Physics 605 – Mechanical Energy
(Read objectives on screen.)
Instructor
No, I haven’t been playing with this toy the whole time you’ve been gone, but it is kind of
hypnotizing, isn’t it? So where were we? Oh yes, we were talking about the relationship between
work and energy and the fact that when work is done on an object, the kinetic energy or potential
energy of the object is increased. That relationship is the basis for calculating both kinetic and
potential energy. Let’s start with kinetic energy. Put your pencils down and watch. I’ll tell you when
to take notes.
(equations on screen)
VO
The basis for calculating kinetic energy is that work is required to put an object in motion. And the
amount of work done on the object will equal its kinetic energy. Here are four equations that you’ve
seen and used before. We’re going to use one at a time to substitute into this basic equation and come
up with the equation for kinetic energy.
First, we’ll use the equation for calculating work and substitute force times displacement for work.
Next, we’ll use Newton’s second law, “F equals m a” and substitute “m times a” for force. Now,
here’s an acceleration equation solving for displacement, so we’ll substitute “1/2 a t2” for "d." Now,
let’s do some rearranging to get “K E equals ½ m times a squared times t squared”. Finally, we’ll
substitute the definition of acceleration, “v over t” for "a". But "a" is squared, so we get v squared
over t squared. The t squared cancel out and we have
“K E = ½ m v2.”
Instructor
Now pick your pencils up and get this important equation in your notes.
This is another one of the equations that will be given to you on quizzes and tests. The reason we
showed you how to derive it is that many students want to know where the one half came from and
why the velocity is squared. You don’t have to derive it, and you don’t have to memorize it. All you
have to do is use it correctly. Let’s try some problems together.
(text on screen)
VO
A shot-putter heaves a 7.2 kg shot with a velocity of 6.9 m/s. What is the kinetic energy of the shot
and how much work did the shot-putter do?
The first problem is a straight plug-and-chug. Kinetic energy will equal one half times 7.2 kg times
6.9 m/s squared. Don’t forget that the number and unit are both squared. The answer, rounded to two
significant digits, is 170 kilogram meter squared per second squared, or 170 joules. Remember, if you
use MKS units, the answer will come out in the proper derived unit. This time it’s the joule.
Part "b" asks how much work was done on the shot. The answer is the same 170 joules. It takes 170
joules of work to give the shot 170 joules of kinetic energy.
(text on screen)
VO
Now you try some kinetic energy problems. Your teacher will stop the tape and put these problems
on the board and give you time to work them. You’ll go over the answers to the first two in your own
classroom and we’ll go over the third one together.
Local Teachers, turn off the tape and give students problem set number one from the facilitator's
guide.
(Pause Tape Now graphic)
(text on screen)
VO
This problem is a little tricky because of the unit for the car’s velocity. For the kinetic energy to be
calculated in joules, the units for mass and velocity must be in the MKS system. Kilogram is the unit
for mass in the MKS system, but kilometers per hour is not. So we need to convert kilometers per
hour to meters per second. This will take two or three conversion factors. We write “how many
meters per second equals 120 km/h?” First, we want to convert the kilometers to meters. The fact is
that one km equals 1,000 meters. Next, we want to get rid of hours and convert to seconds. By now,
you probably know the fact that one hour is 3,600 seconds. If not, go from hours to minutes and then
minutes to seconds. The answer is 33.3 repeating m/s. Since this is not our final answer, it’s not a
good idea to round all the way to two significant digits. Keep at least one or two extra digits so that
your final answer is not affected by your rounding. Or just keep all the digits in your calculator and
go from there.
To calculate kinetic energy, we plug 1,500 kg into mass and 33.33 m/s into velocity. Then we
calculate, remembering the one half and the velocity squared. Our answer, rounded to two significant
digits, is 8.3 X 10 5 joules.
Instructor
We find that velocity has a big effect on kinetic energy, more so than mass. That’s because velocity is
squared in the equation. So doubling an object’s velocity quadruples its kinetic energy.
Now let’s derive the equation for calculating gravitational potential energy. Again, it’s based on the
fact that work and energy are related. Put your pencils down and watch. I’ll tell you when to take
notes.
(equations on screen)
VO
To give an object potential energy, work must be done to raise the object to a certain position. So
we’ll start with the fact that potential energy equals work done on the object to lift it vertically.
Now you already know that work equals force time displacement, but this time, we’ll use a specific
force and a specific displacement. To lift an object straight up, we must apply a force equal to the
object’s weight. And the displacement will be the height. Sometimes you’ll use this equation to
calculate potential energy, but most often we use an equation that includes mass. So let’s go one step
further and use the special case of Newton’s second law for mass and weight. Weight equals mass
times acceleration due to gravity, or "g." This gives us the equation, “P E equals m g h.” This
equation also will be given to you on quizzes and tests.
Instructor
So we have another important equation. This one calculates the gravitational potential energy of an
object, which equals the object’s mass times the acceleration due to gravity times the height.
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And remember that “m times g” is the weight of the object. When mass is given, use “P E equals m g
h,” and when weight is given, use “P E equals weight times height.”
Now, there’s something important you need to know about potential energy. You can’t measure it!
You heard right. You can’t measure it. That’s because potential energy is hidden. The only time you
can measure it is when it goes into or comes out of hiding. In other words, you measure an object’s
change in potential energy. When I lift this book from the table to this position, I change its potential
energy. The “h” in this equation is relative to some frame of reference, like the table, or the floor, or
the ground.
For example, if you told me that you climbed to the top of Stone Mountain, which is 250 meters
above the ground around it, I would say that the height you climbed was 250 meters. But two thirds
of the mountain is hidden beneath the ground, so relative to the actual bottom of the mountain, the
height would be about 750 meters. My point is that when we calculate gravitational potential energy,
we usually assume the frame of reference to be the ground if we’re outside or the floor if we’re inside.
OK. Now that we’ve established the ground rules, so to speak, what’s next? Bet you’ve already
guessed. Example problems.
(text on screen)
VO
In this problem, we are lifting a 2.0 kg book to a shelf that is 1.8 m above the floor. Relative to floor,
the height is 1.8 m. We’ll plug it in to the equation, “potential energy equals mgh” and plug in the
mass and 9.8 m/s2 as "g." The gravitational potential energy equals 35 joules.
What is different about part b? This time we want the change in potential energy from the table to the
shelf, so “h” is 1.8 minus 0.6 m or 1.2 m. Remember that when we subtract, we round to the
estimated column, which is tenths. The gravitational potential energy relative to the table is 2.0
kilograms times 9.8 m/s2 times 1.2 m, or 24 joules.
(text on screen)
VO
Now you try these problems. Your teacher will stop the tape, put these problems on the board and
give you time to work them. We’ll go over the answers when everyone is finished.
Local Teachers, turn off the tape and give students problem set number two from facilitator's guide.
(Pause Tape Now graphic)
(text on screen)
VO
Number one is a straight plug-and-chug problem. The mass is 78 kg, you know that “g” is 9.8 m/s2
squared, and the height is 46 m. That makes the potential energy 3,500 joules.
How much work did the climber do? The answer is the same amount, 3,500 joules.
In the second problem, we are given the weight instead of the mass. But remember that mass times
"g" equals weight. So this problem is even easier than number one. All we do is multiply the weight,
2,400 newtons, times the height, 5.2 m, for an answer of 12,000 j.
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Instructor
Those problems were pretty easy, weren’t they? Let’s see if I can complicate things a little. Don’t
panic. I said a little! But before we get to more complex problems, let’s review what we know about
work and energy.
You’ve already seen how work is related to both potential and kinetic energy. Work done on an
object increases the object’s potential or kinetic energy, and work done by an object decreases that
same energy. Now we need to add another important fact about energy.
Energy can be transformed from one form to another. Kinetic energy can be transformed into
potential energy, and potential energy can change into kinetic energy.
Let’s look at some examples of the relationships between work and energy and between kinetic and
potential energy.
(close-up of hand lifting mass on screen)
VO
When I lift this heavy mass, I’m doing work on the object, and its potential energy increases.
When the ball is released, it starts moving. This means that the potential energy of the object is
changing into kinetic energy.
And if there happens to be an egg in its path, work can be done by the falling mass, decreasing the
object’s kinetic energy.
When I turn the tires of this toy car backwards to wind the spring inside, I’m doing work on it to
increase the car’s potential energy. The work done on the car equals the car’s gain in potential energy.
When I let the car go, this potential energy turns into kinetic energy as the car moves.
When the car knocks over the blocks, work is done by the car and the car loses kinetic energy. The
kinetic energy lost by the car equals the work done by the car on the blocks.
You may have noticed that in each demonstration, we concentrated on one object and its energy
changes. Of course energy changes were happening with the egg and the blocks, too, but we didn’t
want to get into that.
When you look at energy changes, one thing is always true. In fact, it’s a law, the Law of
Conservation of Energy.
(green chalkboard on screen)
VO
The law of conservation of energy states that energy can change form but cannot be created or
destroyed.
In a closed, isolated system of objects, the total mechanical energy, which equals potential energy
plus kinetic energy, remains constant. This second statement is sometimes called the Law of
Conservation of Mechanical Energy.
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Instructor
Here’s a very simple example of the Law of Conservation of Mechanical Energy. It’s a pendulum,
and this is called the bob. We’re going to look at the energy of the bob in three positions. In position
“a” I’ve pulled the bob up but have not released it. When I release it, the bob swings down to its
lowest position, “b,” and then up to its highest position on the other side. That’s “c.” Here’s what I
want you to do. I want you to describe the kinetic energy and potential energy of the bob, using one
of these three words: maximum, minimum, or zero.
(diagram on screen)
VO
Your teacher will give you this diagram and give you time to fill in the blanks. Then you’ll come
back and we’ll talk about your answers.
Local Teachers, turn off the tape and give students problem set number three from the facilitator's
guide.
(Pause Tape Now graphic)
(diagram on screen)
VO
Let’s start with the bob in position "a." Since it has been pulled back to its maximum height and it’s
not moving yet, we can say that its kinetic energy is zero and its potential energy is maximum.
Once the bob starts moving, it loses potential energy as it gains kinetic energy. At the bottom here, its
kinetic energy is maximum. But what about its potential energy? If you said zero, that’s OK.
However, there’s another answer that may be better. The pendulum is suspended over the table,
which is above the floor. So is its potential energy really zero? It depends on your frame of
reference. If you consider the lowest point in its trip to be your reference, then zero is OK. But if
you’re thinking that the table or the floor is your reference, then your answer should be “minimum.”
Either answer would be acceptable, as long as you can explain.
On its way up toward “c,” the bob has lost kinetic energy as it slowed down and gained potential
energy. At position "c," the kinetic energy would be zero again and the potential energy would be
maximum. Mechanical energy has been conserved.
Instructor
Now it’s time to try some conservation of energy problems. All the problems will be based on the
fact that the total mechanical energy remains constant. Here’s the equation we’ll use for every
problem:
“Initial potential energy plus initial kinetic energy equals final potential energy plus final kinetic
energy.” Copy this down and you’ll be ready for the first example problem.
(text on screen)
VO
Let’s work this one together. We start with a 12-kilogram rock at the edge of a 95-meter tall cliff.
And we want to calculate its potential energy and kinetic energy. You know that potential energy
equals “mgh,” so all we do is plug in the 12 kilogram mass times 9.8 m/s2 for “g” and 95 m for “h.”
The rock’s initial potential energy is 11,000 joules.
Since the rock is initially at rest, its initial kinetic energy is zero.
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Then we are told that the rock falls to the ground, and we are asked its final potential energy and
kinetic energy just before it hits. This is where the Law of Conservation of Mechanical Energy comes
in. The law is “initial potential energy plus kinetic energy equals final potential energy plus kinetic
energy.” Let’s plug in what we know. In part a, we calculated the initial potential energy to be
11,000 joules and kinetic energy was zero. At the very bottom of the hill, the ball’s final potential
energy will be zero, and we can calculate its final kinetic energy. Well, that’s easy. It’s 11,000
joules.
Now let’s calculate the final velocity of the rock. Let’s rearrange the equation for kinetic energy to
solve for velocity. We multiply both sides by two and divide by "m." Then we take the square root of
both sides and get “v equals the square root of two times K E divided by m.” We know that the mass
is 12 kg and K E is 11,000 j, so v equals 43 m/s.
Instructor
Did it surprise you that you could use the mass and height of an object to find its final velocity
without using Newton's Second Law and an acceleration equation?
Well, you actually used both! If you’ll remember, we used both to derive the equation for calculating
kinetic energy. So you didn’t have to use them again. It’s a shortcut based on the Law of
Conservation of Mechanical Energy.
(text on screen)
VO
Now you use the Law of Conservation of Mechanical Energy to solve this problem. Your teacher will
give you the problem and time to try it. Then we’ll go over the solution together.
Local Teachers, turn off the tape and give students problem set number four from the facilitator's
guide.
(Pause Tape Now graphic)
(cartoons of diver on screen)
VO
Remember that at any point, the total potential energy plus kinetic energy must remain constant. The
instant the boy steps off the diving board, he has not started falling yet, so his kinetic energy is zero
and his potential energy is 5,800 joules.
When he has fallen half way to the water, he’s lost half of his potential energy and gained that same
amount of kinetic energy. So his potential energy is 2,900 joules and his kinetic energy is 2,900
joules.
At the surface of the water, the swimmer has lost all his potential energy, so his potential energy is
zero, and his kinetic energy is 5,800 joules.
Now for question two, we are given the mass of the swimmer, 62 kilograms. And we’re asked to find
the height of the diving board. We can use the equation for calculating potential energy and rearrange
to solve for "h," which equals “PE divided by m g,” or 5,800 joules divided by 62 kilograms times 9.8
m/s2. The height of the board is 9.5 m.
For part b of this question, we can rearrange the equation for kinetic energy and solve for “v,” just like
we did in the first example. “v equals the square root of 2 times KE divided by m.” When we plug in
the numbers, we get a velocity of 14 meters per second.
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Instructor
Now don’t worry about friction when you solve conservation of mechanical energy problems. You
may always assume that friction is negligible in these problems, just like we consider friction to be
negligible during elastic collisions.
Now that I think about it, we’ve already talked about collisions in terms of conservation of
momentum. Now we need to revisit collisions in terms of kinetic energy.
(student on screen)
VO
Remember that the definition of an elastic collision is one where the objects bounce apart with no loss
of energy. So the system not only conserves momentum, but kinetic energy as well.
(green chalkboard on screen)
During elastic collisions kinetic energy is conserved by the system of objects.
(car collision on screen)
VO
Remember that the definition of an inelastic collision in one where the objects stick together or are
deformed with loss of energy. While momentum is conserved during these and all collisions, kinetic
energy is not.
(green chalkboard on screen)
During inelastic collisions, kinetic energy is not conserved by the system of objects.
Since inelastic collisions involve lots of friction some energy is lost to the surroundings in the form of
heat.
Instructor
Does this mean that the Law of Conservation of Energy is broken during inelastic collisions? No.
Energy is still conserved by the universe, but not by the small system of colliding objects. When heat
escapes, the system is no longer isolated or closed, so the Law of Conservation of Mechanical Energy
no longer applies. Got it?
Now we’re finally ready to keep our promise of explaining this toy. When I pull two balls back and
let them hit, two always fly out together. We wondered why one doesn’t fly out at twice the velocity.
Now you’re ready to help explain why.
(diagrams on screen)
VO
In an earlier program, we decided that momentum would still be conserved if two balls hit and then
one ball were to fly out at twice the velocity. Let’s plug in some simple numbers and do the math.
You know that momentum equals mass times velocity. Let’s say that each ball has a mass of one
kilogram and the two balls together move at one meter per second before they hit. Before the
collision the momentum would be two kilograms times one meter per second, or two-kilogram meters
per second.
If one ball flew out at twice the velocity, that would be one kilogram times two meters per second, or
two kilogram meters per second. Momentum would be conserved. But that never happens.
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And that’s because in an elastic collision, both momentum and kinetic energy are conserved. So let’s
look at kinetic energy, which equals “1/2 mv2.” Before the collision kinetic energy equals one-half
times two kilograms times 1 meter per second, squared or one joule. When two balls fly out after the
collision, the kinetic energy is the same one joule.
But if one ball flew out at twice the velocity, the kinetic energy would equal one half times one
kilogram times two meters per second squared, or two joules. Since energy cannot be created, this
never happens.
Instructor
Have you noticed how similar momentum and kinetic energy are? The formulas for calculating each
involve only mass and velocity. But velocity is much more important to kinetic energy because it is
squared in the equation.
Now, how else do momentum and kinetic energy differ? It’s how they affect objects. A change in
momentum provides an impulse or punch to do things such as knocking objects down. But a change
in kinetic energy actually does work on objects, such as denting fenders or breaking bones.
And it looks like we’ve used our kinetic energy to work on our time for today. We don’t even have
time for your Show What You Know quiz. But don’t worry. Your local teacher has it and will give it
to you when this tape ends.
The last question is a physics challenge to end our study of momentum and energy. You will be
asked which would hurt more: being tackled by a massive, slow moving tackler or a less massive, fast
moving tackler with the same momentum? You might want to plug in some numbers to make
momentum come out the same and then calculate kinetic energy. Talk over what your answers mean.
Oh, and good luck on the unit test. Work hard and efficiently, keep the momentum going, and live up
to your potential. Got it?
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