Know and understand that energy can be
transferred usefully from one form to another, or stored, or dissipated, but
energy cannot be created or destroyed.
This is the law of
conservation of energy.
Another way of expressing this law to say
energy is never lost but transferred between different energy stores
often involving different objects or materials.
However, energy is only useful
if it can be converted from one form to another.
Be careful in using the term energy
loss!
The phrase 'energy loss' is used in the
context of energy transfer when not all the energy is transferred into a
useful form e.g. some energy from a car fuel is 'lost' in the friction of
moving parts, i.e. some chemical energy ends up as heat or sound rather than
kinetic energy to move the car.
Examples  from a suitable
energy source ==> useful form of energy (plus waste in most cases)
Energy is usually transferred by
radiation, heating or doing work in a mechanical sense.
When a gun fires chemical energy
is converted into heat energy, sound energy, light energy and mainly kinetic
energy. When the bullet embeds itself into some material the kinetic energy
of movement is converted into some sound energy, but mainly heat energy.
Some of the energy is wasted but most of the chemical energy stored in
cartridge is converted to the useful kinetic energy of the bullet.
Photovoltaic
solar panels convert light energy into electrical energy. If this
electrical energy is used to charge up a battery then you increase
chemical energy store of the battery.
We
use a large number of electrical devices in the home eg
TV converts electrical energy
into useful light and sound, but some waste heat
A charged mobile phone battery
converts chemical energy into electrical energy, which in turn is converted
into useful light and sound energy. In the charging process you are
increasing the energy store of the mobile phone battery.
Wind
turbines convert kinetic energy into electrical energy
The kinetic energy store of the
wind is decreased, whereas that of the turbine blades is increased,
initially an energy transfer involving the same type of energy
store.
The start of making a cup of tea
is a system. You use an electrical heating element in the kettle
base to boil water by transferring energy to water via the conversion of
electrical energy to heat. The water increases in temperature and
therefore its thermal energy store is increased. The water and heating
element constitute a system.
Whenever an electrical current
flows in a circuit, work is done against the resistance of the wire.
This is all to do with the behaviour of the electrons moving due to
an electrical potential difference.
It doesn't seem the same as say,
pushing a lever to operate the machine, but you are applying a force
to move the lever against a mechanical resistance. In fact the work
done or energy transferred is equal to the force applied x the
distance through which the force operates (see calculations further
down the page).

Work done, power and wasted energy
If an object begins to move or changes
how it moves (change in speed or direction) then a force must be involved.
If a force acts on an object, then work
is being done on the object increasing its energy store  this must involve
an energy transfer.
If an object exerts a force or
transfers energy in some way, work is done by this object and energy is
transferred from its energy store.
The energy can be transferred usefully
i.e. doing useful work, but some energy might (often) be wasted  often
dissipated to the surroundings as heat due to friction.
The energy transferred,
usefully or wastefully, is often referred to, and equal to, work done.
When dealing with the action of a
resultant force acting through a linear distance in the direction of the
force, you can calculate the work done with a very simple formula.
When ever you or a machine moves
something work is done.
Work done (J) = energy
transferred = force (N) x distance (m)
W (J) = F (N) x d (m)
If you apply a force of one
newton through a linear distance of one metre, you do one joule of
work.
one joule = force of one newton x
one metre distance,
1 J = 1 Nm
(note that newton metres =
joules = work done = energy transferred)
The distance must be along the
line of action of the force.
You need an energy source to keep
applying any force for any length of time.
Whenever work is done e.g.
mechanically or electrically, energy is transferred from one energy
store to another.
See
work calculation questions
Power is the rate at which energy
is transferred or 'used up'
In other words power is the
rate of doing work.
The unit of power is the watt,
W.
power (W) = energy
transferred (J) / time taken (s)
P (W) = E (J) / t (s)
A power rating of one watt means
one joule of energy is transferred per second,
or, 1 joule of work is done in 1
second.
The joule is the unit of energy
and the watt is the unit of power (W).
one watt = one joule of energy is
transferred per second,
1 W = 1 J/s
A 1 kW electric heater converts
1000 J of electrical energy ==> heat energy every second
(1 kW = 1000 W =
1000 J/s)
See
work calculation questions
The power of a machine or any
other device is the rate at which it transfers energy.
A powerful machine doesn't
necessary mean a bigger force is generated, it just means a lot
of energy is transferred in a short time (or a lot of work done
in a short time).
Conversely, by using gears
and a low powered electric motor, you can generate a relatively
big force. So, don't confuse power and force  use the terms
appropriately.
See
Turning
forces and moments including gears
However, no mechanical device or
electrical appliance is 'perfect' and energy is lost e.g. by thermal
energy by friction or sound from unwanted vibrations, heat loss from
circuit wires etc., but it is still part of the total work done  it
just isn't all useful!
Therefore you should know and understand that when energy is transferred
only part of it may be usefully transferred as useful energy or
useful work, the rest is ‘wasted’ to the surroundings
You should know and understand that wasted energy is eventually
transferred to the surroundings, which will become warmer AND ...
... the wasted energy
becomes increasingly spread out and so becomes less useful.
More examples of energy
stores, energy transfers and doing work
Doing work in a mechanical sense is just
another set of examples of energy transfer e.g. from one energy store to
another.
This usually involves doing useful work,
but the final energy store destination might have no further use.
For example when you burn fuel in a
car engine the energy changes are as follows.
fuel + oxygen (chemical energy store)
==> hot gases (thermal energy store) ==> car moves (kinetic energy
store)
In the process some of the original
chemical energy store of the system is lost through friction and sound.
BUT, all the energy from the chemical
energy store ends up as heat to warm up the surroundings. Therefore the
thermal energy store of the environment is increased but its now of no
use whatsoever!
The waste of energy increases (but
necessarily!) when you apply the brakes. Work is being done in the
process. The friction effect of the brake pad on the brake disc converts
kinetic energy into heat. The kinetic energy store of the car decreases
(fortunately!) and the thermal energy store of the brake pad + disc
increases. Eventually as the braking system cools down the heat energy
is transferred to increase the thermal energy store of the surroundings
 wasted or degraded energy!
When you lift an object up you are
doing work against the effects of gravity.
The chemical energy stored in your
muscles is decreased as some of it used to lift the objectweight which
on gaining height increases its gravitational potential energy store.
In fact the work you do (J) equals
the weight of the object (N) x vertical distance lifted (m)
When you kick a ball up in the air,
three energy stores should be fairly obvious to you.
chemical energy store of muscles ==>
kinetic energy store of ball ==> gravitational potential energy store of
ball
AND, when the ball falls (ignoring a
bit of lost sound energy)
gravitational potential energy store
of ball ==>`kinetic energy store of ball ==> thermal energy store of the
ground

(1) Examples of simple
mechanical work done calculation questions
AND (2) Examples
of simple power calculation questions
How to solve 'work done' problems.
The formula for work done is quite
simple:
work done in joules = force in newtons x
distance in metres in the line through which the force acts
(1) work done (J) = force (N) x distance
through which force acts (m)
E = F x d
How to solve power problems  just a few
simple 'mechanical' examples here..
Power is the rate at which energy is
transferred, that is the rate at which work is done.
power in watts = (work done = energy
transferred) ÷ time taken
(2) P (W) = E (J) / t (s)
Power P in Watts, Energy E
in Joules, time t in seconds
The power rating of 1 watt is equal to a work rate of 1 joule
of energy transferred per second
See also
FORCES 3. Calculating resultant forces using vector
diagrams and work done gcse physics revision notes
(12) Examples of problem solving using
the 'work done' and 'work rate = power' formulae
In every case watch the units e.g.
W/kW, J/kJ/MJ, min/secs etc.
12
Q1
If you drag a heavy box with a force of 200 N across a floor for 3 m,
(a) what
work is done?
work done = 200 x 3 =
600 J
(b) If you do the job in 5 seconds,
what was your power rating in doing this task?
power (W) = work done (J) / time
taken (s)
your power = 600 / 5 =
120 W
(just over the power of 100 W light bulb!)
12
Q2 (a) If a machine part
does 500 J of work moving linearly 2.5 m, what force was applied by the
machine?
work done = force x distance,
rearranging, force (N) = work done (J) ÷ distance (m)
force = 500 ÷ 2.6 =
200 N
(b) If an electric motor transfers
12.0 kJ of useful energy in 3.0 minutes.
Calculate the power output of the
motor.
work done = energy transferred =
12 x 1000 = 12,000 J, time taken = 3 x 60 = 180 s
power = work done / time = 12,000
/ 180 =
66.7 W (3 sf)
(c) An appliance has a power rating
of 1.2 kW.
How many kJ of energy is
transferred in 8.0 minutes?
1.2 kW = 1200 watts, time = 8
x 60 = 480 seconds
P = E / t, rearranging gives
E = P x t
energy transferred = 1200 x
480 = 576 000 J =
576 kJ
12
Q3 A machine applies a
force of 200 N through a distance of 2.5 m in 2.0 seconds.
What is the power of the machine?
work done = force x distance = 200 x
2.5 = 500 ?
power = work done / time taken = 500
/ 2.0 = 250 J/s = 250 W
12
Q4 A 500 kg express
skyscraper lift moves nonstop up a total height of 100 m.
(a) If the force of gravity = 9.8
N/kg, calculate the weight of the lift.
weight = mass x g
= 500 x 9.8 =
4900 N
(b) Calculate the work done by the
lift motor.
work = force x distance
the
force exerted by the lift motor must be at least equal to the weight
of the lift due to gravity
therefore work done = 4900 x 100
= 490000 J ≡
490 kJ ≡
0.490 MJ
(c) If the lift ascent time is 20
seconds, what is the power of the lift motor?
power = work done / time taken
power = 490000 / 20 =
24500 W ≡
24.5 kW
(d) What assumptions has been made
for calculations (b) and (c)?
Both (b) and (c) calculations ignore
the extra work done in overcoming any forces of friction.
12
Q5 Part of a machine
requires a continuous force of 500 N from a motor to move it in a
linear direction.
(a) How much work is done in moving
it a distance of 50 m?
work done = force x distance =
500 x 50 =
25000 J (25 kJ)
(b) If the power of the machine
is 5.0 kW, how long will it take to move the machine part the 50 m
distance?
power = work done / time taken
time taken = work done / power
time = 25000 / 5000 =
5.0 s (stand clear!)
12
Q6 A machine has to move a conveyor
belt at a rate of 30 m/min.
The system is designed to use 6
kJ/min of electrical energy to move the conveyor belt along.
What is the minimum force the motor
must produce to move the conveyor belt along?
(i) power = rate of energy transfer =
energy transferred (J) / time taken (s)
power = 6000 / 60 = 100 W (100
J/s)
(ii) 30 m/min ≡
30 / 60 = 0.5 m/s
so in one
second, the energy transferred is 100 J and the distance moved is
0.5 m.
work = force x distance
force = work / distance = 100 /
0.5 = 200 N
(iii) If you are smart, you can solve
the problem directly, because both bits of data involved 'per minute'.
So you can just say force = work
/ distance = 6000 / 30 =
200 N !!!
BUT, always work logically in
your own comfort zone, the data might not always be so kind!
12
Q7 A toy model car has
a clockwork motor, whose spring can store 8.75 J of elastic potential
energy.
On release the clockwork motor can
deliver a continuous force of 2.5 N.
How far will the car travel in one
go?
energy store = total work done =
force x distance
distance = energy store / force =
8.75 / 2.5 =
3.5
m
12
Q8 A sliding object
moving across a very rough surface has 5.0 J in its kinetic energy store.
If the object experiences a constant
resistive force of friction of 25 N, calculate in cm how far the object
travels before coming to a halt.
energy transferred = kinetic
energy = resistive force x distance to come to a halt
E = KE = F x d, 5.0 = 25 x
d, d = 5.0/25 =
0.20 m = 20 cm
12
Q9 A small electric
motor uses 120 J of electrical energy in 3 minutes.
Calculate the power of the electric
motor.
Energy transferred = 120 J, time = 3
x 60 = 180 s
P = E / t = 120/180
=
0.67 W (2 sf)
12 Q10
See also
FORCES 3. Calculating resultant forces using vector
diagrams and work done gcse physics revision notes
See also
electrical energy used, power and cost of electricity calculations gcse physics revision notes
Examples of stored energy
calculation questions
How to solve potential energy problems.
(3) How to solve kinetic energy store
problems.
kinetic energy (KE) = 0.5 × mass ×
(speed)^{2}
(3) E_{ke} = ^{1}/_{2}
m v^{2}
Q3.1 A swimmer of mass 70
kg is moving at a speed of 1.4 m/s.
Calculate the kinetic energy of the
swimmer in J (to 2 sf).
E_{ke} = ^{1}/_{2}
m v^{2} = 0.5 x 70 x 1.4^{2} = 68.6 =
69 J (2 sf)
Q3.2 A bullet with a mass
of 10.0 g is travelling with an initial speed of 400 m/s.
(a) What is the bullet's
Initial kinetic energy store?
E_{ke} = ^{1}/_{2}
m v^{2} , 10 g ≡
10/1000 = 0.01 kg
E_{ke} = 0.5 x 0.01 x 400^{2}
E_{ke} =
800 J of kinetic energy
(b) If the bullet embeds
itself in a wooden plank, by how much will the planks thermal energy
store be increased?
All of the 800 J of kinetic
energy will end up as heat. There will be some loss due to friction
(air resistance) and sound energy, but most of the 800 J will end
up as heat energy in the plank.
Q3.3 A car of mass 1200
kg is travelling at a steady speed of 30.0 m/s.
(a) What is the kinetic energy
store of the car?
E_{ke} = ^{1}/_{2}
m v^{2} , 10 g ≡
10/1000 = 0.01 kg
E_{ke} = 0.5 x 1200 x 30^{2}
E_{ke} =
540,000 J of kinetic energy (5.40 x 10^{5}J,
0.540 MJ, 3 s.f.)
(b) If on slowing down the
kinetic energy of the car is halved, what speed is it then doing?
E_{ke} = ^{1}/_{2}
m v^{2} , rearranging: v^{2} = 2E_{ke} / m,
so v = √(2E_{ke} / m), KE = 540,000/2 = 270,000
J
v = √(2 x E_{ke}
/ m) = √(2 x 270,000 / 1200) =
21.2 m/s (3 s.f.)
Q3.4 The chemical
potential energy store of a 100 g rocket, on firing, gives it an initial
kinetic energy store of 200 J.
Calculate the initial vertical
velocity of the rocket.
E_{ke} = ^{1}/_{2}
m v^{2} , rearranging: v^{2} = 2E_{ke} / m,
so v = √(2E_{ke} / m)
m = 100 g ≡
100 / 1000 = 0.1 kg
v =
√(2 x E_{ke} / m) = √(2 x 200 / 0.10) =
63.2 m/s (3 s.f.)
Q3.5 A projectile is
required to impart a kinetic energy blow of 5000 J at a speed of 500 m/s.
What mass of projectile is required?
E_{ke} = ^{1}/_{2}
m v^{2} , rearranging: m = 2 x E_{ke} / v^{2}
m = 2 x E_{ke} / v^{2}
= 2 x 5000 / 500^{2} =
0.040 kg (40.0 g, 3 s.f.)
(4) How to solve gravitational potential
energy store problems
gravitational potential energy
(GPE).= mass × gravitational field strength × height
(4) E_{gpe} = m g h
On the Earth's surface the
gravitational field strength is quoted as g = 9.8 N/kg
Q4.1 A
grandfather clock weight of mass 5500g is raised 135 cm when fully wound up.
Calculate the gain in the weight's
gravitational potential energy store in J (to 3 sf).
5500 g = 5.50 kg,
135 cm = 1.35 m, g = 9.8 N/kg
E_{gpe} = m g h
= 5.50 x 9.8 x 1.35 =
72.8 J (3 s.f.)
Always make sure you have the
correct units for the numbers in the final line of ANY physics
calculation.
Q4.2 An 80.0 kg person
climbs up flights of steps to a vertical gained height of 9.00 m.
(a) Calculate the increase in
the person's gravitational energy store (gravity force = 9.8 N/kg).
E_{gpe} = m g h
GPE store gain =
80 x 9.8 x 9.0 =
7056 =
7060 J (7.06 kJ, 3 s.f.)
(b) What is the work done by
the person in climbing the stairs?
work = force x distance, force =
weight = 90 x 9.8 = 784 J
work = 784 x 9.0
= 7060 J (3
s.f.)
Note: This is the same
answer as (a) because the GPE formula is essentially expressing a
force acting through a specified distance.
It does neglect energy lost
in friction between the person and the steps.
(c) How high would you have to
climb to work off a 45.0 g bar of chocolate with a calorific value of
30.0 kJ/g?
Total energy in the chocolate bar
chemical energy store = 45 x 30 x 1000 = 1350000 J
This will be converted to the
person's gravitational potential energy store, therefore ..
GPE = m g h = 1350000 J
so
h =
1350000 / (m x g) = 1350000 / 784 =
1720 m (to 3 s.f. and higher than Ben Nevis in
Scotland)
Note: A typical active adult
needs 9000 kJ/day. An elderly person with a sedentary lifestyle might
only need 5000 kJ/day. A very active teenager might need 12000 kJ/day.
The calculation does make the point that a single bar of chocolate does
provide a good 'chunk' of you daily energy needs. Of course if you snack
a lot on high calorie foods you will put on weight if it isn't 'burnt
off'! The calorific value of fat is ~37 kJ/g and carbohydrates typically
~17 kJ/g and roast potatoes, which I love, are somewhere in between!
Q4.3 If a 5.00 kg
'weight' is dropped from a height of 10.0 m above the ground, at what speed
will it hit the ground?
The GPE of the 'weight' is easily
calculated from the formula GPE = m g h (and g
= 9.80 N/kg)
GPE = 5 x 9.8 x 10 = 490 J
As the weight falls its gravitational
potential energy store is depleted as its kinetic energy store
increases.
Therefore, at the point of impact,
all the GPE is converted to kinetic energy, so we use the KE equation
to get v.
E_{ke} = ^{1}/_{2}
m v^{2} , rearranging: v^{2} = 2E_{ke} / m,
so v = √(2E_{ke} / m)
v = √(2E_{ke} / m) = √(2 x
490 / 5.0) = 14.0 m/s
(3 s.f.)
Note (i): You can actually
simplify the calculation because you don't actually need to know the
mass of the weight.
initial E_{gpe} = final E_{ke} =
m g h = ^{1}/_{2}
m v^{2} , the m's cancel out,
so g x h = ^{1}/_{2}
x v^{2} and v = √(2 x g x h) = √(2 x 9.8 x 10) =
14.0 m/s (3 s.f.)
Note (i): All these
calculations ignore air resistance, so some KE is lost as heat, so the
final speed is actually just a bit less than the theoretical
calculations above.
Q4.4 ?
See also
FORCES
2. Mass and the effect of gravity force on it  weight, (mention of work done and
GPE
(5) How to solve elastic potential
energy store problems
(5) elastic potential energy = 0.5 × spring constant ×
(extension)^{2,} E_{e} = ^{1}/_{2}
k e^{2}
Q5.1 A spring with spring
constant of 5.00 N/m is stretched for 10.0 cm.
How much energy is stored in the
elastic potential energy store of the spring.
E_{epe} = ^{1}/_{2}
k e^{2}, 10.0 cm ≡
10.0 / 100 = 0.10 m
E_{epe} = 0.5 x 5.0 x 0.10^{2}
E_{epe} =
0.025 J
Q5.2 A car suspension
spring has a spring constant of 2000 N/m.
If the elastic potential energy store
of the spring is 50 J, how far is the spring compressed?
E_{epe} = ^{1}/_{2}
k e^{2}, rearranging gives e^{2} = 2E_{epe} / k,
e = √(2E_{epe} / k) and e = the compression
e = √(2E_{epe} / k) = √(2 x
50 / 2000) = √0.063 =
0.224 m (22.4 cm, 3 s.f.)
Q5.3 It takes 5.0 J
of work to stretch a spring 20 cm.
How much extra work must be done to
stretch it another 20 cm?
(i) You need to work out the spring
constant. 20 cm ≡
20 / 100 = 0.20 m
E_{epe} = ^{1}/_{2}
k e^{2}, rearranging gives k = 2E_{epe} / e^{2}
k = (2 x 5.0) / (0.20 x 0.20) =
250 N/m
(ii) Then work out the total work to
stretch the spring a total of 40 cm.
The total work done on the spring
equals its elastic potential energy store when fully stretched 40 cm
(which is 0.40 m). Since you now know the spring constant, you use
the same equation again, but solving for the total elastic potential
energy.
E_{epe} = ^{1}/_{2}
k e^{2} = 0.5 x 250 x 0.40^{2} =
20 J
(iii) You then subtract (i) from (ii)
to get the extra work done.
Therefore the extra work done =
20  5 =
15 J
Q5.4
See also
Elasticity and energy stored in a spring calculations
Energy resources, and
transfers, work done and
electrical power supply revision notes index
Types of energy store  a comparison with examples explained,
mechanical work done and power calculations
Conservation of energy,
energy transfers, efficiency  calculations and
Sankey diagrams gcse physics
notes
Energy resources & uses, general survey & trends, comparing sources of renewables, nonrenewables
& biofuels
Renewable energy (1) Wind power and
solar power, advantages and disadvantages gcse physics revision
notes
Renewable energy (2) Hydroelectric power and
geothermal power,
advantages and disadvantages physics notes
Renewable energy (3) Wave power and tidal barrage power,
advantages and disadvantages gcse
physics notes
Comparison of methods of generating electricity, 'National Grid' power supply,
mention of small scale supplies
Greenhouse
effect, global warming, climate change,
carbon footprint from fossil fuel burning gcse physics
notes
See also
The Usefulness of Electricity gcse
physics electricity revision notes
aqa gcse
91 physics: Know that a system is an object or group of objects.
There are changes in the way energy is stored when a
system changes. For example: an object projected upwards, a moving object hitting an obstacle, an object accelerated by a constant force, a vehicle slowing down, bringing water to a boil in an electric kettle., Throughout this section on Energy you should be able
to : describe all the changes involved in the way energy
is stored when a system changes and calculate the changes in energy involved when a
system is changed by: heating