Download Calculate the change in kinetic energy

Principles of
Technology/Physics in Context
(PT/PIC)
Chapter 7
Work and Energy 2
Text p. 126 - 132
Key Objectives
At the conclusion of this chapter you’ll be able to:
• Solve problems involving power and work.
• State the equation for calculating kinetic
energy, and solve problems using this
equation.
• State the equation for calculating gravitational
potential energy, and solve problems using this
equation.
• Solve problems that relate changes in kinetic
energy to changes in gravitational potential
energy.
7.3 ENERGY
• When work is done on an object, the
“energy” of the object is changed.
• Energy is a very broad term related to
work, and it has a variety of forms. In this
chapter we will consider two of these
forms, kinetic and potential energy.
7.3 ENERGY
• Together the kinetic energy and the
potential (gravitational and elastic)
energies are called the mechanical energy
of the object.
7.3 ENERGY
• Kinetic Energy
• An object is traveling along a frictionless
horizontal surface.
• A constant force is applied to the object in
the direction of its displacement.
• What is the result of the work done on the
object?
7.3 ENERGY
•
•
•
•
Kinetic Energy
We know from Newton’s second law that
F = ma, therefore W = ma d.
If the force on the object is also constant,
its acceleration is constant and we can
write
7.3 ENERGY
• If we solve this relationship for a d, we find
that
7.3 ENERGY
• We can substitute this solution for a d into
our work relationship:
7.3 ENERGY
• The work done on the object changes a
quantity called kinetic energy that is
related to the mass and the square of the
speed of the object.
7.3 ENERGY
• We define kinetic energy (KE) to be:
• Therefore, work is equal to the change in
kinetic energy.
• W = KEf - KEi = ΔKE.
Assessment Question 1
All of the following statements are TRUE EXCEPT:
A. When work is done on an object, the “energy”
of the object does not change.
B. Energy is a very broad term related to work,
and it has a variety of forms.
C. Together the kinetic energy and the potential
(gravitational and elastic) energies are called
the mechanical energy of the object.
D. Work is equal to the change in kinetic energy.
7.3 ENERGY
• PROBLEM
• A 10-kilogram object subjected to a 20.-newton
force moves across a horizontal, frictionless
surface in the direction of the force.
• Before the force was applied, the speed of the
object was 2.0 meters per second.
• When the force is removed, the object is
traveling at 6.0 meters per second.
• Calculate the initial kinetic energy (KEi)
7.3 ENERGY
• SOLUTION
7.3 ENERGY
• PROBLEM
• A 10-kilogram object subjected to a 20.-newton
force moves across a horizontal, frictionless
surface in the direction of the force.
• Before the force was applied, the speed of the
object was 2.0 meters per second.
• When the force is removed, the object is
traveling at 6.0 meters per second.
• Calculate the final kinetic energy (KEf)
7.3 ENERGY
• SOLUTION
7.3 ENERGY
• PROBLEM
• A 10-kilogram object subjected to a 20.-newton
force moves across a horizontal, frictionless
surface in the direction of the force.
• Before the force was applied, the speed of the
object was 2.0 meters per second.
• When the force is removed, the object is
traveling at 6.0 meters per second.
• Calculate the change in kinetic energy (ΔKE)
7.3 ENERGY
• SOLUTION
Assessment Question 2
•
A 10 kg (m) object subjected to 10 N (F) force
moves across a horizontal, frictionless surface
in the direction of the force.
• Before the force was applied, the speed of the
object was 2 m/s (vi)
• When the force is removed, the object is
traveling at 3 m/s (vf)
Calculate the change in kinetic energy (ΔKE)
ΔKE = KEf - KEi = ½ (m vf 2) – ½ (m vi 2) =
ΔKE = ½( 10 kg∙ (3 m/s)2) - ½( 10 kg∙ (2 m/s)2) =
A. 10 J
B. 15 J
C. 20 J
D. 25 J
7.3 ENERGY
• PROBLEM
• A 10-kilogram object subjected to a 20.-newton
force moves across a horizontal, frictionless
surface in the direction of the force.
• Before the force was applied, the speed of the
object was 2.0 meters per second.
• When the force is removed, the object is
traveling at 6.0 meters per second.
• Calculate work (W)
7.3 ENERGY
• SOLUTION
7.3 ENERGY
• PROBLEM
• A 10-kilogram object subjected to a 20.-newton
force moves across a horizontal, frictionless
surface in the direction of the force.
• Before the force was applied, the speed of the
object was 2.0 meters per second.
• When the force is removed, the object is
traveling at 6.0 meters per second.
• Calculate distance (d)
7.3 ENERGY
• SOLUTION
7.3 ENERGY
Gravitational Potential Energy
• Darya lifts a textbook vertically off a desk
and holds it above her head.
• It is clear that Darya has done work on the
book because she has applied a force
through a distance.
• However the change in the kinetic energy
of the book is zero.
• What did Darya’s work accomplish?
7.3 ENERGY
Gravitational Potential Energy
• In this case, the work overcame the
attraction of the gravitational field, and as
a result the position of the book with
respect to the Earth changed.
• We relate this change to a quantity we call
gravitational potential energy.
7.3 ENERGY
Gravitational Potential Energy
• To calculate the change in the gravitational
potential energy (PE) of the object, we
measure the work done on the object.
7.3 ENERGY
Gravitational Potential Energy
• The force needed to overcome gravity is
simply Fg, which is equal to mg.
• Therefore, since W = Fg∙d, we define the
change in the gravitational potential
energy (ΔPE) as:
7.3 ENERGY
Gravitational Potential Energy
• Here we use Δh, rather than d, to represent
the change in vertical displacement above
the Earth.
7.3 ENERGY
Gravitational Potential Energy
• Since we are dealing with a scalar quantity,
we will not consider the algebraic signs of
the quantities involved.
7.3 ENERGY
Gravitational Potential Energy
• We will simply agree that an object
decreases its gravitational potential energy
as it moves closer to the Earth (and vice
versa).
Assessment Question 3
All of the following statements are TRUE EXCEPT:
A. To increase the gravitational potential energy of
an object work must be performed on it.
B. You have greater gravitational potential energy
in class and less when you are flying in an
airplane.
C. Gravitational Potential Energy is a scalar
quantity.
D. An object decreases its gravitational potential
energy as it moves closer to the Earth
7.3 ENERGY
Gravitational Potential Energy
• PROBLEM
• A 2.00-kilogram mass is lifted to a height of
10.0 meters above the surface of the Earth.
Calculate the change in the gravitational
potential energy of the object.
7.3 ENERGY
Gravitational Potential Energy
• SOLUTION
• Since the object has moved away from the
Earth’s surface, its gravitational potential energy
has increased by 196 joules.
Assessment Question 4
A 12 kg (m) mass is lifted to a height of 15 m (h)
above the surface of the Earth.
g = 9.8 m/s2
• Calculate the change in the gravitational potential
energy (ΔPE) of the object.
ΔPE = mgh = 12 kg∙ 9.8 m/s2∙ 15 m =
A. 520 J
B. 1800 J
C. 2500 J
D. 3200 J
7.3 ENERGY
Gravitational Potential Energy
• For a change in gravitational energy to
occur, there must be a change in the
vertical displacement of an object;
• If it is moved only horizontally, its
gravitational potential energy change is
zero.
7.3 ENERGY
Gravitational Potential Energy
• If an object is moved up an inclined plane,
its potential energy change is measured by
calculating only its vertical displacement;
the horizontal part of its displacement does
not change its potential energy.
Assessment Question 5
All of the following statements are TRUE EXCEPT:
A. For a change in gravitational energy to occur,
there must be a change in the vertical
displacement of an object;
B. If an object is moved only horizontally, its
gravitational potential energy change is zero.
C. If an object is moved up an inclined plane, its
potential energy change is measured by
calculating only its vertical displacement;
D. The horizontal component of an object’s
displacement is needed to calculate its
potential energy.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• Ketan tosses an object upward, and it
returns to the Earth.
• Let’s analyze the motion of this object using
an “energy” point of view.
• For simplicity we will ignore air resistance.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• Ketan’s hand does work on the object.
• The work is transformed into kinetic energy.
As the object rises, we observe that its
speed decreases to zero.
• As a result, the kinetic energy of the object
is decreasing while its potential energy is
increasing.
• This represents a transformation of energy
from kinetic to potential.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• On the downward trip, the speed of the
object increases.
• As the potential energy of the object
decreases, its kinetic energy increases.
• This represents a transformation from
potential to kinetic energy.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• In the system, the sum of potential energy
and kinetic energy (the total mechanical
energy) has been conserved (i.e., is
constant);
• a change in one is accompanied by an
opposite change in the other.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• In the system, the sum of potential energy
and kinetic energy (the total mechanical
energy) has been conserved (i.e., is
constant);
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• This can also be expressed as the law of
conservation of energy, which states that
energy can neither be created nor
destroyed.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• In an ideal mechanical system (a closed
system upon which no friction or other
external forces act) the total mechanical
energy is constant.
Assessment Question 6
All of the following statements are TRUE EXCEPT:
A. According to the law of conservation of energy, friction
forces can be ignored because energy is conserved.
B. According to the law of conservation of energy, a
change in potential or kinetic energy is accompanied by
an opposite change in the other.
C. According to the law of conservation of energy, the sum
of potential energy and kinetic energy (total mechanical
energy) in a system is conserved
D. According to the law of conservation of energy, energy
can neither be created nor destroyed.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• PROBLEM
• A 0.50-kilogram ball is projected vertically
and rises to a height of 2.0 meters above
the ground.
• Calculate: (a) the increase in the ball’s
potential energy
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• SOLUTION
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• PROBLEM
• A 0.50-kilogram ball is projected vertically
and rises to a height of 2.0 meters above
the ground.
• Calculate: (b) the decrease in the ball’s
kinetic energy
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• SOLUTION
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• PROBLEM
• A 0.50-kilogram ball is projected vertically
and rises to a height of 2.0 meters above
the ground.
• Calculate: (c) the initial kinetic energy, and
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• SOLUTION
• (c) At the highest point, the speed of the
ball is zero; therefore its kinetic energy is
zero. As a result, the initial kinetic energy
represents the change in the kinetic
energy of the object.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• PROBLEM
• A 0.50-kilogram ball is projected vertically
and rises to a height of 2.0 meters above
the ground.
• Calculate: (d) the initial speed of the ball.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• SOLUTION
• (c)
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• SOLUTION
• (d)
Assessment Question 7
A 2 kg (m) ball is thrown vertically to a height of 10 m
(h) above the surface of the Earth. g = 9.8 m/s2
Calculate the initial speed (vi) of the ball.
ΔPE = ΔKE = mgh = ½ (m vi 2)
vi = √(2gh) = √(2 ∙ 9.8 m/s2∙ 10 m) =
A. 14 m/s
B. 22 m/s
C. 35 m/s
D. 49 m/s
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• An amusement-park roller coaster is an
example of the interchange of the kinetic
and potential energies of the coaster car.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• As the car falls, its kinetic energy
increases; as it rises, its kinetic energy
decreases, as shown in the diagram.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• Subsequent hills are made shorter and
shorter so that the car will continue to
have kinetic energy as it moves along the
track.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• A simple pendulum, shown in the following
diagram, is another device that illustrates
the transformation between kinetic and
potential energies.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• In the absence of friction, the swing of a
pendulum back and forth will go on
continuously.
• This type of motion, known as simple
harmonic motion (SHM), occurs often in
nature.
• For example, the oscillation of a spring
and the vibration of a tuning fork are
examples of SHM.
Assessment Question 8
All of the following statements are TRUE EXCEPT:
A. An amusement-park roller coaster car falls, its
kinetic energy increases; as it rises, its kinetic
energy decreases,
B. The final hill on a roller coaster track is always
the highest hill on the track and the first hill is
the shortest.
C. In the absence of friction, the swing of a
pendulum back and forth will go on
continuously.
D. A swinging pendulum, the oscillation of a spring
and the vibration of a tuning fork and other
examples of simple harmonic motion (SHM),
occurs often in nature.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• The time needed to complete one full
swing of the pendulum is known as its
period (T).
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• The period of a pendulum (for small angles) is
related to the length of the pendulum (l ) and to
gravitational acceleration (g) according to the
relationship
• Note that the period of a simple pendulum is
independent of the mass of the bob.
Assessment Question 9
Calculate the period (T) of a pendulum with a
length of 1.5 m (l ).
g = 9.8 m/s2
T = 2π√(l /g) =
T = 2∙3.14√(1.5 m/ 9.8 m/s2) =
A. 2.5 s
B. 7.9 s
C. 14 s
D. 29 s
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• PROBLEM
• A pendulum whose bob weighs 12
newtons is lifted a vertical height of 0.40
meter from its equilibrium position.
Calculate: (a) the change in potential
energy between maximum height and
equilibrium height
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• SOLUTION
• (a) We take the potential energy at the
lowest point to be zero. Then:
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• PROBLEM
• A pendulum whose bob weighs 12
newtons is lifted a vertical height of 0.40
meter from its equilibrium position.
Calculate: (b) the gain in kinetic energy
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• SOLUTION
• A pendulum whose bob weighs 12
newtons is lifted a vertical height of 0.40
meter from its equilibrium position.
Calculate: (b) the gain in kinetic energy
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• PROBLEM
• A pendulum whose bob weighs 12
newtons is lifted a vertical height of 0.40
meter from its equilibrium position.
Calculate: (c) the velocity at the
equilibrium point.
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• SOLUTION
• (c) First we calculate
the mass of the bob:
7.3 ENERGY
Interaction of Gravitational Potential and
Kinetic Energy (Conservation of Mechanical Energy)
• SOLUTION
• (c) Then we calculate
the velocity:
Conclusion
• If work is done on an object, the work may
be used to change the object’s kinetic
energy (the energy associated with its
motion), its potential energy (the energy
associated with its position), or its internal
energy (the energy associated with its
atoms and molecules).
Assessment Question 10
•
A.
B.
C.
D.
A pendulum whose bob has a mass of 1 kg (m) is
hangs 7 m (h) from its equilibrium point.
g = 9.8 m/s2
Calculate the change in potential energy (ΔPE)
between maximum height and equilibrium height
ΔPE = PEf-PEi = mgh - 0 =
ΔPE = 1 kg ∙ 7m∙ 9.8 m/s2 =
25 J
48 J
69 J
94 J