Transcript PE = mgh

• Available worksheet, PE, KE, and ME.

First Law of Energy (Thermodynamics)
 All
energy is either kinetic or potential.
Copyright © 2010 Ryan P. Murphy

First Law of Energy (thermodynamics)
 All
energy is either kinetic or potential.
Copyright © 2010 Ryan P. Murphy

First Law of energy (thermodynamics)
 All
energy is either kinetic or potential.
Copyright © 2010 Ryan P. Murphy

First Law of Energy (thermodynamics)
 All
energy is either kinetic or potential.
Copyright © 2010 Ryan P. Murphy

First Law of Energy (thermodynamics)
 All
energy is either kinetic or potential.
Copyright © 2010 Ryan P. Murphy

Potential Energy: (PE) The energy stored
by an object as a result of its position.
Copyright © 2010 Ryan P. Murphy
Potential Enegy (PE)
Kinetic Energy (KE)
Potential Enegy (PE)
Kinetic Energy (KE)
Potential Enegy (PE)
Kinetic Energy (KE)

Potential Energy is the energy of position.
Objects that are elevated have a high
potential energy.
 Kinetic Energy is the energy of motion.
Copyright © 2010 Ryan P. Murphy

Potential Energy is the energy of position.
Objects that are elevated have a high
potential energy.
 Kinetic Energy is the energy of motion.
Copyright © 2010 Ryan P. Murphy

Potential Energy is the energy of position.
Objects that are elevated have a high
potential energy.
 Kinetic Energy is the energy of motion.
Copyright © 2010 Ryan P. Murphy
• Available worksheet, PE, KE, and ME.
• Activity! Please write and plan on sharing
a sentence about PE and KE about the
animation below.
Copyright © 2010 Ryan P. Murphy
• Activity! Please write and plan on sharing
a sentence about PE and KE about the
animation below.
Copyright © 2010 Ryan P. Murphy
• The monkey has potential energy because
of its position in the tree. When she lets go
her potential energy is transferred into the
energy of motion (KE). and
Copyright © 2010 Ryan P. Murphy
• The monkey has potential energy because
of its position in the tree. When he lets go
his potential energy is transferred into the
energy of motion (KE).
Copyright © 2010 Ryan P. Murphy
Copyright © 2010 Ryan P. Murphy
Copyright © 2010 Ryan P. Murphy
Copyright © 2010 Ryan P. Murphy
Copyright © 2010 Ryan P. Murphy
Copyright © 2010 Ryan P. Murphy
• Video Link! (Optional) Energy changes,
Potential and Kinetic Energy.
– http://www.youtube.com/watch?v=Jnj8mc04r9E
• Activity! PE – KE Skateboarder Simulator
• Search Phet Skate Board Demo.
• Download program (Free)
http://phet.colorado.edu/en/simulation/energy
-skate-park
Copyright © 2010 Ryan P. Murphy

PE = mgh
Copyright © 2010 Ryan P. Murphy

PE = mgh
 PE
= Energy (in Joules)
Copyright © 2010 Ryan P. Murphy

PE = mgh
 PE
= Energy (in Joules)
 m = mass (in kilograms)
Copyright © 2010 Ryan P. Murphy

PE = mgh
 PE
= Energy (in Joules)
 m = mass (in kilograms)
 g = gravitational
acceleration of the earth
(9.8 m/s²)
Copyright © 2010 Ryan P. Murphy

PE = mgh
 PE
= Energy (in Joules)
 m = mass (in kilograms)
 g = gravitational
acceleration of the earth
(9.8 m/s²)
 h = height above Earth's
surface (in meters)
Copyright © 2010 Ryan P. Murphy

PE = mgh
 PE
= Energy (in Joules)
 m = mass (in kilograms)
 g = gravitational
acceleration of the earth
(9.8 m/s²)
 h = height above Earth's
surface (in meters)
Learn more about Potential Energy at…
http://www.physicsclassroom.com/clas
s/energy/u5l1b.cfm
Copyright © 2010 Ryan P. Murphy
• Available worksheet, PE, KE, and ME.
• Calculate the potential energy for a 2 kg
basketball dropping from a height of 3.5
meters with a velocity of 9.8 m / sec².
– Find the PE in Joules? PE=mgh
Copyright © 2010 Ryan P. Murphy
• Calculate the potential energy for a 2 kg
basketball dropping from a height of 3.5
meters with a velocity of 9.8 m / s².
– Find the PE in Joules? PE=mgh
Copyright © 2010 Ryan P. Murphy
• Calculate the potential energy for a 2 kg
basketball dropping from a height of 3.5
meters with a velocity of 9.8 m / s².
– Find the PE in Joules? PE=mgh
Copyright © 2010 Ryan P. Murphy
• PE = mgh
m = 2 kg
g = 9.8 m/sec2
h = 3.5 m
Copyright © 2010 Ryan P. Murphy
• PE = mgh
m = 2 kg
g = 9.8 m/sec2
h = 3.5 m
Copyright © 2010 Ryan P. Murphy
• PE = mgh
m = 2 kg
g = 9.8 m/s²
h = 3.5 m
Copyright © 2010 Ryan P. Murphy
• PE = mgh
m = 2 kg
g = 9.8 m/s²
h = 3.5 m
Copyright © 2010 Ryan P. Murphy
• PE = mgh
m = 2 kg
g = 9.8 m/s²
h = 3.5 m
• PE = (2 kg ) (9.8 m/s²) (3.5 m)
Copyright © 2010 Ryan P. Murphy
• PE = mgh
m = 2 kg
g = 9.8 m/s²
h = 3.5 m
• PE = (2 kg ) (9.8 m/s²) (3.5 m)
• PE =
Copyright © 2010 Ryan P. Murphy
• PE = mgh
m = 2 kg
g = 9.8 m/s²
h = 3.5 m
• PE = (2 kg ) (9.8 m/s²) (3.5 m)
• PE = 68.6 Joules
Copyright © 2010 Ryan P. Murphy
• Available worksheet, PE, KE, and ME.
• Calculate the potential energy of a shot put
dropping from a height of 6 meters weighing
5.44 kg with a velocity of 9.8 m/s².
– Find the PE in Joules?
Copyright © 2010 Ryan P. Murphy
• Calculate the potential energy of a shot put
dropping from a height of 6 meters weighing
5.44 kg with a velocity of 9.8 m/s².
– Find the PE in Joules?
Copyright © 2010 Ryan P. Murphy
• Calculate the potential energy of a shot put
dropping from a height of 6 meters weighing
5.44 kg with a velocity of 9.8 m/s².
– Find the PE in Joules? PE=mgh
Copyright © 2010 Ryan P. Murphy
• PE = mgh
m = 5.44 kg
g = 9.8 m/s²
h=6m
Copyright © 2010 Ryan P. Murphy
• PE = mgh
m = 5.44 kg
g = 9.8 m/s²
h=6m
PE = (5.44kg) (9.8m/s²) (6m)
PE =
Copyright © 2010 Ryan P. Murphy
• Answer: PE = 319.87 Joules.
Copyright © 2010 Ryan P. Murphy
• Available worksheet, PE, KE, and ME.
• Calculate the potential energy for a 2500 kg
satellite orbiting at an altitude of 50,000
meters above the surface of the earth if it is
traveling with a velocity of 9.8 m/s². Find PE
in Joules?
– Assume we are using the earth gravity
constant.
• Calculate the potential energy for a 2500 kg
satellite orbiting at an altitude of 50,000
meters above the surface of the earth if it is
traveling with a velocity of 9.8 m/s². Find PE
in Joules?
– Assume we are using the earth gravity
constant.
• Calculate the potential energy for a 2500 kg
satellite orbiting at an altitude of 50,000
meters above the surface of the earth if it is
traveling with a velocity of 9.8 m/s². Find PE
in Joules? PE=mgh
– Assume we are using the earth gravity
constant.
• Calculate the potential energy for a 2500 kg
satellite orbiting at an altitude of 50,000
meters above the surface of the earth if it is
traveling with a velocity of 9.8 m/s². Find PE
in Joules? PE=mgh
– Assume we are using the earth gravity
constant.
• PE = mgh
m = 2500 kg
g = 9.8 m/s²
h = 50,000m
Copyright © 2010 Ryan P. Murphy
• PE = mgh
m = 2500 kg
g = 9.8 m/s²
h = 50,000m
Copyright © 2010 Ryan P. Murphy
• PE = mgh
m = 2500 kg
g = 9.8 m/s²
h = 50,000m
• PE = (2500 kg) (9.8 m/s²) (50,000 m)
Copyright © 2010 Ryan P. Murphy
• PE = mgh
m = 2500 kg
g = 9.8 m/s²
h = 50,000m
• PE = (2500 kg) (9.8 m/s²) (50,000 m)
• PE = ?
Copyright © 2010 Ryan P. Murphy
• Or PE = 1,225,000,000 Joules
Copyright © 2010 Ryan P. Murphy
• Or PE = 1,225,000,000 Joules
• Can you put it into scientific notation?
Copyright © 2010 Ryan P. Murphy
• Or PE = 1,225,000,000 Joules
9
• Can you put it into scientific notation?
Copyright © 2010 Ryan P. Murphy
• Or PE = 1,225,000,000 Joules
9
• Can you put it into scientific notation?
• PE = 1.225 x 109 Joules
Copyright © 2010 Ryan P. Murphy

Kinetic energy
Copyright © 2010 Ryan P. Murphy

Kinetic energy
 The energy that matter has because of its
motion and mass.
Copyright © 2010 Ryan P. Murphy

Kinetic energy
 The energy that matter has because of its
motion and mass.
 Where m = mass of object (kg).
Copyright © 2010 Ryan P. Murphy

Kinetic energy
 The energy that matter has because of its
motion and mass.
 Where m = mass of object (kg).
 v = speed of object.
Copyright © 2010 Ryan P. Murphy

Kinetic energy
 The energy that matter has because of its
motion and mass.
 Where m = mass of object (kg).
 v = speed of object.
 KE = Energy in Joules.
Copyright © 2010 Ryan P. Murphy
• Kinetic energy
– The
energy shows
that matter
has kinetic
because
of its
This
equation
that the
energy
and
of motion
an object
is mass.
proportional to the square of
its
speed. m
For= amass
twofold
increase
– Where
of object
(kg).in speed,
the
energy
will increase by a factor of
– v kinetic
= speed
of object.
four.
– KE = Energy in Joules.
Copyright © 2010 Ryan P. Murphy
• Kinetic energy
– The
energy shows
that matter
has kinetic
because
of its
This
equation
that the
energy
and
of motion
an object
is mass.
proportional to the square of
its
speed. m
For= amass
twofold
increase
– Where
of object
(kg).in velocity,
the
energy
will increase by a factor of
– v kinetic
= speed
of object.
four.
– KE = Energy in Joules.
Copyright © 2010 Ryan P. Murphy

Kinetic energy
-
Copyright © 2010 Ryan P. Murphy
Copyright © 2010 Ryan P. Murphy
Kinetic Energy
Copyright © 2010 Ryan P. Murphy
Kinetic Energy
Copyright © 2010 Ryan P. Murphy
• Amount of KE depends on both the
objects mass and its velocity / (speed).
Copyright © 2010 Ryan P. Murphy
• Amount of KE depends on both the
objects mass and its velocity / (speed).
Copyright © 2010 Ryan P. Murphy
• Amount of KE depends on both the
objects mass and its velocity / (speed).
Copyright © 2010 Ryan P. Murphy
• Amount of KE depends on both the
objects mass and its velocity / (speed).
Copyright © 2010 Ryan P. Murphy
• Amount of KE depends on both the
objects mass and its velocity / (speed).
Copyright © 2010 Ryan P. Murphy
• Available worksheet, PE, KE, and ME.
• What is the kinetic energy of a 10 kilogram
cannon ball traveling at 50 meters per
second?
• m = 10 kg
• v = 50 m/s
Copyright © 2010 Ryan P. Murphy
• What is the kinetic energy of a 10 kilogram
cannon ball traveling at 50 meters per
second?
• m = 10 kg
• v = 50 m/s
Copyright © 2010 Ryan P. Murphy
• What is the kinetic energy of a 10 kilogram
cannon ball traveling at 50 meters per
second?
• m = 10 kg
• v = 50 m/s
Copyright © 2010 Ryan P. Murphy
• Don’t forget your order of operations.
Copyright © 2010 Ryan P. Murphy
• Don’t forget your order of operations.
• PEMDAS
Copyright © 2010 Ryan P. Murphy
• Don’t forget your order of operations.
• PEMDAS
• For KE, you must do exponents (E) before
multiplying (M).
Copyright © 2010 Ryan P. Murphy
• Don’t forget your order of operations.
• PEMDAS
• For KE, you must do exponents (E) before
multiplying (M).
Copyright © 2010 Ryan P. Murphy
• Don’t forget your order of operations.
• PEMDAS
• For KE, you must do exponents (E) before
multiplying (M).
Copyright © 2010 Ryan P. Murphy
• KE = 0.5 times 10 kg times (50) ² Joules
Copyright © 2010 Ryan P. Murphy
• KE = 0.5 times 10 kg times (50) ² Joules
• KE = 0.5 times 10 kg times 2,500 Joules
Copyright © 2010 Ryan P. Murphy
• KE = 0.5 times 10 kg times (50) ² Joules
• KE = 0.5 times 10 kg times 2,500 Joules
Copyright © 2010 Ryan P. Murphy
• KE = 0.5 times 10 kg times (50) ² Joules
• KE = 0.5 times 10 kg times 2,500 Joules
Copyright © 2010 Ryan P. Murphy
• KE = 0.5 times 10 kg times (50) ² Joules
• KE = 0.5 times 10 kg times 2,500 Joules
• KE = 5 kg times 2,500 Joules
Copyright © 2010 Ryan P. Murphy
•
•
•
•
KE = 0.5 times 10 kg times (50) ² Joules
KE = 0.5 times 10 kg times 2,500 Joules
KE = 5 kg times 2,500 Joules
KE =
Copyright © 2010 Ryan P. Murphy
•
•
•
•
KE = 0.5 times 10 kg times (50) ² Joules
KE = 0.5 times 10 kg times 2,500 Joules
KE = 5 kg times 2,500 Joules
KE = 12,500 Joules
Copyright © 2010 Ryan P. Murphy
•
•
•
•
KE = 0.5 times 10 kg times (50) ² Joules
KE = 0.5 times 10 kg times 2,500 Joules
KE = 5 kg times 2,500 Joules
KE = 12,500 Joules
Copyright © 2010 Ryan P. Murphy
• Available worksheet, PE, KE, and ME.
• What is the kinetic energy of a .142 kilogram
baseball traveling at 45 meters per second?
• m = .142 kg
• v = 45 m/s
Copyright © 2010 Ryan P. Murphy
• What is the kinetic energy of a .142 kilogram
baseball traveling at 45 meters per second?
• m = .142 kg
• v = 45 m/s
Copyright © 2010 Ryan P. Murphy
• What is the kinetic energy of a .142 kilogram
baseball traveling at 45 meters per second?
• m = .142 kg
• v = 45 m/s
Copyright © 2010 Ryan P. Murphy
• KE = 0.5 times .142 kg times (45) ² Joules
Copyright © 2010 Ryan P. Murphy
• KE = 0.5 times .142 kg times (45) ² Joules
• KE = 0.5 times .142 kg times 2,025 Joules
PEMDAS
Copyright © 2010 Ryan P. Murphy
• KE = 0.5 times .142 kg times (45) ² Joules
• KE = 0.5 times .142 kg times 2,025 Joules
Copyright © 2010 Ryan P. Murphy
• KE = 0.5 times .142 kg times (45) ² Joules
• KE = 0.5 times .142 kg times 2,025 Joules
• KE = .071 kg times 2,025 Joules
Copyright © 2010 Ryan P. Murphy
•
•
•
•
KE = 0.5 times .142 kg times (45) ² Joules
KE = 0.5 times .142 kg times 2,025 Joules
KE = .071 kg times 2,025 Joules
KE =
Copyright © 2010 Ryan P. Murphy
•
•
•
•
KE = 0.5 times .142 kg times (45) ² Joules
KE = 0.5 times .142 kg times 2,025 Joules
KE = .071 kg times 2,025 Joules
KE = 143.775 Joules
Copyright © 2010 Ryan P. Murphy
•
•
•
•
KE = 0.5 times .142 kg times (45) ² Joules
KE = 0.5 times .142 kg times 2,025 Joules
KE = .071 kg times 2,025 Joules
KE = 143.775 Joules
Copyright © 2010 Ryan P. Murphy

Mechanical Energy (ME): Energy due to
position and motion.
-
Copyright © 2010 Ryan P. Murphy

Mechanical Energy (ME): Energy due to
position and motion.
 Sum
of potential and kinetic energies,
includes heat and friction. PE + KE = ME
Copyright © 2010 Ryan P. Murphy
• Available worksheet, PE, KE, and ME.
• A ski jumper moving down the hill had a
Potential Energy of 6,500 Joules, and a
Kinetic Energy of 10,500 Joules.
– What is her Mechanical Energy?
• A ski jumper moving down the hill had a
Potential Energy of 6,500 Joules, and a
Kinetic Energy of 10,500 Joules.
– What is her Mechanical Energy?
• A ski jumper moving down the hill had a
Potential Energy of 6,500 Joules, and a
Kinetic Energy of 10,500 Joules.
– What is her Mechanical Energy?
• A ski jumper moving down the hill had a
Potential Energy of 6,500 Joules, and a
Kinetic Energy of 10,500 Joules.
– What is her Mechanical Energy?
ME = PE + KE
• A ski jumper moving down the hill had a
Potential Energy of 6,500 Joules, and a
Kinetic Energy of 10,500 Joules.
– What is her Mechanical Energy?
ME = PE + KE
ME = 6,500 J + 10,500 J
• A ski jumper moving down the hill had a
Potential Energy of 6,500 Joules, and a
Kinetic Energy of 10,500 Joules.
– What is her Mechanical Energy?
ME = PE + KE
ME = 6,500 J + 10,500 J
ME =
• A ski jumper moving down the hill had a
Potential Energy of 6,500 Joules, and a
Kinetic Energy of 10,500 Joules.
– What is her Mechanical Energy?
ME = PE + KE
ME = 6,500 J + 10,500 J
ME = 17,000 Joules.
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. The
run into the vault was 8.3 m/s and they
weighed 77 kilograms.
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. The
run into the vault was 8.3 m/s and they
weighed 77 kilograms. KE= ½ m * V²
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. The
run into the vault was 8.3 m/s and they
weighed 77 kilograms. KE= ½ m * V2
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
“The homework isn’t
color coded.”
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. The
run into the vault was 8.3 m/s and they
weighed 77 kilograms. KE= ½ m * V2
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
“The homework isn’t
color coded.”
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. The
run into the vault was 8.3 m/s and they
weighed 77 kilograms. KE= ½ m * V²
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. The
run into the vault was 8.3 m/s and they
weighed 77 kilograms. KE= ½ m * V²
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
KE
KE
KE
KE
= ½ m * V²
=
=
=
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. The
run into the vault was 8.3 m/s and they
weighed 77 kilograms. KE= ½ m * V²
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
KE
KE
KE
KE
= ½ m * V²
= .5* 77 kg * 8.3 m/s
=
=
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. The
run into the vault was 8.3 m/s and they
weighed 77 kilograms. KE= ½ m * V²
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
KE
KE
KE
KE
= ½ m * V²
= .5* 77 kg * 8.3 m/s
= .5* 77 kg * 68.89 m/s
=
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. The
run into the vault was 8.3 m/s and they
weighed 77 kilograms. KE= ½ m * V²
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
KE
KE
KE
KE
=
=
=
=
½ m * V²
.5* 77 kg * 8.3 m/s
.5* 77 kg * 68.89 m/s
2652.2 Joules
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh (9.8 m/s²)
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh (9.8 m/s²)
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh (9.8 m/s²)
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
“Organize
your work
please.”
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh (9.8 m/s²)
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
PE= mgh
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh (9.8 m/s²)
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
PE= mgh
PE = 77 kg* 9.8 m/s² * 3 m
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh (9.8 m/s²)
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
PE= mgh
PE = 77 kg* 9.8 m/s² * 3 m
PE = 2263.8 Joules
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh (9.8 m/s²)
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
PE= mgh
PE = 77 kg* 9.8 m/s² * 3 m
PE = 2263.8 Joules
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh (9.8 m/s²)
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
PE= mgh
PE = 77 kg* 9.8 m/s² * 3 m
PE = 2263.8 Joules
KE = 2652.2 Joules
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh (9.8 m/s²)
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
PE= mgh
PE = 77 kg* 9.8 m/s² * 3 m
PE = 2263.8 Joules
KE = 2652.2 Joules
-388.4 Joules for heat,
sound, and other losses.
• Please calculate the potential energy of a
pole-vaulter at the top of their vault. Their
height was 3 meters and they weighed 77
kilograms. PE= mgh (9.8 m/s²)
– (Assume all energy in the vault was transformed into
potential energy to make this question easier.)
PE= mgh
PE = 77 kg* 9.8 m/s² * 3 m
PE = 2263.8 Joules
KE = 2652.2 Joules
-388.4 Joules for heat,
sound, and other losses.
• Activity! Please make a roller coaster on a
page in your science journal.
– Color Key Areas with high potential energy
and kinetic energy.
Copyright © 2010 Ryan P. Murphy
• Activity! Please make a roller coaster on a
page in your science journal.
– Color Key Areas with high potential energy
and kinetic energy.
Copyright © 2010 Ryan P. Murphy
• Activity! Please make a roller coaster on a
page in your science journal.
– Color Key Areas with high potential energy
and kinetic energy.
Copyright © 2010 Ryan P. Murphy

Centripetal Force: A force that makes a body
follow a curved path.
Copyright © 2010 Ryan P. Murphy

Centripetal Force: A force that makes a body
follow a curved path.
Copyright © 2010 Ryan P. Murphy
• Video! Centripetal Force
– http://www.youtube.com/watch?v=XWCBk9Vl-rc
• Gravity from the mass of the sun keeps the
earth from heading out into space.
Copyright © 2010 Ryan P. Murphy
• Gravity from the mass of the sun keeps the
earth from heading out into space.
Copyright © 2010 Ryan P. Murphy
• Gravity from the mass of the sun keeps the
earth from heading out into space.
Copyright © 2010 Ryan P. Murphy
• The World of the Hammer Throw. Centripetal
and Centrifugal Force
– http://www.youtube.com/watch?v=tB00eDfTNhs
• Activity (Optional) Funky foam tube roller
coaster.
– Use ½ inch foam pipe insulation cut in half,
duct tape to connect the tubes and anchor, cup
to catch at end, and marbles.
• Create a one page visual of a roller
coaster with drawings.
– Name your coaster.
– Create a not to scale visual that will be
achievable with the materials provided by
teacher.
– Class will vote to choose a model and build
the coaster.
– Calculate the PE and KE.
– Find the mass of the marble.
– Measure the height of the coaster.
– Calculate the velocity.
• Distance / meters divided by seconds and direction
• Create a one page visual of a roller
coaster with drawings.
– Name your coaster.
– Create a not to scale visual that will be
achievable with the materials provided by
teacher.
– Class will vote to choose a model and then
build the coaster.
– Calculate the PE and KE.
– Find the mass of the marble.
– Measure the height of the coaster.
– Calculate the velocity.
• Distance / meters divided by seconds and direction
• Academic Link! (Optional)
– http://www.youtube.com/watch?v=BSWl_Zj-CZs
• F=MA, PE, KE and more ramp activity.
– Available Sheet
• Activity! Kinetic and Potential Energy +
Newton’s Laws F=MA.
Copyright © 2010 Ryan P. Murphy
• Activity! Kinetic and Potential Energy +
Newton’s Laws F=MA.
Copyright © 2010 Ryan P. Murphy
• Please create this spreadsheet in your journal.
• Truck (D Battery) Car (AA Batter) – Cup (Parked Car)
Ramp
Height
Parked car
One
Washer
Parked Car Parked Car
Two Washer Three
Washers
Lowest
AA –Car_________
(Distance of
Parked Car)
D – Truck________
Middle
AA –Car___________
AA –Car_________
(Distance of
Parked Car)
D – Truck__________
D – Truck________
Highest
AA –Car___________
AA –Car_________
(Distance of
Parked Car)
D – Truck__________
D – Truck________
Make Prediction
after data
collection,
AA –Car_________
D – Truck________
Copyright © 2010 Ryan P. Murphy
Set-up of the activity. The height can change
by placing the rectangular block on its various
sides.
Ramp start line
Height
D
5cm
gap
Plastic Cup
Washers
1-3
AA
Meter Stick to measure distance
cup “parked car” travels after hit.
Copyright © 2010 Ryan P. Murphy
• Conduct trials with small car (AA Battery) with
one and three washers and the three different
heights, measuring the distance the parked
car traveled after hit in cm.
• Repeat with Truck / D Battery.
– Do not do medium height as we will predict later.
Copyright © 2010 Ryan P. Murphy
• F=MA, PE, KE and more ramp activity.
– Available Sheet
• F=MA, PE, KE and more ramp activity.
– Available Sheet
• Based on your data, make a prediction for
the distance the parked car should travel
for both the small car (AA) and truck (D)
on your spreadsheets for medium height
with two washers.
Copyright © 2010 Ryan P. Murphy
• Based on your data, make a prediction for
the distance the parked car should travel
for both the small car (AA) and truck (D)
on your spreadsheets for medium height
with two washers.
– Run some trials afterward to see if your
prediction is correct.
Copyright © 2010 Ryan P. Murphy
Increase in Friction /
Mass to move.
Increase
in Friction
/ Mass to
move.
• F=MA, PE, KE and more ramp activity.
– Available Sheet
• F=MA, PE, KE and more ramp activity.
– Available Sheet
• How did the height of the ramp affect the
movement of the parked car?
– Use potential energy and kinetic energy in
your response.
– Measure the height of the ramp, mass of the
batteries, and determine the Potential
Energy. PE=mgh
Copyright © 2010 Ryan P. Murphy
• How did the resistance to force (washers)
affect the movement of the parked car?
Copyright © 2010 Ryan P. Murphy
• How did the height of the ramp affect the
movement of the parked car?
Copyright © 2010 Ryan P. Murphy
• How did the height of the ramp affect the
movement of the parked car?
– Increasing the height of the ramp
increased the batteries potential energy.
Copyright © 2010 Ryan P. Murphy
• How did the height of the ramp affect the
movement of the parked car?
– Increasing the height of the ramp
increased the batteries potential energy.
This increase of potential energy created
an increase in kinetic energy /
(Acceleration) which caused the parked
car to move further (force).
Copyright © 2010 Ryan P. Murphy
• How did the resistance to force (washers)
affect the movement of the parked car?
Copyright © 2010 Ryan P. Murphy
• How did the resistance to force (washers)
affect the movement of the parked car?
– The more mass added to the parked car
(washers) decreased the distance it
traveled after being struck.
Copyright © 2010 Ryan P. Murphy
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity2
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/sec
– Velocity 3 m/sec
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/sec
– Velocity 3 m/sec
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
Organize your work!
– Gravity = 9.8 m/s²
PE= mgh
PE = ____ * ___ * ____
– Velocity 3 m/s
PE =
Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
Organize your work!
PE= mgh
PE = ____ * ___ * ____
PE =
Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m) PE=mgh
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m) PE=mgh
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
PE= .148kg * 9.8 m/s² * .06m
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m) PE=mgh
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
PE= .148kg * 9.8 m/s² * .06m
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m) PE=mgh
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
PE= .148kg * 9.8 m/s² * .06m
PE = .087 Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
• PE = .087 Joules
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
• PE = .087 Joules
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
KE=1/2 m * velocity²
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
• PE = .087 Joules
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
KE=1/2 m * velocity²
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
• PE = .087 Joules
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
KE=1/2 m * velocity²
KE=.5 * .148 * 3 m/s²
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
• PE = .087 Joules
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
KE=1/2 m * velocity²
KE=.5 * .148 * 3 m/s²
KE=.5 * .148 * 9 m/s²
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
• PE = .087 Joules
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
KE=1/2 m * velocity²
KE=.5 * .148 * 3 m/s²
KE=.5 * .148 * 9 m/s²
KE = .666 Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
• PE = .087 Joules
KE = .666 Joules
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
KE=1/2 m * velocity²
KE=.5 * .148 * 3 m/s²
KE=.5 * .148 * 9 m/s²
KE = .666 Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
• PE = .087 Joules
KE = .666 Joules
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
KE=1/2 m * velocity²
KE=.5 * .148 * 3 m/s²
KE=.5 * .148 * 9 m/s²
KE = .666 Joules
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
• PE = .087 Joules
KE = .666 Joules
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
KE=1/2 m * velocity²
KE=.5 * .148 * 3 m/s²
KE=.5 * .148 * 9 m/s²
KE = .666 Joules
Mechanical Energy (ME) =
• Find the Mechanical Energy of the large D
battery hitting the parked car from the highest
position.
• PE = mgh
KE = ½ mass * velocity²
• PE = .087 Joules
KE = .666 Joules
– D Battery mass = 148 g (.148kg)
– Height = 6 cm (.06m)
– Gravity = 9.8 m/s²
– Velocity 3 m/s
– Answer in Joules
KE=1/2 m * velocity²
KE=.5 * .148 * 3 m/s²
KE=.5 * .148 * 9 m/s²
KE = .666 Joules
Mechanical Energy (ME) = .753 Joules
• Question on homework: Describe three ways
potential energy of position as well as
potential chemical energy are combined with
kinetic energy to generate kinetic electrical
energy.
Copyright © 2010 Ryan P. Murphy
• Question on homework: Describe three ways
potential energy of position as well as
potential chemical energy are combined with
kinetic energy to generate kinetic electrical
energy.
Copyright © 2010 Ryan P. Murphy
• Hydropower : Potential energy turned into
kinetic energy of motion turned into kinetic
electrical energy.
Copyright © 2010 Ryan P. Murphy
• Hydropower : Potential energy turned into
kinetic energy of motion turned into kinetic
electrical energy.
Copyright © 2010 Ryan P. Murphy
• Hydropower gave rise to early industry.
– One of our earliest ways to harness energy.
Copyright © 2010 Ryan P. Murphy
• Hydropower gave rise to early industry.
– One of our earliest ways to harness energy.
Potential Energy
Copyright © 2010 Ryan P. Murphy
• Hydropower gave rise to early industry.
– One of our earliest ways to harness energy.
Potential Energy
Transfer to
Kinetic
Energy
Copyright © 2010 Ryan P. Murphy
• In Dinowrig, Wales. Water is pumped from
the lower lake to the upper lake when
electricity is low in demand.
• During times high electrical demand, the
stored potential energy flows downhill to
generate electricity (Kinetic).
• During times high electrical demand, the
stored potential energy flows downhill to
generate electricity (Kinetic).
• During times high electrical demand, the
stored potential energy flows downhill to
generate electricity (Kinetic).
• Kinetic energy to kinetic electrical energy
Copyright © 2010 Ryan P. Murphy
• Gravity turns potential energy in tides, into
kinetic energy (flowing tides) into kinetic
electrical energy.
Copyright © 2010 Ryan P. Murphy
• Geothermal
Copyright © 2010 Ryan P. Murphy
• Geothermal -Kinetic energy heat, turns water
into steam, water rises and runs a turbine to
generate electrical energy.
Copyright © 2010 Ryan P. Murphy
• Geothermal -Kinetic energy heat, turns water
into steam, water rises and runs a turbine to
generate electrical energy.
Copyright © 2010 Ryan P. Murphy
• Geothermal -Kinetic energy heat, turns water
into steam, water rises and runs a turbine to
generate kinetic electrical energy.
Copyright © 2010 Ryan P. Murphy
• Steam / Coal and wood burning electric
plant
• Nuclear energy – Nuclear reactions
generate kinetic electrical energy using
water, steam, and a turbine.
• When you lift an object, chemical energy
(a form of potential energy) stored in the
chemicals obtained from your digested
food is converted into the mechanical
energy (kinetic) used to move your arm
and the object upward and into heat given
off by your body.
Copyright © 2010 Ryan P. Murphy
• When you lift an object, chemical energy
(a form of potential energy) stored in the
chemicals obtained from your digested
food is converted into the mechanical
energy (kinetic). Which is then used to
move your body. Heat is produced
Copyright © 2010 Ryan P. Murphy
• When you lift an object, chemical energy
(a form of potential energy) stored in the
chemicals obtained from your digested
food is converted into the mechanical
energy (kinetic). Which is then used to
move your body. Heat is produced
Copyright © 2010 Ryan P. Murphy
• When you lift an object, chemical energy
(a form of potential energy) stored in the
chemicals obtained from your digested
food is converted into the mechanical
energy (kinetic). Which is then used to
move your body. Heat is produced
Copyright © 2010 Ryan P. Murphy
• When you lift an object, chemical energy
(a form of potential energy) stored in the
chemicals obtained from your digested
food is converted into the mechanical
energy (kinetic). Which is then used to
move your body. Heat is released.
Copyright © 2010 Ryan P. Murphy