Work, Energy and Power

1Work by a constant force

For a constant force, W = F·s = Fs cosθ, where θ is the angle between the force and the displacement. Work is a scalar with a sign, and the sign lives entirely in cosθ: positive when the force has a component along the motion, negative when it opposes, exactly zero when it is perpendicular. A normal force on a block sliding along the floor, and the tension holding a stone in a horizontal circle, both do zero work however large they are.

Each force is charged separately. The weight, the normal reaction, the applied pull and friction each get their own W, and only afterwards are they added.

Figure. Four forces act; only two do work. N is perpendicular to the displacement and mg is perpendicular to it too, so each contributes exactly zero — no matter that mg is as large as the rope's pull. F, N and mg are drawn to one scale; the 4 N friction arrow is not, because at that scale it would be invisible. The rope contributes its horizontal component 50 cos 37° = 40 N, friction 4 N against it, and the net 36 N over 10 m is the 360 J the crate ends up carrying.

2Work by a variable force

When the force changes along the path, W = ∫F·dr — which on a force-displacement graph is the signed area under the curve. Area above the axis counts positive, area below it negative, and the net work is their difference, not their sum.

The spring is the case worth knowing by heart. To hold a spring stretched by x you must apply F = kx, so the area under that straight line from 0 to x is ½kx². Stretching further, from x₁ to x₂, costs ½k(x₂² − x₁²) — the difference of the squares, never k times the difference of the extensions.

Figure. The applied force rises linearly, so the work to reach any extension is the area of a triangle: ½ × x × kx. The strip between 10 cm and 20 cm is a trapezium of area ½(20 + 40) × 0.10 = 3.0 J, three times the triangle of area ½ × 20 × 0.10 = 1.0 J beneath the first 10 cm.

3The work-energy theorem

The net work done on a particle equals the change in its kinetic energy: W_net = ΔK = ½mv² − ½mu². In an inertial frame it holds for every force — constant, variable, conservative, frictional — and it needs no knowledge of the time taken or of the shape of the path, which is exactly why it is faster than Newton's laws whenever the question gives you a distance and asks for a speed.

Kinetic energy and momentum are not independent: K = p²/2m. So K goes as the square of p, and a small fractional change in one is twice the fractional change in the other, ΔK/K ≈ 2Δp/p.

Figure. The curve is a parabola, so equal steps in momentum are not equal steps in energy: the same 20% rise in p that costs 44% more energy near the middle of the curve would cost a far larger absolute amount further out. This is why doubling a car's speed quadruples its braking distance.

Using the theorem

1. List every forceInclude the ones you expect to vanish — the normal reaction and any force perpendicular to the path contribute zero, but say so rather than forgetting them.
2. Add the works with signsFriction and any resisting force enter negative; the driving force positive. This sum is W_net.
3. Equate to ΔKSet W_net = ½mv² − ½mu². The time and the shape of the path never appear.

4Conservative forces and potential energy

A force is conservative when the work it does between two points is the same along every path — equivalently, when its work round any closed loop is zero, ∮F·dr = 0. Only then can you define a potential energy U at each point, with ΔU = −W by the force, and recover the force from it as F = −dU/dx in one dimension.

Gravity near the ground gives U = mgh, an ideal spring gives U = ½kx². Friction gives nothing: its work depends on how long the path was, and going out and back does not return what it took. Potential energy is only ever defined up to an additive constant, so choose the zero level wherever it makes the arithmetic shortest — only differences appear in any answer.

Figure. Two routes from A to B. Gravity does exactly +mgh on each, because only the vertical drop enters — on path 2 the horizontal leg contributes nothing at all. Friction would not behave like this: it charges by the length of the path, and path 2 is longer than path 1, so a rough surface would take more from a block sent the long way round.

5Reading a potential-energy curve

Given U(x), the force is minus the slope: F = −dU/dx. A particle is therefore always pushed downhill on the U graph, towards lower potential energy. Where the slope is zero the force is zero and the particle is in equilibrium — but the kind of equilibrium is decided by the curvature, not the slope. A valley, d²U/dx² > 0, is stable: displace the particle and the slope pushes it back. A hill, d²U/dx² < 0, is unstable: displace it and the slope drives it further away.

Draw the horizontal line U = E across the graph and you read the motion off directly. K = E − U, so the particle can only exist where the curve lies below the line; where the curve meets the line, K = 0 and the particle turns round. Between two such turning points it is trapped, and oscillates.

Figure. Both marked extrema have zero slope, so the force vanishes at both — yet one is a trap and the other is a knife edge. The dashed line is the particle's total energy: the vertical gap down to the curve is its kinetic energy, which is largest at the bottom of the valley and falls to zero at the two turning points. Outside them the curve rises above the line, K would have to be negative, and the particle simply cannot go there.

Reading the curve

1. Flat points firstSolve dU/dx = 0. Every solution is an equilibrium position, and there is no force there whatever the value of U itself.
2. Then the curvatureEvaluate d²U/dx² at each: positive is a valley and stable, negative is a hill and unstable, zero needs a closer look.
3. Then the energy lineDraw U = E. The intersections are turning points, and the region where U < E is the only region the particle can reach.

6Conservation, and what friction takes

When only conservative forces do work, E = K + U is the same number at every instant, and a problem that would need the whole trajectory collapses into two snapshots: write K + U at the start, write it at the end, set them equal.

Non-conservative forces break that equality by exactly the work they do: W_nc = ΔK + ΔU = ΔE. For friction the work is −f times the length of path actually travelled — not the displacement, and not the straight-line distance. Every centimetre counts, including the centimetres travelled while a spring is being compressed.

Figure. Distance is measured along the floor from the foot of the track, where the block carries all 100 J as kinetic energy. Over the first 3 m friction alone acts, and the kinetic energy falls along a straight line of slope −4 J per metre. From 3 m on, the spring takes over: the kinetic curve dives while the spring curve climbs to meet it. At every position the two curves sum to 100 − 4s, not to 100 — the shortfall is what friction has already taken.

The energy audit

1. Fix the two instantsChoose a start and an end where you know the most — 'at rest', 'at maximum compression', 'just leaving the surface'.
2. List the storesKinetic, gravitational mgh, spring ½kx². Pick one datum for h and use it at both instants.
3. Charge the rest as workAnything that does not store — friction, air drag, an applied push — enters as W_nc, computed over the real path length.

7Power

Power is the rate of doing work, P = dW/dt, and for a force acting on a body moving with velocity v it is P = F·v — a dot product again, so a force perpendicular to the motion delivers no power just as it does no work. Average power is total work over total time; instantaneous power uses the velocity at that instant, and the two agree only when the rate is steady.

The F·v form carries a consequence worth internalising: at a fixed power the available force falls as the speed rises. A vehicle working at constant power therefore accelerates ever more feebly as it speeds up, and from rest its speed grows as v = √(2Pt/m) — proportional to √t, not to t.

Figure. At a fixed 2.5 kW the force available is a hyperbola in the speed: every doubling of v halves F. Four times the speed, a quarter of the force — which is why a machine rated by its power runs out of acceleration long before it runs out of engine, and why top speed is set by where this falling curve meets the rising drag.

8Collisions

During the brief contact of a collision the internal forces are enormous and any external force — gravity, friction — contributes negligible impulse. So momentum is conserved in every collision, elastic or not. Kinetic energy is not: it is conserved only when the collision is perfectly elastic.

The coefficient of restitution grades the rest: e = (speed of separation)/(speed of approach) along the line of impact, running from e = 1 for perfectly elastic down to e = 0, where the bodies move off together. Two standard results are worth carrying: in a head-on elastic collision equal masses simply exchange velocities, and in a perfectly inelastic collision the energy lost is ½μ(u₁ − u₂)², where μ = m₁m₂/(m₁ + m₂) is the reduced mass.

Figure. The arrows are drawn to scale: the second is a third of the first, because the momentum 12 kg m s⁻¹ is unchanged while the moving mass has trebled. Momentum survives the collision intact; two-thirds of the kinetic energy does not, and goes into heat and permanent deformation of the two bodies.

Solving any collision

1. Momentum firstWrite m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ along the line of impact. This equation is always available.
2. Then restitutionAdd v₂ − v₁ = e(u₁ − u₂). Two linear equations, two unknowns — no quadratic to solve, whatever e may be.
3. Energy only if askedCompute K before and after separately and subtract. Never assume K is conserved unless you are told e = 1.