Energy, Work and Power

Syllabus coverage

understand the concept of the work done by a force, and calculate the work done by a constant force when its point of application undergoes a displacement not necessarily parallel to the force, using $W = F d \cos θ$ ; the scalar product is not required
understand the concepts of gravitational potential energy and kinetic energy, and use the appropriate formulae
understand and use the relationship between the change in energy of a system and the work done by the external forces, and use in appropriate cases the principle of conservation of energy, including cases where the motion may not be linear, such as a child on a smooth curved slide
use the definition of power as the rate at which a force does work, and use the relationship between power, force and velocity for a force acting in the direction of motion, including average power as work done divided by time taken, and $P = F v$
solve problems involving, for example, the instantaneous acceleration of a car moving on a hill against a resistance

Part of the AS & A Level Maths syllabus 9709.

Work done against friction is force times distance along the slope, while the gain in gravitational potential energy depends only on the vertical height risen. On the hill the engine works against the component

m g \sin θ

of the weight down the slope.

Work done by a force

A force does work when it moves its point of application. If a constant force $F$ moves a body through a displacement $d$ in a straight line, and the angle between the force and the displacement is $θ$ , then the work done is $W = F d \cos θ$ . The unit is the joule: 1 J is the work done by a force of 1 N moving a body 1 m in the direction of the force.

The formula is a statement about the component of the force in the direction of the displacement. A force at 60° to the direction of motion does only half as much work as the same force along it, because $\cos 60^{\circ} = \frac{1}{2}$ . An equivalent reading, useful when the displacement is not along the force, is force times the component of the displacement in the direction of the force - the two give the same number.

The angle $θ$ is always the angle between the force and the displacement, never the angle between the force and something else in the diagram. Three cases cover most questions: $θ = 0$ , so the force is along the motion and $W = F d$ ; $θ = 90^{\circ}$ , so the force is perpendicular to the motion and $W = 0$ ; and $θ = 180^{\circ}$ , so the force opposes the motion and $W = - F d$ .

A negative value of work means the force is taking energy out of the body rather than putting it in. Friction and air resistance always do negative work, which is why a body sliding on a rough surface loses kinetic energy.

Angle $θ$	Work done	What it means
0°	$F d$	the force drives the motion
90°	0	the force is perpendicular to the motion
180°	$- F d$	the force opposes the motion
between 0° and 90°	$F d \cos θ$	the force drives the motion, but only partly

A normal reaction does no work on a body moving along a surface, because it is perpendicular to the displacement. Neither does the weight of a body moving horizontally. A force does no work only when its component along the motion is zero.

Kinetic energy

Kinetic energy is the energy a body has because of its motion: $K E = \frac{1}{2} m v^{2}$ where $m$ is the mass in kg and $v$ is the speed in $m s^{- 1}$ . It is measured in joules, and it is never negative because $v^{2}$ is never negative.

The formula follows from $F = m a$ and $v^{2} = u^{2} + 2 a s$ . From $v^{2} = u^{2} + 2 a s$ , $a s = \frac{v^{2} - u^{2}}{2}$ , so the work done by the resultant force is $F s = m a s = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$ : the work done by the resultant force is the change in kinetic energy.

Kinetic energy depends on the square of the speed, so doubling the speed multiplies the kinetic energy by four. That is why a car's braking distance at 60 mph is about four times its braking distance at 30 mph: the brakes must remove four times as much energy, and they do work at a roughly constant rate.

A body at rest has zero kinetic energy. If a question says a particle 'starts from rest', then $u = 0$ and the initial kinetic energy is zero, which usually removes a term from the energy equation.

a 2 kg body moving at $3 m s^{- 1}$ has $\frac{1}{2} \times 2 \times 9 = 9 J$
a change in kinetic energy is measured in J and can be negative
the mass never cancels from a kinetic-energy calculation on its own, but it often cancels when it meets $m g h$

Speed, not velocity, goes into

\frac{1}{2} m v^{2}

- kinetic energy is a scalar and is always positive, even when the body is moving in the negative direction.

Gravitational potential energy

Gravitational potential energy is the energy a body has because of its height. Raising a mass $m$ through a vertical height $h$ against its weight $m g$ takes work $m g h$ , and that energy is stored: $G P E = m g h$ measured from whichever level you choose as zero.

The zero level is your choice, and only changes in potential energy ever appear in an equation, so the choice never affects an answer. It is tidiest to put the zero at the lowest point of the motion, so that every height is positive.

The height $h$ is always the vertical height, never the distance along a slope. A block moving 5 m up a plane inclined at 30° rises $5 \sin 30^{\circ} = 2.5 m$ , so it gains $m g \times 2.5$ joules, not $m g \times 5$ .

A body falling a vertical distance $h$ loses $m g h$ of potential energy, and this does not depend on the path. Falling, sliding down a smooth slope and sliding down a curved slide all lose the same energy for the same vertical drop.

Height is measured vertically. Using the distance along a slope in

m g h

is the commonest single error in this topic, and it makes the answer too big by a factor of

\frac{1}{\sin θ}

The work-energy principle

The work done by the resultant force on a body equals the change in its kinetic energy: $W_{resultant} = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$ . This is the work-energy principle, and it is just $F = m a$ integrated along the motion.

The wider statement, and the one the syllabus asks for, is about all the external forces together: the work done by the external forces equals the change in the total energy of the system, kinetic plus potential. Written out for a body moving from a height $h_{1}$ to a height $h_{2}$ : $W_{external} = (\frac{1}{2} m v_{2}^{2} + m g h_{2}) - (\frac{1}{2} m v_{1}^{2} + m g h_{1})$ . The quantity in brackets is the mechanical energy of the body.

It is often easier to think in terms of separate forces. The weight is built into the potential-energy terms, so it should not also be put into $W_{external}$ . Every other force - an applied pull, a tension, a resistance, friction - does work that goes into $W_{external}$ , with a minus sign when the force opposes the motion.

So for a body sliding on a rough surface, the loss in mechanical energy equals the work done against friction. That energy has not disappeared - it has gone into heating the surfaces.

no resistance: mechanical energy is constant, and the equation is $\frac{1}{2} m u^{2} + m g h_{1} = \frac{1}{2} m v^{2} + m g h_{2}$
with resistance: $\frac{1}{2} m u^{2} + m g h_{1} - F d = \frac{1}{2} m v^{2} + m g h_{2}$ , where $F d$ is the work done against the resistance over the distance $d$ actually travelled
with friction on a horizontal surface: $\frac{1}{2} m u^{2} - μ m g d = \frac{1}{2} m v^{2}$

Work done against friction is

F \times

the distance along the surface, not the vertical height. On a slope the two are different, and mixing them up is the second commonest error here.

Conservation of energy

When the only forces doing work are the weight and forces perpendicular to the motion, no mechanical energy is lost, and the principle of conservation of energy says that kinetic plus potential energy stays constant. The equation is $\frac{1}{2} m u^{2} + m g h_{1} = \frac{1}{2} m v^{2} + m g h_{2}$ and the mass appears in every term, so it cancels: the answer never depends on the mass of the body.

That cancellation is the reason a heavy sledge and a light one reach the same speed at the bottom of the same smooth slope, and the same speed as a stone dropped from the same height.

The syllabus singles out motion that is not in a straight line - a child on a smooth curved slide, for example. The energy method still works, because only the overall change in height matters: the shape of the slide affects the direction of the normal reaction but never the work it does, which is zero throughout, since the reaction is perpendicular to the surface at every point. So the speed at the bottom is $v = \sqrt{2 g h}$ for a vertical drop $h$ , whatever the slide looks like.

When a resistance is present the energy is not lost but transferred: initial mechanical energy = final mechanical energy + energy dissipated against resistance.

The principle fails as stated if you forget a force that is doing work - a driving force, a pull, or a resistance. List every force doing work along the path before writing the equation.

the mass cancels when only weight and reactions act, so do not panic if the question never gives it
the normal reaction does no work, so a smooth curved path is no harder than a straight one
resistance does negative work: $m g h$ becomes less than the gain in kinetic energy, by exactly $F \times$ distance travelled

Energy is conserved in every case; what changes is whether the mechanical part of it is conserved. With friction, mechanical energy falls and thermal energy rises by the same amount.

Power

Power is the rate at which work is done, and it is measured in watts. One watt is one joule per second, $1 W = 1 J s^{- 1}$ . Average power over an interval is $P = \frac{work done}{time taken}$ and for a force acting in the direction of motion at a given instant, $P = F v$ where $F$ is the force and $v$ the speed at that instant.

The two formulas answer different questions. $P = \frac{W}{t}$ is about an interval of time and is used when the question gives a distance, a time and asks for a rate - an engine lifting a load, a pump raising water. $P = F v$ is about a single instant, and it is used when the question gives a speed and asks for a force, or a force and asks for a speed.

At constant speed on a level road the driving force equals the total resistance, so $P = R v$ , where $R$ is the total resistance. At higher speed the same power gives a smaller driving force, which is why a car cannot accelerate as hard at 70 mph as it can at 30 mph.

Watch the units. Engine powers are often given in kW, so $9 k W = 9000 W$ and $40 k W = 40000 W$ . Converting is a step worth writing down, because forgetting it changes the answer by a factor of a thousand.

The maximum speed is the speed at which the driving force has fallen, through $P = F v$ , to balance the resistance.

average power: $P = \frac{W}{t}$ , with $W$ the work done in that time
instantaneous power: $P = F v$ for a force along the motion
constant speed on a level road: $P = R v$
maximum speed on a hill: $P = (R + m g \sin θ) v_{max}$

P = F v

needs the force along the motion. If the force acts at an angle, use its component in the direction of motion,

F \cos θ

Cars on hills and against resistance

A car on a hill has three forces along the line of the slope: the driving force $F$ up the hill, the resistance $R$ down the hill, and the component of the weight $m g \sin θ$ down the hill. Newton's second law along the slope gives $F - R - m g \sin θ = m a$ .

The driving force is usually not given directly. It comes from the engine power through $P = F v$ , so $F = \frac{P}{v}$ at the instant the car is travelling at speed $v$ . That is the step that makes these questions work: power and speed give the force, and the force with $F = m a$ gives the acceleration.

At constant speed on the hill the acceleration is zero, so $F = R + m g \sin θ$ : the engine is supplying exactly enough force to overcome the resistance and hold the car up the slope. The work the engine does against gravity is not wasted - it is stored as potential energy.

For the instantaneous acceleration, put the driving force from $P = F v$ into $F - R - m g \sin θ = m a$ and solve. As the car speeds up, $F = \frac{P}{v}$ falls, so the acceleration falls with it; that is why the acceleration a question asks for is the acceleration at one particular speed.

The resistance is modelled as a constant force in this component, so use the value the question gives.

find the driving force from $F = \frac{P}{v}$
resolve along the slope: $F - R - m g \sin θ = m a$
constant speed: $a = 0$ , so $F = R + m g \sin θ$ and $P = (R + m g \sin θ) v$
power in watts, force in newtons, speed in $m s^{- 1}$ - convert kW before you start

Do not put the weight into the equation of motion as well as using

m g \sin θ

. The component down the slope is the only part of the weight that acts along the motion, and

m g \cos θ

is balanced by the normal reaction.

Choosing between energy and $F = m a$

Both methods are correct and both earn full marks; the marks come from using one of them properly. Energy methods are quicker whenever the question gives speeds, heights and distances and asks for a speed, a distance or an amount of energy. A curved path, a change of direction, or a question that never mentions the forces are all signals to use energy.

Use $F = m a$ when the question asks for a force, an acceleration or a time, or when a resistance changes during the motion. Energy methods cannot give you a time, because they compare two instants and say nothing about what happened between them.

Many questions need both: a car climbing a hill uses $P = F v$ for the force and $F = m a$ for the acceleration.

Whichever you choose, write down the principle you are using before substituting numbers. A line such as 'loss in potential energy = gain in kinetic energy + work done against resistance' is worth a mark on its own, and it is the line that shows the marker which equation you meant to write.

speeds, heights, distances, energies: use energy
forces, accelerations, times: use $F = m a$
state the energy balance in words first, then substitute
check that every force doing work appears exactly once

Energy methods never give a time. If the last line of the question asks for a time, you will need the constant-acceleration formulae with an acceleration you have already found - or found from

F = m a

The quantities and their units

Every number in this topic carries a unit, and the units are the quickest check on a calculation. Work and every form of energy are in joules, power is in watts, force is in newtons, mass in kilograms and speed in metres per second.

Two conversions are worth knowing by heart: $1 k W = 1000 W$ and $1 M J = 10^{6} J$ . An answer in kW is fine if the question asks for it, but mixing kW and W in the same calculation is not.

A watt is a joule per second, so power is an energy divided by a time - exactly what $P = \frac{W}{t}$ says.

Quantity	Symbol	Unit	Formula
work done by a force	$W$	J	$F d \cos θ$
kinetic energy	$K E$	J	$\frac{1}{2} m v^{2}$
gravitational potential energy	$G P E$	J	$m g h$
power	$P$	W	$\frac{W}{t}$ or $F v$
force	$F$	N	$m a$
speed	$v$	$m s^{- 1}$	$\frac{P}{F}$ when $P = F v$

If an answer comes out a thousand times too big or too small, you have almost certainly left a power in kW or a mass in grams. Check the units before you check the algebra.

Key terms you must be able to define

Work done by a force: The energy transferred by a force moving its point of application: $W = F d \cos θ$ , where $θ$ is the angle between the force and the displacement.

Joule: The unit of work and energy. One joule is the work done by a force of 1 N moving a body 1 m in the direction of the force.

Kinetic energy: The energy a body has because of its motion: $\frac{1}{2} m v^{2}$ , measured in joules. It is never negative.

Gravitational potential energy: The energy a body has because of its height above a chosen zero level: $m g h$ , with $h$ the vertical height.

Mechanical energy: Kinetic energy plus gravitational potential energy. It is constant when no resistance or friction acts.

Work-energy principle: The work done by the resultant force equals the change in kinetic energy of the body.

Principle of conservation of energy: Energy is neither created nor destroyed, only transferred. When only the weight does work, kinetic plus potential energy is constant.

Work done against resistance: The energy dissipated by a force opposing the motion: the resistance times the distance travelled along the surface.

Zero level: The height chosen as the zero of gravitational potential energy. Only changes in potential energy matter, so the choice is free.

Power: The rate at which work is done: $P = \frac{W}{t}$ on average, and $P = F v$ for a force acting in the direction of motion.

Watt: The unit of power: one joule per second, $1 W = 1 J s^{- 1}$ . $1 k W = 1000 W$ .

Average power: The total work done divided by the time taken, for the whole interval.

Instantaneous power: The rate of working at a particular instant, given by $P = F v$ with the force and the speed at that instant.

Driving force: The force the engine of a vehicle supplies along the direction of motion. It is found from $F = \frac{P}{v}$ when the power is known.

Resistance: A force opposing motion, such as air resistance. It does negative work, removing mechanical energy from the body.

Maximum speed: The speed at which the driving force has fallen, through $P = F v$ , to equal the total resistance, so the acceleration is zero.

Efficiency: The useful energy or power output as a fraction of the total input. It has no units and is often given as a percentage.

Applied

At AS & A Level the marks sit in the setup: choosing the model, stating the variable and writing the equation before any working. These show the shape of a full-mark answer, then you do the algebra yourself.

Work done at an angle

A force of $40 N$ acts at 60° to the direction in which a body moves $6 m$ . The work done is $40 \times 6 \times \cos 60^{\circ} = 40 \times 6 \times 0.5 = 120 J$ . Only the component of the force along the motion does work; the perpendicular component does none.

Work done against friction

A box of mass $10 k g$ is pulled $8 m$ across rough horizontal ground by a horizontal force of $50 N$ against a constant resistance of $30 N$ . The pull does $50 \times 8 = 400 J$ of work, the resistance does $- 30 \times 8 = - 240 J$ , so the gain in kinetic energy is $400 - 240 = 160 J$ , and the speed after 8 m is $\sqrt{\frac{2 \times 160}{10}} = 5.66 m s^{- 1}$ .

A smooth curved slide

A child of mass $30 k g$ slides from rest down a smooth slide whose top is $2 m$ above the bottom. The normal reaction is perpendicular to the slide at every point, so it does no work, and $m g h = \frac{1}{2} m v^{2}$ gives $10 \times 2 = \frac{1}{2} v^{2}$ , so $v = \sqrt{40} = 6.32 m s^{- 1}$ . The shape of the slide and the mass of the child both cancel out.

Power at a steady speed

A car travels at a constant $25 m s^{- 1}$ along a level road against a total resistance of $800 N$ . At constant speed the driving force equals the resistance, so the power developed is $P = F v = 800 \times 25 = 20000 W = 20 k W$ .

Acceleration on a hill

A car of mass $1000 k g$ moves up a hill inclined at $θ$ to the horizontal, where $\sin θ = 0.08$ , at $15 m s^{- 1}$ . The engine works at $30 k W$ and the resistance is $400 N$ . The driving force is $\frac{30000}{15} = 2000 N$ , and the component of the weight down the hill is $1000 \times 10 \times 0.08 = 800 N$ . Then $2000 - 400 - 800 = 1000 a$ , so $a = 0.8 m s^{- 2}$ .

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What is the work done by a constant force

F

moving a body a distance

d

at an angle

θ

to the force?

W = F d \cos θ

joules. Only the component of the force along the displacement does work.

Q2How much work does a force perpendicular to the motion do?

None, because

\cos 90^{\circ} = 0

. That is why a normal reaction does no work on a body sliding along a surface.

Q3What is the kinetic energy of a body of mass

m

moving at speed

v

\frac{1}{2} m v^{2}

joules. It depends on the square of the speed, so doubling the speed quadruples the kinetic energy.

Q4What is the gain in gravitational potential energy when a mass

m

rises a vertical height

h

m g h

joules. The height is always vertical, never the distance along a slope.

Q5State the work-energy principle.

The work done by the resultant force equals the change in kinetic energy:

W = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}

Q6How do you write the work done by the external forces in terms of energy?

Work done by the external forces equals the change in kinetic energy plus the change in gravitational potential energy.

Q7A child slides from rest down a smooth curved slide. Why is the speed at the bottom independent of the shape of the slide?

The normal reaction is perpendicular to the slide everywhere and so does no work. Only the vertical drop matters, and

m g h = \frac{1}{2} m v^{2}

gives

v = \sqrt{2 g h}

Q8A sledge slides down a rough slope. How do you use energy to find its speed at the bottom?

m g h = \frac{1}{2} m v^{2} + F d

, where

F

is the friction and

d

is the distance slid along the slope.

Q9Define power.

The rate at which work is done:

P = \frac{W}{t}

on average, measured in watts.

1 W = 1 J s^{- 1}

Q10What is the relationship between power, force and speed?

P = F v

, for a force acting in the direction of motion. It is an instantaneous relationship, at one speed.

Q11A car moves at constant speed along a level road. What power is it developing?

P = R v

, where

R

is the total resistance, because the driving force equals the resistance when the acceleration is zero.

Q12How do you find the driving force of a car from its engine power?

F = \frac{P}{v}

, with the power in watts. The driving force falls as the speed rises for a fixed power.

Q13Write the equation of motion for a car climbing a hill against a resistance.

F - R - m g \sin θ = m a

, where

F

is the driving force up the hill and

R

the resistance.

Q14When can energy methods not be used on their own?

When the question asks for a time, or for a force or acceleration that changes during the motion. Energy methods compare two instants only.

0 of 14 marked as known

Exam technique for this topic

Write down the formula you are using and the value of $g$ before substituting. In this component take $g = 10 m s^{- 2}$ unless the question says otherwise.
In $W = F d \cos θ$ , the angle is the one between the force and the displacement. If the force is at 60° to the horizontal and the body moves horizontally, then $θ = 60^{\circ}$ , and the work is $F d \cos 60^{\circ}$ .
In $m g h$ the height is vertical. Moving 5 m up a plane at 30° raises the body $5 \sin 30^{\circ} = 2.5 m$ , so the gain in potential energy is $m g \times 2.5$ .
State your energy balance in words before the algebra - 'loss in potential energy = gain in kinetic energy plus work done against the resistance' - then write it out. The line itself often earns a mark.
A smooth curved path is handled by heights alone, because the normal reaction does no work; that is a signal to use energy methods.
Convert kW to W before using $P = F v$ , and check whether the question wants a force, an acceleration or a power. Give final answers to three significant figures with the unit.

Why students lose marks here

A student wroteThe block moves 5 m up a plane at 30°, so its gain in potential energy is

m g \times 5

Why it loses marks

h

m g h

is the vertical height, not the distance along the slope. The block rises

5 \sin 30^{\circ} = 2.5 m

The answer that scoresGain in potential energy

= m g \times 5 \sin 30^{\circ} = 2.5 m g

. The distance along the plane is only used for the work done against friction.

A student wroteA force of 50 N at 60° to the motion over 8 m does

50 \times 8 = 400 J

of work.

Why it loses marksOnly the component of the force along the motion does work, and that component is

50 \cos 60^{\circ} = 25 N

. The other component is perpendicular to the displacement and does no work.

The answer that scores

W = F d \cos θ = 50 \times 8 \times \cos 60^{\circ} = 200 J

A student wroteThe normal reaction pushes the block up the slope, so it does work as well as the weight.

Why it loses marksThe normal reaction is perpendicular to the surface, and the block moves along the surface, so the angle between the force and the displacement is 90°.

\cos 90^{\circ} = 0

The answer that scoresThe normal reaction does no work. On a smooth slide it does not appear in the energy equation at all, which is why a curved smooth slide is no harder than a straight one.

A student wroteA sledge slides down a rough slope, so

m g h = \frac{1}{2} m v^{2}

Why it loses marksFriction has done work against the motion, so some of the potential energy has gone into heating the surfaces. The speed at the bottom is less than

\sqrt{2 g h}

The answer that scores

m g h = \frac{1}{2} m v^{2} + F d

, where

F

is the friction and

d

is the distance slid along the slope. The missing energy is exactly

F d

A student wroteA car develops 30 kW at 15 m/s, so the driving force is

30 \times 15 = 450 N

Why it loses marks

P = F v

rearranges to

F = \frac{P}{v}

, not

P v

. The power must also be in watts, so 30 kW is 30000 W.

The answer that scores

F = \frac{30000}{15} = 2000 N

. Multiplying instead of dividing gives 450 N, less than a quarter of the force actually needed, so the size of the answer alone should send you back to the formula.

A student wroteThe engine works at 9 kW, so the power is 9 W.

Why it loses marksA kilowatt is a thousand watts. Leaving the power in kW changes every force and acceleration that follows from it.

The answer that scores

9 k W = 9000 W

. Convert before using

P = F v

P = \frac{W}{t}

, and keep one unit for the whole calculation.

Syllabus coverage

Work done by a force

Kinetic energy

Gravitational potential energy

The work-energy principle

Conservation of energy

Power

Cars on hills and against resistance

Choosing between energy and $F = m a$

The quantities and their units

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here

Energy, Work and Power

Syllabus coverage

Work done by a force

Kinetic energy

Gravitational potential energy

The work-energy principle

Conservation of energy

Power

Cars on hills and against resistance

Choosing between energy and F=ma

The quantities and their units

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here

Choosing between energy and $F = m a$