Download I. Newton`s Laws of Motion

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Dear Studious Student,
This chapter called, “Newton’s Laws for NEWBIES” is meant to help your academic career in courses such as Physics
Academic or Physics Honors with any high school teacher. (It is actually best with a Professor at Montgomery High School)
This is not for an AP class, not is it to bore to tears. It is not 84,903,285,902 pages that sound smart (although we could have
made ourselves sound smart if we wanted to), but more importantly, we wanted you, the reader to actually be able to read and
decipher every piece of information given. Every single bullet point, lab, demonstration, and diagram has a purpose of
teaching, not repeating, so read carefully! Good Luck!
Best Wishes,
Jina, Robi, Yang
P.S. Of course every chapter or book comes with a list…our list is a list of rules you should follow so that your feet do get
flattened by a car, skateboard, or any means of transportation with wheels. Also that you don’t break a part of your body
from trying to do an experiment or lab. Okay, honestly, the most important reason for us to give you a list of rules is so that we,
the company don’t get sued for having a dangerous or faulty lab.
So here they are:
READ CAREFULLY!
1.
Always have an adult (teacher, parent, guardian, etc) to supervise the lab, experiment, demonstration, etc.
2.
Always wear goggles and gloves for labs that involve toxic chemicals, an engine, battery-powered motor, or any heavy
objects.
3.
When dealing with especially sharp, heavy, or slippery objects, always wear appropriate clothing. (Ex: When dealing
with rocks, wear sneakers, not flip-flops)
4.
Be patient with group members-do not get aggravated as you may create greater experimental uncertainty because of
bursts of rage
5.
Treat all objects and materials with respect-they have to last a long time
6.
No fooling around! This will create hysteria and chaos, which results in greater amounts of experimental uncertainties,
and digression.
I, ___________________________ read and understand the safety procedure required of “Newton’s Laws for NEWBIES”.
X__________________________________________
(Parent/Guardian)
X_________________________________________
(Student)
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Chapter: Dynamics Unit using free-body diagrams
Objectives:
 To know and understand all three of Newton’s Laws and how to apply them to force problems
 To be able to draw and use free body diagrams with motion diagrams to describe the forces acting on an object
One day, after school, your mother asked you to go purchase a gallon of milk from Shoprite, and since you have completed
all your homework, you decided to walk there. Just one block before you reached the supermarket, however, you observed a police
chase: A ’74 Chevy Impala sped down the corner towards a police blockade at approximately 80 miles per hour. The Chevy made a
sharp left turn to avoid the police blockade. Unfortunately, one of the police cars behind the Chevy wasn’t able to stop in time, and it
rammed into another police car that’s part of the blockade. You observed that the blockade car, which was positioned sideways,
stopped the chasing police car and caused it to bounce back a few meters as well. The blockade police car was also pushed back a few
meters from the force of the moving police car crashing into the blockade. As a very studious physics student, you quickly jotted down
some notes about what forces were present in the situation, on a post-it. The next day you asked the physics professor regarding the
crash. He answered, “The moving police car was stopped by the police blockade because of Newton’s First Law, the police car was
able to accelerate to catch up with the Chevy Impala because of Newton’s Second Law, and the police that was moving bounced back
after hitting the police blockade because of Newton’s Third Law, understand?” You still did not have a clear idea about what
Newton’s Laws really were and how forces behaved in phenomena such as this one, so you asked your professor for more help. He
answered by saying, “Why don’t you go read the chapter, ‘Newton’s Laws for NEWBIES’ to learn more?”
So, the next week, you came back to physics class knowing everything there is to know about forces. What you learned
was…
I.
Newton’s Laws of Motion
Newton’s First Law
→Any object in motion will stay in motion unless acted upon by an external unbalanced
force.
→Any object at rest will stay at rest unless acted upon by an external unbalanced force.
Demonstration: Newton’s First Law
Need: Bowling ball, bowling pins, wall, 1 person
1.
2.
3.
4.
5.
Go to any large open space with a wall at the end.
1 person push the bowling ball straight down the space until it reaches the wall
Observe and record the results of the approximate direction and motion of the ball.
Repeat steps 1 and 2 but this time, instead of having a wall, roll it into open space without a wall
Repeat step 3
1.
2.
3.
4.
Go to any large open space
Set up the bowling pins at one end and set a person with a bowling ball on the other
1 person push the bowling ball straight down the space until it hits the pins on the other side
Observe and record the results of the motion of the bowling pins
The student should have come to the conclusion that if any object is in motion and comes into contact with another outside force, then
the motion and direction of that object will be altered. Also, when the object did not have another external force acting upon it, it
should have continued to stay that constant motion. The second part of the demonstration shows the second part of Newton’s First law.
The student should have observed that the bowling pins stayed at rest until it was hit with the bowling ball. This demonstration shows
that any object in motion stays in motion unless acted upon by an outside force, and any object at rest will stay at rest unless acted
upon by an external unbalanced force.
Newton’s Second law
→fnet is dependent on the mass and the acceleration of the object.
→fnet=m*a
Lab: Newton’s Second Law
Need: 1 heavy and 1 light friend, A scooter or skateboard,
open space with little friction, stop watch, calculator, yardstick
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This picture of Newton
under a tree and an apple
falling from the tree shows
Newton’s Second law
because the apple
accelerated (from 0-rest to
a greater speed) and it has
a mass. There is the force
of the apple dropping from
the tree (force of the earth)
and that can be figured out
with the equation f=ma.)
1.
2.
3.
4.
5.
6.
7.
Find a large open space
Measure the distance to be traveled
Get the light friend to sit on the bike or skateboard.
Push your light friend along carefully. Try to keep a constant force. The friend must be still within the scooter or skateboard.
Record observations of the force exerted on the person and the acceleration of the person using the stopwatch and yardstick.
Repeat steps 1-5 except with the heavier person.
Analyze how the acceleration and force changed.
You should have found that when you started off walking, you had to walk quicker and quicker, and eventually run, to keep your force
constant. This is a demonstration of constant force needs a constant acceleration.
This shows that no matter the mass, or weight of
the object, the acceleration is always the same for
all objects.
The first part (left side) is 100 kg and the second
part (10kg) is 10 kg but they still accelerate at the
same rate.
This was calculated using the equation from
Newton’s Second Law.
Newton’s Third law
→Every action has an equal and opposite reaction
Demonstration: Newton’s Third Law
This picture shows both
of the two men pulling
with the same force. This
is because it is a picture
so it can be inferred that
they are not moving.
Need: Two sets of skates, two people (of similar weight), a flat surface, yardstick
1.
2.
3.
4.
5.
6.
Find a large open space
Put the skates on the two people
Mark a line on the floor
Both people stand on the line (toe-side on the line) facing each other
Push gently against each other (but do not do any more work other than that)
Measure the distance each person traveled from the starting line.
If you are the same weight you will move about the same distance backwards. Newton’s third law states that there will be an equal and
opposite reaction. The actual being pushed the other way is the opposite part and the equal part is when their distance apart is equal.
Experiment: Bocce Ball
Need: 1 Bocce Ball, 1 wall, 1 flat surface (as close to frictionless as possible), and 1 person
1.
2.
3.
4.
Set the Bocce Ball at one end of the surface opposite the wall, which is stationary on the ground.
1 person push the ball (do work on it) to move it straight in front nearing the wall
Observe and record results of the behavior of the ball.
Think about which of Newton’s Laws are applied.
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The ball moved so All three of Newton’s laws should have been applied. You should have gotten that the ball bounced off the wall a
bit. This shows Newton’s Third Law. (Read Third law explanation.) Also, The First Law can be shown because it came close to a stop
(the roundness of the object creates flaws). The wall was the interference in constant acceleration. Any object that is moving can be
applied to the equation a=f/m because it has mass and acceleration.
 LET’S REVIEW
Fan Carts Demonstration
Need: Two fan carts, one wall (of some sort), a flat surface, 1 calculator, 1 timer, 1 yardstick, 2 people, speed radar, tape, different
masses
Newton’s First Law
1.
2.
3.
Place the cart in an open space with a wall at the end of its path
Turn the fan on and observe the motion of the car as it travels towards the wall
Observe what happens to the car when it comes into contact in the wall. Record your answer
This demonstration represents Newton’s first law by showing how an object in motion (the cart) will stay
in motion until acted upon by an outside force (the wall).
Newton’s Second Law Part 1: How is mass related to acceleration:
1.
2.
3.
4.
5.
6.
7.
8.
Put the cart on the ground and mark the starting point with tape.
Set the speed radar a certain distance away (ex. 10 meters) and mark that with tape too.
Have a person hold a stopwatch
Turn the fan cart on and start stop watch simultaneously
Once the cart passes the speed radar, stop the stopwatch.
Find the acceleration by using equation a = ∆v/t
Observe and record your results
Repeat steps 1-7, but start adding more and more weight to the cart each time. Record the acceleration each time
Notice how the acceleration of the cart gets less and less as more and more mass is added to it. This is because according to the
equation a=fnet/m, acceleration and the mass are inversely related.
Newton’s Second Law Part 2: How changing the force is related to acceleration:
1.
2.
3.
4.
5.
6.
7.
8.
Put the cart on the ground and mark the starting point with tape.
Set the speed radar a certain distance away (ex. 10 meters) and mark that with tape too.
Have a person hold a stopwatch
Turn the fan cart on and start stop watch simultaneously
Once the cart passes the speed radar, stop the stopwatch.
Find the acceleration by using equation a = ∆v/t
Observe and record your results.
Repeat steps 1-7, but put more batteries into the cart so that there is more force pushing the cart each time.
Notice how the acceleration of the cart increases as more and more batteries are added resulting in greater forces pushing the cart.
This is because according to the equation a=fnet/m, acceleration and the forces are directly related.
Newton’s Third Law
1.
2.
3.
4.
Put one fan cart on one side of the flat surface and the other fan cart on the other side (approx 30cm/12in/1ft or a small
increment of measure but bigger than ~6 inches), facing the first cart.
Turn on one of the two fan carts
Observe and record direction and behavior of the moving fan cart (fan cart 1) and the one that was originally at rest (fan cart
2).
Measure the distance that fan cart 2 traveled
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The distance traveled by fan cart 1 and fan cart 2 should have been equal. (If not, there were some experimental uncertainty.) This
shows that the force of fan cart 1 was equal to the reaction of fan cart 2. Although their speeds when colliding were different, because
of Newton’s Third Law, they traveled the same distance by the end.
Newton’s Laws Practice Problems
(g=9.8m/s/s)
1.
In an objectless projectile path, a rocket was launched straight up. If the rocket stops its engine after exiting
the Earth’s effective gravitational field, what will happen to the rocket? How does this have to do with
Newton’s Laws?
2.
The steel cables of an elevator are in a pulley system of two ropes. Each rope can exert a force of 1000N.
You and your friends have a combined weight of 200kg. Considering gravity of the Earth accelerates your at
9.8 m/s/s, will all of you be able to get in together? Show your work.
3.
(Following question 2.) Your trip upstairs has been hit by an unfortunately turn of events: The elevator break fails and a man
‘squeezed’ into the elevator. And, he’s as massive as you and your friends combined. What will happen? Give details. (Hint:
Ignore the mass of the elevator.)
4.
In photography class you observed a photo of a raindrop plunged into a pond, and making magnificent
sprinkles. Why didn’t the drop of water enter the pond swiftly, but instead, part of it bounced back?
Which Newton’s Laws can you relate this to?
5.
True or False: Because a feather is so small, when one hits the ground, it just settles there without ever bouncing around like a
ball does.
6.
Bob has the ability to teleport. If he teleports to the top of a mountain, side of a hill, or the bottom of the sea, and stand there, in
what direction is the mountain, hill and the sea floor pushing him?
7.
John took a scale into an elevator going down. He observed that when the elevator moves, his mass changes. Why does this
phenomena occur? Is John really losing weight?
8.
While driving, Mr. Grieco notices a bug hitting the windshield of his car. He had just learned about Newton’s Third in class that
day. He knew that the bug hit the windshield and the windshield hit the bug. Which of the two forces is greater: the force on the
bug or the force on the windshield?
(Answers are shown on the next page, but don’t look before solving the problem!)
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Answers to Newton’s Laws Practice Problems
1.
The rocket will be going at the same velocity that it exited Earth’s pull. Objects at rest stays at rest, objects at motion stays in
motion unless acted upon by another force.
2.
Yes, 9.8x200=1960N, since there are two steel cables in the pulley system, it can support 2000N. (1000Nx2)
3.
Since the mass of the system is doubled: (2000-(1960x2))/400= -4.8=4.8m/s/s downwards
(2000-(9.8)(2x200kg)/(2x2000kg)
4.
The droplet exerts a force to the surface of the pond, and the pond exerted an equal and opposite force back to the droplet. For
every force, there’s an equal and opposite force., which is Newton’s Third Law.
5.
False, it’s just too little to observe. For every force, there’s an equal and opposite force., which is Newton’s Third Law
6.
The force always pushes Bob upward because the only other force on Bob is the force of the earth, which is pulling down on
him.
7.
The scale measures the normal force. When the elevator accelerates downwards the force of gravity is reduced. This therefore
reduces the normal force, and the measurements on the scale.
8.
This is a trick question! Newton’s third Law states that all actions have equal and opposite reaction.
Show work here:
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Free Body Diagrams
How do free-body diagrams model motion?
Free-body diagrams show all the forces that are acting on an object. They also show what directions the forces are going in
and how big they are.
This is an example of a force diagram. The dot in
the middle represents the object, which the forces
are acting upon. The arrows represent the forces
and the directions they are going in. The longer
the arrow, the greater the force. This object is
either at rest or moving at a constant velocity,
because all the arrows are equal in length.
When there are two arrows going in opposite directions, the forces they represent are going against each other. If those two
arrow are equal in length (the forces are of the same magnitude), then they cancel each other out and the object is either at rest OR
traveling at a constant speed. The force diagram shown above is an example of an object at rest or traveling at a constant speed
because all the arrows are of the same length (the vertical arrows cancel each other out and the horizontal arrows cancel each other
out.)
When an arrow in one direction is longer than the arrow going in the opposite direction, it shows that the first force is greater
than the second. Therefore, the object accelerates in the direction of the longer arrow. The diagram shown below is an example of this.
In this force diagram, the force that are
pulling/pushing on the object upward and
downward are the same, so they cancel
each other out. However, the force
pushing/pulling the object to the right is
greater than the force pulling/pushing the
object to the left. Therefore, the object is
accelerating towards the right.
How to interpret free-body diagrams:
Once you know how to draw free-body diagrams, it is pretty easy to interpret free-body diagrams that have already been
drawn for you. First, look at the arrows that represent the forces that are acting upon the object. Are any of them longer than the one
opposite them? If this is the case, then the object is accelerating in the direction of the longer arrow. If the forces are all equal, then the
object is either at rest or moving at a constant velocity (you’ll need a motion diagram to determine this).
Example problem: How is this object moving?
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Answer: The arrows going to the right and
left are equal in length, so those two forces
cancel each other out. However, the arrow
going up is longer than the arrow going
down. Therefore, the object is accelerating in
the upward direction.
Limitations of Free Body Diagrams:
Believe it or not, free body diagrams aren’t perfect. When the arrows are equal in length, you don’t know whether the object
is at rest or moving at a constant velocity. Also, if one arrow is longer than the arrow opposite it, you don’t know whether the object is
moving faster in the direction of the longer arrow or slowing down in the direction of the shorter arrow.
This is why you need motion diagrams to truly be able to interpret free-body diagrams.
Without the motion diagram on the
bottom, you wouldn’t be able to tell
if this object was at rest or moving
at a constant speed. However, the
motion diagram shows you that the
object is indeed moving at a
constant speed to the right.
Acceleration = change in force/mass
Newton’s second law states that F = ma, or a = F/m where F = change in force, m = mass, and a = acceleration. How does
this work? Here is an experiment that shows how Newton’s second law works.
Newton’s Second Law Part 1: How is mass related to acceleration:
1.
2.
3.
4.
5.
6.
7.
8.
Put the cart on the ground and mark the starting point with tape.
Set the speed radar a certain distance away (ex. 10 meters) and mark that with tape too.
Have a person hold a stopwatch
Turn the fan cart on and start stop watch simultaneously
Once the cart passes the speed radar, stop the stopwatch.
Find the acceleration by using equation a = ∆v/t
Observe and record your results
Repeat steps 1-7, but start adding more and more weight to the cart each time. Record the acceleration each time
Notice how the acceleration of the cart gets less and less as more and more mass is added to it.
Newton’s Second Law Part 2: How changing the force is related to acceleration:
1.
2.
3.
4.
5.
6.
7.
8.
Put the cart on the ground and mark the starting point with tape.
Set the speed radar a certain distance away (ex. 10 meters) and mark that with tape too.
Have a person hold a stopwatch
Turn the fan cart on and start stop watch simultaneously
Once the cart passes the speed radar, stop the stopwatch.
Find the acceleration by using equation a = ∆v/t
Observe and record your results.
Repeat steps 1-7, but put more batteries into the cart so that there is more force pushing the cart each time.
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Notice how the acceleration of the cart increases as more and more batteries are added resulting in greater forces pushing the cart.
Why this representation is useful to enhance understanding:
A free-body diagram combined with a motion diagram is very useful because it helps you understand the forces acting on the
object. They give you a visual of what is happening to an object and why it is moving the way it is. If you just describe the forces
acting on an object in words, it can get really confusing. However, a free-body diagram combined with a motion diagram let’s you see
in a glance what is happening to an object. It makes everything clearer and more understandable.
Force
Fearth (Also known as gravity)
Fnormal
Description of the Common Force
The force of gravity is the force with which the earth’s core
attracts an object towards itself. All objects on earth have the
force of gravity, which is, directed usually in the negative
direction towards the center of the earth.
Fgrav = m * g
g = acceleration of gravity = 9.8 m/s2 (on Earth)
m = mass (in kg)
The normal force is the support force exerted upon an object,
which is in contact with another still object. (Think of any object
sitting on a table or surface of some sort)
Exemplary Problems: Here are a few examples of problems that you will have to be able to solve.
1.
Elevator Problem: A 100-kg person stands on a scale in an elevator. What does the scale read, in Newtons, when
a.
b.
c.
d.
e.
the elevator is at rest?
the elevator is rising at a constant speed?
the elevator is falling at a constant speed?
the elevator is accelerating upwards at 5 m/s/s?
the elevator is accelerating downwards at 5 m/s/s?
How to answer this question:
a.
Since the elevator is at rest, nothing is interfering with the force of the person on the scale (his weight), or the force of the scale
on the person (the scale reading). You can simply find the answer by using the equation F = ma from Newton’s second law,
where F = the person’s weight in newtons, m = the person’s mass (100 kg), and a = the acceleration due to gravity (9.8 m/s/s).
So using this equation, the person’s weight in newtons can be found like this: F scale on person = 100 kg x 9.8 m/s/s = 980 N.
b.
Once again, the only acceleration you have in this system is the acceleration due to gravity (9.8 m/s/s). Use Newton’s second
law equation just like you did in the above problem, and your answer should still be 980 N.
c.
Same as the previous 2 problems. The acceleration of the elevator is still zero, so the “a” in the equation is still 9.8 m/s/s. Your
answer should still be 980 N.
d.
Now it gets a little tricky because the elevator is accelerating upwards and the scale is accelerating along with it. Therefore, the
force of the scale on the person should be greater than when the elevator is not accelerating. Now you use the equation F = W +
ma where “F” is the scale reading (force of scale on the person), “W” is the weight of the person when the elevator is not
accelerating, “m” is the person’s mass, and “a” is the acceleration of the elevator. Your work should look like this:
F = 980N + (100)(5 m/s/s)
F = 980N + 500N = 1480N. That is the scale reading.
e.
You solve this problem with the same equation as in problem “d”. However, since the elevator is now accelerating downward,
the acceleration of the elevator is negative (we assumed that the upward direction was positive). Therefore, the acceleration of
the elevator in this scenario is -5 m/s/s. So using the equation above, your work should look like this:
F = 980N + (100 kg)(-5 m/s/s)
F = 980N – 500N = 480N
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2. Falling object problem:
A 9800N piano is tossed off the Empire State building by an angry toddler. Find the mass of the piano if air resistance is
negligible (use 9.8 m/s/s for acceleration due to gravity). Draw an label a free-body diagram and a motion diagram for this
scenario.
How to do this problem:
This problem is pretty simple because all you need to do is use the equation from Newton’s second law. In this case, the force is
equal to the weight of the piano (9800N). The downward acceleration is 9.8 m/s/s. All you need to do is solve for the mass like
this:
9800N = m x 9.8 m/s/s
m = 9800N/9.8 m/s/s
m = 1000 kg
Motion Diagram:
Piano
Force of earth on
piano (9800N)
3. Horizontal drag problem:
A 150 kg man is being dragged along a highway by an 18 wheeler and is accelerating at 20 m/s/s. The force of the surface of
the road against the man (friction) is 5N. How much force is the truck exerting on the man? Draw and label a force-body
diagram and a motion diagram for this problem.
How to do this problem:
Once again, you need to use Newton’s second law to solve this problem. Since we are trying to find the force of the truck,
lets call it F1. The friction of the road opposes the force of the truck on the man so you need to subtract that from F1. Your
equation should look like the following:
(F1 – 5N)/150 kg = 20 m/s/s
F1 – 5N = 3000N
F1 = 3005N. This is how much force the truck is exerting on the man to make him accelerate at 20 m/s/s.
Motion Diagram:
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Force of road
on man
Man
Force of surface of road on
man
Force of truck on man
Force of earth
on man
4. Pulley Problem:
Two blocks are suspended above the ground by a massless frictionless pulley. Block 1 has a mass of 5 kg and Block 2 has a
mass of 10 kg. Draw and label a free-body diagram and a motion diagram for each block and then find the acceleration for both
blocks.
1
2
How to do this problem:
You first need to find the weight (force of earth on block) of each block so you can use Newton’s second law. To do this,
simply multiply the mass of each block by the acceleration due to gravity (9.8 m/s/s). So the weight of Block 1 is 49N and the
weight of Block 2 is 98N. Now you can use the equation from Newton’s second law: F = ma, where F is the weight of the block
(or the force of the earth on the block). However, the “m” in this equation is actually the sum of the masses of both blocks
because they are both attached to the pulley. Also, since Block 1 is lighter than Block 2, it will accelerate upwards while Block
2 will accelerate downwards. You are trying to solve for a, so you equation should be a = F/m. Here’s how you should do you
work:
Block 1
a = 49N/(5 kg + 10 kg)
= 49N/15 kg
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= 3.27 m/s/s upwards
Block 2:
a = 98N/(5 kg + 10 kg)
= 98N/15 kg
= 6.53 m/s/s downwards
Motion Diagram for Block 1 and Block 2:
Force of Block
1 on Block 2
Force of block 2
on block 1
Block 2
Block 1
Force of
earth on
block 1
Force of earth on
Block 2
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Extra Practice Problems
1.
Tom and Billy are playing tug of war. Tom is stronger, and is exerting more force on the rope than Billy is. Draw a
qualitative free-body diagram with a motion diagram for the rope.
2.
Gerardo is standing on a scale in an elevator. His mass is 75 kg and the elevator is accelerating upward at 3 m/s/s. Find
Gerardo’s weight in pounds (hint: 1 lb = 4.5N).
3.
A cat that was thrown out of an airplane is falling through very thick air. The air resistance is 2N. If the cat is accelerating
downwards at 6 m/s/s and has a mass of 9 kg, find the weight of the cat (force of earth on cat) in Newtons. Provide a labeled
free-body diagram and motion diagram for this problem.
4.
A bag of lead weighing 100N and a bag of feathers weighing 2N are thrown off a building. Calculate the acceleration for
each bag. Which has the greater acceleration?
5.
A man and an empty bucket are suspended above the ground by a massless frictionless pulley. If the man has a mass of 20 kg
and the empty bucket has a mass of 1 kg, how much mass has to be added to the bucket for the man to accelerate upwards at
2 m/s/s?
6.
True or False: George badly bruised another physics student. Since George wasn’t hurt at all, he exerted more force than the
other student who exerted none.
7.
A stunt car went racing off a cliff that’s 1,000 meters high. The car is being pulled by the Earth with 10,000N, and propelled
by its rocket engine with twice that force. Where will the stunt car first touch the ground? (The car starts at 10m/s, Hint:
Distance=Vit+0.5a(t^2))
8.
Bob was tending his mule one day, he bend over to pick up its hay, then, from one meter away, the mule delivered a powerful
kick to his behind. The mule’s hooves accelerated at 40m/s/s, and it took half a second for the mule to send the punishment.
If Bob’s derrière is 0.9 meters above ground, how far will Bob go? (Bob weight in at 55kg, assume the kick was parallel to
the ground landing Bob on his butt, ignore friction, it’s a big mule)
Show Work Here:
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Answers to practice problems:
1.
Rope
Force of
Billy on
rope
Force of Tom
on rope
.
2.
F (earth on Gerardo) = 75 kg x 9.8 m/s/s = 735N
F (scale on Gerardo) = W + ma
F (scale on Gerardo) = 735N + (75 kg)(3 m/s/s)
= 735N + 225N
= 960N = 213.3 lbs
3.
FBD/Motion Diagram:
Force of air
on cat
Cat
Force of
earth on cat
Let F1 = force of earth on cat
a = F1-F2/m
6 m/s/s = (F1- 2N)/9 kg
54N = F1 – 2N
56N = F1
The cat’s weight is 56 N
4.
Both have the same acceleration because the acceleration due to gravity is the same for all objects (9.8 m/s/s). This was
supposed to be a trick question.
5. a = F/(m1 + m2)
Let F = weight of man, m1 = mass of man, m2 = mass of what the bucket needs to be
F = 20 kg x 9.8 m/s/s = 196N
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2 m/s/s = 196N/(20 kg + m2)
2 m/s/s x (20kg + m2) = 196N
20kg + m2 = 98kg
m2 = 78 kg.
You need to add 77 kg to the empty 1kg bucket.
6.
False, the bruised physics student gave George equal and opposite forces due to Newton’s third law.
7.
10,000/9.8=mass=1020.4kg, acceleration of engine=20,000/1020.4=19.6 (twice gravity), to solve for time:
1,000=0.5*9.8*(t^2), t=14.3sec, D=10*14.3+0.5*19.6*(14.3^2)=2147meters before touch down.
8.
0.5*40=displacement of Bob at 20m/s. To find time: 1=0.5*9.8(t^2)=0.45sec. 20*0.45=Distance=9meters.
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