Download Physics 402 – Newton`s Second Law (Read objectives on screen

Physics 402 – Newton’s Second Law
(Read objectives on screen.)
Instructor
In the last program, we introduced Newton’s Laws of Motion. We spent a lot of time on the first law.
Do you remember another name for it?
If you said the law of inertia, you’re right. Remember that this law says that objects at rest tend to
stay at rest and objects in motion stay in uniform motion. But that’s only true under what condition?
It’s true only when there is no net force acting on the object.
(graphic on screen)
And what does the "net" mean? It means that there could be many forces acting on an object in
different directions so that they add up to zero.
At the very end the last program, we introduced part of Newton’s second law of motion, the Law of
Acceleration. Do you know how this one differs from the first law?
(Graphic changes)
This time, there is a net force exerted on our object.
And the law states that the object will accelerate in the direction of that net force. We’ll always use a
big, fat arrow like this to show the direction of the acceleration. It will come in handy when we work
problems involving several forces. But one thing is certain. The acceleration will be in the same
direction as the net force, even if the object is moving in the opposite direction. Remember that
acceleration describes change in motion, not the motion itself.
But there’s more to Newton’s second law than what we told you before. And it has to do with the
effects of mass and force on acceleration. To find out how these three variables are related, let’s do a
simple experiment.
(lab carts on screen)
VO
We’ll use the spring-loaded exploder on this cart to exert a force on another cart. When the pin is
tapped, the exploder will be released. To increase the force exerted, we’ll push the rod in to the
second notch, compressing the spring more.
To get a general idea of how much the cart accelerates, we’ll measure the time it takes the cart to go
from one end of the track to the other. The greater the acceleration of the cart at the beginning, the
less time it will take to travel a set distance.
(table on screen)
Here’s a data table for Part One of our experiment. You do not need to copy it. We’ll fill in the chart
on the screen, and you can make conclusions on the spot. All we want is enough data to make a
conclusion about the relationship between force exerted and acceleration. So we’ll keep the mass of
the cart constant, vary the force, and see what happens to acceleration.
(students on screen)
Here goes trial one, with the exploder set to the first notch, which provides a small force.
The time for trial one is 4.15 seconds.
Now, we increase the force to be exerted on our cart for trial two.
The time for this trial is 1.41 seconds.
(table on screen)
Look at the times in our data table, and make a conclusion about the relationship between the net
force exerted on an object and the object’s acceleration. Use the times to help you determine whether
acceleration has increased or decreased, but don’t include the word, time, in your conclusion. Tell
your teacher.
(Pause Tape Now graphic)
Instructor
Now a physics surfer would just look at this last column, see that the numbers decreased, and
conclude that as force increases, acceleration decreases. But we aren’t physics surfers, are we?
(table on screen)
Diving deeper into the situation, we see that as time decreases, acceleration must be increasing
because the cart moves faster. This means that the acceleration of an object increases when the net
force increases.
If we did a more quantitative experiment, we would discover the same thing Newton discovered:
Acceleration is directly proportional to net force.
Now let’s see how mass and acceleration are related.
(students on screen)
VO
This time we’ll use the first notch of the spring exploder for both trials, keeping the force exerted on
our cart constant.
We can use the data from trial one because in it we set the exploder on the first notch. Remember that
the time was 1.41 seconds for that trial.
Now we’ll add a heavy rider to the cart, increasing the mass. We’ll keep the force constant and repeat
the experiment.
For this trial, the time is 1.75 seconds.
(table on screen)
Look at the data for Part two of the lab, and tell your teacher how you think mass and acceleration are
related. Remember, no surfing allowed. Think about what the data tells you.
(Pause Tape Now graphic)
Instructor
If surfing led you to the wrong conclusion last time, maybe you were more careful this time. As the
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mass of our cart increased, it took more time for the trip. This means that acceleration decreased.
(text on screen)
So acceleration must be inversely proportional to mass.
Now we can complete our statement of Newton’s Second Law of Motion.
(green chalkboard on screen)
VO
You already have the first part of the law in your notes. When a net force is exerted on an object, the
object accelerates in the direction of the net force.
Now add the second part. The acceleration of the object is directly proportional to the net force
exerted and inversely proportional to mass of the object.
Instructor
Now that’s a lot of writing for one law, isn’t it? Let’s make things easy by changing our two
proportions into one equation. Here it is.
(text on screen)
Since force and acceleration are on opposite sides of the equals sign, they are directly proportional.
And since mass and acceleration are on the same side, they are inversely proportional.
All we need to add in order to complete the law is the fact that the acceleration is in the same direction
as the net force. You can add it in writing like this.
Or we could simply put arrows over force and acceleration in the equation, like this. It shows that
both force and acceleration are vectors and must be in the same direction. So that’s Newton’s second
law, the law of acceleration, in a nutshell.
Now we need to check out the units, since they must be the same on both sides of the equation. We
know that the MKS unit for mass is the kilogram, and acceleration is meters per second squared.
So, in fundamental units, the unit for force is the kilogram meter per second squared. Of course, we
call this derived unit by a nickname, the newton. Let’s get all this in your notes.
(text on screen)
VO
Mathematically, Newton’s second law is stated as “net force equals mass times acceleration.” The
arrows show that force and acceleration are vectors in the same direction.
The unit for force is the newton, which is defined as the force required to accelerate a mass of one
kilogram one meter per second squared.
In fundamental units, a newton is a kilogram meter per second squared. The dot shows that the units
are multiplied.
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Instructor
Do you need to memorize the law? Absolutely! But that will come naturally after you use it in a few
problems. Do you need to memorize the fundamental units for the newton? No way. If you are asked
for this information, just remember “F = ma.” You should already know the units for mass and
acceleration, so you can figure it out.
Remember, don’t memorize anything that you can figure out. Memorizing without thinking is
skimming the surface. No surfing allowed.
Now, with every mathematical law, there comes a set of problems. So let’s dive into some second
law problems. Your teacher will pause the tape and give you some easy problems to solve. The last
one will be more difficult, but you can do it if you use what you already know. After everyone has
finished, we’ll come back and go over the problems with you.
(text on screen)
VO
Local Teachers: Turn off tape and give students problem set number one from facilitator's guide.
(Pause Tape Now graphic)
(text on screen)
VO
1. The first problem is a straight plug and chug one. We know mass and acceleration and want to
solve for net force. So we just plug into the equation, “F=ma,” and get our answer, 2,100
kilogram meter per second squared. That’s a newton. So our answer is 2,100 N. It is
understood that both acceleration and net force are in the forward direction.
2. In the second problem, we know net force and acceleration and we want to solve for the mass
of the rock. We rearrange the equation to solve for mass, and then plug the numbers in a chug
out the answer. If the units for force and acceleration are in the MKS system, you can trust
that the unit for mass will be kilograms, even though you can’t see them cancel because or the
derived unit, the newton. The answer is 1.1 m/s2.
3. Number three is a two step problem, but each step should be familiar to you. The trick to
solving this kind of problem is recognizing the fact that you need two steps and two different
equations. One step will involve “F=ma,” and the other requires an acceleration equation.
This will help. When you recognize this kind of problem, divide your data into two parts.
Actually draw a line and label one side, “F=ma” and the other “acceleration equation.” Mass
and net force will go on this side, and time will go on the other side. You also know that
initial velocity is zero, which goes on the acceleration equation side.
If you know two out of the three F, m, and a, you can solve for the third. If not, you’ll have to
use this equation later. This time, we know mass and net force and can calculate acceleration.
The answer is that acceleration equals 2.8 m/s2. But we’re not there yet, because we haven’t
answered the question, “how far.” Now that we know the value of “a,” we’ll pick an
acceleration equation using “a,” “t,” and “d,” starting from rest.
We’ll use “d = ½ a t2” and solve for “d.” The answer, rounded to two significant digits, is 360
meters.
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Instructor
Always re-read a problem before you leave it, to be sure that you’ve answered the question.
Sometimes you can get so involved in the first step that you’ll go on after you get that first answer.
And you’d regret it later.
(text on screen)
Now look at this problem. Don’t copy it. Just decide whether you have enough information to solve
it or not. Can you solve it in your head?
If you multiplied 48 times 3.0 and got an answer of 140 newtons, you are indeed a physics surfer, and
a wrong one at that!!! Look at the problem again.
Is 48 newtons the mass of the bowling ball? The newton is not a unit of mass. It’s a unit for force.
Let’s look at a diagram of the forces acting on the ball.
( force diagram on screen)
The weight of the ball is the force of gravity pulling the ball straight down.
But the floor pushes upward on the ball, supporting it. These two forces cancel out, so there is no net
force acting on the ball in a vertical direction.
Because the bowler pushes the ball forward, across the floor, the net force acting on the ball is
horizontal. And the acceleration will be in that same direction. So we can’t use the weight of the ball
as a force in this problem. What we need is the mass of the ball. How can we find it? That’s another
matter. We’ll come back to this problem after we explore the relationship between mass and weight.
It turns out that the relationship is based on Newton’s second law of motion, “F = ma.” (F equals m a.)
Watch this.
(green chalkboard on screen)
VO
The relationship between mass and weight can be found in a special case of Newton’s
Second law. In free fall, the net force acting on an object is the object’s weight. And the acceleration
will be acceleration due to gravity, or “g.”
In most places on earth, the value of “g” is nine point eight zero meters per second squared.
So to convert mass to weight, we multiply by nine point eight zero meters per second squared.
And to convert weight to mass, we rearrange the equation and divide by nine point eight zero meters
per second squared. It’s as simple as that.
Instructor
Do you have another equation to memorize? NO! It’s still “F=ma.” Just remember that when weight
and mass are involved, "a" becomes "g." And you’ve done enough work with “g” to know that its
value is nine point eight zero meters per second squared.
(text on screen)
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VO
Your local teacher is going to put this problem set on the board for you to copy and work. When you
are finished, we’ll go over the answers.
Local Teachers: Turn off tape and give students problem set number two from facilitator's guide.
(Pause Tape Now graphic)
(text on screen)
VO
Here are the answers to the problems you worked.
For number one, you simply multiply mass times 9.8 m/s2 . That gives us weights of 18,000 newtons
for "a" and 1.2 newtons for "b."
In number two, we rearrange to solve for mass. Dividing by 9.8 m/s2, we get masses of 1.4 kilograms
and 69 kilograms.
Now let’s look at number three. On earth, 75 kilograms equals 740 newtons, rounded to two
significant digits. On the moon, we use 1.6 m/s2 for "g" and that gives us a weight of 120 newtons.
Remember that mass does not change, but weight depends on the force of gravity or the value of "g."
(bowling ball problem on screen)
I think we’re ready to go back to our bowling ball problem now, don’t you? When you read a
problem, make a list of the data. And pay attention to the units. Remember that we want no naked
numbers. We want to solve for net force, and we know acceleration and weight. But to use “F=ma,”
we need mass. So we’ll quickly convert 48 newtons to mass by dividing by 9.8 m/s2 . You don’t
even have to do the math. In fact, it’s probably better if you just leave it as mass equals 48 newtons
divided by 9.8 m/s2.
Now we’re ready to plug and chug. Net force will equal 48 divided by 9.8 m/s2 times 3.0 m/s2, or 15
newtons.
(text on screen)
VO
Try this one. Your teacher will pause the tape and give you the problem and enough time to work the
it. Then we’ll come back and go over the solution. Remember to watch your units!
(Pause Tape Now graphic)
(text on screen)
VO
Let’s start with part "a." We’re given the net force and asked to find acceleration. Whenever force
and acceleration are involved, look for a way to use “F=ma.” All we need now is mass. But we’re
given the weight of the ball, not the mass. No problem. We’ll just use the special case of “F=ma,”
“weight equals m g.” So mass will be weight divided by "g," which is 9.80 m/s2 . The mass of the
ball is 1.1 N divided by 9.80 m/s2), which is 0.11 kg.
Now we can go to “F=ma” and solve for acceleration, which is acceleration equals force divided by
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mass. Force is 19 N and mass is 0.11 kg, so acceleration equals 170 m/s2. Did you realize that a
baseball accelerates that much?
Now for Part b. You should recognize this one as an acceleration equation problem. Distance is
3.3m, initial velocity is zero, and we want to solve for final velocity. Now we know that the
acceleration of the ball from part a is 170 m/s2 and we need it. We’ll choose acceleration equation
number three, which is final velocity squared equals two times "a" times "d." So final velocity is the
square root of two times 170 m/s2 times 3.3 m. The answer is 33 m/s. This is about 74 miles per
hour, which is not a fast ball in the major leagues.
Instructor
It’s time to try some problems like this one and the one with the bowling ball on your own. Your
teacher has some practice problems for you. Pay attention to the units and separate your data if the
problem involves more than one step. You’ll go over the solutions in your class and then have a quiz.
When we come back, you’ll finally find out why this bowling ball and this marble would fall at the
same rate in an airless environment. So practice your problem solving and I’ll go bowling. See you
later.
VO
But right now, it’s time to …Show What You Know!
Jot down your choice for each question. After the program, your local teacher will go over the correct
answers with you.
(Read Show What You Know questions on screen)
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