Download Plan for March 2010

LeaPS Workshop
March 12, 2010
Morehead Conference Center
Morehead, KY
Word Bank:
Acceleration, mass, inertia, weight, gravity, work, heat, kinetic energy, potential energy,
closed systems, open systems, energy conservation, Newton’s laws of motion.
Learning targets:
I can show that forces come in pairs that are equal and opposite.
I recognize that changing motion can be speeding up, slowing down, or changing
direction.
I can represent changes in motion using arrows.
I can calculate acceleration from data about changes in motion over time.
I can recognize that any change in motion is a consequence of unbalanced forces.
I know that a change in motion (acceleration) is always in the direction of net force.
I can show that acceleration is directly proportional to net force.
I know that the mass of an object is related to its inertia, its tendancy to maintain its
current state of motion.
I know that the mass of an object is not the same as its weight; weight being the
attraction between objects due to the force of gravity between them.
I recognize the need to use low-friction objects to study the laws of motion.
I can show that the acceleration of an object depends on both it mass and thenet force
acting on it.
I know that work performed on an object equals the force applied times its displacement.
I recognize that work is a type of kinetic energy.
I can calculate the kinetic energy of an object in motion knowing its mass and velocity.
I can differentiate between open and closed systems with respect to energy transfers.
I know that while energy is conserved in a closed system, it may be converted between
kinetic energy, potential energy, or heat.
.
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•
Newton’s Third Law
In our work in January, we found that forces always occur in pairs:
You bumped hands with a groupmate. You felt a force in a direction
toward you and your groupmate felt a force toward her. By measuring the
size of these forces we would find that they are equal in size; they are
clearly oppositely directed. Forces always occur between bodies and
always occur in pairs, for example you are pushing on your groupmate
and s/he is pushing on you. We cannot have one force without the other.
This law of physics is stated as:
In an interaction between body A and body B, body A
exerts a force on body B and body B exerts an equal and
opposite force on body A.
These equal and opposite forces of an interaction never act on the same
body, because an interaction is between bodies.
This insight is not only a description of interactions but is also Newton’s
Third Law. You may have heard this stated (but less clearly so) as: “To
every action there is always an equal and opposite reaction.”
Consider a collision between two billiard balls: one moving directly at another
one not initially moving. What happens with each ball? Does the moving ball
change its motion? Does the stationary ball change its motion? Which one
experiences a force? How do you know? What are the directions of all of the
forces? How do you know? Observe the demonstration and discuss these
questions in your group.
When two magnets interact, one magnet experiences a force and the other
magnet experiences an equal and opposite force.
If one magnet is very large or heavy
and the other is small or light, both
magnets will experience the force
equal and opposite, but the lighter one
will respond more because it has less
mass.
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When a small car runs into a large truck, the force of the truck on the car is equal
and opposite to the force of the car on the truck. The forces are equal and
opposite, but the responses of the two bodies are different because the masses
are different.
•
Forces and changing motion
From our work in February, we showed that changing motion can be described
as changing velocity; continuously changing motion means continuously
changing velocity. All changing motion involves changing velocity. If there is a
changing velocity, there is a change in v, ∆v, occurring as time passes along;
after a very small time v(final) must be different from v(initial)
∆v = v (final) – v(initial) = vf – vi
I find it easier to think about rearranging and writing vf = vi + ∆v. In other words,
we need to add ∆v to vi to get vf . There are three change-in-velocity scenarios
to consider: when an object is speeding up, when it is slowing down, and when it
is changing direction:
For increasing v (speeding up)
vi
∆v
When v is increasing, then ∆v is
in the same direction as vi,
corresponding to a increase in v.
vf
For decreasing v (slowing down)
vf
∆v
Note that the direction of ∆v is
opposite to that of vi and
corresponds to a reduction of
v: vf is less than that of vi
vi
For changing direction
What is the ∆v associated with changing direction? We know that we must have
a ∆v,
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and vf = vi + ∆v. When we change direction, the direction of the velocity changes
and we velocity representations that look like this:
vi
vf
∆v
Note that if there is any ∆v, then there must be a matching unbalanced force. In
other words:
unbalanced forces → net force → changing motion → ∆v
Furthermore, the direction of the ∆v is always in the same direction as the net
force:
∆v
Net force
As you suspected all along, any change in motion (speeding up, slowing down,
or changing direction), is an acceleration. Changing motion is motion with a ∆v,
so acceleration is associated with ∆v and in fact the definition of acceleration is:
a = acceleration = ∆v/∆t, the change in velocity per unit of time
Acceleration and ∆v are always in the same direction. Whenever you want to
know the direction of the acceleration, determine the ∆v and you have it. The
units of v and hence ∆v are for most applications meters per second or m/s.
Hence, the units for acceleration are meters per second x second or m/s2.
Now we can see that net force produces acceleration, following the logic
unbalanced forces → net force → changing motion → ∆v
→ a
And just as with velocity, the direction of acceleration is the same as the direction
of the net force.
This connection between unbalanced forces and acceleration tells us that
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NET FORCE IS PROPORTIONAL TO ACCELERATION
If we are pushing on a body and increase net force then we get a larger
acceleration. The symbol ∝ means proportional. Therefore, can write this
proportionality as
F(net)
∝
acceleration
The net force€
and acceleration are directly proportional. Doubling the net force
causes double the acceleration etc. We can write this in equation form including
an as yet unspecified constant as
€
F(net) = (constant) acceleration
or
F(net) = (Constant)a
•
How does mass fit into this?
Motion and Inertia
The idea that a force is involved in motion is not terribly surprising. The value is
in the details: an applied force causes acceleration (the continual change in
motion of an object), the absence of a net force is necessary to maintain a
constant speed, and deviance from these rules may be accounted for if we
consider friction as a somewhat hidden force.
The next topic we will deal with, the role of inertia, is at first glance, on the
surface not tremendously shocking. Inertia refers to the tendency of an object
to maintain its current state of motion. Therefore, an object at rest will remain
at rest until put into motion by an applied force. A cart in motion will maintain its
motion in a straight line unless a net force speeds it up, slows it down, or
changes its direction. (This is called Newton’s First Law of Motion.)
As with the effect of force, the role or importance of inertia is in the details. In
this section, you will investigate why some objects resist motion more than
others, and how inertia can be incorporated into the general theory of motion.
What characteristic of an object seems to determine how much force is needed
to accelerate it?
Imagine needing to accelerate two vehicles - a small car and a large truck
(consider this under low-friction conditions). For which vehicle do you think it
would be easier to change the speed continually from at-rest to a speed of
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12mph with a given force – the small car or the large truck? Discuss this in your
groups.
It is obvious to most people that if there is more stuff, then more force will be
required to cause a continual change in speed. But suppose we double the
amount of stuff. Will that mean that twice the net force is needed to accelerate
double the stuff? Scientists have named the “stuff” about which we’re talking
mass.
What Is Mass? Philosophers of science have great debates about the definition
of mass. Many equate mass with weight, assuming that more stuff weighs more.
While true on the surface of the earth, this is not true in outer space, where
things become “weightless”. The weight of an object is due to the attraction its
mass feels to other objects with mass. People have observed that one piece of
mass attracts another piece of mass. Objects on earth have weight because
their relatively small mass interacts with the mass of the earth and is attracted to
the earth. We can call that pull its weight. In outer space, because things are
not near an object with large mass such as the earth, they do not have this pull
and hence do not have weight. The nuances of mass and weight are beyond the
scope of our studies here, although you may wish to investigate this in another
context. For now we will rely on our intuitive understanding of mass as some
“amount of stuff” that affects how much an object resists changes in its motion.
Mass measures inertia, the resistance to changes in motion
a) Recall the experiments with the person on the skateboard. Suppose we had a
volunteer that had much less mass than the original person. Imagine how the
observations of the change would be different if we kept the pushing force the
same.
Note: We have to keep the pushing force the same!! Often we make assumptions
about an experiment and lose track of what is being controlled. Here the pushing
force needs to be kept constant; the experiment has no meaning if that is not the
case.
Suppose we had a volunteer who had much more mass than the original person.
Imagine how the observation of the change in motion would be different if we
kept the pushing force the same.
Describe your ideas about the observed motion for original person on the cart, a
much lighter person on the cart and a much heavier person on the cart always
keeping the pushing force the same.
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b) Think about a soccer ball being kicked so as to change its motion (slowing,
speeding, changing direction). Suppose we replaced the soccer ball with a
bowling ball. What do you think you would observe as a result of this increase in
mass? Discuss this with your group, then observe the demonstration.
[Group discussion about the effect of mass just observed]
What do you think you would observe about changes in the motion of a low
friction cart if we added weights with about the same amount of stuff (mass) as
the cart itself? We will do this activity by approximately doubling the amount of
stuff compared to the plain cart. You may note that the cart with more mass may
be pushing down harder on the table and maybe have a higher friction force.
This would be true for objects with measurable friction, like the person on the
blanket, however with good low friction devices, the friction remains essentially
zero for all reasonable added masses. In reality, the friction is never zero but for
all practical purposes it can be ignored. If we apply the same net force for both,
how would the motion of the “friction-free” cart with about double the stuff
compare with that of cart alone?
We will do the experiment with the cart mass alone, more mass and cart, and
finally even more mass and cart and do it with a medium force and a large force.
For the activities in this section each group will need:
•
1 spring scale
like we used before
•
1 low friction cart or 4 inch car on which we can place additional mass like
we used previously
•
Weights for increasing mass, each block should have about the same
mass as the car/truck so that adding one doubles mass and adding two
triples the mass
•
LabQuests (LQ), fully charged
You will need to investigate how to interpret LabQuest velocity data. In this
experiment, you will be applying a constant force and observing a change in
velocity. Based on the earlier discussion about acceleration, you can take the
change in velocity and divide it by the time interval during that velocity change to
calculate the acceleration. Discuss in your group exactly what data is needed.
Then try it a few times to get the sense of how to do it efficiently.
 Check with facilitators that you are carrying out this operation
correctly.
At first, try this activity without using the LQ, just to get a feel of what is going on.
With no added weights pull the cart or block with a medium force and observe
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the change in motion qualitatively. Record your feel for it such as: slowly
increasing in velocity or rapidly increasing in velocity, remembering that you want
to keep force constant. Then add weights equal to the mass of the cart and
repeat the experiment with the same medium force. Then add additional weights
and repeat with the same medium force. As mass is increased is the change in
motion greater or lesser? Record your observations in writing here and in the
table below.
Now do the same set of experiments except exert a larger force with the springscale, keeping the force equal for the different masses. Record your results in
writing here and in the table below. Is the change in motion for the higher force
greater, lesser or the same as that for the medium force for all three cases? You
may need to repeat some of your medium force experiments to make the
comparisons.
Complete the following chart as a way of summarizing your observations:
Table for Cart Experiments, circle appropriate term or make entry
Cart mass
Medium force exerted
Change in motion is:
very small, small,
moderate, large, very large
Larger force exerted
Change in motion is:
very small, small,
moderate, large, very large
Double mass
Change in motion is:
very small, small,
moderate, large, very large
Change in motion is:
very small, small,
moderate, large, very large
Triple mass
Change in motion is:
very small, small,
moderate, large, very large
Change in motion is:
very small, small,
moderate, large, very large
No added weights
Next, carry out these activities in a quantitative manner using the LabQuests.
 CHECK write down your conclusions from this activity as to how
changing mass affects the ability of a constant net force to change
motion in a straight line. Share your conclusions with a facilitator.
Most people observe that as mass increases, for the same net force the
change in motion is less; conversely as the mass is decreased for the
same net force applied to the body the change of motion is greater. Do
your observations agree with this? Do you believe this?
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There are two linked observations here critical to understanding the laws of
motion:
You have verified the law of proportionality between the applied force and the
resulting “speeding up.” Acceleration is directly proportional to the applied
force; that is, the acceleration doubles when the applied force doubles.
The acceleration also depends upon the mass of the carts, however. When
there is twice as much mass, the acceleration is halved. We say that
acceleration is inversely proportional to the mass, when a constant net force
is applied.
The data you collected imply a mathematical relationship among net force, mass,
and acceleration. These parameters can be combined in an equation form like
this:
acceleration
∝ Net F/m
where net F is the net force and m is the mass. Note this equation tells us that
for a larger m while keeping the net F constant, the acceleration is less. For a
larger net force while keeping the mass constant results in an acceleration that is
€
larger.
With units of kilograms for mass, newtons for force and meters per second2 for
acceleration, we can write this equation as a = net F/m or if we multiply both of
the equation by m we get net F = ma.
net F = ma
Congratulations! Through a study of motion, you have uncovered
Newton’s Second Law, which essentially says that the changing motion,
or acceleration, of an object depends directly on the net force applied
to it and inversely on how massive the object is. That it took over
2000 years of research and the development of calculus to derive
these laws should give you an idea of exactly how difficult these
understandings are. And yet you did it in just a morning’s worth of
work!
Congratulations!!!!!!!!!!, but we are not through.
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 Work and Heat
Mechanical work is defined in terms of force and the displacement
through which the force acts. When the direction of the force and the
direction of the displacement are the same the relationship is Work =
(Force)x(displacement) often written as (Force)(distance) or Fd.
Force direction (F)
Distance (d)
W = Fd, when force and displacement in same direction.
Note that when the force and displacement are not in the same direction, the
equation for calculating the work is a little more complicated. We will not work
with these cases, so we are limited to studying situations where the force and
displacement are in the same direction.
Think back to one of our first experiments – pushing a person on the
skateboard. We applied a force of about 40 pounds as measured by our
weight scale. If we had used a scale calibrated in newtons (N) we would
have had 178 N where the conversion is 1 pound = 4.45 newtons.
(40 pounds)x(4.45newtons/pound) = 178 newtons
Notice the unit “pounds” cancels out in the equation
(40 pounds)x(4.45newtons/pound) = 178 newtons
Kinetic energy (KE) is the energy associated with motion and it can be
calculated with the defining equation
Kinetic energy =KE= 1/2 m v2, where KE is measured in joules, m is
mass in kilograms and v is the speed in meters per second.
The most basic statement about energy in mechanics is that
The work done by all forces on an object is equal to the
CHANGE in kinetic energy of the object.
For our pushing on the “friction-free” skateboard the forces add to form
the net force; in our case the net force is equal to the force of our push
since there is essentially no friction force (and there are no unbalanced
forces in the vertical direction).
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Calculate the work done when we pushed the person a displacement of 1
meter. The units in your answer will be newton x meter or newton meter.
One newton meter is equal to a joule.
As you may recall we stopped taking data when we had pushed the
person 747 cm or 7.47 m. Calculate the work done by the pushing force
when we pushed the person 7.47 meters.
Discuss the following in your group: If the net force on an object is zero, because
all the forces are balanced and there is no change in motion, what total work is
done? What change in velocity would you expect? What change in kinetic
energy would you expect? Does this agree with what you would expect using
Newton's Second Law? When the total or net work done on an object is zero, is
the kinetic energy of that object conserved, i.e. does it remain constant?
Energy Transformations
The problem just completed is a simple example of energy conservation. In this
case kinetic energy is conserved because the final kinetic energy is equal to the
initial kinetic energy.
If you studied heat you know that it is energy in transit. Heat is energy that
transfers from a hotter (higher temperature) body to a colder body. If a heating
experiment is done in an insulated container, all the energy that leaves the hotter
body goes to the colder body and energy in the insulated container is conserved.
One can keep track of the heating and cooling and confirm this.
You have likely thought about other forms of energy like chemical energy, light
energy etc. The principle of energy conservation can be extended to many other
cases, although we will not prove or demonstrate it here. For example, if an
object like a 1 kg block of wood is sliding on a table with a speed of 2 m/s it has
kinetic energy in an amount that you can calculate. Discuss what you think
happens to this kinetic energy as the block slides to a stop.
There are many applications of the energy conservation principle throughout
physics and all of science. It is one of the cornerstones of our scientific view of
the world. If there are no external sources or losses of energy from a
system, when all forms of energy are considered for the system, the total
energy of the system remains constant. We call a system for which energy
does not go out or in and is only transformed within and transferred
between things in the system, a CLOSED or ISOLATED SYSTEM. Energy
can change forms, but the total amount of energy in a closed system
remains constant
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What is potential energy?
Here are some basic definitions, phrased informally:
Kinetic energy = The energy something has because of its motion. The heavier
or faster something is, the more kinetic energy it has.
Gravitational potential energy = The energy “stored” in an object because it’s
been lifted. The energy is “potential” because it has the potential to turn into
kinetic energy—for instance, if the object gets dropped. The heavier or higher an
object is, the more potential energy is stored up. There are other types of
potential energy as well such as chemical energy.
Consider and discuss the following scenario:
A child pushes a loaded wagon up a hill, starting slowly but gradually
getting faster and faster. Is the wagon gaining kinetic energy, gravitational
potential energy, both, or neither? Explain.
Suppose the wagon gains a total of 50 joules of energy. According to
energy conservation, energy is never created nor destroyed. But the
wagon just gained 50 joules! What is meant by this statement? Does this
scenario contradict conservation of energy? Explain why or why not.
When you say something like “a jogger just burned 100 calories” of energy, what
does that mean? Where exactly do those 100 calories come from? In other
words, what form of energy is depleted by 100 calories? Hint: This is as much a
biology or chemistry question as it is a physics question.
Roughly speaking, when work is done on a body, the mechanical energy—the
kinetic and potential energy—are changed an amount equal to the work done.
How much work did the child do on the wagon?
 CHECK Make sure that you have written full responses to all the questions and
discussed your ideas with your groupmates before doing the check.
Lifting Objects; Work Done
In this problem, you’ll use the formal definition of work to figure out some stuff.
Recall W = Fd, with force and displacement in the same direction.
A student holds a book of mass m in her hand and raises the book
vertically at constant speed. Make a drawing to illustrate the
motion of the book, showing its position at regular time intervals;
this diagram is sometimes called a Motion diagram, especially if the
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positions at equally separated times are shown. For example, an object moving
at a constant speed could be represented with dots at equally spaced times.
Suppose the student does 25 joules of work lifting the book.
1. Does the book (lifted at constant speed) gain potential energy, kinetic
energy, or both? Explain.
Is the potential energy gained by the book greater than, less than, or
equal to 25 joules? Explain.
2.
 Check. Make sure that you have written full responses to all the questions
and discussed your ideas with your groupmates before doing the check.
Pushing on a wall
In this section we’ll clarify the meaning of work.
A student pushes hard enough on a wall that she breaks a sweat. The wall,
however, does not move. Does the student do any work on the wall? Answer
using:
a. your intuition
b. work as defined above
Apparently, some reconciliation is needed. We’ll lead you through it.
In this scenario, does the student give the wall any energy?
Does the student expend energy (i.e., use up chemical energy stored in her
body)?
If the energy “spent” by the student doesn’t go into the wall’s mechanical
energy, where does it go? Is it just gone, or is it transformed into something
else? Hint: How do you feel when you’ve expended lots of energy?
Intuitively, when you push on a wall, are you doing useful work or are you
“wasting energy”?
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A student says,
“In everyday life, ‘doing work’ means the same thing as ‘expending
energy.’ But in physics, work corresponds more closely to the intuitive
idea of useful work, work that accomplishes something, as opposed to just
wasting energy. That’s why it’s possible to expend energy without doing
work in the physics sense.”
In what ways do you agree or disagree with the student’s analysis?
Let’s check for consistency between intuition and definitions. Does the student’s
idea that “doing work” means “expending energy” mesh with the rough definition
of work as the mechanical (kinetic and potential) energy given to something?
 Check. Make sure that you have written full responses to all the questions
and discussed your ideas with your groupmates before doing the check.
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