Work-Energy Theorem and Power Calculator: From Variable Force to Watts

Push a stalled car and the effort feels obvious, but physics wants a precise number for what that effort actually accomplished. Push with a steady 300 N for 2 meters and you've done exactly 600 J of work — full stop, whether the push took two seconds or twenty. Speed only enters the picture when you ask a different question: how fast was that work delivered? That question has its own unit, the watt, and its own formula. This is where work stops and power begins.

This page is both a work-energy theorem calculator and an interactive power calculator. Drag a force curve and it computes the work done by a force that changes as the object moves along with it. Watch that same number feed straight into the work-energy theorem and come out the other side as a change in speed. Then switch over to a motor lifting a crate, where the same idea splits into two separate readings: the power being delivered right now, and the power that's been delivered on average since the lift began.

What Is the Work-Energy Theorem?

The work-energy theorem says the net work done on an object — by every force acting on it, added together — equals its change in kinetic energy:

$W_{net} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

It isn't a separate law you have to memorize on faith. It falls straight out of Newton's second law. Start from $F=ma$ and integrate both sides over the displacement:

$W_{net} = \int F\,dx = \int ma\,dx$

Acceleration is $dv/dt$, and swapping the integration variable from position to velocity using $dx = v\,dt$ collapses the whole thing into something clean:

$\int ma\,dx = \int m\frac{dv}{dt}(v\,dt) = \int mv\,dv = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

Mass and acceleration disappear and kinetic energy falls out on its own. That's the whole theorem, derived in four lines.

If you've used our Kinetic and Potential Energy Visualiser, you've already watched kinetic and potential energy trade places as a ball rolls down a ramp — energy conservation viewed from the outside, two reservoirs filling and draining. This page looks at the same physics from the inside: how work actually gets done, one newton and one meter at a time, and how fast that work can be delivered.

How Do You Calculate Work Done by a Constant Force?

When a force stays constant while an object moves in a straight line, work is just force times distance times the cosine of the angle between them:

$W = Fd\cos\theta$

The cosine is doing real work here. Push a box straight forward ($\theta = 0°$) and every newton contributes fully. Push down on it at an angle while it slides horizontally and only the horizontal component of your push, $F\cos\theta$, moves it along its path — the vertical component just presses it into the floor and does nothing useful. Push perpendicular to the motion entirely ($\theta = 90°$) and you do zero work, no matter how hard you push. That's why carrying a suitcase down a level hallway, however tiring, technically does zero work on the suitcase.

Quick example: push a crate 5 m with a steady 20 N force applied 30° above horizontal: $W = 20 \times 5 \times \cos 30° \approx 86.6$ J.

How Do You Calculate Work Done by a Variable Force?

Real forces rarely hold still. A spring pushes back harder the more you compress it. Friction shifts as the normal force changes. A rocket's thrust tapers off as it burns fuel. None of these fit into $W = Fd\cos\theta$, because there's no single F to plug in — force is a function of position, not a fixed number.

The fix is calculus. Chop the displacement into tiny steps, treat force as roughly constant across each one, add up all the little bits of work, and let the steps shrink to zero:

$W = \int_{x_i}^{x_f} F(x)\,dx$

Graphically, this integral has an unusually simple meaning.

How Do You Find Work from a Force-vs-Displacement Graph?

Plot force on the vertical axis and displacement on the horizontal axis, and the work done is exactly the area under the curve between the start and end points — area above the axis counts as positive work, area below it counts as negative work. "Find the work done using the area under the graph" is one of the most common instructions on AP Physics 1 and A-Level force-and-motion questions. When the curve isn't a clean shape you can integrate by hand, break it into trapezoids between known points and sum their areas — the trapezoidal rule — which is exactly how the calculator below produces its live readout.

Interactive Variable-Force Work Calculator

Drag the five green points below to shape a force-versus-displacement curve — steep, flat, dipping negative, whatever you like. The shaded region is the live-computed work integral, summed trapezoid by trapezoid exactly as described above. Flatten every point to the same height and the curve collapses into the constant-force case from the last section, with the readout confirming $W = Fd$. The angle slider then peels off the along-displacement component of that work, and the mass and initial-speed sliders feed the result straight into the work-energy theorem to show the resulting change in speed.

Try it: set every point to the same height and the dashed "equivalent constant force" line lands exactly on your curve — when the force really is constant, its average is its value everywhere.

How Is Power Related to Work? Instantaneous vs Average Power

Work tells you how much energy changed hands. It says nothing about how long that took — and often, how long it took is what actually matters. A car engine and a hand crank could each do the same 50,000 J of work getting a car rolling; nobody would call the two equivalent. Power is the rate at which work is done, and it comes in two flavors worth keeping straight.

Average power over an interval is the total work divided by the total time it took:

$P_{avg} = \frac{W}{t}$

Instantaneous power is the rate of work at one specific moment. Since $dW = F\,dx$ over a tiny displacement, dividing by $dt$ gives:

$P = \frac{dW}{dt} = F\frac{dx}{dt} = Fv$

Instantaneous power is force times the velocity at that instant; average power looks at the whole journey. The two only agree at every moment if force and velocity both stay constant throughout — the instant either one changes, they diverge.

What Is the SI Unit of Power?

Power is measured in watts (W), where $1\text{ W} = 1\text{ J/s} = 1\text{ kg·m}^2/\text{s}^3$ — one joule of work done every second. For comparison, one mechanical horsepower is about 746 W, so a 1 hp motor delivers roughly the same work per second as seven and a half 100-watt light bulbs running at once.

How Do You Calculate Instantaneous Power from Force and Velocity?

Multiply the force by the velocity at that exact moment. Both are vectors, so strictly speaking only the component of force along the direction of motion counts: $P = Fv\cos\theta$. Go back to the suitcase from a couple sections ago — pull it along level ground by a strap held at an angle instead of walking it flat, and only the horizontal component of your pull, $F\cos\theta$, does any work as the suitcase moves forward. Multiply that component by the suitcase's speed and you get the power actually being delivered to its motion; the vertical part of your pull, however hard you're gripping it, contributes exactly zero, because it never does any work in the first place.

Interactive Power Calculator: A Motor Lifting a Crate

A winch motor applies a constant upward force to lift a crate straight up, starting from rest. Adjust the crate's mass, the force the motor applies, and how much time has passed — the gauge shows the instantaneous power being delivered right now, with the amber line marking the average power delivered since the lift began.

Notice something: as long as the motor's force stays constant and the crate starts from rest, the instantaneous power is always exactly double the average power at any given moment. That's not a coincidence specific to this scenario — it falls straight out of the kinematics. Starting from rest under constant acceleration, velocity grows in a straight line from zero, so the average velocity over any interval is always half the velocity reached at the end of it. Average power depends on the average velocity over that interval; instantaneous power depends on the velocity right at the end of it. Since the average is always exactly half the final value here, the power ratio comes out the same way — instantaneous is double, every time.

Worked Examples for Physics Exams

Example 1: Work done by a changing push

A warehouse worker pushes a 10 kg crate across the floor. The push isn't steady — they lean in harder to get it moving, hold a steady push once it's rolling, then ease off as they let it coast: 5 N at the start, ramping up to 15 N by 2.5 m, up to 30 N by 5 m, holding at 30 N until 7.5 m, then easing to 10 N by the full 10 m. (This is the exact default curve loaded in the calculator above.) Breaking it into its four straight-line segments and summing trapezoid areas:

$W = \underbrace{\tfrac{1}{2}(5+15)(2.5)}_{25} + \underbrace{\tfrac{1}{2}(15+30)(2.5)}_{56.25} + \underbrace{\tfrac{1}{2}(30+30)(2.5)}_{75} + \underbrace{\tfrac{1}{2}(30+10)(2.5)}_{50} = 206.25\text{ J}$

Check it in the simulator: load the default curve (or press Reset curve) at θ = 0° — the readout shows W ≈ 206.3 J, matching this by-hand calculation to rounding.

Example 2: Applying the work-energy theorem

That same crate wasn't quite starting from rest — it was already coasting at 2 m/s when the worker's push took over. With $W_{net} = 206.25$ J acting on the 10 kg crate:

$\frac{1}{2}(10)v_f^2 = \frac{1}{2}(10)(2)^2 + 206.25$

$5v_f^2 = 20 + 206.25 = 226.25 \quad\Rightarrow\quad v_f = \sqrt{45.25} \approx 6.73\text{ m/s}$

Check it in the simulator: with the default curve, θ = 0°, mass = 10 kg and v₀ = 2 m/s — all defaults — the theorem readout shows v_f ≈ 6.73 m/s.

Example 3: Power output of a lifting motor

A winch applies a constant 600 N to lift a 50 kg crate straight up. The crate's weight is $mg = 50 \times 9.8 = 490$ N, so the net upward force is $600 - 490 = 110$ N and the acceleration is $a = 110/50 = 2.2\text{ m/s}^2$. After 4 seconds:

$v = at = 2.2 \times 4 = 8.8\text{ m/s} \qquad d = \tfrac{1}{2}at^2 = \tfrac{1}{2}(2.2)(4)^2 = 17.6\text{ m}$

Instantaneous power at that moment: $P = Fv = 600 \times 8.8 = 5{,}280$ W. Total work done over those 4 seconds: $W = Fd = 600 \times 17.6 = 10{,}560$ J, so average power is $P_{avg} = W/t = 10{,}560/4 = 2{,}640$ W — exactly half the instantaneous value, as the relationship above predicts.

Check it in the simulator: the power calculator's defaults are mass = 50 kg, force = 600 N, elapsed time = 4 s — exactly this example.