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Science 10 – Chapter B2
ENERGY TRANSFORMATIONS
B2.1 – Types of Energy
Types of Energy
 There are many types of energy, and often in nature
and in mechanical devices, we see a transformation
of energy from one type to another
 As we will see, in any transformation, energy is never
created or destroyed
 Energy can be transferred from one object to another
or transformed from one type to another
 While there are many types of energy, many forms
can be classified as either potential or kinetic energy
Categories of Energy
Potential Energy
Kinetic Energy
 Energy that is stored in readiness
 Potential energy cannot be seen
 Energy of motion
 The faster an object
or observed until it is
transformed into another form
 Examples of potential energy
include



Chemical energy – such as the E stored
in food before you eat it, or the E
stored in a battery
Gravitational energy – the E stored in
objects held above the ground that
have the potential to fall
Elastic energy – the E stored in a
compressed spring or a stretched
elastic
moves, the higher its Ek
 Because particles within
substances are always
moving, and warmer
objects have fastermoving particles,
thermal energy (heat) is
one type of kinetic
energy
Nuclear energy
 Another type of potential energy, nuclear energy is
energy stored in the nucleus of an atom
 It can be released during a nuclear reaction, such as:


Radioactive decay
Nuclear fusion – when the nuclei of two atoms fuse to make
one larger atom


This is the type of reaction that occurs in the sun to produce solar
energy
Nuclear fission – when one atom breaks apart to form two or
more smaller atoms

This is the type of reaction that occurs in a nuclear power plant
and in a nuclear bomb
Energy conversions
 An energy conversion refers to the transformation of
energy from one form to another
 In each example below, a device converts one type of
energy input into a different type of energy output




A flashlight converts chemical potential energy into
electromagnetic (light) energy
A windmill converts kinetic energy into electrical energy
A car engine converts chemical potential energy into kinetic
energy
A solar panel converts electromagnetic (solar) energy into
electrical energy
Volta Pile – the first battery
 one notable energy conversion was a
device invented by Alessandro Volta
called the Volta pile
 the Volta Pile converted the chemical
potential energy of different metals into
electrical energy and was, therefore, the
first battery
Joule’s experiments
 James Prescott Joule performed two experiments to
demonstrate energy transformations:



Experiment #1: showed that a falling object (that had potential
energy) can create heat (thermal energy)
Experiment #2: showed that a moving object (that had kinetic
energy) can create heat
Because of his work on energy, the unit we use to quantify it is called
the joule
 Joule’s work came at an important time – working at the
start of the Industrial Revolution, people were very
interested in designing machines that could perform
useful work – that is, transform one type of energy into
mechanical energy
Joule’s experiments
 As the weights fall, the paddles
inside the chamber turn
 The paddles agitate the water,
increasing its temperature
 See an animation
B2.2 – Potential Energy
What is potential energy?
 Recall, potential energy is energy stored in readiness
 It can only be observed once it is transformed into
some other type of energy


E.g. Gravity cannot be observed until you allow an object to fall
– Ep(grav) is converted into Ek
E.g. Chemical energy in a fuel cannot be observed, but once the
fuel burns in your gas tank the car moves – Ep(chem) is
converted into Emechanical
Gravitational potential energy
 An object has Ep(grav) if it has the potential to fall
 That is, if its position is above the Earth’s surface
 The greater the mass of the object, or the higher it is,
the greater its Ep(grav)
 The formula for gravitational potential energy is:

Ep(grav) = mgh, where
m = the mass of the object, in kg
 g = the acceleration of the object due to gravity = 9.81 m/s2
 h = the height of the object, in m


Like all types of energy, Ep(grav) is measured in joules (J)
Why no negatives?
 Notice, the value given for ‘g’ is 9.81 m/s2 but does
not include the negative. Why is this, if the object is
accelerating downwards?
 Two reasons:


the height of the object should technically be – as well, since it
will be traveling in a – direction; the two –’s would cancel out
energy is a scalar quantity, which means we are only interested
in its magnitude, not its direction
Example:
 A 80kg diver handstands on a
10m diving platform. Before
diving, how much
gravitational potential
energy does the diver have?
 Ep = mgh
= (80kg)(9.81 m/s2)(10m)
= 7848 J
= 7.8 x 103 J
Weight vs. mass
 Recall from Ch1,
 the mass of an object refers to the amount of matter it has, and
is measured in kg.
 the weight of an object refers to the force that the Earth is
applying to that object as a result of gravity.

Further, since F = ma, we can create a formula to describe this
force:
 𝐹𝑔
= 𝑚𝑔, where
= the force of gravity, in N
 m = mass of the object, in kg
 𝑔 = the acceleration of the object due to gravity, - 9.81 m/s2
 𝐹𝑔
Other types of potential energy
 As mentioned, there are other types of potential
energy, but unlike Ep(grav), they do not require a
special formula
 Whenever work is done on an object to give it
potential energy, its Ep can be calculated using a
formula we’ve already seen: W = Fd, where



W = the work done on the object and therefore its potential E
F = the force applied to the object, in N
d = the distance over which the force was applied, in m
Example:
 An athlete stretches a resistance band
0.20m with a force of 50N. How much
elastic potential energy does the tubing
have?
 Ep = W = Fd
= (50N)(0.20m)
= 10 J
B2.3 – Kinetic Energy & Motion
Calculating kinetic energy
 Anytime an object is in motion, it has kinetic energy
 Heat is also a form of kinetic energy, since the
particles of a warmer substance move faster than
those of a cooler substance
 The formula for calculating Ek is

Ek = ½ mv2, where
m = the mass of the object, in kg
 v = the speed of the object, in m/s

How do we calculate Ek if the object is accelerating?
 Notice that the formula includes ‘v’ for the speed of
the object, but what happens if an object is not
traveling at a constant speed, that is, if it is
accelerating?
 The ‘v’ in this formula refers to the object’s
instantaneous speed, that is, the speed the object is
going at a specific moment


Therefore, even if the object is speeding up or slowing down,
we are only interested in the Ek at one moment
E.g. What is the Ek of the diver right before he hits the water?
A closer look at the formula & an example
 How do we type this in to our calculator? A few
choices:



Ek = ½ mv2
Ek = 0.5 mv2
𝐸𝑘 =
𝑚𝑣 2
2
 Notice, the v is squared!
 Example: A 0.300-kg ball is pushed horizontally at a
speed of 20.0m/s. Calculate the kinetic energy of the
ball at the moment it starts to move.

Ek = ½ mv2 = (0.5)(0.300 kg)(20.0m/s)2 = 60.0 J
Solving for other values
 This is the toughest formula to manipulate for the other
values. Let’s look at it.
 Solving for mass:
 𝐸𝑘 =
𝑚𝑣 2
2
so
2Ek = mv2
so
𝑚=
2𝐸𝑘
𝑣2
𝑣2 =
2𝐸𝑘
𝑚
 * you will need to put your v2 in brackets
 Solving for velocity:
𝑚𝑣 2
2
so
2Ek = mv2
so

𝐸𝑘 =

However we need to isolate ‘v’, not v2 so we have to take the square root

𝑣=

* to avoid making any errors, try to calculate this all in one step
2𝐸𝑘
𝑚
Example
 A 80-kg diver has 7848 J of kinetic energy right
before she strikes the water. What is her speed at the
moment before she strikes the water?
 𝐸𝑘 =
𝑚𝑣 2
2
𝑣=
2(7848 𝐽)
80 𝑘𝑔

so 𝑣 =
2𝐸𝑘
𝑚
= 14 m/s
Why not negative? We asked for speed, not velocity, so we are
not interested in her direction.
B2.4 – Mechanical Energy
Energy conversions
 Recall, in any energy transformation, energy cannot
be created or destroyed


As an object moves upward or downward, its total energy
remains constant, even though its Ek or Ep(grav) might be
changing
This total energy is referred to as the mechanical energy, or Em
 We can even calculate Em since, at any point in an
object’s motion, it is the total of the Ep and the Ek

E m = Ep + Ek
Converting between Ep and Ek
 When an object is falling from a height, it has Ep the
entire time, until the moment it strikes the ground

During the fall, its height is decreasing, so its Ep is steadily
decreasing until it becomes zero (at the ground)
 Similarly, it has Ek the moment it starts falling
 During the fall, it is accelerating, so since is speed is
increasing, its Ek is steadily increasing
 Though, as it falls, its Ep is  and its Ek is , its Em
remains constant
Example:




At the top of the diving board, the diver has
a maximum amount of gravitational
potential energy, but as he is not yet falling,
no kinetic energy
As the diver falls, his height above the
Earth decreases, so as he nears the water,
his potential energy decreases
As he falls, his velocity is increasing
because gravity causes objects to accelerate
as they fall. As he speeds up, his kinetic
energy increases
The moment the diver strikes the water’s
surface, he is at “ground level” and no
longer has potential energy. However, he
has reached his maximum velocity, so his
kinetic energy is at a maximum.
Calculating Em
 Calculating mechanical energy requires you to
analyze both the position and the movement of the
object



based on its position above the Earth, you can calculate its Ep
based on its speed, you can calculate Ek
its Em will be the sum of the two
Example:
 A roller coaster and its passengers have a combined
mass of 1000 kg. Halfway down one slope, the
coaster is 20.0m off the ground and traveling at 5.00
m/s. What is the mechanical energy of the coaster at
that point?
Em
= Ep + Ek
= mgh + ½ mv2
= (1000kg)(9.81 m/s2)(20.0m) + ½ (1000kg)(5.00m/s)2
= 196 200 J + 12 500 J
= 208 700 J
= 2.09 x 105 J
Solving for one of the other variables:
 The most challenging type of question here would be
if a question gives you the Em but asks you to solve
for one of the other variables (the height or the
speed).
 Rearranging this formula looks tricky, but uses the
same skills you’ve already learned.
Rearranging the formula
 Solving for height:
Em = mgh + ½ mv2
Em – ½
mv2
= mgh
 Solving for speed:
Em = mgh + ½ mv2
Em – mgh = ½ mv2
2(Em – mgh) = mv2
Em – ½ mv2 = h
mg
2(Em – mgh) = v2
m
2(𝐸𝑚 −𝑚𝑔ℎ)
𝑚
=v
The Law of Conservation of Energy
 Recall, according to the Law of Conservation of
Energy, energy cannot be created or destroyed
 In the context of mechanical energy,



the total energy, Em remains constant, but also
the object’s maximum Ep (at the top) is equal to the object’s
maximum Ek (at the bottom)
So if a question gives us enough information to find Ep at the
top, we can calculate Ek at the bottom, and vice versa
Example
 An 80.0-kg diver has 7848 J of potential energy
when she is standing on the 10m platform.





a) how much kinetic energy does she have right before she
strikes the water?
b) how fast is she going right before she strikes the water?
c) how much mechanical energy does she have at the top
before she starts the dive?
d) how much mechanical energy does she have at the bottom
right before she strikes the water?
e) how much mechanical energy does she have when she is
halfway down (at 5.0m from the water)?
Example - solution
 An 80.0-kg diver has 7848 J of potential energy
when she is standing on the 10m platform.

a) how much kinetic energy does she have right before she
strikes the water?


Ek(bottom) = Ep(top) = 7848 J
b) how fast is she going right before she strikes the water?

𝐸𝑘 =

𝑣=
𝑚𝑣 2
2
so 𝑣 =
2(7848 𝐽)
80 𝑘𝑔
2𝐸𝑘
𝑚
= 14 m/s
Example - solution
 An 80.0-kg diver has 7848 J of potential energy
when she is standing on the 10m platform.

c) how much mechanical energy does she have at the top
before she starts the dive?
Em = Ep + Ek before she starts to move, her Ek is zero
 Em = 7848 J + 0 J = 7848 J


d) how much mechanical energy does she have at the bottom
right before she strikes the water?
Em = Ep + Ek as she is at her lowest height, her Ep is zero
 Em = 0 J + 7848 J = 7848 J


e) how much mechanical energy does she have when she is
halfway down (at 5.0m from the water)?

At any point, her Em remains constant, at 7848 J
For falling or rising objects:
 Therefore, for objects who are both at a position
above the Earth, and moving, you can set these two
formulas equal to each other.
 This means, a question could tell you how fast the
object was moving at the bottom, and ask you for its
height at the top, or vice versa
For falling or rising objects:
 Ep(top) = Ek(bottom)
 mgh = ½ mv2
*Notice that the masses cancel out
 gh = ½ v2
 This formula will not be given to you, only the
formulas for Ep and Ek
 It is up to you, based on the context of the question,
to realize that you have to set these two formulas
equal to each other.
Given… solve for…
 Given the height at the
top, solve for the speed at
the bottom



Ep(top) = Ek(bottom)
mgh = ½ mv2
gh = ½ v2
𝑣=
2𝑔ℎ
 Given the speed at the
bottom, solve for height
at the top



Ep(top) = Ek(bottom)
mgh = ½ mv2
gh = ½ v2 or gh = v2
2
𝑣2
ℎ=
2𝑔
Example:
 A barbell is lifted from the ground to a height of
2.00m off the ground. At what speed did the
weightlifter lift the barbell off the ground?
 *only one number is given to us in this question. You
don’t need the mass, because it will cancel out, and
you always know ‘g’.



Ep(top) = Ek(bottom)
mgh = ½ mv2
gh = ½ v2
𝑣=
2𝑔ℎ =
𝑚
2(9.81 2 )(2.00𝑚)=
𝑠
6.26 m/s
Example:
 A 0.125 kg arrow is shot directly upwards. If it leaves
the bow travelling 11m/s, how high does it fly?



ℎ=
Ep(top) = Ek(bottom)
mgh = ½ mv2
gh = ½ v2 or gh = v2
2
𝑣2
2𝑔
𝑚
=
(11 𝑠 )2
𝑚
2(9.81 2 )
𝑠
= 6.2 m
How do I know which formula to use?
 If a question gives you energy (a number in joules)
you will likely need Em = Ep + Ek
 If a question gives you a height and asks for speed, or
a speed and asks for a height, you will use Ep(top) =
Ek(bottom)
One more example
 Though not an example of an object rising or falling,
a pendulum is another great example of a constant
transformation between Ek and Ep.
Pendulums
 another example of a conversion between Ep and
Ek



at position A, the pendulum is not moving, but is at a
maximum height  maximum Ep, minimum Ek
at position B, the pendulum is moving at a maximum speed,
but is the closest to the ground  minimum Ep, maximum Ek
at position C, the pendulum is at the same position as position
A
B2.5 – Energy Conversions
Evidence of Energy Conversions
 as we’ve seen, an energy conversion is the
transformation of energy from one type to another
 different results can occur as a result:




motion – an object can move
a change in position – an object can rise or fall
a change in shape – an elastic could stretch, or a spring
compress
a change in temperature – an object could heat up
Energy conversions in nature
 Two major energy conversions occur in nature, and
will be addressed in the Biology Unit:


photosynthesis is the conversion of solar energy into
chemical potential energy and is performed by plants
cellular respiration is the conversion of chemical potential
energy into useful energy, e.g. energy of motion, thermal
energy, etc. and is performed by all living organisms
Energy conversions to produce electricity
 There are a number of methods used in modern
society to produce electricity


These methods don’t create energy, but rather convert it from
one form to another
Different types of power plants (e.g. hydro-electric, coalburning, nuclear)
use different input energies (a water reservoir, coal, radioactive
uranium)
 to cause water to move,
 which turns a turbine,
 which creates an electrical current
 and produces the output energy: electrical energy

Hydro-electric power stations
 Uses the gravitational
potential energy of water
that is in a position
above the power plant
 when the water falls, it
gains kinetic energy
 this fast-moving water
makes the turbine spin
and produces electricity
Hydro-electric power plant
ADVANTAGES:
 relies on a renewable
resource
 clean and efficient
 no air pollution
 little maintenance
required
DISADVANTAGES:
 water reservoirs are large
and unsightly
 re-routing water to go
through the plant could
negatively impact the
ecosystem
Coal-burning power plants
 also known as thermoelectric power plants
 convert chemical energy stored in fossil fuels into
electric energy
 when the fuel is burned, the chemical energy is
released
 this is transformed into thermal energy which is used
to heat steam
 this fast-moving steam makes the turbine turn,
producing electricity
Coal-burning power plant
Coal-burning power plant
ADVANTAGES:
 relatively inexpensive
 waste heat can be used to
heat surrounding
buildings
 in Alberta, fossil fuels are
still readily available
DISADVANTAGES:
 relies on a nonrenewable resource
 pollutes the air with
greenhouse gases
 obtaining fossil fuels
from the ground can
damage the environment
Nuclear power plant
 atoms of uranium are split by nuclear fission
 splitting atoms releases a tremendous amount of
thermal energy
 this thermal energy is used to make steam, which
turns a turbine, and again, produces electricity
Nuclear power plant
Nuclear power plant
ADVANTAGES:
 do not produce harmful
greenhouse gases
 very efficient: a small
amount of uranium
produces an enormous
amount of electricity
DISADVANTAGES:
 nuclear waste is
dangerous to any living
thing exposed to it for
100s of years
 facilities for storing
nuclear waste are
expensive
Solar cells
 An alternative to a
conventional power
plant, solar cells produce
electrical energy directly
from solar energy
without the need of a
rotating turbine
Hydrogen fuel cells
 A hydrogen fuel cell is another alternative to fossil
fuels to produce electricity
 It uses the formation reaction of water,

H2(g) + O2(g)  H2O(l) to produce electricity
 Because the product is water, fuel cells do not
produce harmful carbon compounds
 Unfortunately, due to lack in availability of pure
hydrogen, a high risk of explosion, and the large size
of these fuel cells, they are not yet a viable alternative
to gasoline engines