Energy Forms and Transformations

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Transcript Energy Forms and Transformations

Lesson 1
unit 4
Energy
 Energy can be defined as the capacity to work or to
accomplish a task.
 Example: burning fuel allows an engine to do the work
of moving a car.
Forms of Energy
 Chemical Potential Energy
 In chemical reactions, new molecules are formed and
chemical potential energy is released or absorbed.
Sound Energy
 Produced by vibrations; the energy travels by waves
through a material to the receiver.
Radiant Energy
 Components of the electromagnetic spectrum have
characteristics of waves, such as wavelengths,
frequencies, and energies; they travel in a vacuum at the
speed of light (3.0 x 108 m/s).
Nuclear Energy
 The nucleus of every atom contains energy.


Nuclear fission – The breaking down of atoms
Nuclear fusion – the joining of atoms
Electrical Energy
 Electrons in a electric circuit can transfer energy to the
components of the circuit.
Thermal Energy
 The more rapidly atoms and molecules move, the
greater their total thermal energy.
Gravitational Potential Energy
 A raised object has stored energy due to its position
above some reference level
Kinetic Energy
 Energy of motion, every moving object has this energy
Elastic Potential Energy
 Is stored in objects that are stretched or compressed.
Energy Transformations
 The 9 forms of energy listed above are able to change from
one to another; this change is called energy
transformation. A energy transformation equation can
be used to summarize the changes in a transformation.
Example: a microwave
 Electrical energy  Radiant Energy  Thermal Energy
Energy-Transformation Technology
 A device used to transform energy for a specific
purpose.
 Work – The energy transferred to an object by a force
applied over a measured distance.
 As the force or the distance increases, the work also
increases.
 W is the work done on the object
 F is the magnitude of the applied force in the direction
of the displacement
 ∆d is the magnitude of the displacement
 Because the applied force and displacement are in the
same direction there are no signs needed making work
a scalar quantity, it has magnitude but no direction.
 Work is measured in Newton-meters (N•m)
 One Newton-meter is referred to as a joule
Example 1
 A store employee exerts a horizontal applied force of
magnitude 44 N on a set of carts. How much work is
done by the employee when pulling the carts 15m?
Express the answer in joules and kilojoules.
 F = 44 N
 ∆d = 15 m
 W=?
 W = F ∆d
 W = ( 44 N) x ( 15 m)
 W = 6.6 x 102 J
 Therefore, the work done by the employee in
pulling the cars is 6.6 x 102 J, or 0.66 kJ.
Negative Work
 If the force is opposite in direction to the
displacement, the work done is negative.
 W = -F∆d
 Consider the previous example but this time a second
employee exerts a horizontal force in the opposite
direction to the 44 N force.
 Since the force is in the opposite direction to the
displacement, the work done by the second employee
on the carts is negative.
 Now the total work done by the two employees on the
carts is the sum of the positive and negative values.
Work done by the 44 N force
Work done by the 14 N force
Total work done by the forces on the carts
 Negative work can also occurs with kinetic friction
because the force of kinetic friction always happens in
the direction opposite to the direction of motion of the
object.
 W = -FK∆d
 -FK is the magnitude of the force of kinetic friction
Example 2
 A toboggan carrying two children (total mass 85kg)
reaches its maximum speed at the bottom of a hill. It
then glides to a stop 21 m along a horizontal surface.
The coefficient of friction between the toboggan and
the snowy surface is 0.11.
A. Calculate the magnitude of the force of kinetic
friction acting on the toboggan
B. Calculate the work done by the force of kinetic
friction on the toboggan
 m = 85 kg
 g = 9.8 N/kg
 µk = 0.11
 FK = ?
 FK = µkFN
 FK = µkmg
 FK = (0.11)(85 kg) (9.8 N/kg)
 FK = 92 N
 The magnitude of the kinetic force is 92 N
 Calculate the work done by the force of kinetic friction on
the toboggan
 ∆d = 21 m
 W=?
 W = -FK∆d
 W = -(92 N) (21 m)
 W = - 1.9 x 103 J
Therefore, the work done by the kinetic friction on the
toboggan is -1.9kJ
 The work done by friction has been transformed into
thermal energy. This is observed as a increase in
temperature.
Work done in raising objects
 To lift an object to a higher position, an upward force
must be applied against the downward force of gravity
acting on the object.
 If the force applied and the displacement are both
vertically upward and no acceleration occurs, the work
done by the upward force is positive and calculated
with the formula:
 W = F∆d.
Work done in raising objects
 The force in this case is equal in magnitude to the
weight of the object or the force of gravity on the
object.
 F = mg
Example 3
 A bag of groceries of mass 8.1 kg is raised vertically at a
slow, consistent velocity from the floor to a countertop,
for a distance of 92 cm. Calculate
A. The force needed to raise the bag of groceries at a
constant velocity
B. The work done on the bag of groceries by the upward
force.
 m = 8.1 kg
 g = 9.8 N/kg
F = ?
 F = mg
 F = (8.1 kg) (9.8 N/kg)
 F = 79 N
 Therefore, the force needed is 79 N.
 The work done on the bag of groceries by the upward
force.
 ∆d = 92 cm = 0.92
 W =?
 W = F∆d
 W = (79 N) (0.92 m)
 W = 73 J
 Therefore, the work done on the bag of groceries
by the upward force is 73 J.
Zero Work
 Sometimes an object can experience a force, a
displacement, or both, yet no work is done on the object.
 Example: Holding a box in your hands. The box is not
moving so no work is done so the displacement is zero.
 A puck on an air hockey table is moving but it does not have
force acting parallel to the movement as friction is negligible.
Work and springs
 When a spring or rubber band is stretched, the more
difficult it becomes to stretch. If a typical spring was
stretched and then graphed the force applied to it
against the stretch experienced by the sting, the line
would be straight.
 The area under the triangle can be calculated by the
equation
 this yields the work done by the force used to stretch
the spring by an amount ∆x. The slope of the line, is
found by using the equation

or
 this represents the force constant of the spring, k. The
force constant represents the stiffness of the spring.
Questions
1.
Using an energy transformation equation, show how
energy is transformed for each of the following:
1.
2.
A hotdog grilled on an outdoor BBQ
A truck acceleration on a highway
2. A Truck does 3.2 kJ of work pulling horizontally on a
car to move 1.8 m horizontally in the direction of the
force. Calculate the magnitude of the force.
3. A store clerk moves a 4.4 kg box of soap at a constant
velocity along a shelf by pushing it with a horizontal
force of magnitude 8.1 N. The clerk does 5.9 J of work
on the box. How far did the box move?
 A 150 g text book is lifted from the floor to the shelf 2.0
m above. Calculate
 The force needed to lift the book without acceleration.
 The work done by the force on the book to lift it up to a shelf.
 An Electric forklift is capable of doing 4.0 x 105 J of
work on a 4.5 x 103 kg load, What height can the truck
lift the load?
 A student pushes against a large tree with a force of
magnitude 250 N, but the tree does not move. How
much work has the student done on the tree?
 Draw a graph of a spring that has the following data
Calculate the work done in stretching of
the spring represented in the graph you
just drew.
a) 0.12 m
b) 0.24 m
Force
(N)
5
Stretch
(m)
0.06
10
0.12
15
0.18
20
0.24