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ENERGY
Chapter 12 Section 3
Warm-up
• Name different types of energy and their daily
uses.
What is Energy?
• Energy can be described as the ability to do
work.
• Look at the picture below where do you
believe energy is.
Potential Energy
• The energy that an object has because of the
position, shape, or condition of the object.
• There are two types of potential energy:
elastic and gravitational
Elastic Potential Energy
• The energy stored in any type of stretched or
compressed elastic material.
• Name different objects that would have elastic
potential energy.
Gravitational Potential Energy
• Energy that is stored in the gravitational field
is known as gravitational potential energy.
• Depends on both mass and height.
• Let’s look at the equation used for
gravitational potential energy
Gravitational Potential Energy
Equation
• grav.PE= mass x free-fall acceleration x height
Or in other terms
• PE= mgh
• PE= Potential energy, m= mass, g= free-fall
acceleration, h= height
Example
• A 65 kg rock climber ascends a cliff. What is
the climber’s gravitational potential energy at
a point 35m above the base of the cliff?
How to solve this problem
• Step 1: State the equation you are going to
use.
• PE=mgh
How to solve problem cont.
• Step 2: List the known and unknown values
given in the equation
• Known values
• mass= 65 kg
• height= 35 m
• free-fall acceleration g= 9.8m/s²
• Unknown value- gravitational potential
energy
How to solve the problem cont.
• Step 3: Plug in the values and solve.
• PE= (65 kg)(9.8m/s²)(35 m)
• PE= 2.2x 10⁴ kg x m²/s² = 2.2 x 10⁴ J
Practice Problems
• Try problem 1 and 2 on your own. Follow the
steps of the example given above.
Warm up
• A book weighing 8.0 kg is on a table 75cm
high. What is the gravitational potential
energy of the book?
Warm-up Solved
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•
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Step 1: State equation we are going to use
PE = mgh
Step 2: State known and unknown values
Known values
mass = 8.0kg g= 9.8m/s² h= 75cm
Unknown values
Potential Energy
Warm –up cont.
• Step 3: Plug in values and solve
• We have to convert 75cm to meters
• 75cm/ 100 = 0.75m
• PE = (8.0kg)(9.8m/s²)(.75m)
• PE = 58.8 J
Kinetic Energy
• The energy of a moving object due to the
object’s motion
• Just like with potential energy the amount of
kinetic energy is dependant on the objects
mass as well as its speed
Kinetic Energy Equation
• Kinetic Energy = ½ mass x velocity²
Or in other words
• KE = ½ mv²
• KE= kinetic energy m= mass v= velocity
Example Problem
• What is the kinetic energy of a 44 kg cheetah
running at 31 m/s?
How to solve the problem
• Step 1: Write out the equation you are going
to use
KE= ½ m x v²
How to solve the problem cont.
• Step 2: Write out the known and unknown
values
• m= 44kg
• v= 31 m/s
• Unknown
• KE= ?
How to solve the problem cont.
• Step 3: Plug in values and solve
• KE= ½ (44)(31)²
• KE= 2.1 x 10⁴ J
Mechanical Energy
• The amount of work an object can do
because of the object’s kinetic and potential
energies
• Remember all objects contain both kinetic and
potential energy
Mechanical Energy can change forms
• Just like kinetic and potential energy
mechanical energy can change forms as well.
• Changes from mechanical to nonmechanical
energy due to the presence of friction and air
resistance
• Nonmechanical energy is a special form of
kinetic or potential energy
Conservation of Energy
Section 12.4
Conservation of Energy
• Imagine you are riding the rollercoaster
pictured below.
• Where is the energy coming from?
The Law of Conservation of Energy
• Energy can not be created or destroyed.
• What does that mean for the energy that is
found around us.
Energy does not appear or disappear
• Energy works in systems.
• Two types of systems : closed and open
Closed System
• When the flow of energy into and out of a
system is small enough to be ignored is called
a closed system
• Based on the information above what is an
example of a closed system.
Open System
• Occur when the energy exchange occurs
within the space that surrounds them
• Why is a car an example of an open system
Efficiency
• A quantity usually expressed as a percentage
that measures the ratio of useful work
output to work input
• efficiency = useful work output/ work input
• Since this is expressed as a percentage we
must multiply the answer by 100.
Efficiency Problem
• A sailor uses a rope and an old squeaky pulley
to raise a sail that weighs 140 N. He finds that
he must do 180 J of work on the rope in order
to raise the sail by 1 m (doing 140 J of work on
the sail). What is the efficiency of the pulley?
Express your answer as a percentage.
Efficiency Problem
• Step 1: State the equation
• efficiency = useful work output/ work imput
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•
•
•
Step 2: State known and unknown values
Known work input- 180 J
Useful work output- 140 J
Unknown values- efficiency percentage
Efficiency Problem
• Step 3: Solve
• 140/180= .78
• .78 x 100 = 78 %
Practice Problems
• Try pg 407 # 1 and pg 408 # 7 for practice on
your own