IB DP Physics · Theme A · Space, time and motion

Work, energy and power.

A.3 · Work done by a force, mechanical energy and its conservation, power, efficiency and fuels
Standard level and higher level

SL + HL · 8 hours

The journey

Four parts.

Part 01

Work

A force transfers energy

Part 02

Kinetic and potential energy

Three kinds of mechanical energy

Part 03

Conservation

With and without resistive forces

Part 04

Power and efficiency

Rates, Sankey diagrams, energy density

A.3 has no additional higher level content: every slide is for SL and HL.

A.3 WORK, ENERGY AND POWER · SL + HL

Part 01 · SL + HL

Work.

Energy is conserved. Work is how a force moves it from one place to another

A.3 WORK, ENERGY AND POWER · SL + HL

The principle behind the whole topic

Conservation of energy.

Energy cannot be created or destroyed. It can only be transferred or changed from one form to another
The total energy of an isolated system stays constant

Mechanical energy

Kinetic and potential

Of moving bodies, raised masses and stretched springs

Transfer by a force

Work

A force acting through a distance

Other forms

Thermal, chemical, nuclear …

Where “lost” mechanical energy goes, and where fuels store it

A.3 WORK, ENERGY AND POWER · SL + HL

Why energy is so useful

Forces and accelerations change from moment to moment and have directions. Energy is a single number that stays the same from start to finish. If you know the energy of a system at the beginning and can account for what enters and leaves, you know its energy at the end, without following the motion in between. Many problems that are long with Newton’s laws take two lines with energy.

Choosing the system

“Conserved” applies to a system that nothing enters or leaves. A falling ball alone is not isolated: the Earth pulls on it. The ball and the Earth together are, so their total energy is constant. Whenever energy seems to appear or vanish, the system has been drawn too small.

Energy is a scalar, measured in joules. There are no components to resolve, which is much of its appeal.

Transferring energy

Work done by a force.

\(W = Fs \cos \theta\)

For a constant force
\(\theta\): the angle between the force and the displacement
Work done = energy transferred. Unit: \(1 \,\mathrm{J} = 1 \,\mathrm{N} \,\mathrm{m}\)

PULLING A BLOCK

A.3 WORK, ENERGY AND POWER · SL + HL

Two ways to read the equation

\((F \cos \theta) \times s\): the component of the force along the motion, times the distance moved. Or \(F \times (s \cos \theta)\): the whole force, times the distance moved in the direction of the force. They give the same number. Use whichever is easier to see in the diagram.

What the rest of the force does

The component at right angles to the motion, \(F \sin \theta\), does no work. In the diagram it reduces the normal force, but the block stays on the table: it moves no distance in the direction of this component.

Work in physics is not effort

Holding a heavy bag still, you tire, but the force on the bag moves through no distance, so no work is done on the bag. Your muscles are transferring energy internally, which is a different matter.

Leaving out \(\cos \theta\) when the force is not along the motion, or using \(\sin \theta\) because the angle was given from the vertical.

The sign of the work

Positive, negative and zero work.

\(\theta\) less than 90°

Positive work

The force has a component along the motion. It gives energy to the body

Example: gravity on a falling ball

\(\theta\) equal to 90°

No work

The force is at right angles to the motion

Examples: the normal force on a sliding block, the centripetal force in a circle

\(\theta\) more than 90°

Negative work

The force has a component against the motion. It takes energy from the body

Example: friction, \(\cos 180^{\circ} = -1\)

A.3 WORK, ENERGY AND POWER · SL + HL

Worked example

Pulling a sledge across snow.

Rope at \(30^{\circ}\) to the ground · tension \(F = 120 \,\mathrm{N}\) The sledge moves \(s = 25 \,\mathrm{m}\) \(W = Fs \cos \theta\) \(W = 120 \times 25 \times \cos 30^{\circ}\) \(W = \class{res}{2.6 \times 10^{3} \,\mathrm{J}}\)

Friction 104 N, at constant velocity \(W_\mathrm{f} = F_\mathrm{f} s \cos 180^{\circ}\) \(W_\mathrm{f} = 104 \times 25 \times (-1) = \class{res}{-2.6 \times 10^{3} \,\mathrm{J}}\) Total work: zero

A.3 WORK, ENERGY AND POWER · SL + HL

When the force is not constant

Work from a force–displacement graph.

Work done = the area under a force–displacement graph
Constant force: a rectangle, \(F \times s\)
Varying force: count squares, or split into simple shapes
Use the component of force along the displacement

A VARYING FORCE

A.3 WORK, ENERGY AND POWER · SL + HL

Why the area

Cut the displacement into thin strips. Across one strip the force hardly changes, so the work done is the height of the strip times its width: its area. Adding the strips gives the whole area under the curve. The same argument gave displacement from a velocity–time graph in A.1 and impulse from a force–time graph in A.2.

Check the axes first

Area under force against displacement is work, in joules. Area under force against time is impulse, in newton seconds. The two graphs can look alike and mean quite different things.

Counting squares: work out what one square is worth, for example \(5 \,\mathrm{N} \times 0.2 \,\mathrm{m} = 1 \,\mathrm{J}\), then count, taking part squares in pairs.

The story · Manchester, 1845

Background story · not assessed

Joule’s paddle wheel.

Photograph of James Prescott Joule as an older man with a white beard, seated beside a table — **James Prescott Joule**, brewer and amateur scientist

Engraving of Joule’s apparatus: a weight hanging from a string that passes over a pulley and turns a spindle with paddles inside a closed vessel of water, with a thermometer in the lid — A **falling weight** turns paddles in water (engraving of 1869)

Heat was thought to be a fluid, separate from mechanical energy
Joule let a falling weight churn water and measured the tiny rise in temperature
The same work always gave the same heating
His 1850 value is within 1% of the modern one

Work and heat are both transfers of energy. The unit carries his name.

Portrait: from Henry Roscoe, 1906, public domain. Apparatus: Harper’s New Monthly Magazine, 1869, public domain.

A.3 WORK, ENERGY AND POWER · SL + HL

Quick check

Work.

01A 20 N force moves a box 3.0 m in its own direction. Work done?

\(20 \times 3.0 = 60 \,\mathrm{J}\)

02The same force acts at 60° to the motion. Work done?

\(60 \cos 60^{\circ} = 30 \,\mathrm{J}\)

03How much work does the normal force do on a sliding block?

None: it is at right angles to the motion

04Friction of 8.0 N acts over 5.0 m. Work done by friction?

\(-40 \,\mathrm{J}\)

05What does the area under a force–displacement graph give?

The work done

06Is work a scalar or a vector?

A scalar

PRESS → TO REVEAL EACH ANSWER

Part 02 · SL + HL

Kinetic and potential energy.

The three kinds of mechanical energy, each found from the work done

A.3 WORK, ENERGY AND POWER · SL + HL

Energy of motion

Kinetic energy.

\(E_\mathrm{k} = \dfrac{1}{2} m v^{2} = \dfrac{p^{2}}{2m}\)

Depends on the square of the speed
A scalar: never negative, whatever the direction
\(p = mv\) is the momentum

A resultant force \(F\) acts over a distance \(s\), from rest \(W = Fs = mas\) \(v^{2} - u^{2} = 2as\) with \(u = 0\): \(as = \dfrac{v^{2}}{2}\) \(W = \dfrac{1}{2} m v^{2}\)

A.3 WORK, ENERGY AND POWER · SL + HL

What the square means

Double the speed and the kinetic energy is four times as large. A car at 100 km h⁻¹ has four times the kinetic energy it has at 50 km h⁻¹, and with the same braking force it needs four times the distance to stop. Speed matters far more than mass.

The momentum form

Put \(v = \frac{p}{m}\) into \(\frac{1}{2} m v^{2}\) and you get \(\frac{p^{2}}{2m}\). It is the quickest route when a question gives momentum, as after a collision or an explosion. Two bodies with the same momentum do not have the same kinetic energy: the lighter one has more. That is why the bullet, not the rifle, carries nearly all the energy.

Squaring the whole of \(\frac{1}{2} m v\). Only the speed is squared.

Work done by the resultant force

Work and the change in kinetic energy.

The work done by the resultant force equals the change in kinetic energy
\(F s = \dfrac{1}{2} m v^{2} - \dfrac{1}{2} m u^{2}\)
No times, no accelerations: just the start and the end

A 1200 kg car at \(20 \,\mathrm{m} \,\mathrm{s}^{-1}\) · braking force 6000 N \(W_\mathrm{brakes} = \Delta E_\mathrm{k} = E_{\mathrm{k,f}} - E_{\mathrm{k,i}}\) \(-F s = 0 - \dfrac{1}{2} m u^{2}\) so \(s = \dfrac{m u^{2}}{2F}\) \(s = \dfrac{1200 \times 20^{2}}{2 \times 6000} = \class{res}{40 \,\mathrm{m}}\)

A.3 WORK, ENERGY AND POWER · SL + HL

The same answer the long way

With Newton’s second law, take the direction of motion as positive: \(a = \frac{-6000}{1200} = -5.0 \,\mathrm{m} \,\mathrm{s}^{-2}\), then \(v^{2} - u^{2} = 2as\) gives \(s = \frac{0 - 20^{2}}{2 \times (-5.0)} = 40 \,\mathrm{m}\). The energy method reaches it in one step, and it still works when the force is not constant, where the equations of motion do not.

Why braking distance goes as speed squared

From \(s = \frac{m u^{2}}{2F}\): at 40 m s⁻¹ the same car needs 160 m. Where the energy goes: the brake disks get hot. The brakes have done −240 kJ of work on the car, and 240 kJ of thermal energy has appeared.

Use the resultant force. If the car were also on a slope, the component of its weight along the slope would do work too.

Energy of position

Gravitational potential energy.

\(\Delta E_\mathrm{p} = mg \Delta h\)

The work done against gravity in raising a mass
Only the change in height matters, not the route
Valid close to the Earth’s surface, where \(g\) is constant

TWO ROUTES TO THE SAME HEIGHT

A.3 WORK, ENERGY AND POWER · SL + HL

Why the route does not matter

Lifted straight up, the block needs a force \(mg\) over a distance \(\Delta h\): work \(mg \Delta h\). On a smooth ramp at angle \(\theta\), the force needed is smaller, \(mg \sin \theta\), but the distance along the ramp is longer, \(\frac{\Delta h}{\sin \theta}\). Multiply them and \(\sin \theta\) cancels: \(mg \Delta h\) again. A ramp trades force for distance; it does not save energy.

Why only changes

The equation gives a change in potential energy, not an amount. You may choose any level as zero, the floor or the bench or the ground outside, because only differences in height affect the motion. Choose any convenient zero, but always use \(\Delta h = h_\mathrm{f} - h_\mathrm{i}\). A fall gives \(\Delta h < 0\) and \(\Delta E_\mathrm{p} < 0\), whatever zero you choose.

Far from the surface \(g\) is not constant, and \(E_\mathrm{p} = -\frac{G M m}{r}\) must be used: D.1, higher level.

Energy stored in a spring

Elastic potential energy.

\(E_\mathrm{H} = \dfrac{1}{2} k (\Delta x)^{2}\)

\(k\): spring constant. \(\Delta x\): extension or compression
The force grows from 0 to \(k \Delta x\): the average force is half of that
The same for stretching and for squashing

FORCE AGAINST EXTENSION

A.3 WORK, ENERGY AND POWER · SL + HL

Where the half comes from

The force needed to stretch a spring is not constant. It starts at zero and rises steadily to \(k \Delta x\) (Hooke’s law, A.2). The work done is the area under the graph, a triangle: \(\frac{1}{2} \times \Delta x \times k \Delta x\). Or think of the average force, \(\frac{1}{2} k \Delta x\), acting over the distance \(\Delta x\).

The square again

Double the extension and the stored energy is four times as large. Pulling a bowstring back twice as far stores four times the energy.

Using \(F \times \Delta x\) with the final force. That gives twice the right answer: it is the rectangle, not the triangle.

\(\Delta x\) is measured from the natural length of the spring, not from wherever it happened to start.

Worked example

A spring launcher.

\(k = 200 \,\mathrm{N} \,\mathrm{m}^{-1}\) · \(\Delta x = 0.15 \,\mathrm{m}\) · ball 0.050 kg \(E_\mathrm{H} = \dfrac{1}{2} k (\Delta x)^{2}\) \(E_\mathrm{H} = \dfrac{1}{2} \times 200 \times 0.15^{2} = \class{res}{2.25 \,\mathrm{J}}\) \(\Delta E_\mathrm{k} + \Delta E_\mathrm{H} = \dfrac{1}{2} m v^{2} - E_{\mathrm{H,i}} = 0\) \(v = \sqrt{\dfrac{2 \times 2.25}{0.050}} = \class{res}{9.5 \,\mathrm{m} \,\mathrm{s}^{-1}}\)

RELEASED FROM RESTPRESS → TO PLAY

A.3 WORK, ENERGY AND POWER · SL + HL

Why energy is the only sensible method here

The force of the spring on the ball falls as the spring expands, so the acceleration is not constant and the equations of motion cannot be used. Energy does not care how the force varies on the way: all 2.25 J stored in the spring ends up as kinetic energy of the ball, if friction and the mass of the spring are ignored.

Fired straight up instead

The same 2.25 J would become gravitational potential energy at the top: \(\Delta h = \frac{E_\mathrm{H}}{mg} = \frac{2.25}{0.050 \times 9.8} = 4.6 \,\mathrm{m}\) above the release point.

Compress the spring twice as far and the ball leaves at twice the speed: the energy is four times as large, and speed goes as the square root of energy.

Quick check

Kinetic and potential energy.

01A 2.0 kg mass moves at 3.0 m s⁻¹. Kinetic energy?

\(\frac{1}{2} \times 2.0 \times 3.0^{2} = 9.0 \,\mathrm{J}\)

02Its speed doubles. What happens to its kinetic energy?

Four times as large

03A 5.0 kg mass is raised 2.0 m. Gain in potential energy?

\(5.0 \times 9.8 \times 2.0 = 98 \,\mathrm{J}\)

04A spring with \(k = 400 \,\mathrm{N} \,\mathrm{m}^{-1}\) is stretched 0.10 m. Energy stored?

\(\frac{1}{2} \times 400 \times 0.10^{2} = 2.0 \,\mathrm{J}\)

05A body has momentum 6.0 kg m s⁻¹ and mass 3.0 kg. Kinetic energy?

\(\frac{6.0^{2}}{2 \times 3.0} = 6.0 \,\mathrm{J}\)

06What equals the work done by the resultant force?

The change in kinetic energy

PRESS → TO REVEAL EACH ANSWER

Part 03 · SL + HL

Conservation of mechanical energy.

When it holds, when it does not, and where the missing energy has gone

A.3 WORK, ENERGY AND POWER · SL + HL

Without resistive forces

Mechanical energy is conserved.

A CART ON A FRICTIONLESS TRACKKINETIC + POTENTIAL = CONSTANT

Total mechanical energy = \(E_\mathrm{k} + E_\mathrm{p} + E_\mathrm{H}\). With no frictional or resistive forces, it stays the same.

A.3 WORK, ENERGY AND POWER · SL + HL

What to watch

As the cart drops, the teal bar shrinks and the apricot bar grows by exactly the same amount. At the bottom of the dip nearly all the energy is kinetic and the cart is at its fastest. On the second hill, which is lower than the first, the cart still has some kinetic energy left at the top, so it gets over. It could never climb a hill higher than the one it started from.

The speed depends only on the height

The shape of the track between two points makes no difference to the speed at the second one. Two tracks from the same height to the same level, one steep and one gentle, deliver the cart at the same speed, though not after the same time.

Work, in this picture, is the amount of energy changed from one mechanical form to another: gravity does positive work on the way down, turning potential energy into kinetic.

Worked example

A roller coaster drops 20 m.

From rest at the top · no resistive forces \(\Delta E_\mathrm{k} + \Delta E_\mathrm{p} = 0\) \(\dfrac{1}{2} m v^{2} + mg \Delta h = 0,\quad \Delta h = -20 \,\mathrm{m}\) \(v = \sqrt{-2 g \Delta h} = \sqrt{-2 \times 9.8 \times (-20)}\) \(v = \class{res}{20 \,\mathrm{m} \,\mathrm{s}^{-1}}\)

The mass cancels: every car reaches the same speed
The shape of the track does not matter
If it starts at \(5.0 \,\mathrm{m} \,\mathrm{s}^{-1}\): \(v = \sqrt{u^{2} - 2 g \Delta h} = 20.4 \,\mathrm{m} \,\mathrm{s}^{-1}\)

A.3 WORK, ENERGY AND POWER · SL + HL

A method for every problem of this kind

Choose the system and two moments, initial and final. With no transfer to the surroundings, write \(\Delta E_\mathrm{total} = 0\). For this frictionless track, \(\Delta E_\mathrm{k} + \Delta E_\mathrm{p} = 0\). Every change is final minus initial: the kinetic-energy change is positive and the potential-energy change is negative. Substitute with those signs before rearranging.

Why starting at 5.0 m s⁻¹ adds so little

Energies add, not speeds. The starting kinetic energy, proportional to \(5.0^{2} = 25\), is small beside the \(-2 g \Delta h = 392\) gained in the fall. The final speed is \(\sqrt{25 + 392} = 20.4\), barely more than before.

Using the equations of motion along a curved track. The acceleration is not constant there. Energy is the right tool.

When mechanical energy is not conserved

With frictional and resistive forces.

Friction and drag do negative work: mechanical energy falls
Change in mechanical energy = work done by the non-conservative forces
The “lost” energy becomes thermal energy

\(\Delta E_\mathrm{total} = 0\), including thermal energy.

A 70 kg skier descends 30 m, from rest, and arrives at \(18 \,\mathrm{m} \,\mathrm{s}^{-1}\) \(W = \Delta E_\mathrm{mech} = \dfrac{1}{2} m v^{2} + mg \Delta h\) \(W = \dfrac{1}{2} \times 70 \times 18^{2} + 70 \times 9.8 \times (-30)\) \(W = 11\,340 - 20\,580 = \class{res}{-9.2 \times 10^{3} \,\mathrm{J}}\)

A.3 WORK, ENERGY AND POWER · SL + HL

Reading the answer

The skier started with 20 580 J of potential energy and finished with 11 340 J of kinetic energy. With \(\Delta h = -30 \,\mathrm{m}\), the change in mechanical energy is \(-9240 \,\mathrm{J}\), the work done by friction and air resistance. Including the thermal-energy rise gives \(\Delta E_\mathrm{total} = \Delta E_\mathrm{mech} + \Delta U_\mathrm{thermal} = 0\).

Finding the force

If the slope is 120 m long, the average resistive force is \(\frac{9240}{120} = 77 \,\mathrm{N}\). Work done by friction depends on the length of the path, unlike the work done by gravity, which depends only on the height. That difference is what “non-conservative” means.

Conservative forces (gravity, the force of a spring) store energy that can be got back. Non-conservative forces (friction, drag) turn it into thermal energy that cannot.

Quick check

Conservation of mechanical energy.

01Name the three kinds of mechanical energy.

Kinetic, gravitational potential, elastic potential

02When is the total mechanical energy conserved?

When there are no frictional or resistive forces

03A ball is dropped 5.0 m. Speed at the bottom, ignoring drag?

\(\sqrt{2 \times 9.8 \times 5.0} = 9.9 \,\mathrm{m} \,\mathrm{s}^{-1}\)

04Does that speed depend on the mass of the ball?

No: the mass cancels

05A block loses 50 J of potential energy and gains 35 J of kinetic. Work done by friction?

\(-15 \,\mathrm{J}\)

06Where has that energy gone?

To thermal energy of the surfaces

PRESS → TO REVEAL EACH ANSWER

Part 04 · SL + HL

Power and efficiency.

How fast energy is transferred, how much of it is useful, and how much a fuel holds

A.3 WORK, ENERGY AND POWER · SL + HL

The rate of doing work

Power.

\(P = \dfrac{\Delta W}{\Delta t} = Fv\)

The rate of doing work, or of transferring energy
Unit: the watt, \(1 \,\mathrm{W} = 1 \,\mathrm{J} \,\mathrm{s}^{-1}\)
\(P = Fv\) for a force along the velocity

A lift raises 800 kg through 15 m in 12 s \(P = \dfrac{\Delta W}{\Delta t} = \dfrac{mg \Delta h}{\Delta t} = \dfrac{800 \times 9.8 \times 15}{12} = \class{res}{9.8 \,\mathrm{kW}}\) A car at a steady \(30 \,\mathrm{m} \,\mathrm{s}^{-1}\) against 600 N of drag \(P = Fv = 600 \times 30 = \class{res}{18 \,\mathrm{kW}}\)

A.3 WORK, ENERGY AND POWER · SL + HL

Where \(P = Fv\) comes from

In a time \(\Delta t\) a body moving at speed \(v\) covers a distance \(v \Delta t\). A force \(F\) along the motion does work \(F v \Delta t\). Divide by \(\Delta t\): \(P = Fv\). It gives the power at one instant, even while the speed is changing.

Why cars have a top speed

At a steady speed the driving force equals the resistive force, and the engine’s power is \(Fv\). Drag rises steeply with speed, roughly as \(v^{2}\), so the power needed rises roughly as \(v^{3}\). Doubling the top speed would take about eight times the power.

Energy = power × time. A kilowatt-hour is 1 kW for one hour: \(1000 \times 3600 = 3.6 \times 10^{6} \,\mathrm{J}\). It is a unit of energy, not of power.

The story · Birmingham, 1780s

Background story · not assessed

Watt and the horse.

Painted portrait of James Watt, a gray-haired man in a dark coat, seated with a drawing of an engine in front of him — **James Watt**, painted by Carl Frederik von Breda in 1792

A large steam engine in a museum, with a wooden beam overhead, a tall vertical cylinder and a big spoked flywheel — A **Boulton & Watt engine** of 1785, now in Sydney

In 1765 Watt saw how to stop the steam engine wasting most of its fuel: a separate condenser
His customers were replacing horses, so he needed a way to compare
He settled on 33 000 foot-pounds per minute for one horse: about 746 W
In 1882 it was proposed to name the unit of power the watt

An engine is sold by its power, not its energy.

Portrait: Carl Frederik von Breda, 1792, public domain. Engine: David Maciulaitis, CC BY 2.0.

A.3 WORK, ENERGY AND POWER · SL + HL

How much is useful?

Efficiency and Sankey diagrams.

\(\eta = \dfrac{E_\mathrm{output}}{E_\mathrm{input}} = \dfrac{P_\mathrm{output}}{P_\mathrm{input}}\)

The useful output as a fraction of the total input
No unit. Always less than 1
Sankey diagram: the width of each arrow shows the energy

A PETROL ENGINE\(\eta = 0.30\)

A.3 WORK, ENERGY AND POWER · SL + HL

Reading a Sankey diagram

Energy flows from left to right. The input arrow is 100 units wide. The useful output carries straight on; the wasted energy peels away, usually downwards. The widths of all the outputs must add up to the width of the input: that is conservation of energy, drawn to scale. Here \(30 + 70 = 100\), and the efficiency is \(\frac{30}{100} = 0.30\), or 30%.

Why it is never 1

Some energy always goes to thermal energy through friction, electrical resistance or hot exhaust gases. That energy still exists, but it has spread out into the surroundings and cannot be gathered up to do useful work. B.4 explains at higher level why engines that run on heat have a limit well below 100%.

When drawing one: choose a scale such as 1 mm for 1 J, keep the widths in proportion and label each arrow with its form of energy.

Giving an efficiency greater than 1, or 100%. If that happens, the input and output have been swapped.

Worked example

The efficiency of an electric motor.

Input power 500 W It lifts 20 kg at a steady \(1.5 \,\mathrm{m} \,\mathrm{s}^{-1}\) \(P_\mathrm{output} = Fv = mgv\) \(P_\mathrm{output} = 20 \times 9.8 \times 1.5 = 294 \,\mathrm{W}\) \(\eta = \dfrac{P_\mathrm{output}}{P_\mathrm{input}} = \dfrac{294}{500} = \class{res}{0.59}\)

At a steady speed the lifting force equals the weight
As a percentage: 59%
Wasted: \(500 - 294 = 206 \,\mathrm{W}\), as thermal energy

A.3 WORK, ENERGY AND POWER · SL + HL

How much energy a fuel holds

Energy density of fuels.

Specific energy: energy released per unit mass, in J kg⁻¹
Energy density: energy released per unit volume, in J m⁻³
High values: less fuel to carry and to store

Hydrogen: the most energy per kilogram, very little per cubic meter.

Fuel (approximate values)	MJ kg⁻¹	MJ m⁻³
Wood, dry	16	\(1 \times 10^{4}\)
Petrol	46	\(3.4 \times 10^{4}\)
Natural gas, as a gas	55	38
Hydrogen, as a gas	142	12
Uranium-235, fission	\(8 \times 10^{7}\)	\(1.5 \times 10^{12}\)

A.3 WORK, ENERGY AND POWER · SL + HL

Why both numbers matter

An aircraft cares most about mass: every kilogram of fuel has to be lifted. A car or a ship cares about volume as well: the fuel has to fit in a tank. Petrol does well on both counts, which is a large part of why it has been so hard to replace.

The trouble with gases

A kilogram of hydrogen releases three times the energy of a kilogram of petrol, but at atmospheric pressure it fills about 11 cubic meters. To be useful as a fuel it must be compressed to hundreds of atmospheres or cooled to a liquid at −253 °C, and both take energy and heavy tanks.

Nuclear fuel

Fission releases energy from the nucleus, not from the electrons, so the scale is different altogether: one kilogram of uranium-235 gives as much as about two thousand tonnes of petrol.

Energy released = specific energy × mass, or energy density × volume. Check which of the two a question gives you by its unit.

Worked example

How much coal does a power station burn?.

Electrical output 500 MW · efficiency 0.35 Specific energy of coal \(30 \,\mathrm{MJ} \,\mathrm{kg}^{-1}\) \(P_\mathrm{input} = \dfrac{P_\mathrm{output}}{\eta} = \dfrac{500}{0.35} = 1430 \,\mathrm{MW}\) Mass each second \(= \dfrac{P_\mathrm{input}}{\text{specific energy}}\) \(= \dfrac{1430 \times 10^{6}}{30 \times 10^{6}} = \class{res}{48 \,\mathrm{kg} \,\mathrm{s}^{-1}}\)

Power is energy each second, so this gives mass each second
In a day: \(48 \times 86\,400 = 4.1 \times 10^{6} \,\mathrm{kg}\)
About 4000 tonnes: several long trains of coal, every day

A.3 WORK, ENERGY AND POWER · SL + HL

Quick check

Power and efficiency.

01600 J of work is done in 4.0 s. Power?

\(\frac{600}{4.0} = 150 \,\mathrm{W}\)

02A boat moves at 5.0 m s⁻¹ against 400 N of drag. Engine power?

\(400 \times 5.0 = 2.0 \,\mathrm{kW}\)

03A lamp takes in 60 W and gives out 9 W of light. Efficiency?

\(\frac{9}{60} = 0.15\)

04What does the width of an arrow on a Sankey diagram show?

The amount of energy, or power

052.0 kg of fuel with specific energy 45 MJ kg⁻¹ is burned. Energy released?

90 MJ

06What is the unit of energy density?

J m⁻³

PRESS → TO REVEAL EACH ANSWER

Summary

Work, energy and power on one slide.

Work

\(W = Fs \cos \theta\)

A transfer of energy
Area under a force–displacement graph

Kinetic energy

\(E_\mathrm{k} = \frac{1}{2} m v^{2} = \frac{p^{2}}{2m}\)

Work by the resultant force changes it

Potential energy

\(mg \Delta h\) and \(\frac{1}{2} k (\Delta x)^{2}\)

Gravitational, near the surface
Elastic, in a spring

Conservation

Mechanical energy constant

If there are no resistive forces
Otherwise the change is the work they do

Power

\(P = \frac{\Delta W}{\Delta t} = Fv\)

The rate of transferring energy, in watts

Efficiency and fuels

\(\eta = \frac{E_\mathrm{output}}{E_\mathrm{input}}\)

Sankey diagrams show it to scale
Energy density: J m⁻³

A.3 WORK, ENERGY AND POWER · SL + HL

Summary

Key words.

WorkForce × distance along the force

Kinetic energyEnergy of a moving body

Gravitational potential energyEnergy of a raised mass

Elastic potential energyEnergy of a stretched spring

Mechanical energyKinetic + potential

Non-conservative forceFriction or drag: path matters

PowerRate of transferring energy

EfficiencyUseful output as a fraction of input

Sankey diagramArrow widths show energy

Specific energyEnergy per kilogram of fuel

Energy densityEnergy per cubic meter of fuel

Isolated systemNothing enters or leaves

A.3 WORK, ENERGY AND POWER · SL + HL