WORK, ENERGY AND POWER 2010
Definition and Mathematics of Work
When a force acts upon an object to cause a displacement of the object, it is said that work was done
upon the object. There are three key ingredients to work - force, displacement, and cause. In order for a
force to qualify as having done work on an object, there must be a displacement and the force must
cause the displacement. There are several good examples of work which can be observed in everyday
life - a horse pulling a plow through the field, a father pushing a grocery cart down the aisle of a grocery
store, a freshman lifting a backpack full of books upon her shoulder, a weightlifter lifting a barbell above
his head, an Olympian launching the shot-put, etc. In each case described here there is a force exerted
upon an object to cause that object to be displaced.
Read the following five statements and determine whether or not they represent examples of
Statement
Answer with
Explanation
A teacher applies a force to a wall and becomes exhausted.
A book falls off a table and free falls to the ground.
A waiter carries a tray full of meals above his head by one arm straight across the
room at constant speed. (Careful! This is a very difficult question which will be
discussed in more detail later.)
A rocket accelerates through space Mathematically, work can be expressed by the following equation.
W= F.ΔX COS Θ
where F is the force, ΔX is the displacement, and the angle (theta) is defined as the angle between the
force and the displacement vector. Perhaps the most difficult aspect of the above equation is the angle
"theta." The angle is not just any 'ole angle, but rather a very specific angle. The angle measure is
defined as the angle between the force and
the displacement. To gather an idea of its
meaning, consider
the following three
scenarios.

Scenario A: A force acts
rightward upon an object as it is
displaced rightward. In such an
instance, the force vector and
the displacement vector are in
the same direction. Thus, the
angle between F and d is 0
degrees.
WORK, ENERGY AND POWER 2010

Scenario B: A force acts leftward upon an object which is displaced rightward. In
such an instance, the force vector and the displacement vector are in the opposite
direction. Thus, the angle between F and d is 180 degrees.

Scenario C: A force acts upward on an object as it is displaced rightward. In such
an instance, the force vector and the displacement vector are at right angles to each
other. Thus, the angle between F and d is 90 degrees.
To Do Work, Forces Must Cause Displacements
Let's consider Scenario C above in more detail. Scenario C
involves a situation similar to the waiter who carried a tray full of
meals above his head by one arm straight across the room at constant speed. It
was mentioned earlier that the waiter does not do work upon
the tray as he carries it across the room. The force supplied by
the waiter on the tray is an upward force and the displacement
of the tray is a horizontal displacement. As such, the angle
between the force and the displacement is 90 degrees. If the
work done by the waiter on the tray were to be calculated, then the results would be 0. Regardless of
the magnitude of the force and displacement, F*d*cosine 90 degrees is 0 (since the cosine of 90
degrees is 0). A vertical force can never cause a horizontal displacement; thus, a vertical force does not
do work on a horizontally displaced object!!
It can be accurately noted that the waiter's hand did push forward on the tray for a brief period of time
to accelerate it from rest to a final walking speed. But once up to speed, the tray will stay in its straightline motion at a constant speed without a forward force. And if the only force exerted upon the tray
during the constant speed stage of its motion is upward, then no work is done upon the tray. Again, a
vertical force does not do work on a horizontally displaced object.
The equation for work lists three variables - each variable is associated with
one of the three key words mentioned in the definition of work (force,
displacement, and cause). The angle theta in the equation is associated with
the amount of force which causes a displacement. As mentioned in a previous unit,
when a force is exerted on an object at an angle to the horizontal, only a part
of the force contributes to (or causes) a horizontal displacement. Let's consider
the force of a chain pulling upwards and rightwards upon Fido in order to drag
Fido to the right. It is only the horizontal component of the tension force in the
chain which causes Fido to be displaced to the right. The horizontal component
is found by multiplying the force F by the cosine of the angle between F and d.
In this sense, the cosine theta in the work equation relates to the cause factor - it selects the portion of
the force which actually causes a displacement.
WORK, ENERGY AND POWER 2010
The Meaning of Theta
When determining the measure of the angle in the work equation, it is
important to recognize that the angle has a precise definition - it is the
angle between the force and the displacement vector. Be sure to avoid
mindlessly using any 'ole angle in the equation. A common physics lab
involves applying a force to displace a cart up a ramp to the top of a
chair or box. A force is applied to a cart to displace it up the incline at
constant speed. Several incline angles are typically used; yet, the force
is always applied parallel to the incline. The displacement of the cart is
also parallel to the incline. Since F and d are in the same direction, the
angle theta in the work equation is 0 degrees. Nevertheless, most
students experienced the strong temptation to measure the angle of incline and use it in the equation.
Don't forget: the angle in the equation is not just any 'ole angle. It is defined as the angle between the
force and the displacement vector.
The Meaning of Negative Work
On occasion, a force acts upon a moving object to hinder a displacement. Examples might include a car
skidding to a stop on a roadway surface or a baseball runner sliding to a stop on the infield dirt. In such
instances, the force acts in the direction opposite the objects motion in order to slow it down. The force
doesn't cause the displacement but rather hinders it. These situations involve what is commonly called
negative work. The negative of negative work refers to the numerical value which results when values of
F, d and theta are substituted into the work equation. Since the force vector is directly opposite the
displacement vector, theta is 180 degrees. The cosine(180 degrees) is -1 and so a negative value results
for the amount of work done upon the object. Negative work will become important (and more
meaningful) as we begin to discuss the relationship between work and energy
For each diagram, state the numerical value that should be substituted in for θ when using the
work equation W = F cos θ. ΔX
. Refer to the following information for the next question.
A block is being pushed across a frictionless surface by a force at 30º
WORK, ENERGY AND POWER 2010
What is θ when calculating the work done by F?
Refer to the following information for the next question.
A block is sliding across a horizontal surface which is not frictionless.
What is θ when calculating the work done by F?
Refer to the following information for the next question.
A book is being held at a constant height and moved at a constant speed across a room.
WORK, ENERGY AND POWER 2010
What is θ when calculating the work done by the person carrying the
book?
Refer to the following information for the next question.
A block is being dragged up a 20º incline by a rope that acts at an angle of 15º to the incline's surface.
What is θ when calculating the work done by the tension in the rope?
Calculating the Amount of Work Done by Forces
1.
Apply the work equation to determine the amount of work done by the applied force in each of
the three situations described below.
WORK, ENERGY AND POWER 2010
2. On many occasions, there is more than one force acting upon an object. A free-body diagram is a
diagram which depicts the type and the direction of all the forces acting upon an object. The following
descriptions and their accompanying free-body diagrams show the forces acting upon an object. For
each case, indicate which force(s) are doing work upon the object. Then calculate the work done by
these forces.
Free-Body
Diagram
Forces
Doing Work
Amount of
Work Done
on the
by Each Force
Object
A 10-N forces is applied to push a block across a friction free surface for a
displacement of 5.0 m to the right.
A 10-N frictional force slows a moving block to a stop after a displacement
of 5.0 m to the right.
A 10-N force is applied to push a block across a frictional surface at
constant speed for a displacement of 5.0 m to the right.
An approximately 2-kg object is sliding at constant speed across a friction
free surface for a displacement of 5 m to the right.
An approximately 2-kg object is pulled upward at constant speed by a 20-N
force for a vertical displacement of 5 m.
3. Before beginning its initial descent, a roller coaster car is always pulled up the first hill to a high initial
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height. Work is done on the car (usually by a chain) to achieve this initial height. A coaster designer is
considering three different incline angles at which to drag the 2000-kg car train to the top of the 60meter high hill. In each case, the force applied to the car will be applied parallel to the hill. Her critical
question is: which angle would require the most work? Analyze the data, determine the work done in
each case, and answer this critical question.
Angle
Force
Distance
35 deg1.12 x 104 N 105 m
a.
Work (J)
45 deg1.39 x 104 N 84.9 m
b.
55 deg1.61 x 104 N 73.2 m
c.
4. Ben Travlun carries a 200-N suitcase up three flights of stairs (a height of 10.0 m) and then pushes it
with a horizontal force of 50.0 N at a constant speed of 0.5 m/s for a horizontal distance of 35.0 meters.
How much work does Ben do on his suitcase during this entire motion?
5. A force of 50 N acts on the block at the angle shown in the diagram. The block moves a horizontal
distance of 3.0 m. How much work is done by the applied force?
6. How much work is done by an applied force to lift a 15-Newton block 3.0 meters vertically at a
constant speed?
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7. A student with a mass of 80.0 kg runs up three flights of stairs in 12.0 sec. The student has gone a
vertical distance of 8.0 m. Determine the amount of work done by the student to elevate his body to this
height. Assume that her speed is constant.
8. Calculate the work done by a 2.0-N force (directed at a 30° angle to the
vertical) to move a 500 gram box a horizontal distance of 400 cm across a
rough floor at a constant speed of 0.5 m/s. (HINT: Be cautious with the
units.)
9. A tired squirrel (mass of 1 kg) does push-ups by applying a force to elevate its center-of-mass by 5
cm. Estimate the number of push-ups which a tired squirrel must do in order to do a approximately 5.0
Joules of work.
Potential Energy
An object can store energy as the result of its position. For example, the heavy heavy ball of a
demolition machine is storing energy when it is held at an elevated position. This stored energy of
position is referred to as potential energy. Similarly, a drawn bow is able to store energy as the result of
its position. When assuming its usual position (i.e., when not drawn), there is no energy stored in the
bow. Yet when its position is altered from its usual equilibrium position, the bow is able to store energy
by virtue of its position. This stored energy of position is referred to as potential energy. Potential
energy is the stored energy of position possessed by an object.
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Gravitational Potential Energy
The two examples above illustrate the two forms of potential
energy to be discussed in this course - gravitational potential
energy and elastic potential energy. Gravitational potential energy
is the energy stored in an object as the result of its vertical
position or height. The energy is stored as the result of the
gravitational attraction of the Earth for the object. The
gravitational potential energy of the massive ball of a demolition machine is dependent on two variables
- the mass of the ball and the height to which it is raised. There is a direct relation between gravitational
potential energy and the mass of an object. More massive objects have greater gravitational potential
energy. There is also a direct relation between gravitational potential energy and the height of an object.
The higher that an object is elevated, the greater the gravitational potential energy. These relationships
are expressed by the following equation:
PEgrav = mass * g * height
PEgrav = m * g * h
In the above equation, m represents the mass of the object, h represents the height of the object and g
represents the acceleration of gravity (9.8 m/s/s on Earth).
To determine the gravitational potential energy of an object,
a zero height position must first be arbitrarily assigned.
Typically, the ground is considered to be a position of zero
height. But this is merely an arbitrarily assigned position
which most people agree upon. Since many of our labs are
done on tabletops, it is often customary to assign the
tabletop to be the zero height position. Again this is merely
arbitrary. If the tabletop is the zero position, then the
potential energy of an object is based upon its height
relative to the tabletop. For example, a pendulum bob swinging to and from above the table top has a
potential energy which can be measured based on its height above the tabletop. By measuring the mass
WORK, ENERGY AND POWER 2010
of the bob and the height of the bob above the tabletop, the potential energy of the bob can be
determined.
Since the gravitational potential energy of an object is directly proportional to its height above the zero
position, a doubling of the height will result in a doubling of the gravitational potential energy. A tripling
of the height will result in a tripling of the gravitational potential energy.
Use this principle to determine the blanks in the following diagram. Knowing that the potential
energy at the top of the tall platform is 50 J, what is the potential energy at the other positions
shown on the stair steps and the incline?
Check Your Understanding
Check your understanding of the concept of potential energy by answering the following questions.
When finished, click the button to view the answers.
1. A cart is loaded with a brick and pulled at constant speed
along an inclined plane to the height of a seat-top. If the mass of
the loaded cart is 3.0 kg and the height of the seat top is 0.45
meters, then what is the potential energy of the loaded cart at
the height of the seat-top?
WORK, ENERGY AND POWER 2010
2. If a force of 14.7 N is used to drag the loaded cart (from previous question) along the incline for a
distance of 0.90 meters, then how much work is done on the loaded cart?
Note that the work done to lift the loaded cart up the inclined plane at constant speed is equal to the
potential energy change of the cart. This is not coincidental!
Kinetic Energy
Kinetic energy is the energy of motion. An object which has motion - whether it be vertical or
horizontal motion - has kinetic energy. There are many forms of kinetic energy - vibrational (the energy
due to vibrational motion), rotational (the energy due to rotational motion), and translational (the
energy due to motion from one location to another). To keep matters simple, we will focus upon
translational kinetic energy. The amount of translational kinetic energy (from here on, the phrase kinetic
energy will refer to translational kinetic energy) which an object has depends upon two variables: the
mass (m) of the object and the speed (v) of the object. The following equation is used to represent the
kinetic energy (KE) of an object.
where m = mass of object
v = speed of object
This equation reveals that the kinetic energy of an object is directly
proportional to the square of its speed. That means that for a twofold
increase in speed, the kinetic energy will increase by a factor of four. For a
threefold increase in speed, the kinetic energy will increase by a factor of
nine. And for a fourfold increase in speed, the kinetic energy will increase by a factor of sixteen. The
kinetic energy is dependent upon the square of the speed. As it is often said, an equation is not merely a
recipe for algebraic problem-solving, but also a guide to thinking about the relationship between
quantities.
Kinetic energy is a scalar quantity; it does not have a direction. Unlike
velocity, acceleration, force, and
momentum, the kinetic energy of an object is completely described by magnitude alone. Like work and
potential energy, the standard metric unit of measurement for kinetic energy is the Joule. As might be
implied by the above equation, 1 Joule is equivalent to 1 kg*(m2.S-2)
WORK, ENERGY AND POWER 2010
Check Your Understanding
Use your understanding of kinetic energy to answer the following questions. Then click the button to
view the answers.
1. Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s.
2. If the roller coaster car in the above problem were moving with twice the speed, then what would be
its new kinetic energy?
3. Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12
000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then what is her speed?
4. A 900-kg compact car moving at 60 Km/hr has approximately 320 000 Joules of kinetic energy.
Estimate its new kinetic energy if it is moving at 30 Km/hr. (HINT: use the kinetic energy equation as a
"guide to thinking.")
Mechanical Energy
In a previous part , it was said that work is done upon an object
whenever a force acts upon it to cause it to be displaced. Work is
a force acting upon an object to cause a displacement. In all
instances in which work is done, there is an object which supplies
the force in order to do the work. If a World Civilization book is
lifted to the top shelf of a student locker, then the student
supplies the force to do the work on the book. If a plow is
displaced across a field, then some form of farm equipment
(usually a tractor or a horse) supplies the force to do the work on
the plow. If a pitcher winds up and accelerates a baseball towards
home plate, then the pitcher supplies the force to do the work on
the baseball. If a roller coaster car is displaced from ground level to the top of the first drop of a roller
coaster ride, then a chain driven by a motor supplies the force to do the work on the car. If a barbell is
WORK, ENERGY AND POWER 2010
displaced from ground level to a height above a weightlifter's head, then the weightlifter is supplying a
force to do work on the barbell. In all instances, an object which possesses some form of energy
supplies the force to do the work. In the instances described here, the objects doing the work (a
student, a tractor, a pitcher, a motor/chain) possess chemical potential energy stored in food or fuel
which is transformed into work. In the process of doing work, the object which is doing the work
exchanges energy with the object upon which the work is done. When the work is done upon the object,
that object gains energy. The energy acquired by the objects upon which work is done is known as
mechanical energy.
Mechanical energy is the energy which is possessed by an object due
to its motion or due to its position. Mechanical energy can be either
kinetic energy (energy of motion) or potential energy (stored energy
of position). Objects have mechanical energy if they are in motion
and/or if they are at some position relative to a zero potential
energy position (for example, a brick held at a vertical position
above the ground or zero height position). A moving car possesses
mechanical energy due to its motion (kinetic energy). A moving
baseball possesses mechanical energy due to both its high speed
(kinetic energy) and its vertical position above the ground (gravitational potential energy). A World
Civilization book at rest on the top shelf of a locker possesses mechanical energy due to its vertical
position above the ground (gravitational
potential energy). A barbell lifted high above a weightlifter's
head possesses mechanical energy due to its vertical position above the ground (gravitational potential
energy). A drawn bow possesses mechanical energy due to its stretched position (elastic potential
energy).
Mechanical Energy as the Ability to Do Work
An object which possesses mechanical energy is able to do work. In
fact, mechanical energy is often defined as the ability to do work.
Any object which possesses mechanical energy - whether it be in
the form of potential energy or kinetic energy - is able to do work.
That is, its mechanical energy enables that object to apply a force to another object in order to cause it
to be displaced.
Numerous examples can be given of how an object with mechanical energy can harness that energy in
order to apply a force to cause another object to be displaced. A classic example involves the massive
wrecking ball of a demolition machine. The wrecking ball is a massive object which is swung backwards
to a high position and allowed to swing forward into building structure or other object in order to
demolish it. Upon hitting the structure, the wrecking ball applies a force to it in order to cause the wall of
the structure to be displaced. The diagram below depicts the process by which the mechanical energy of
a wrecking ball can be used to do work.
WORK, ENERGY AND POWER 2010
A hammer is a tool which utilizes mechanical energy to do work. The
mechanical energy of a hammer gives the hammer its ability to apply a
force to a nail in order to cause it to be displaced. Because the hammer
has mechanical energy (in the form of kinetic energy), it is able to do work
on the nail. Mechanical energy is the ability to do work.
Another example which illustrates how mechanical
energy is the ability of an object to do work can
be seen any evening at your local bowling alley. The mechanical energy of a
bowling ball gives the ball the ability to apply a force to a bowling pin in order to
cause it to be displaced. Because the massive ball has mechanical energy (in the
form of kinetic energy), it is able to do work on the pin. Mechanical energy is the
ability to do work.
A dart gun is still another example of how mechanical energy
of an object can do work on another object. When a dart gun
is loaded and the springs are compressed, it possesses
mechanical
energy. The
mechanical energy of the compressed springs give the springs
the ability to apply a force to the dart in order to cause it to
be displaced. Because of the springs have mechanical energy
(in the form of elastic potential energy), it is able to do work on
the dart. Mechanical energy is the ability to do work.
A common scene in some parts of the countryside is a "wind
farm." High speed winds are used to do work on the blades of
a turbine at the so-called wind farm. The mechanical energy
WORK, ENERGY AND POWER 2010
of the moving air give the air particles the ability to apply a force and cause a displacement of the
blades. As the blades spin, their energy is subsequently converted into electrical energy (a nonmechanical form of energy) and supplied to homes and industries in order to run electrical appliances.
Because the moving wind has mechanical energy (in the form of kinetic energy), it is able to do work on
the blades. Once more, mechanical energy is the ability to do work.
The Total Mechanical Energy
As already mentioned, the mechanical energy of an object can be the result of its motion (i.e.,
kinetic
energy) and/or the result of its stored energy of position (i.e., potential energy). The total amount of
mechanical energy is merely the sum of the potential energy and the kinetic energy. This sum is simply
referred to as the total mechanical energy (abbreviated TME).
ME = PE + KE
The diagram below depicts the motion of Li Ping Phar (esteemed Chinese ski jumper) as she glides down
the hill and makes one of her record-setting jumps.
The total mechanical energy of Li Ping Phar is the sum of the potential and kinetic energies. The two
forms of energy sum up to 50 000 Joules. Notice also that the total mechanical energy of Li Ping Phar is
a constant value throughout her motion. There are conditions under which the total mechanical energy
will be a constant value and conditions under which it will be a changing value. Energy
. For now,
merely remember that total mechanical energy is the energy possessed by an object due to either its
motion or its stored energy of position. The total amount of mechanical energy is merely the sum of
these two forms of energy. And finally, an object with mechanical energy is able to do work on another
object.
Power
The quantity
work has to do with a force causing a displacement. Work has
nothing to do with the amount of time that this force acts to cause the
displacement. Sometimes, the work is done very quickly and other times the
work is done rather slowly. For example, a rock climber takes an abnormally
long time to elevate her body up a few meters along the side of a cliff. On
the other hand, a trail hiker (who selects the easier path up the mountain)
might elevate her body a few meters in a short amount of time. The two
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people might do the same amount of work, yet the hiker does the work in considerably less time than
the rock climber. The quantity which has to do with the rate at which a certain amount of work is done is
known as the power. The hiker has a greater power rating than the rock climber.
Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using
the following equation.
The standard metric unit of power is the Watt. As is implied by the
equation for power, a unit of power is equivalent to a unit of work
divided by a unit of time. Thus, a Watt is equivalent to a
Joule/second. For historical reasons, the horsepower is
occasionally used to describe the power delivered by a machine.
One horsepower is equivalent to approximately 750 Watts.
Most machines are designed and built to do work on objects. All machines are typically described by a
power rating. The power rating indicates the rate at which that machine
can do work upon other objects. Thus, the power of a machine is the
work/time ratio for that particular machine. A car engine is an example of
a machine which is given a power rating. The power rating relates to how
rapidly the car can accelerate the car. Suppose that a 40-horsepower
engine could accelerate the car from 0 mi/hr to 60 mi/hr in 16 seconds. If this were the case, then a car
with four times the horsepower could do the same amount of work in one-fourth the time. That is, a
160-horsepower engine could accelerate the same car from 0 mi/hr to 60 mi/hr in 4 seconds. The point
is that for the same amount of work, power and time are inversely proportional. The power equation
suggests that a more powerful engine can do the same amount of work in less time.
A person is also a machine which has a power rating. Some people are more power-full than others.
That is, some people are capable of doing the same amount of work in less time or more work in the
same amount of time. A common physics lab involves quickly climbing a flight of stairs and using mass,
height and time information to determine a student's personal power. Despite the diagonal motion along
the staircase, it is often assumed that the horizontal motion is constant and all the force from the steps
are used to elevate the student upward at a constant speed. Thus, the weight of the student is equal to
the force which does the work on the student and the height of the
staircase is the upward displacement. Suppose that Ben Pumpiniron
elevates his 80-kg body up the 2.0 meter stairwell in 1.8 seconds. If this
were the case, then we could calculate Ben's power rating. It can be
assumed that Ben must apply a 800-Newton downward force upon the
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stairs to elevate his body. By so doing, the stairs would push upward on Ben's body with just enough
force to lift his body up the stairs. It can also be assumed that the angle between the force of the stairs
on Ben and Ben's displacement is 0 degrees. With these two approximations, Ben's power rating could
be determined as shown below.
Ben's power rating is 871 Watts. He is quite a horse.
The expression for power is work/time. And since the expression for work is force*displacement, the
expression for power can be rewritten as (force*displacement)/time. Since the expression for velocity is
displacement/time, the expression for power can be rewritten once more as force*velocity. This is shown
below.
This new equation for power reveals that a powerful machine is
both strong (big force) and fast (big velocity). A powerful car
engine is strong and fast. A powerful piece of farm equipment is
strong and fast. A powerful weightlifter is strong and fast. A
powerful linemen on a football team is strong and fast. A
machine which is strong enough to apply a big force to cause a
displacement in a small mount of time (i.e., a big velocity) is a
powerful machine.
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Check Your Understanding
Use your understanding of work and power to answer the following questions. When finished, click the
button to view the answers.
1. Two physics students, Will N. Andable and Ben Pumpiniron, are in the
weightlifting room. Will lifts the 100-pound barbell over his head 10 times in
one minute; Ben lifts the 100-pound barbell over his head 10 times in 10
seconds. Which student does the most work? ______________ Which
student delivers the most power? ______________ Explain your answers.
2. During a physics lab, Jack and Jill ran up a hill. Jack is twice as massive as Jill; yet Jill ascends the
same distance in half the time. Who did the most work? ______________ Who delivered the most
power? ______________ Explain your answers.
3. A tired squirrel (mass of approximately 1 kg) does
push-ups by applying a force to elevate its center-ofmass by 5 cm in order to do a mere 0.50 Joule of
work. If the tired squirrel does all this work in 2
seconds, then determine its power.
4. When doing a chin-up, a physics student lifts her 42.0-kg body a
distance of 0.25 meters in 2 seconds. What is the power delivered by the
student's biceps?
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5. Your household's monthly electric bill is often expressed in kilowatt-hours. One kilowatt-hour is the
amount of energy delivered by the flow of l kilowatt of electricity for one hour. Use conversion factors to
show how many joules of energy you get when you buy 1 kilowatt-hour of electricity.
6. An escalator is used to move 20 passengers every minute from the first floor of a department store to
the second. The second floor is located 5.20 meters above the first floor. The average passenger's mass
is 54.9 kg. Determine the power requirement of the escalator in order to move this number of
passengers in this amount of time.
Internal vs. External Forces
There are a variety of ways to categorize all the types of forces. In a previous unit, it was mentioned that all
the types of forces can be categorized as contact forces or as action-at-a-distance forces. Whether a
force was categorized as an action-at-a-distance force was dependent upon whether or not that type of
force could exist even when the objects were not physically
touching. The force of gravity, electrical forces, and magnetic
forces were examples of forces which could exist between two
objects even when they are not physically touching. In this
lesson, we will learn how to categorize forces based upon
whether or not their presence is capable of changing an object's
total mechanical energy. We will learn that there are certain types of
forces, which when present and when involved in doing work on
objects will change the total mechanical energy of the object. And there are other types of forces which
can never change the total mechanical energy of an object, but rather can only transform the energy of
an object from potential energy to kinetic energy (or vice versa). The two categories of forces are
referred to as internal forces and external forces.
Forces can be categorized as internal forces or external forces. There are many sophisticated and worthy
ways of explaining and distinguishing between internal and external forces. Many of these ways are
commonly discussed at great length in physics textbooks. For our purposes, we will simply say that
external forces include the applied force, normal force, tension force, friction force, and air resistance
force. And for our purposes, the internal forces include the gravity forces, magnetic force, electrical
force, and spring force.
Internal Forces
External Forces
Fgrav
Fspring
Fapp
Ffrict
Fair
Ftens
Fnorm
WORK, ENERGY AND POWER 2010
The importance of categorizing a force as being either internal or external is related to the ability of that
type of force to change an object's total mechanical energy when it does work
upon an object. When net work is done upon an object by an external force, the
total mechanical energy (KE + PE) of that object is changed. If the work is positive
work, then the object will gain energy. If the work is negative work, then the
object will lose energy. The gain or loss in energy can be in the form of potential
energy, kinetic energy, or both. Under such circumstances, the work which is done
will be equal to the change in mechanical energy of the object. This principle will be discussed in great detail
later in this lesson. Because external forces are capable of changing the total mechanical energy of an
object, they are sometimes referred to as nonconservative forces.
When the only type of force doing net work upon an object is an internal force (for example,
gravitational and spring forces), the total mechanical energy (KE + PE) of that object remains constant. In such
cases, the object's energy changes form. For example, as an object is "forced" from a high elevation to a
lower elevation by gravity, some of the potential energy of that object is transformed into kinetic
energy. Yet, the sum of the kinetic and potential energies remain constant. This is referred to as energy
conservation and will be discussed in detail later in this lesson. When the only forces doing work are internal forces,
energy changes forms - from kinetic to potential (or vice versa); yet the total amount of mechanical is
conserved. Because internal forces are capable of changing the form of energy without changing the
total amount of mechanical energy, they are sometimes referred to as conservative forces.
In the following descriptions, the only forces doing work upon the objects are internal forces gravitational and spring forces. Thus, energy is transformed from KE to PE (or vice versa) while
the total amount of mechanical energy is conserved. Read each description and indicate whether energy
is transformed from KE to PE or from PE to KE. Click the mouse to check your answers.
Description of Motion
1.
A ball falls from a height of 2 meters in
the absence of air resistance.
2.
A skier glides from location A to location
B across a friction free ice.
KE to PE or PE to KE?
Explain.
WORK, ENERGY AND POWER 2010
3.
A baseball is traveling upward towards a
man in the bleachers.
4.
A bungee cord begins to exert an upward
force upon a falling bungee jumper.
The spring of a dart gun exerts a force on
5. a dart as it is launched from an initial rest
position.
NOTE: Perhaps at this time you might find it useful to review the lessons on kinetic energy and potential
energy.
When work is done by external forces (nonconservative forces), the total mechanical energy of the
object is altered. The work that is done can be positive work or negative work depending on whether the
force doing the work is directed opposite the object's motion or in the same direction as the object's
motion. If the force and the displacement are in the same direction, then positive work is done on the
object. If positive work is done on an object by an external force, then the object gains mechanical
energy. If the force and the displacement are in the opposite direction, then negative work is done on
the object; the object subsequently loses mechanical energy.
The following descriptions involve external forces (friction, applied, normal, air resistance and
tension forces) doing work upon an object. Read the description and indicate whether the object
gained energy (positive work) or lost energy (negative work). (NOTE: If this is part is difficult, review the
section on work.) Then, indicate whether the gain or loss of energy resulted in a change in the object's
kinetic energy, potential energy, or both. Click the buttons to view answers.
Description
Megan drops the ball and hits an awesome forehand. The racket is moving
horizontally as the strings apply a horizontal force while in contact with the ball.
A tee ball player hits a long ball off the tee. During the contact time between
ball and bat, the bat is moving at a 10 degree angle to the horizontal.
Rusty Nales pounds a nail into a block of wood. The hammer head is moving
horizontally when it applies force to the nail.
+ or Work?
Change
PE or
KE or
Both?
WORK, ENERGY AND POWER 2010
The frictional force between highway and tires pushes backwards on the tires of
a skidding car.
A diver experiences a horizontal reaction force exerted by the blocks upon her
feet at start of the race.
A weightlifter applies a force to lift a barbell above his head at constant speed.
Note that in the five situations described above, a horizontal force can never change the potential
energy of an object. Horizontal forces cannot cause vertical displacements. The only means by which an
external or nonconservative force can contribute to a potential energy change is if the force has a
vertical component. Potential energy changes are the result of height changes and only a force with a
vertical component can cause a height change.
Analysis of Situations Involving External Forces
The previous part of the Lesson discussed the relationship between work and energy change. Whenever work
is done upon an object by an external force, there will be a change in the total mechanical energy of the
object. If only internal forces are doing work (no work done by external forces), there is no change in
total mechanical energy; the total mechanical energy is said to be conserved. Because external forces
are capable of changing the total mechanical energy of an object, they are sometimes referred to as
nonconservative forces. Because internal forces do not change the total mechanical energy of an object,
they are sometimes referred to as conservative forces. In this part of the Lesson , we will further
explore the quantitative relationship between work and energy.
The quantitative relationship between work and mechanical energy is expressed by the following
equation:
MEi + Wext = MEf
The equation states that the initial amount of total mechanical energy (MEi) plus the work done by
external forces (Wext) is equal to the final amount of total mechanical energy (MEf). A few notes should be
WORK, ENERGY AND POWER 2010
made about the above equation. First, the mechanical energy can be either potential energy (in which
case it could be due to springs or gravity) or kinetic energy. Given this fact, the above equation can be
rewritten as
KEi + PEi + Wext = KEf + PEf
The second note which should be made about the above equation is that the work done by external
forces can be
a positive or a negative work term. Whether the work term takes on a positive or a
negative value is dependent upon the angle between the force and the motion. Recall from previous
Lesson that the work is dependent upon the angle between the force and the displacement vectors. If the
angle is 180 degrees as it occasionally is, then the work term will be negative. If the angle is 0 degrees,
then the work term will be positive.
The above equation is expresses the quantitative relationship
between work and energy. This equation will be the basis for the rest
of this unit. It will form the basis of the conceptual aspect of our
study of work and energy as well as the guiding force for our
approach to solving mathematical problems. A large slice of the world
of motion can be understood through the use of this relationship
between work and energy.
To begin our investigation of the work-energy relationship, we will investigate situations involving work
being done by external forces (nonconservative forces). Consider a weightlifter who applies an upwards
force (say 1000 N) to a barbell to displace it upwards a given distance (say 0.25 meters) at a constant
speed. The initial energy plus the work done by the external force equals the final energy. If the barbell
begins with 1500 Joules of energy (this is just a made up value) and the weightlifter does 250 Joules of
work (F*d*cosine of angle = 1000 N*0.25 m*cosine 0 degrees = 250 J), then the barbell will finish with
1750 Joules of mechanical energy. The final amount of mechanical energy (1750 J) is equal to the initial
amount of mechanical energy (1500 J) plus the work done by external forces (250 J).
WORK, ENERGY AND POWER 2010
Now consider a baseball catcher who applies a rightward force (say 6000 N) to a leftward moving
baseball to bring it from a high speed to a rest position over a given distance (say 0.10 meters). The
initial energy plus the work done by the external force equals the final energy. If the ball begins with
605 Joules of energy (this is just another made up value), and the catcher does -600 Joules of work
(F*d*cosine of angle = 6000 N*0.10 m*cosine 180 degrees = -600 J), then the ball will finish with 5 Joules
of mechanical energy. The final energy (5 J) is equal to the initial energy (605 J) plus the work done by
external forces (-600 J).
WORK, ENERGY AND POWER 2010
Now consider a car which is skidding from a high speed to a lower speed. The force of friction between
the tires and the road exerts a leftward force (say 8000 N) on the rightward moving car over a given
distance (say 30 m). The initial energy plus the work done by the external force equals the final energy.
If the car begins with 320 000 Joules of energy (this is just another made up value), and the friction
force does -240 000 Joules of work (F*d*cosine of angle = 8000 N*30 m*cosine 180 degrees = -240 000
J), then the car will finish with 80 000 Joules of mechanical energy. The final energy (80 000 J) is equal
to the initial energy (320 000 J) plus the work done by external forces (-240 000 J).
As a final example, consider a cart being pulled up an inclined plane at constant speed by a student
during a Physics lab. The applied force on the cart (say 18 N) is directed parallel to the incline to cause
the cart to be displaced parallel to the incline for a given displacement (say 0.7 m). The initial energy
plus the work done by the external force equals the final energy. If the cart begins with 0 Joules of
energy (this is just another made up value), and the student does 12.6 Joules of work (F*d*cosine of angle
= 18 N*0.7 m*cosine 0 degrees = 12.6 J), then the cart will finish with 12.6 Joules of mechanical
energy. The final energy (12.6 J) is equal to the initial energy (0 J) plus the work done by external
forces (12.6 J).
WORK, ENERGY AND POWER 2010
In each of these examples, an external force does work upon an object over a given distance to change
the total mechanical energy of the object. If the external force (or nonconservative force) does positive
work, then the object gains mechanical energy. The amount of energy gained is equal to the work done
on the object. If the external force (or nonconservative force) does negative work, then the object loses
mechanical energy. The amount of mechanical energy lost is equal to the work done on the object. In
general, the total mechanical energy of the object in the initial state (prior to the work being done) plus
the work done equals the total mechanical energy in the final state.
The work-energy relationship presented here can be combined with the
expressions for potential and kinetic energy to solve complex problems.
Like all complex problems, they can be made simple if first analyzed from
a conceptual viewpoint and broken down into parts. In other words, avoid
treating work-energy problems as mere mathematical problems. Rather,
engage your mind and utilize your understanding of physics concepts to approach the problem. Ask
"What forms of energy are present initially and finally?" and "Based on the equations, how much of each
form of energy is present initially and finally?" and "Is work being done by external forces?" Use this
approach on the following three practice problems. After solving, click the button to view the answers.
Practice Problem #1
A 1000-kg car traveling with a speed of 25 m/s skids to a stop. The car experiences an 8000 N
WORK, ENERGY AND POWER 2010
force of friction. Determine the stopping distance of the car.
Practice Problem #2
At the end of the Shock Wave roller coaster ride, the 6000-kg train of cars (includes passengers)
is slowed from a speed of 20 m/s to a speed of 5 m/s over a distance of 20 meters. Determine
the braking force required to slow the train of cars by this amount.
Practice Problem #3
A shopping cart full of groceries is sitting at the top of a 2.0-m hill. The cart begins to roll until it
hits a stump at the bottom of the hill. Upon impact, a 0.25-kg can of peaches flies horizontally
out of the shopping cart and hits a parked car with an average force of 500 N. How deep a dent is made
in the car (i.e., over what distance does the 500 N force act upon the can of peaches before bringing it
to a stop)?
WORK, ENERGY AND POWER 2010
All three of the above problems have one thing in common: there is a force which does work over a
distance in order to remove mechanical energy from an object. The force acts opposite the object's
motion (angle between force and displacement is 180 degrees) and thus does negative work. Negative work
results in a loss of the object's total amount of mechanical energy. In each situation, the work is related to the kinetic
energy change.
Analysis of Situations in Which Mechanical Energy is Conserved
It has previously been mentioned that there is a relationship between work and mechanical energy change.
Whenever work is done upon an object by an external force (or nonconservative force), there will be a change in
the total mechanical energy of the object. If only internal forces are doing work (no work done by external
forces), then there is no change in the total amount of mechanical energy. The total mechanical energy
is said to be conserved. In this part of Lesson 2, we will further explore the quantitative relationship
between work and mechanical energy in situations in which there are no external forces doing work.
The quantitative relationship between work and the two forms of mechanical energy is expressed by the
following equation:
KEi + PEi + Wext = KEf + PEf
The equation illustrates that the total mechanical energy (KE + PE) of the object is changed as a result
of work done by external forces. There are a host of other situations in which the only forces doing work
are internal or conservative forces. In such situations, the total mechanical energy of the object is not
changed. The external work term cancels from the above equation and mechanical energy is conserved.
The previous equation is simplified to the following form:
KEi + PEi = KEf + PEf
In these situations, the sum of the kinetic and potential energy is everywhere the same. As the potential
energy is increased due to the stretch/compression of a spring or an increase in its height above the
earth, the kinetic energy is decreased due to the object slowing down. As the potential energy is
decreased due to the return of a spring to its rest position or a decrease in height above the earth, the
kinetic energy is increased due to the object speeding up. We would say that energy is transformed or
changes its form from kinetic energy to potential energy (or vice versa); yet the total amount present is
Application and Practice Questions
The Lesson has thus far focused on how to analyze motion situations using the work and energy
relationship. The relationship could be summarized by the following statements:
There is a relationship between work and mechanical energy change. Whenever work is done
upon an object by an external or nonconservative force, there will be a change in the total
mechanical energy of the object. If only internal forces are doing work (no work done by
external forces), there is no change in total mechanical energy; the total mechanical energy is
said to be "conserved." The quantitative relationship between work and the two forms of
mechanical energy is expressed by the following equation:
WORK, ENERGY AND POWER 2010
KEi + PEi + Wext = KEf + PEf
Now an effort will be made to apply this relationship to a variety of motion scenarios in order to test our
understanding.
Check Your Understanding
Use your understanding of the work-energy theorem to answer the following questions. Then click the
button to view the answers.
1. Consider the falling and rolling motion of the ball in the following two resistance-free situations. In
one situation, the ball falls off the top of the platform to the floor. In the other situation, the ball rolls
from the top of the platform along the staircase-like pathway to the floor. For each situation, indicate
what type of forces are doing work upon the ball. Indicate whether the energy of the ball is conserved
and explain why. Finally, fill in the blanks for the 2-kg ball.
2. If frictional forces and air resistance were acting upon the falling ball in #1 would the kinetic energy
of the ball just prior to striking the ground be more, less, or equal to the value predicted in #1?
WORK, ENERGY AND POWER 2010
Use the following diagram to answer questions #3 - #5. Neglect the affect of resistance forces.
3. As the object moves from point A to point D across the surface, the sum of its gravitational potential
and kinetic energies ____.
a. decreases, only
b. decreases and then increases
c. increases and then decreasesd. remains the same
4. The object will have a minimum gravitational potential energy at point ____.
a. A
b. B
c. C
d. D
e. E
5. The object's kinetic energy at point C is less than its kinetic energy at point ____.
a. A only
b. A, D, and E
c. B only
d. D and E
6. Many drivers education books provide tables which relate a car's braking distance to the speed of the
car (see table below). Utilize what you have learned about the stopping distance-velocity relationship to
complete the table.
7. Some driver's license exams have the following question.
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A car moving 50 km/hr skids 15 meters with locked brakes. How far will the car skid with locked brakes
if it is moving at 150 km/hr?
8. Two baseballs are fired into a pile of hay. If one has twice the speed of the other, how much farther
does the faster baseball penetrate? (Assume that the force of the haystack on the baseballs is constant).
9. Use the law of conservation of energy (assume no friction) to fill in the blanks at the various marked
positions for a 1000-kg roller coaster car.
10. If the angle of the initial drop in the roller coaster diagram above were 60 degrees (and all other
factors were kept constant), would the speed at the bottom of the hill be any different? Explain.
11. Determine Li Ping Phar's (a mass of approximately 50 kg) speed at locations B, C, D and E.
12. An object which weighs 10 N is dropped from rest from a height of 4 meters above the ground.
When it has free-fallen 1 meter its total mechanical energy with respect to the ground is ____.
a. 2.5 J
b. 10 J
c. 30 J
d. 40 J
WORK, ENERGY AND POWER 2010
13. During a certain time interval, a 20-N object free-falls 10 meters. The object gains _____ Joules of
kinetic energy during this interval.
a. 10
b. 20
c. 200
d. 2000
14. A rope is attached to a 50.0-kg crate to pull it up a frictionless incline at constant speed to a height
of 3-meters. A diagram of the situation and a free-body diagram is shown below. Note that the force of
gravity has two components (parallel and perpendicular component); the parallel component balances
the applied force and the perpendicular component balances the normal force.
Of the forces acting upon the crate, which one(s) do work upon it?
Based upon the types of forces acting upon the system and their classification as internal or external
forces, is energy conserved? Explain.Calculate the amount of work done upon the crate.
Inclined Planes
An object placed on a tilted surface will often slide down the surface.
The rate at which the object slides down the surface is dependent upon
how tilted the surface is; the greater the tilt of the surface, the faster
the rate at which the object will slide down it. In physics, a tilted
surface is called an inclined plane. Objects are known to accelerate
down inclined planes because of an unbalanced force. To understand
this type of motion, it is important to analyze the forces acting upon an
object on an inclined plane. The diagram at the right depicts the two
forces acting upon a crate which is positioned on an inclined plane
(assumed to be friction-free). As shown in the diagram, there are
always at least two forces acting upon any object that is positioned on
an inclined plane - the force of gravity and the normal force. The force of
WORK, ENERGY AND POWER 2010
gravity (also known as weight) acts in a downward direction; yet the normal force acts in a direction perpendicular to the
surface (in fact, normal means "perpendicular").
The first peculiarity of inclined plane problems is that the normal force is not directed in the direction
which we are accustomed to. Up to this point in the course, we have always seen normal forces acting in
an upward direction, opposite the direction of the force of gravity. But this is only because the objects
were always on horizontal surfaces and never upon inclined planes. The truth about normal forces is not
that they are always upwards, but rather that they are always directed perpendicular to the surface that
the object is on.
The task of determining the net force acting upon an object on an inclined plane is a difficult manner
since the two (or more) forces are not directed in opposite directions. Thus, one (or more) of the forces
will have to be resolved into perpendicular components so that they can be easily added to the other
forces acting upon the object. Usually, any force directed at an angle to the horizontal is resolved into horizontal and
vertical components. However, this is not the process that we will pursue with inclined planes. Instead, the
process of analyzing the forces acting upon objects on inclined planes will involve resolving the weight
vector (Fgrav) into two perpendicular components. This is the second peculiarity of inclined plane
problems. The force of gravity will be resolved into two components of force - one directed parallel to
the inclined surface and the other directed perpendicular to the inclined surface. The diagram below
shows how the force of gravity has been replaced by two components - a parallel and a perpendicular
component of force.
The perpendicular component of the force of gravity is directed opposite the normal force and as such
balances the normal force. The parallel component of the force of gravity is not balanced by any other
force. This object will subsequently accelerate down the inclined plane due to the presence of an
unbalanced force. It is the parallel component of the force of gravity which causes this acceleration. The
WORK, ENERGY AND POWER 2010
parallel component of the force of gravity is the net force.
The task of determining the magnitude of the two components of the force of gravity is a mere manner
of using the equations. The equations for the parallel and perpendicular components are:
In the absence of friction and other forces (tension, applied, etc.), the acceleration of an object on an
incline is the value of the parallel component (m*g*sine of angle) divided by the mass (m). This yields
the equation
(in the absence of friction and other force.
In the presence of friction or other forces (applied force, tensional
forces, etc.), the situation is slightly more complicated. Consider
the diagram shown at the right. The perpendicular component
offorce still balances the normal force since objects do not
accelerate perpendicular to the incline. Yet the frictional force
must also be considered when determining the net force. As in all
net force problems, the net force is the vector sum of all the forces.
That is, all the individual forces are added together as vectors. The
perpendicular component and the normal force add to 0 N. The
parallel component and the friction force add together to yield 5 N.
The net force is 5 N, directed along the incline towards the floor.
The above problem (and all inclined plane problems) can be simplified through a useful trick known as
"tilting the head." An inclined plane problem is in every way like any other net force problem with the
sole exception that the surface has been tilted. Thus, to transform the problem back into the form with
which you are more comfortable, merely tilt your head in the same direction that the incline was tilted.
Or better yet, merely tilt the page of paper (a sure remedy for TNS - "tilted neck syndrome" or "taco
neck syndrome") so that the surface no longer appears level. This is illustrated below.
WORK, ENERGY AND POWER 2010
Once the force of gravity has been resolved into its two components and the inclined plane has been
tilted, the problem should look very familiar. Merely ignore the force of gravity (since it has been
replaced by its two components) and solve for the net force and acceleration.
As an example consider the situation depicted in the diagram at the right. The free-body diagram shows
the forces acting upon a 100-kg crate which is sliding down an inclined plane. The plane is inclined at an
angle of 30 degrees. The coefficient of friction between the crate and the incline is 0.3. Determine the
net force and acceleration of the crate.
Begin the above problem by finding the force of gravity acting upon the crate and the components of
this force parallel and perpendicular to the incline. The force of gravity is 980 N and the components of
this force are Fparallel = 490 N (980 N • sin 30 degrees) and Fperpendicular = 849 N (980 N • cos30 degrees).
Now the normal force can be determined to be 849 N (it must balance the perpendicular component of the weight
vector). The force of friction can be determined from the value of the normal force and the coefficient of
friction; Ffrict is 255 N (Ffrict = "mu"*Fnorm= 0.3 • 849 N). The net force is the vector sum of all the forces.
The forces directed perpendicular to the incline balance; the forces directed parallel to the incline do not
balance. The net force is 235 N (490 N - 255 N). The acceleration is 2.35 m/s/s (Fnet/m = 235 N/100
kg).
Practice
The two diagrams below depict the free-body diagram for a 1000-kg roller coaster on the first drop of
two different roller coaster rides. Use the above principles of vector resolution to determine the net force
and acceleration of the roller coaster cars. Assume a negligible affect of friction and air resistance. When
done, click the button to view the answers.
WORK, ENERGY AND POWER 2010
The affects of the incline angle on the acceleration of a roller coaster (or any object on an incline) can be
observed in the two practice problems above. As the angle is increased, the acceleration of the object is
increased. The explanation of this relates to the components which we have been drawing. As the angle
increases, the component of force parallel to the incline increases and the component of force
perpendicular to the incline decreases. It is the parallel component of the weight vector which causes
the acceleration. Thus, accelerations are greater at greater angles of incline. The diagram below depicts
this relationship for three different angles of increasing magnitude.
Roller coasters produce two thrills associated with the initial drop down a steep incline. The thrill of
acceleration is produced by using large angles of incline on the first drop; such large angles increase the
value of the parallel component of the weight vector (the component which causes acceleration). The
thrill of weightlessness is produced by reducing the magnitude of the normal force to values less than
their usual values. It is important to recognize that the thrill of weightlessness is a feeling associated
WORK, ENERGY AND POWER 2010
with a lower than usual normal force. Typically, a person weighing 700 N will experience a 700 N normal
force when sitting in a chair. However, if the chair is accelerating down a 60-degrees incline, then the
person will experience a 350 Newton normal force. This value is less than normal and contributes to the
feeling of weighing less than one's normal weight - i.e., weightlessness.
Check Your Understanding
The following questions are intended to test your understanding of the mathematics and concepts of
inclined planes. Once you have answered the question, click the button to see the answers.
1. Two boys are playing ice hockey on a neighborhood street. A stray puck travels across the frictionfree ice and then up the friction-free incline of a driveway. Which one of the following ticker tapes (A, B,
or C) accurately portrays the motion of the puck as it travels across the level street and then up the
driveway?
Explain your answer.
2. Little Johnny stands at the bottom of the driveway and kicks a soccer ball. The ball rolls northward up
the driveway and then rolls back to Johnny. Which one of the following velocity-time graphs (A, B, C, or
D) most accurately portrays the motion of the ball as it rolls up the driveway and back down?
Explain your answer.
WORK, ENERGY AND POWER 2010
3. A golf ball is rolling across a horizontal section of the green on the 18th hole. It then encounters a
steep downward incline (see diagram). Friction is involved. Which of the following ticker tape patterns
(A, B, or C) might be an appropriate representation of the ball's motion?
Explain why the inappropriate patterns are inappropriate.
4. Missy dePenn's eighth frame in the Wednesday night bowling league was a disaster. The ball rolled off
the lane, passed through the freight door in the building's rear, and then down the driveway. Millie
Meater (Missy's teammate), who was spending every free moment studying for her physics test, began
visualizing the velocity-time graph for the ball's motion. Which one of the velocity-time graphs (A, B, C,
or D) would be an appropriate representation of the ball's motion as it rolls across the horizontal surface
and then down the incline? Consider frictional forces.
5. Three lab partners - Olive N. Glenveau, Glen Brook, and Warren Peace - are discussing an incline
problem (see diagram). They are debating the value of the normal force. Olive claims that the normal
force is 250 N; Glen claims that the normal force is 433 N; and Warren claims that the normal force is
500 N. While all three answers seem reasonable, only one is correct. Indicate which two answers are
wrong and explain why they are wrong.
WORK, ENERGY AND POWER 2010
6. Lon Scaper is doing some lawn work when a 2-kg tire escapes from his wheelbarrow and begins
rolling down a steep hill (a 30° incline) in San Francisco. Sketch the parallel and perpendicular
components of this weight vector. Determine the magnitude of the components using trigonometric
functions. Then determine the acceleration of the tire. Ignore resistance force.
Finally, determine which one of the velocity-time graph would represent the motion of the tire as it rolls
down the incline.
Explain your answer.
7. In each of the following diagrams, a 100-kg box is sliding down a frictional surface at a constant
speed of 0.2 m/s. The incline angle is different in each situation. Analyze each diagram and fill in the
blanks.
Diagram A
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Diagram B
Energy Conservation on an Incline
Consider an ordinary lab cart loaded with bricks and accelerating down an inclined plane. How could
work and energy be utilized to analyze the motion of the loaded cart? Would the total mechanical energy
of the cart be altered in the process of rolling down the incline? Or would the total mechanical energy of
the cart merely be conserved?
Of course the answers to these questions begin by determining whether or not there are any external
forces doing work upon the loaded cart. If external forces do work upon the cart, the total mechanical
energy of the cart is not conserved; the initial amount of mechanical energy is not the same as the final
amount of mechanical energy. On the other hand, if external forces do not do work upon the loaded
cart, then the total mechanical energy is conserved; that is, mechanical energy is merely transformed
from the form of potential energy to the form of kinetic energy while the total amount of the two forms
remains unchanged.
In the case of the cart rolling down the incline, there are three external forces (the normal force, the
force of friction and air resistance) and one internal force (the force of gravity). The normal force does
not do work upon the cart because it acts in a direction perpendicular to the direction of motion. In such
instances, the angle between F and d is 90 degrees and the work done by the force is 0 Joules. The
force of friction does not do work upon the cart because it acts upon the wheels of the cart and actually
does not serve to displace either the cart nor the wheels. The friction force only serves to help the
wheels turn as the cart rolls down the hill. Friction only does work upon a skidding wheel. Finally, the
force of air resistance does do work upon the cart; air resistance does negative work upon the cart since
it acts in a direction opposite the direction of the cart's motion. Sometimes referred to as a dissipative
force, air resistance contributes to a loss in the total amount of mechanical energy possessed by the
cart. Subsequently, it would be expected that there would be a small amount of energy loss as the cart
rolls down the hill from an elevated position to a position just above the ground.
Due to the difficulty in measuring air resistance forces and due to the small amount of existing Fair in
situations in which a streamlined object moves at relatively low speeds, the affect of air resistance is
WORK, ENERGY AND POWER 2010
often neglected. If air resistance is neglected, then it would be expected that the total mechanical
energy of the cart would be conserved. The animation below depicts this phenomenon (in the absence of
air resistance).
As the cart rolls down the hill from its elevated position, its mechanical energy is transformed from
potential energy to kinetic energy. At a height just above the ground, the majority of the energy is in
the form of kinetic energy. This is to say, energy due to vertical position has been transformed into
energy due to motion. In fact, if air resistance can be neglected, then the amount of potential energy
loss equals the amount of kinetic energy gained. Calculations can be performed to illustrate that the
potential energy lost by the 3.0-kg cart in the above animation is equal to the kinetic energy gained.
Assume that the height of the cart changes from 0.40 meters to 0.05 meters and the speed changes
from 0 m/s to 2.62 m/s.
Work, Energy and Power: Problem Set Overview
This set of 32 problems targets your ability to use equations related to work and power, to calculate the
kinetic, potential and total mechanical energy, and to use the work-energy relationship in order to
determine the final speed, stopping distance or final height of an object. The more difficult problems are
color-coded as blue problems.
Work
Work results a force acts upon an object to cause a displacement (or a motion) or in some instances, to
hinder a motion. Three variables are of importance in this definition - force, displacement, and the
extent to which the force causes or hinders the displacement. Each of these three variables find their
way into the equation for work. That equation is:
Work = Force • Displacement • Cosine(theta)
W = F • d • cos(theta)
Since the standard metric unit of force is the Newton and the standard meteric unit of displacement is
WORK, ENERGY AND POWER 2010
the meter, then the standard metric unit of work is a Newton•meter, defined as a Joule and abbreviated
with a J.
The most complicated part of the work equation and work calculations is the meaning of the angle theta
in the above equation. The angle is not just any stated angle in the problem; it is the angle between the
F and the d vectors. In solving work problems, one must always be aware of this definition - theta is the
angle between the force and the displacement which it causes. If the force is in the same direction as
the displacement, then the angle is 0 degrees. If the force is in the opposite direction as the
displacement, then the angle is 180 degrees. If the force is up and the displacement is to the right, then
the angle is 90 degrees. This is summarized in the graphic below.
Power
Power is defined as the rate at which work is done upon an object. Like all rate quantities, power is a
time-based quantity. Power is related to how fast a job is done. Two identical jobs or tasks can be done
at different rates - one slowly or and one rapidly. The work is the same in each case (since they are
identical jobs) but the power is different. The equation for power shows the importance of time:
Power = Work / time
P=W/t
The unit for standard metric work is the Joule and the standard metric unit for time is the second, so the
standard metric unit for power is a Joule / second, defined as a Watt and abbreviated W. Special
attention should be taken so as not to confuse the unit Watt, abbreviated W, with the quantity work,
also abbreviated by the letter W.
Combining the equations for power and work can lead to a second equation for power. Power is W/t and
work is F•d•cos(theta). Substituting the expression for work into the power equation yields P =
F•d•cos(theta)/t. If this equation is re-written as
P = F • cos(theta) • (d/t)
one notices a simplification which could be made. The d/t ratio is the speed value for a constant speed
motion or the average speed for an accelerated motion. Thus, the equation can be re-written as
P = F • v • cos(theta)
where v is the constant speed or the average speed value. A few of the problems in this set of problems
will utilize this derived equation for power.
WORK, ENERGY AND POWER 2010
Mechanical, Kinetic and Potential Energies
There are two forms of mechanical energy - potential energy and kinetic energy.
Potential energy is the stored energy of position. In this set of problems, we will be most concerned with
the stored energy due to the vertical position of an object within Earth's gravitational field. Such energy
is known as the gravitational potential energy (PEgrav) and is calculated using the equation
PEgrav = m•g•h
where m is the mass of the object (with standard units of kilograms), g is the acceleration of gravity
(9.8 m/s/s) and h is the height of the object (with standard units of meters) above some arbitraily
defined zero level (such as the ground or the top of a lab table in a physics room).
Kinetic energy is defined as the energy possessed by an object due to its motion. An object must be
moving to possess kinetic energy. The amount of kinetic energy (KE) possessed by a moving object is
dependent upon mass and speed. The equation for kinetic energy is
KE = 0.5 • m • v2
where m is the mass of the object (with standard units of kilograms) and v is the speed of the object
(with standard units of m/s).
The total mechanical energy possessed by an object is the sum of its kinetic and potential energies.
Work-Energy Connection
There is a relationship between work and total mechanical energy. The relationship is best expressed by
the equation
TMEi + Wnc = TMEf
In words, this equations says that the initial amount of total mechanical energy (TMEi) of a system is
altered by the work which is done to it by non-conservative forces (Wnc). The final amount of total
mechanical energy (TMEf) possessed by the system is equivalent to the initial amount of energy (TMEi)
plus the work done by these non-conservative forces (Wnc).
The mechanical energy possessed by a system is the sum of the kinetic energy and the potential energy.
Thus the above equation can be re-arranged to the form of
KE + PE + W
= KE + PE
i
i
nc
f
f
0.5 • m • vi2 + m • g • hi + F • d • cos(theta) = 0.5 • m •
vf2 + m • g • hf
The work done to a system by non-conservative forces (Wnc) can be described as either positive work or
WORK, ENERGY AND POWER 2010
negative work. Positive work is done on a system when the force doing the work acts in the direction of
the motion of the object. Negative work is done when the force doing the work opposes the motion of
the object. When a positive value for work is substituted into the work-energy equation above, the final
amount of energy will be greater than the initial amount of energy; the system is said to have gained
mechanical energy. When a negative value for work is substituted into the work-energy equation above,
the final amount of energy will be less than the initial amount of energy; the system is said to have lost
mechanical energy. There are occasions in which the only forces doing work are conservative forces
(sometimes referred to as internal forces). Typically, such conservative forces include gravitational
forces, elastic or spring forces, electrical forces and magnetic forces. When the only forces doing work
are conservative forces, then the Wnc term in the equation above is zero. In such instances, the system
is said to have conserved its mechanical energy.
The proper approach to work-energy problem involves carefully reading the problem description and
substituting values from it into the work-energy equation listed above. Inferences about certain terms
will have to be made based on a conceptual understanding of kinetic and potential energy. For instance,
if the object is initially on the ground, then it can be inferred that the PEi is 0 and that term can be
canceled from the work-energy equation. In other instances, the height of the object is the same in the
initial state as in the final state, so the PEi and the PEf terms are the same. As such, they can be
mathematically canceled from each side of the equation. In other instances, the speed is constant during
the motion, so the KEi and KEf terms are the same and can thus be mathematically canceled from each
side of the equation. Finally, there are instances in which the KE and or the PE terms are not stated;
rather, the mass (m), speed (v), and height (h) is given. In such instances, the KE and PE terms can be
determined using their respective equations. Make it your habit from the beginning to simply start with
the work and energy equation, to cancel terms which are zero or unchanging, to substitute values of
energy and work into the equation and to solve for the stated unknown.
Habits of an Effective Problem-Solver
An effective problem solver by habit approaches a physics problem in a manner that reflects a collection
of disciplined habits. While not every effective problem solver employs the same approach, they all have
habits which they share in common. These habits are described briefly here. An effective problemsolver...

...reads the problem carefully and develops a mental picture of the physical
situation. If needed, they sketch a simple diagram of the physical situation to help visualize
it.

...identifies the known and unknown quantities in an organized manner, often times
recording them on the diagram iteself. They equate given values to the symbols used to
represent the corresponding quantity (e.g., m = 1.50 kg, vi = 2.68 m/s, F = 4.98 N, t =
0.133 s, vf = ???).

...plots a strategy for solving for the unknown quantity; the strategy will typically
center around the use of physics equations be heavily dependent upon an understaning of
physics principles.

...identifies the appropriate formula(s) to use, often times writing them down. Where
needed, they perform the needed conversion of quantities into the proper unit.

...performs substitutions and algebraic manipulations in order to solve for the
unknown quantity.
Work, Energy and Power: Problem Set
WORK, ENERGY AND POWER 2010
Problem 1
Renatta Gass is out with her friends. Misfortune occurs and Renatta and her friends find themselves
getting a workout. They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel
station. Determine the work done on the car.

Problem 2
Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and
rightwards with a force of 22.9 Newtons at an angle of 35 degrees above the horizontal to drag his
backpack a horizontal distance of 129 meters to the right. Determine the work (in Joules) done upon the
backpack.
Problem 3
Lamar Gant, U.S. powerlifting star, became the first man to deadlift five times his own body weight in
1985. Deadlifting involves raising a loaded barbell from the floor to a position above the head with
outstretched arms. Determine the work done by Lamar in deadlifting 300 kg to a height of 1.9 m above
the ground.
Problem 4
Sheila has just arrived at the airport and is dragging her suitcase to the luggage check-in desk. She pulls
on the strap with a force of 190 N at an angle of 35° to the horizontal to displace it 45 m to the desk.
Determine the work done by Sheila on the suitcase.

Problem 5
While training for breeding season, a 380 gram male squirrel does 32 pushups in a minute, displacing its
center of mass by a distance of 8.5 cm for each pushup. Determine the total work done on the squirrel
while moving upward (32 times).

Problem 6
During the Powerhouse lab, Jerome runs up the stairs, elevating his 102 kg body a vertical distance of
2.29 meters in a time of 1.32 seconds at a constant speed.
a. Determine the work done by Jerome in climbing the stair case.
b. Determine the power generated by Jerome.
Problem 7
A new conveyor system at the local packaging plan will utilize a motor-powered mechanical arm to exert
an average force of 890 N to push large crates a distance of 12 meters in 22 seconds. Determine the
power output required of such a motor.
Problem 8
The Taipei 101 in Taiwan is a 1667-foot tall, 101-story skyscraper. The skyscraper is the home of the
world’s fastest elevator. The elevators transport visitors from the ground floor to the Observation Deck
on the 89th floor at speeds up to 16.8 m/s. Determine the power delivered by the motor to lift the 10
WORK, ENERGY AND POWER 2010
passengers at this speed. The combined mass of the passengers and cabin is 1250 kg.

Problem 9
The ski slopes at Bluebird Mountain make use of tow ropes to transport snowboarders and skiers to the
summit of the hill. One of the tow ropes is powered by a 22-kW motor which pulls skiers along an icy
incline of 14° at a constant speed. Suppose that 18 skiers with an average mass of 48 kg hold onto the
rope and suppose that the motor operates at full power.
a. Determine the cumulative weight of all these skiers.
b. Determine the force required to pull this amount of weight up a 14° incline at a constant speed.
c. Determine the speed at which the skiers will ascend the hill.

Problem 10
The first asteroid to be discovered is Ceres. It is the largest and most massive asteroid is our solar
system’s asteroid belt, having an estimated mass of 3.0 x 1021 kg and an orbital speed of 17900 m/s.
Determine the amount of kinetic energy possessed by Ceres.

Problem 11
A bicycle has a kinetic energy of 124 J. What kinetic energy would the bicycle have if it had …
a. … twice the mass and was moving at the same speed?
b. … the same mass and was moving with twice the speed?
c. … one-half the mass and was moving with twice the speed?
d. … the same mass and was moving with one-half the speed?
e. … three times the mass and was moving with one-half the speed?
Problem 12
A 78-kg skydiver has a speed of 62 m/s at an altitude of 870 m above the ground.
a. Determine the kinetic energy possessed by the skydiver.
b. Determine the potential energy possessed by the skydiver.
c. Determine the total mechanical energy possessed by the skydiver.

Problem 13
Li Ping Phar, the esteemed Chinese ski jumper, has a mass of 59.6 kg. He is moving with a speed of
23.4 m/s at a height of 44.6 meters above the ground. Determine the total mechanical energy of Li Ping
Phar.
Problem 14
Chloe leads South’s varsity softball team in hitting. In a game against New Greer Academy this past
weekend, Chloe slugged the 181-gram softball so hard that it cleared the outfield fence and landed on
Lake Avenue. At one point in its trajectory, the ball was 28.8 m above the ground and moving with a
speed of 19.7 m/s. Determine the total mechanical energy of the softball.

WORK, ENERGY AND POWER 2010
Problem 15
Olive Udadi is at the park with her father. The 26-kg Olive is on a swing following the path as shown.
Olive has a speed of 0 m/s at position A and is a height of 3.0-m above the ground. At position B, Olive
is 1.2 m above the ground. At position C (2.2 m above the ground), Olive projects from the seat and
travels as a projectile along the path shown. At point F, Olive is a mere picometer above the ground.
Assume negligible air resistance throughout the motion. Use this information to fill in the table.
Position
Height (m)
A
3.0
B
1.2
C
2.2
F
0

Audio Guided Solution

Show Answer
PE (J)
KE (J)
TME (J)
Speed (m/s)
0.0
Position
Height (m)
PE (J)
KE (J)
TME (J)
Speed (m/s)
A
3.0
760
0
760
0.0
B
1.2
310
460
760
5.9
C
2.2
560
200
760
4.0
F
0
0
760
760
7.7
Problem 16
Suzie Lavtaski (m=56 kg) is skiing at Bluebird Mountain. She is moving at 16 m/s across the crest of a
ski hill located 34 m above ground level at the end of the run.
a. Determine Suzie's kinetic energy.
b. Determine Suzie's potential energy relative to the height of the ground at the end of the run.
c. Determine Suzie's total mechanical energy at the crest of the hill.
d. If no energy is lost or gained between the top of the hill and her initial arrival at the end of the run,
then what will be Suzie's total mechanical energy at the end of the run?
e. Determine Suzie's speed as she arrives at the end of the run and prior to braking to a stop.

Problem 17
Nicholas is at The Noah's Ark Amusement Park and preparing to ride on The Point of No Return racing
slide. At the top of the slide, Nicholas (m=72.6 kg) is 28.5 m above the ground.
a. Determine Nicholas' potential energy at the top of the slide.
b. Determine Nicholas's kinetic energy at the top of the slide.
c. Assuming negligible losses of energy between the top of the slide and his approach to the bottom of
WORK, ENERGY AND POWER 2010
the slide (h=0 m), determine Nicholas's total mechanical energy as he arrives at the bottom of the slide.
d. Determine Nicholas' potential energy as he arrives at the bottom of the slide.
e. Determine Nicholas' kinetic energy as he arrives at the bottom of the slide.
f. Determine Nicholas' speed as he arrives at the bottom of the slide.

Problem 18
Ima Scaarred (m=56.2 kg) is traveling at a speed of 12.8 m/s at the top of a 19.5-m high roller coaster
loop.
a. Determine Ima's kinetic energy at the top of the loop.
b. Determine Ima's potential energy at the top of the loop.
c. Assuming negligible losses of energy due to friction and air resistance, determine Ima's total
mechanical energy at the bottom of the loop (h=0 m).
d. Determine Ima's speed at the bottom of the loop.

Problem 19
Justin Thyme is traveling down Lake Avenue at 32.8 m/s in his 1510-kg 1992 Camaro. He spots a police
car with a radar gun and quickly slows down to a legal speed of 20.1 m/s.
a. Determine the initial kinetic energy of the Camaro.
b. Determine the kinetic energy of the Camaro after slowing down.
c. Determine the amount of work done on the Camaro during the deceleration.

Problem 20
Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders
for customers. He fills a pitcher full of Cola, places it on the counter top and gives the 2.6-kg pitcher a
8.8 N forward push over a distance of 48 cm to send it to a customer at the end of the counter. The
coefficient of friction between the pitcher and the counter top is 0.28.
a. Determine the work done by Pete on the pitcher during the 48 cm push.
b. Determine the work done by friction upon the pitcher .
c. Determine the total work done upon the pitcher .
d. Determine the kinetic energy of the pitcher when Pete is done pushing it.
e. Determine the speed of the pitcher when Pete is done pushing it.

Problem 21
The Top Thrill Dragster stratacoaster at Cedar Point Amusement Park in Ohio uses a hydraulic launching
system to accelerate riders from 0 to 53.6 m/s (120 mi/hr) in 3.8 seconds before climbing a completely
vertical 420-foot hill.
a. Jerome (m=102 kg) visits the park with his church youth group. He boards his car, straps himself in
and prepares for the thrill of the day. What is Jerome's kinetic energy before the acceleration period?
b. The 3.8-second acceleration period begins to accelerate Jerome along the level track. What is
WORK, ENERGY AND POWER 2010
Jerome's kinetic energy at the end of this acceleration period?
c. Once the launch is over, Jerome begins screaming up the 420-foot, completely vertical section of the
track. Determine Jerome's potential energy at the top of the vertical section. (GIVEN: 1.00 m = 3.28 ft)
d. Determine Jerome's kinetic energy at the top of the vertical section.
e. Determine Jerome's speed at the top of the vertical section.

Problem 22
Paige is the tallest player on South's Varsity volleyball team. She is in spiking position when Julia gives
her the perfect set. The 0.226-kg volleyball is 2.29 m above the ground and has a speed of 1.06 m/s.
Paige spikes the ball, doing 9.89 J of work on it.
a. Determine the potential energy of the ball before Paige spikes it.
b. Determine the kinetic energy of the ball before Paige spikes it.
c. Determine the total mechanical energy of the ball before Paige spikes it.
d. Determine the total mechanical energy of the ball upon hitting the floor on the opponent's side of the
net.
e. Determine the speed of the ball upon hitting the floor on the opponent's side of the net.
Problem 23
According to ABC's Wide World of Sports show, there is the thrill of victory and the agony of defeat. On
March 21 of 1970, Vinko Bogataj was the Yugoslavian entrant into the World Championships held in
former West Germany. By his third and final jump of the day, heavy and persistent snow produced
dangerous conditions along the slope. Midway through the run, Bogataj recognized the danger and
attempted to make adjustments in order to terminate his jump. Instead, he lost his balanced and
tumbled and flipped off the slope into the dense crowd. For nearly 30 years thereafter, footage of the
event was included in the introduction of ABC's infamous sports show and Vinco has become known as
the agony of defeat icon.
a. Determine the speed of 72-kg Vinco after skiing down the hill to a height which is 49 m below the
starting location.
b. After descending the 49 m, Vinko tumbled off the track and descended another 15 m down the ski hill
before finally stopping. Determine the change in potential energy of Vinko from the top of the hill to the
point at which he stops.
c. Determine the amount of cumulative work done upon Vinko's body as he crashes to a halt.

Problem 24
Nolan Ryan reportedly had the fastest pitch in baseball, clocked at 100.9 mi/hr (45.0 m/s) If such a
pitch had been directed vertically upwards at this same speed, then to what height would it have
traveled?

Problem 25
In the Incline Energy lab, partners Anna Litical and Noah Formula give a 1.00-kg cart an initial speed of
2.35 m/s from a height of 0.125 m above the lab table. Determine the speed of the cart when it is
located 0.340 m above the lab table.
WORK, ENERGY AND POWER 2010

Problem 26
In April of 1976, Chicago Cub slugger Dave Kingman hit a home run which cleared the Wrigley Field
fence and hit a house located 530 feet (162 m) from home plate. Suppose that the 0.145-kg baseball
left Kingman's bat at 92.7 m/s and that it lost 10% of its original energy on its flight through the air.
Determine the speed of the ball when it cleared the stadium wall at a height of 25.6 m.

Problem 27
Dizzy is speeding along at 22.8 m/s as she approaches the level section of track near the loading dock of
the Whizzer roller coaster ride. A braking system abruptly brings the 328-kg car (rider mass included) to
a speed of 2.9 m/s over a distance of 5.55 meters. Determine the braking force applied to Dizzy's car.

Problem 28
A 6.8-kg toboggan is kicked on a frozen pond, such that it acquires a speed of 1.9 m/s. The coefficient
of friction between the pond and the toboggan is 0.13. Determine the distance which the toboggan
slides before coming to rest.

Problem 29
Connor (m=76.0 kg) is competing in the state diving championship. He leaves the springboard from a
height of 3.00 m above the water surface with a speed of 5.94 m/s in the upward direction.
a. Determine Connor's speed when he strikes the water.
b. Connor's body plunges to a depth of 2.15 m below the water surface before stopping. Determine the
average force of water resistance experienced by his body.

Problem 30
Gwen is baby-sitting for the Parker family. She takes 3-year old Allison to the neighborhood park and
places her in the seat of the children's swing. Gwen pulls the 1.8-m long chain back to make a 26° angle
with the vertical and lets 14-kg Allison (swing mass included) go. Assuming negligible friction and air
resistance, determine Allison's speed at the lowest point in the trajectory.

Problem 31
Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near
Bluebird Lake. She approaches a long embankment inclined upward at 16° above the horizontal. As she
slides up the embankment, she encounters a coefficient of friction of 0.128. Determine the height to
which she will travel before coming to rest.

Problem 32
Matthew starts from rest on top of 8.45 m high sledding hill. He slides down the 32-degree incline and
across the plateau at its base. The coefficient of friction between the sled and snow is 0.128 for both the
hill and the plateau. Matthew and the sled have a combined mass of 27.5 kg. Determine the distance
WORK, ENERGY AND POWER 2010
which Matthew will slide along the level surface before coming to a complete stop.
Work done by a net forces
If we look at the forces on an object being pulled across a table's surface there would be three: F, the
applied force, N, the normal or supporting force supplied by the table, and mg, its weight or the
gravitational force of attraction to the earth.
The normal force and the object's weight are in static equilibrium (they are balanced forces), the applied
force, F, is an unbalanced force and will result in the object being accelerated across the top of the
table's surface in the same direction as the force. This acceleration will change the object's velocity and
subsequently its kinetic energy. We say that this applied force is doing work on the object. The amount
of work done by F is directly proportional to the distance through which the force is applied as it pulls
the object across the table's surface.
net Workdone = (F
net)
cos Ө. ΔX
By using Newton's second law, F net = ma, our equation becomes
net Workdone = (ma) cos Ө. ΔX but cosӨ is 1 because 𝜽 is 00
net Workdone = m(a) ΔX
Remembering the kinematics equation vf2 = vo2 + 2as and solving for "as" let's our equation become
net Workdone = m[½(vf2 - vo2)]
net Workdone = ½(mvf2 - mvo2)
net Workdone = ½mvf2 - ½mvo2
net Wdone = ΔKE
WORK, ENERGY AND POWER 2010
The relationship we just derived is called the energy-work theorem..
This statement tells us that when an external force does work on an object it will change the object's
kinetic energy; that is, it will cause the object to either gain or lose speed. When more than one force is
acting on an object, all forces that are either parallel or antiparallel to the direction the object moves will
do work. If the object's velocity remains constant, that just means that the work done by opposing
forces (for example, a forward applied force, F, and an opposing force, friction) are equal.
Note that if F and s are perpendicular to each other no work is done on the object. In our example of
the block being dragged across the table, neither the normal force nor the weight would do any work on
the block since they act at right angles to the direction of the block's motion. Another example would be
when a satellite is being held in circular orbit by the force of gravity. Note that since the satellite's speed
and orbital radius remain constant, no energy is being changed; therefore, no work is being done on the
satellite.
Work done by conservative forces
Work done by conservative forces, or path-independent forces, results in changes in the object's
potential energy.
Let's use gravity an example of a CONSERVATIVE FORCE (or path-independent force). Remember that
the changes in an object's potential energy only depend on comparing its starting position and its ending
position, not on whether it does or does not pass through various points in-between. The block's final
change in potential energy is the same whether it follows the path with the intermediate stops B, C and
D or whether it is directly taken from A to E. The height of the post is the same.
When you observe an object falling, it loses potential energy (height) while it gains kinetic energy
(speed). That is, in the absence of another external, non-conservative force, such as friction,
pushing/pulling, or tensions in strings, the total amount of potential energy before the fall equals the
total amount of kinetic energy after the fall and the energy-work theorem is restated as the Law of
Conservation of Energy:
Workdone = ΔKE
Workdone conservative force= - ΔPE
- ΔPE = ΔKE
- (PEf - PEo) = KEf - KEo
- PEf + PEo = KEf - KEo
KEo+ PEo = KEf + PEf
For projectiles in freefall this statement of conservation of energy can be used to compare the energies
at two different locations (A and B) in its trajectory:
PEA + KEA = PEB + KEB