Work, kinetic and potential energy, the work–energy theorem, and power

Work is force times the distance moved along the force, W = FΔd cos θ, in joules. The work–energy theorem says the net work on an object equals its change in kinetic energy, W_net = ΔKE = ½mv² − ½mv₀². When only gravity and springs do work, that becomes conservation of mechanical energy: KE + PE stays constant, and the route the object took stops mattering. Power is energy per second. Below, one hill worked by energy and then by forces so you can see the same answer appear twice, the lines students get wrong, and the formula that stops working far from Earth.

Work: force along the motion, and the cos θ that is often 0 or −1

The work formula is W = FΔd cos θ, where θ is the angle between the force and the displacement. One joule is one newton acting through one metre along its own direction, 1 J = 1 N·m.

The cos θ is where the physics hides, and in most school problems it takes one of three values.

• cos 0° = 1. The force points along the motion. A push on a box that moves the way you push does positive work, W = FΔd.
• cos 90° = 0. The force is perpendicular to the motion and does no work at all. The normal force on a box sliding down a slope is perpendicular to the slope, so it does nothing to the box's energy. Carrying a bag across a level floor at steady speed does no work on the bag, because the upward force on it is at right angles to the sideways motion.
• cos 180° = −1. The force points against the motion. Friction on a sliding box does negative work, W = −fΔd, which is energy leaving the box.

If a question gives you a force at some other angle, resolve it: only the component along the displacement does work.

The work–energy theorem, in one line and one picture

The theorem: W_net = ΔKE = ½mv² − ½mv₀². The net work on an object, from all forces together, equals the change in its kinetic energy. The kinetic energy formula is KE = ½mv², with m in kg and v in m/s giving joules.

The picture is of a box pushed across a floor: the push does work, the floor's friction undoes some of it, and whatever is left over shows up as the box moving faster. The theorem is a shortcut because it skips two things the force route needs, the acceleration and the time. Prefer it whenever the question gives distances and speeds and never asks what time anything happened at.

A small case. A 2.0 kg block starts from rest and is pushed with a net force of 6.0 N over 3.0 m. How fast is it moving?

• W_net = 6.0 N × 3.0 m = 18 J (the net work, cos 0° = 1)
• ½ × 2.0 kg × v² − 0 = 18 J (the theorem; it started from rest)
• v² = 18 J ÷ 1.0 kg = 18 m²/s² (dividing)
• v = √18 = 4.2 m/s (the root; two significant figures from the data)

No acceleration found, no time found. The force route gives the same 4.2 m/s in three lines instead of two, with a = 3.0 m/s² and v² = 2 × 3.0 × 3.0 in between.

Potential energy and conservation of mechanical energy

Near Earth's surface, gravitational potential energy is PE_g = mgh, where h is the height above a reference level you choose once and keep. The choice does not matter, because only changes in PE ever enter a calculation. What matters is not changing it halfway through a problem. A stretched or compressed spring stores PE_s = ½kx², with k the spring constant and x the stretch from its natural length.

Gravity and springs are conservative forces: the work they do depends only on where the object starts and ends, not on the path. When they are the only forces doing work, the sum KE + PE is the same at every point of the motion. That is conservation of mechanical energy, and it is why a roller coaster's speed at the bottom of a drop depends on the height of the drop and nothing else about the track's shape.

Friction is the leak. It does negative work, W_f = −fd, and the energy it removes becomes heat in the surfaces. The bookkeeping line that handles every case is:

KE₀ + PE₀ + W_f = KE + PE

Start with the energy you had, subtract what friction took, and what remains is what you have. With no friction, W_f = 0 and the line is plain conservation. Write this one line at the top of every energy problem before any number goes in.

Worked example: a 60.0 kg skier, a 25.0 m hill, then flat snow with μₖ = 0.10

A 60.0 kg skier starts from rest at the top of a 25.0 m hill. The slope is smooth enough to treat as frictionless, an idealisation stated so you know it is one. At the bottom the snow is flat, with μₖ = 0.10 between skis and snow. Find the skier's speed at the bottom of the hill and how far along the flat they slide before stopping.

Givens: m = 60.0 kg, h = 25.0 m, g = 9.8 m/s², μₖ = 0.10 on the flat only. The reference height for PE is the flat snow. Three significant figures.

Line 1, potential energy at the top.

• PE₀ = mgh = 60.0 kg × 9.8 m/s² × 25.0 m = 14,700 J (kg × m/s² × m = N·m = J)

Line 2, speed at the bottom, by energy. No friction on the slope, so KE₀ + PE₀ = KE + PE with KE₀ = 0 and PE = 0 at the bottom.

• 0 + 14,700 J = ½ × 60.0 kg × v² + 0 (the ledger line with the numbers in)
• v² = 14,700 J ÷ 30.0 kg = 490 m²/s² (dividing)
• v = √490 = 22.1 m/s (the root)

The slope's shape, steep or gentle, straight or curved, never entered. That is conservation of energy doing its job. The normal force did no work on the slope, which is why it does not appear.

Line 3, friction on the flat. On flat snow the normal force equals the weight, which is the free-body diagram you drew for a box on a level floor.

• f = μₖN = μₖmg = 0.10 × 60.0 kg × 9.8 m/s² = 58.8 N (kinetic friction)

Line 4, the stopping distance, by the work–energy theorem. Friction is the only force doing work on the flat, and its work is negative.

• W_f = −f × d = −58.8 N × d (friction opposes the motion, cos 180° = −1)
• −58.8 d = 0 − 14,700 J (the theorem: net work equals final KE minus initial KE)
• d = 14,700 ÷ 58.8 = 250 m (dividing; J ÷ N = m)

Line 5, the check by forces. The same flat, done the long way.

• a = −f/m = −58.8 N ÷ 60.0 kg = −0.98 m/s², or −μₖg (Newton's second law; the minus means slowing)
• v² = v₀² + 2aΔd → 0 = 490 − 2 × 0.98 × d = 490 − 1.96 d (the kinematics equation v² = v₀² + 2aΔd, which leaves out the time nobody asked for)
• d = 490 ÷ 1.96 = 250 m (the same)

The same 250 m appeared twice from two different routes. If your two routes disagree, one of them has a wrong line, and now you know which pair of lines to compare.

Line 6, the shortcut worth noticing. Set the two ends of the whole trip equal in symbols: mgh at the top, nothing at the end, and friction's work in between.

• mgh = μₖmgd (all the potential energy ends up as friction's work)
• d = h/μₖ = 25.0 m ÷ 0.10 = 250 m (the masses and the g cancel)

The mass cancelled. Say to yourself why: a heavier skier has more energy to lose and proportionally more friction to lose it to. Chi and colleagues (1994) found that students who explained a worked example's steps to themselves, as you have just been asked to, understood the material measurably better than students who only read it twice. The sentence "why does the mass cancel" is the self-explanation this page is built around.

Power: the same energy, how fast

Power is the rate of transferring energy, P = W/t = ΔE/t, in watts; one watt is one joule per second. At a constant speed, where the force applied balances the force resisting, it is also P = Fv, force times speed, because W/t = FΔd/t = Fv.

The same 60.0 kg student climbs 3.0 m of stairs in 4.0 s.

• ΔPE = mgh = 60.0 kg × 9.8 m/s² × 3.0 m = 1,764 J (the energy gained)
• P = 1,764 J ÷ 4.0 s = 441 W, about 440 W (the power; J/s = W)

For scale, one horsepower is 746 W, so the climb was about 0.59 horsepower. Efficiency is useful energy out divided by energy in, written as a percentage, and it is never over 100: a motor that is 80 % efficient turns 80 J of every 100 J of electrical energy into mechanical work and the rest into heat.

The three wrong lines a marker sees

1. v instead of v² in the kinetic energy, or the ½ forgotten. The symptom is a speed that is absurd: solving 60.0 × v = 14,700 gives 245 m/s, and 60.0 × v² = 14,700 gives 15.7 m/s where the right answer is 22.1. A skier at 245 m/s is a sign to look at the formula, not the arithmetic.
2. The normal force's work counted on the slope, or friction's work given a positive sign. The normal force is perpendicular to the slope and does no work at all. Friction's work is negative, always, because friction opposes the motion. The symptom of a positive W_f is a skier who gains energy from the snow.
3. Reference heights mixed. PE measured from the top in one line and from the bottom in the next. Choose the reference once, write "h = 0 at the flat" at the top of the page, and never move it.

A fourth, briefer: power written in joules. Power is joules per second, watts. A number in joules is an energy, and a question asking "how powerful" wants a rate.

A sibling to try cold

Cover the example. A 70.0 kg skier starts from rest on a frictionless 16.0 m hill and slides onto flat snow with μₖ = 0.080. Find the speed at the bottom and the stopping distance. Write the energy ledger line, KE₀ + PE₀ + W_f = KE + PE, before any number.

Check

• PE₀ = 70.0 × 9.8 × 16.0 = 10,976 J, about 1.10 × 10⁴ J
• v = √(2gh) = √(2 × 9.8 × 16.0) = √313.6 = 17.7 m/s
• d = h/μₖ = 16.0 ÷ 0.080 = 200 m

If you got 200 m by the shortcut, check that you could also get it the long way: f = 0.080 × 70.0 × 9.8 = 54.9 N, and 10,976 ÷ 54.9 = 200 m. The shortcut is only safe once you have seen it come out of the ledger.

When mgh stops being enough

PE = mgh assumes g is the same at every height, which is true to within a fraction of a percent for a hill and false for a satellite. Far from Earth, the gravitational potential energy is PE_g = −GMm/r, with its zero set at infinite distance, so every bound object has negative potential energy. Near the surface the curve of −GMm/r is almost a straight line, and mgh is the first step up that line: the same shape as the gravitation formula with one fewer power of r. That is idea 15 of the course, and the orbit, where −GMm/r is needed is on the gravitation page; this one keeps its example near the ground.

Where this sits in the course

This page covers ideas 6, 7 and 8 of 25 in SPH4U, the first half of strand C: Work and kinetic energy, Conservation of mechanical energy and Power and efficiency, with idea 15, Gravitational potential energy and potential, in strand D named just above as the place mgh is replaced. Before idea 6 is idea 5, Relative motion in two dimensions, the last of the kinematics ideas.

After idea 8 is idea 9, Impulse and momentum, where the course says "conserved" of a second quantity, and this time the conserved thing is a vector with a sign. Impulse and momentum, the idea after power is written the same way as this page. The SPH4U course map shows all 25 ideas in teaching order, and every idea of the three science courses has a page like this one.

Work and energy on one screen

• Three formulas. W = FΔd cos θ; KE = ½mv²; PE_g = mgh near the surface, PE_s = ½kx².
• The theorem. W_net = ΔKE. A speed from forces and distances, with no time found.
• The ledger line. KE₀ + PE₀ + W_f = KE + PE. Write it first; friction's work is negative.
• Energy or forces. Energy when no time is asked for; forces when it is. The same answer should appear from both.
• Power. P = W/t = Fv, in watts. Efficiency is out over in, never above 100 %.
• The three wrong lines. v for v² or the ½ dropped; the normal force's work counted or friction's work positive; the reference height moved mid-problem.
• The habit. Say why the mass cancelled before you move on; how to study physics is built on deriving each line and explaining it to yourself.