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).

Wasted energy
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

Examples of simple
mechanical work done 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
work done (J) = force (N) x distance
through which force acts (m)
Q1.1
If you drag a heavy box with a force of 200 N across a floor for 3 m, what
work is done?
work done = 200 x 3 =
600 J
Q1.2 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
Q1.3
Examples
of simple power calculation questions
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
P (W) = E (J) / t (s)
1 watt is equal to a work rate of 1 joule
per second
Q2.1 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
Q2.2
See also
electrical energy used, power and cost of electricity calculations
Examples of stored energy
calculation questions
How to solve potential energy problems.
How to solve kinetic energy store
problems.
Q3.1 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
Intial 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.
Q3.2 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 (5.4 x 10^{5}J,
0.54 MJ)
(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.)
Q3.3 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.)
Q3.4 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 g)
How to solve gravitational potential
energy store problems
Q4.1 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 J (7.06 kJ)
(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
= 7056 J
Note: This is the same
answer as (a) because the GPE formula is essentially expressing a
force acting through a specified distance.
(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.
Q4.2 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
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.
Q4.3
See also
FORCES
2. Mass and the effect of gravity force on it  weight, (mention of work done and
GPE
How to solve elastic potential
energy store problems
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, comparison with examples explained,
energy stores, mechanical work done & power calculations
Energy resources and their uses  a general survey
more details in
gcse physics revision notes below
Renewable energy (1) Wind power and
solar power, advantages and disadvantages
Renewable energy (2) Hydroelectric power and
geothermal power, advantages and disadvantages
Renewable energy (3) Wave power and tidal power, advantages and
disadvantages
Biofuels, renewables and nonrenewables, advantages and disadvantages
Generating electricity and the 'National Grid' power supply,
mention of small scale supplies
Energy transfer and efficiency  calculations and
Sankey diagrams
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