration you need a kinematic equation or an average speed, worked out separately. The same limit applies to direction: both speeds enter squared, so the equation is blind to which way the body was travelling.
When should I use the work energy theorem instead of SUVAT?
Use it when you are given a force and a distance but no time, and especially when the force is not constant. SUVAT needs one steady acceleration to exist, which a fading spring or a variable drag does not provide. Where both apply they agree to fifteen decimal places, so the real difference between them is scope, not accuracy.
Why do the simulator's two work readouts sometimes not add up?
Because each readout is rounded on its own before it is printed. The quantities add up perfectly; the strings on screen need not. A 30 N push over 20.0 m on a 1.0 kg block shows 600.0 J and -49.1 J beside a net of 551.0 J, and 600.0 minus 49.1 is 550.9. Trust the physics rather than the last printed digit.
Explore this topic interactively Work Energy Theorem Simulator (interactive lab) Work Energy Theorem Calculator
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