Work, kinetic energy, potential energy, and power

The same speed can hide an energy transfer

Imagine a small cart sliding rightward along a horizontal track while stretching a spring fixed to a stationary anchor. Treat the cart as a 2 kg particle: its translation is the only motion we calculate. It starts at x = 0 with velocity +3 m/s. For this teaching model, a drive pushes rightward with 15 N, sliding resistance acts leftward with 3 N, and an ideal spring pulls leftward with force magnitude (6 N/m)x. The coordinate x also measures extension from the spring’s unstretched state.

Follow only the rightward part from x = 0 to x = 4 m. The cart finishes at the same speed, 3 m/s, yet the stretched spring stores 48 J and the drive has supplied 60 J. Those statements can all hold together. We will account for the work of each force and the energy converted out of mechanical form, rather than treating final speed as a complete energy record. These are exact stipulated conditions, not measurements or a tested elastic range.

Start with signed arithmetic, substitution and units, as taught in the published Units, standards, and physical constants and Algebra, functions, graphs, and logarithms lessons. Forces and energy quantities are introduced here. Vectors and vector spaces provides an optional route for the dot-product example. On a first reading, use the straight force line and its triangular areas; Derivatives, Taylor series, and local approximation and Integrals and multivariable calculus provide optional return paths for the rate and integral statements.

Name the force and the point it acts on

For a constant force along one fixed axis, its work is W = FₓΔx, where Fₓ is the signed force component and Δx is the displacement of its point of application. A rightward 15 N force acting through a rightward 4 m displacement does +60 J of work. A leftward 3 N force over that same displacement does −12 J. Reversing the coordinate axis reverses both signs in the product, so it preserves the work.

In several dimensions, constant-force work is the dot product W = F·Δr: multiply matching perpendicular components and add. Equivalently, when both vectors are nonzero, W = |F||Δr|cosθ, with θ the angle between the force and displacement. If either vector is zero, the dot product gives zero work without assigning an angle. Work is a signed scalar, not a force or a vector. Its unit is the joule: 1 J = 1 N·m = 1 kg·m²/s². A force perpendicular to each local displacement does zero work along that path.

A force can be present without doing work on the chosen object: its application point might remain stationary. That statement does not imply that a person maintaining the force uses no biological energy. For an extended object that rotates or deforms, the application point matters; do not substitute the object’s overall displacement automatically. Our particle approximation deliberately avoids those additional motions.

Add the work before changing the speed

In this fixed inertial frame, the classical kinetic energy of a constant-mass particle is K = (1/2)m|v|². It is nonnegative and depends on speed squared. Our 2 kg cart at 3 m/s has K = (1/2)(2 kg)(3 m/s)² = 9 J. Doubling the speed at the same mass multiplies K by four. Changing the reference frame can change the velocity and kinetic energy, so keep one frame throughout the comparison.

The work–kinetic-energy theorem states Wnet = Kfinal − Kinitial. Here Wnet is the sum of the work of every force on the particle, rather than the drive’s work alone. A positive net contribution raises K; a negative one lowers it. Zero net work over a whole interval establishes equal endpoint kinetic energies, without establishing constant speed between them.

When a force varies along a one-dimensional path, work adds its signed contributions over displacement: W = ∫Fₓ dx. This is the signed area of a force–position graph. It is not the area of a force–time graph. The integral is a compact form of adding progressively smaller force-times-displacement contributions; our straight-line case can be calculated exactly with triangle arithmetic.

Stored energy belongs to an interaction

A conservative force has work determined by the initial and final configurations, within its stated model. Represent that interaction with potential energy U so that its work is −ΔU. The potential’s zero is a chosen reference: adding the same constant everywhere changes neither ΔU nor the force. In one dimension, the force is −dU/dx when that derivative exists.

For an ideal linear spring, Fspring = −kx and U = (1/2)kx² with U = 0 at its unstretched state. The stiffness k has units N/m. In our model k = 6 N/m, so the spring’s work from 0 to 4 m is −48 J and its potential increase is +48 J. The spring is treated as massless and its anchor stays fixed. This U describes the cart–spring configuration; it is not an extra kinetic energy of the isolated cart.

For a separate near-surface gravity model with constant downward acceleration magnitude g, a height increase Δh changes gravitational potential by mgΔh. That is energy of the body–Earth interaction. Constant g is an approximation for a sufficiently limited height range, not a formula for every distance from Earth. Sliding resistance generally requires a separate transfer or conversion term, rather than a single position-only potential valid for every path and direction.

Compare force area with energy height

Let q = x/(1 m), the dimensionless number of metres of extension. Adding the three horizontal forces gives Fnet = (12 − 6q) N. Its straight line starts at +12 N, crosses zero at x = 2 m and ends at −12 N. The upper plot’s positive triangle contributes +(1/2)(2 m)(12 N) = +12 J; the negative triangle contributes −12 J. Their sum is zero net work from 0 to 4 m.

Work from the start to a position x is (12q − 3q²) J, giving K = (9 + 12q − 3q²) J. Spring potential is U = 3q² J. Adding them gives mechanical energy Emech = K + U = (9 + 12q) J. The lower plot shows those three energies on one common joule scale. Its height gives energy; its horizontal axis gives position. The force and energy panels have different vertical quantities and scales.

At x = 2 m, K reaches 21 J and the net force is zero. The cart is moving at √21 m/s, so zero force at that event is not rest. Rewriting K as [21 − 3(q − 2)²] J shows that it stays between 9 J and 21 J on 0–4 m. Thus the selected rightward branch remains moving throughout this interval. The supplied functions, not the sparse plotted markers alone, define these curves.

Keep kinetic and mechanical balances distinct

For the cart particle, the three works over 0–4 m are +60 J from the drive, −12 J from sliding resistance and −48 J from the spring. Their sum is 0 J, so K ends at its initial 9 J. Omitting the spring or counting only positive work would give the wrong final speed.

For the cart–spring mechanical balance, the spring’s conservative transfer is included as U. Now Δ(K + U) = Wdrive + Wresistance = 60 J − 12 J = 48 J. The mechanical total rises from 9 J to 57 J. It is not conserved in this driven, resistive case. Do not count the spring’s −48 J work again on that right-hand side after including its +48 J potential change on the left.

The simplified ledger allocates 12 J of the drive’s input to conversion out of mechanical energy. A fuller cart-and-surface model can follow thermal and internal energy, but this calculation does not decide how that energy is divided between materials or give their temperatures. Mechanical energy changing is compatible with conservation of energy for an appropriately complete system. Choose the boundary and include its transfers before invoking conservation.

A transfer and its rate answer different questions

Average power over a nonzero duration Δt is the selected work or energy transfer divided by that duration. Instantaneous power is its local rate. For a force on our particle, P = F·v; along this axis it is Fₓv. One watt is one joule per second. The letter W can denote work in a formula and the unit symbol W denotes watt; the quantity’s context and units distinguish them.

At the cart’s start, the drive’s power is (15 N)(3 m/s) = 45 W, resistance contributes −9 W, and the unstretched spring contributes 0 W. Net power is 36 W, the local rate of increase of K. At x = 2 m, the net force and net power are zero while the cart still moves and the drive still transfers energy. The separate force powers need not vanish when their sum does.

The net work from 0 to 4 m is zero, but the drive’s work is 60 J. Its average power would require the journey’s duration as well. Equal endpoint speeds or a work total alone does not supply that duration. Energy fixes a speed magnitude under the model; direction and timing need additional motion information.

Check what this balance describes

Our formulas describe a constant-mass particle in one fixed inertial frame, with a stipulated ideal spring and fixed resistance on a rightward sliding branch. The horizontal track makes weight and normal force perpendicular to the particle’s displacement, so they do zero work here. Neither force is universally incapable of doing work. A moving support, rotation, deformation or a different path needs its own application-point and energy accounting.

Use the model only on 0–4 m. Continuing beyond it, reversing motion or claiming that a real spring stays linear would require new assumptions and evidence. The classical expression (1/2)m|v|² is also limited to nonrelativistic mechanics. The work theorem gives a kinetic-energy change; it does not by itself certify every force model, every energy form or a complete trajectory.

Worked examples

Try it yourself

1. At x = 1 m in the main cart model, find net force, kinetic energy, spring potential and mechanical energy.

   Show the answer and reasoning

   Fnet = +6 N, K = 18 J, U = 3 J and K + U = 21 J.

   Use q = 1 in the declared force and energy functions. The energy sum includes the conservative spring interaction; it is distinct from the net force and from kinetic energy alone.

2. The main cart has zero net force at x = 2 m. Is it at rest there?

   Show the answer and reasoning

   No. Its K is 21 J and its rightward velocity is +√21 m/s.

   For m = 2 kg, speed is √(2K/m) = √21 m/s at this event. The positive branch was specified and remains moving. Zero net force gives zero acceleration at that event, rather than zero velocity.

3. A separate 2 kg particle moves at 6 m/s in the same frame. What is its kinetic energy, compared with the main cart’s initial 9 J?

   Show the answer and reasoning

   Its K is 36 J, four times 9 J.

   K = (1/2)(2 kg)(6 m/s)² = 36 J. Speed doubled while mass stayed fixed, so the squared speed multiplied energy by four.

4. Replace every spring potential value U by U + 5 J. What is its increase from 0 to 4 m, and does the spring force change?

   Show the answer and reasoning

   The increase remains 48 J and the spring force is unchanged.

   The shifted endpoints are 5 J and 53 J, whose difference is 48 J. A constant offset has zero derivative, so −dU/dx is preserved. Absolute mechanical values shift too, but their changes do not.

5. At x = 4 m, the main cart still moves rightward at 3 m/s. Find the instantaneous powers of the drive, resistance and spring, and their sum.

   Show the answer and reasoning

   Drive +45 W, resistance −9 W and spring −72 W; net power is −36 W.

   The forces are +15 N, −3 N and −24 N. Multiply each by +3 m/s. A locally decreasing K at the final event is compatible with equal initial and final K over the whole earlier interval.

6. Repeat the separate 12 J ideal lift in 2 s instead of 4 s, preserving its endpoint and no-loss conditions. What changes in lifting work and average input power?

   Show the answer and reasoning

   Work stays +12 J; average input power rises from 3 W to 6 W.

   The same gravitational potential change and equal endpoint speeds require the same input energy. Dividing that energy by half the duration doubles average power. This does not mean that the two instantaneous power histories are known or identical.

What to carry forward

• Work uses the selected force and its application-point displacement; its sign is preserved by a consistent axis reversal.
• For the stated classical constant-mass particle, net work equals kinetic-energy change.
• Potential represents a conservative interaction, with a fixed but arbitrary zero reference.
• Mechanical conservation requires the chosen system’s transfer conditions; a changing mechanical total is not proof that energy vanished.
• Force–position area gives work, energy height gives energy, and power describes a transfer rate.