Chapter 6 Applications of Newton’s Laws. Midterm # 1 Average = 16.35 Standard Deviation = 3.10...
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Transcript of Chapter 6 Applications of Newton’s Laws. Midterm # 1 Average = 16.35 Standard Deviation = 3.10...
![Page 1: Chapter 6 Applications of Newton’s Laws. Midterm # 1 Average = 16.35 Standard Deviation = 3.10 Well done! 2.](https://reader036.fdocuments.us/reader036/viewer/2022081514/56649dc85503460f94abe2d0/html5/thumbnails/1.jpg)
Chapter 6
Applications of Newton’s Laws
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Midterm # 1
Average = 16.35
Standard Deviation = 3.10
Well done!
2
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Who Wins?
a) I need to know a lot more about angles, masses of the line men, etc.
b) I could answer this if only I had the initial velocity
c) I would also need to know what planet this happened on.
d) 90 N, upfield.
e) This isn’t the Tech game, is it?
Kevin Parks (5’6”, 90kg) takes a handoff from Michael Rocco, but the hole closes as he hits the line. He is pinned between to O-linemen behind him, and 3 D-linemen ahead... and everyone is pushing, lifting, grabbing, and kicking. His acceleration is 1 m/s2 upfield.
What is the net force on KP?
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If you know the acceleration of a
body, you know the direction of
the net force. If you also know
the mass, then you know the
magnitude.
Who Wins?Kevin Parks (5’6”, 90kg) takes a handoff from Michael Rocco, but the hole closes as he hits the line. He is pinned between to O-linemen behind him, and 3 D-linemen ahead... and everyone is pushing, lifting, grabbing, and kicking. His acceleration is 1 m/s2 upfield.
What is the net force on KP?
a) I need to know a lot more about angles, masses of the line men, etc.
b) I could answer this if only I had the initial velocity
c) I would also need to know what planet this happened on.
d) 90 N, upfield.
e) This isn’t the Tech game, is it?
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Will It Budge?a) moves to the left, because the force of
static friction is larger than the applied force
b) moves to the right, because the applied force is larger than the static friction force
c) the box does not move, because the static friction force is larger than the applied force
d) the box does not move, because the static friction force is exactly equal the applied force
e) The answer depends on the value for μk.
Tm
Static friction (s= 0.4)
A box of weight 100 N is at rest on a floor where μs = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?
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The static friction force has a
maximum of sN = 40 N. The
tension in the rope is only 30 N.
So the pulling force is not big
enough to overcome friction.
Will It Budge?a) moves to the left, because the force of
static friction is larger than the applied force
b) moves to the right, because the applied force is larger than the static friction force
c) the box does not move, because the static friction force is larger than the applied force
d) the box does not move, because the static friction force is exactly equal the applied force
e) The answer depends on the value for μk.
Follow-up: What happens if the tension is 35 N? What about 45 N?
Tm
Static friction (s= 0.4)
A box of weight 100 N is at rest on a floor where μs = 0.4. A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?
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Springs
Hooke’s law for springs states that the force increases with the amount the spring is stretched or compressed:
The constant k is called the spring constant.
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SpringsNote: we are discussing the force of the spring on the mass. The force of the spring on the wall are equal, and opposite.
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Springs and TensionA mass M hangs on spring 1, stretching it length L1
Mass M hangs on spring 2, stretching it length L2
Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch?
S1
S2
a) (L1 + L2) / 2b) L1 or L2, whichever is smallerc) L1 or L2, whichever is biggerd) depends on which order the springs are attachede) L1 + L2
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Springs and TensionA mass M hangs on spring 1, stretching it length L1
Mass M hangs on spring 2, stretching it length L2
Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch?
S1
S2
W
Fs=T
a) (L1 + L2) / 2b) L1 or L2, whichever is smallerc) L1 or L2, whichever is biggerd) depends on which order the springs are attachede) L1 + L2
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Springs and TensionA mass M hangs on spring 1, stretching it length L1
Mass M hangs on spring 2, stretching it length L2
Now spring1 and spring2 are connected end-to-end, and M1 is hung below. How far does the combined spring stretch?
Spring 1 supports the weight.Spring 2 supports the weight.Both feel the same force, and stretch the same distance as before.
S1
S2
W
Fs=T
a) (L1 + L2) / 2b) L1 or L2, whichever is smallerc) L1 or L2, whichever is biggerd) depends on which order the springs are attachede) L1 + L2
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Recall -- Instantaneous acceleration
Velocity vector is always in the direction of motion; acceleration vector can points in the direction velocity is changing:
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Circular Motion and Centripetal Force
This force may be provided by the tension in a string, the normal force, or friction, or....
An object moving in a circle must have a force acting on it; otherwise it would move in a straight
line! centripetal forceThe net force must have a component pointing to the center of the circle
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Circular Motion and Centripetal AccelerationAn object moving in a circle must have a force
acting on it; otherwise it would move in a straight line.If the speed is constant, the direction of the acceleration (which is due to the net force) is towards the center of the circle.The magnitude of this
centripetal component of the force is given by:
For circular motion problems, it is often convenient to choose coordinate axes with one pointing along the direction of this centripetal force
aa
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Circular Motion
An object may be changing its speed as it moves in a circle; in that case, there is a tangential acceleration as well:
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Missing Link
A Ping-Pong ball is shot into a
circular tube that is lying flat
(horizontal) on a tabletop.
When the Ping-Pong ball
leaves the track, which path
will it follow?
a b c
d e
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Missing Link
Once the ball leaves the tube, there is no
longer a force to keep it going in a circle.
Therefore, it simply continues in a straight line,
as Newton’s First Law requires!
A Ping-Pong ball is shot into a
circular tube that is lying flat
(horizontal) on a tabletop.
When the Ping-Pong ball
leaves the track, which path
will it follow?
Follow-up: What physical force provides the centripetal force?
a b c
d e
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Tetherballa) toward the top of the pole
b) toward the ground
c) along the horizontal component of the tension force
d) along the vertical component of the tension force
e) tangential to the circle
In the game of tetherball,
the struck ball whirls
around a pole. In what
direction does the net
force on the ball point?
W
T
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The vertical component of the
tension balances the weight. The
horizontal component of tension
provides the necessary centripetal
force that accelerates the ball to
towards the pole and keeps it
moving in a circle.
W T
W
T
a) toward the top of the pole
b) toward the ground
c) along the horizontal component of the tension force
d) along the vertical component of the tension force
e) tangential to the circle
In the game of
tetherball, the struck
ball whirls around a
pole. In what direction
does the net force on
the ball point?
Tetherball
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Examples of centripetal force
whenno friction is needed to hold the track!
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A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in the figure. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?
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A hockey puck of mass m is attached to a string that passes through a hole in the center of a table, as shown in the figure. The hockey puck moves in a circle of radius r. Tied to the other end of the string, and hanging vertically beneath the table, is a mass M. Assuming the tabletop is perfectly smooth, what speed must the hockey puck have if the mass M is to remain at rest?
necessary centripetal force:
Only force on puck is tension in the string!
To support mass M, the necessary tension is:
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Circular motion and apparent weight
This normal force is the apparent, or perceived, weight
Car at the bottom of a dip
Note: at any specific point, any curve can be approximated by a portion of a circle
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Going in Circles I
a) N remains equal to mg
b) N is smaller than mg
c) N is larger than mg
d) none of the above
You’re on a Ferris wheel moving in a
vertical circle. When the Ferris wheel
is at rest, the normal force N exerted
by your seat is equal to your weight
mg. How does N change at the top of
the Ferris wheel when you are in
motion?
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Going in Circles I
a) N remains equal to mg
b) N is smaller than mg
c) N is larger than mg
d) none of the above
You’re on a Ferris wheel moving in a
vertical circle. When the Ferris wheel
is at rest, the normal force N exerted
by your seat is equal to your weight
mg. How does N change at the top of
the Ferris wheel when you are in
motion?
You are in circular motion, so there
has to be a centripetal force
pointing inward. At the top, the
only two forces are mg (down) and
N (up), so N must be smaller than
mg. Follow-up: Where is N larger than mg?
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Vertical circular motion
vertical (down)
vertical (up)
C
A
B horizontal
Centripetal acceleration must be
Condition for falling: N=0
at C:
(note: before falling, apparent weight is in the opposite direction to true weight!)
So, as long as:
at the top, then N>0 and pointing down.
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Chapter 7
Work and Kinetic Energy
http://people.virginia.edu/~kdp2c/downloads/WorkEnergySelections.html
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Work Done by a Constant ForceThe definition of work, when the force is parallel to the displacement:
SI unit: newton-meter (N·m) = joule, J
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Atlas holds up the world. Sisyphus pushes his
rock up a hill.
Who does more work?(a)
(b)
Working Hard... or Hardly Working
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Atlas holds up the world. Sisyphus pushes his
rock up a hill.
(a)
(b)
Working Hard... or Hardly Working
With no displacement, Atlas does no work
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Forces not along displacement
If the force is at an angle to the displacement:
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Friction and Work I
a) friction does no work at all
b) friction does negative work
c) friction does positive work
A box is being pulled
across a rough floor
at a constant speed.
What can you say
about the work done
by friction?
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f
N
mg
Displacement
Pull
Friction acts in the opposite direction to
the displacement, so the work is
negative. Or using the definition of
work (W = F (Δr)cos ), because =
180º, then W < 0.
Friction and Work I
a) friction does no work at all
b) friction does negative work
c) friction does positive work
A box is being pulled
across a rough floor
at a constant speed.
What can you say
about the work done
by friction?
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Can friction ever
do positive work? a) yes
b) no
Friction and Work II
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Can friction ever
do positive work? a) yes
b) no
Consider the case of a box on the back of a pickup truck. If the
box moves along with the truck, then it is actually the force of
friction that is making the box move. ... but .... the friction isn’t
really doing the work, it is only transmitting the forces to the box,
while work is done by the truck engine.
Friction and Work II
My view: the language is confusing so I’m not interested in arguing the point.
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Convenient notation: the dot productThe work can also be written as the dot product of the force and the displacement:
vector “dot” operation: project one vector onto the other
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Force and displacement
The work done may be positive, zero, or negative, depending on the angle between the force and the displacement:
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Sum of work by forces = work by sum of forces
If there is more than one force acting on an object, the work done by each force is the same as the work done by the net force:
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Units of Work
Lifting 0.5 L H2O up 20 cm = 1 J
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In a baseball game, the
catcher stops a 90-mph
pitch. What can you say
about the work done by
the catcher on the ball?
a) catcher has done positive work
b) catcher has done negative work
c) catcher has done zero work
Play Ball!
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In a baseball game, the
catcher stops a 90-mph
pitch. What can you say
about the work done by
the catcher on the ball?
a) catcher has done positive work
b) catcher has done negative work
c) catcher has done zero work
The force exerted by the catcher is opposite in direction to the displacement of the ball, so the work is negative. Or using the definition of work (W = F (Δr)cos ), because = 180º, then W < 0. Note that the work done on the ball is negative, and its speed decreases.
Play Ball!
Follow-up: What about the work done by the ball on the catcher?
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Tension and Work
a) tension does no work at all
b) tension does negative work
c) tension does positive work
A ball tied to a string is
being whirled around in
a circle with constant
speed. What can you
say about the work
done by tension?
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Tension and Work
a) tension does no work at all
b) tension does negative work
c) tension does positive work
A ball tied to a string is
being whirled around in
a circle with constant
speed. What can you
say about the work
done by tension?
v
T
No work is done because the force
acts in a perpendicular direction to
the displacement. Or using the
definition of work (W = F (Δr)cos ),
= 90º, then W = 0.
Follow-up: Is there a force in the direction of the velocity?
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Work by gravity
Fg a
h
A ball of mass m drops a distance h. What is the total work done on the ball by gravity?
N
Fg
A ball of mass m rolls down a ramp of height h at an angle of 45o. What is the total work done on the ball by gravity?
h
a
θ
Fgx = Fg sinθ
h = L sinθ
W = Fd = Fgx L = (Fg sinθ) (h / sinθ)
W = Fg h = mgh
W = Fd = Fgx h
W = mgh
Path doesn’t matter when asking “how much work did gravity do?”
Only the change in height!
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Motion and energyWhen positive work is done on an object, its speed increases; when negative work is done, its speed decreases.
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Kinetic Energy
As a useful word for the quantity of work we have done on an object, thereby giving it motion, we define the kinetic energy:
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Work-Energy Theorem
Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy.
(True for rigid bodies that remain intact)
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Lifting a Book
You lift a book with your hand
in such a way that it moves up
at constant speed. While it is
moving, what is the total work
done on the book?
a) mg r
b) FHAND r
c) (FHAND + mg) r
d) zero
e) none of the above
mg
r FHAND
v = consta = 0
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Lifting a Book
You lift a book with your hand
in such a way that it moves up
at constant speed. While it is
moving, what is the total work
done on the book?
The total work is zero because the net force
acting on the book is zero. The work done
by the hand is positive, and the work done
by gravity is negative. The sum of the two
is zero. Note that the kinetic energy of the
book does not change either!
a) mg r
b) FHAND r
c) (FHAND + mg) r
d) zero
e) none of the above
mg
r FHAND
v = consta = 0
Follow-up: What would happen if FHAND were greater than mg?
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By what factor does the
kinetic energy of a car
change when its speed
is tripled?
a) no change at all
b) factor of 3
c) factor of 6
d) factor of 9
e) factor of 12
Kinetic Energy I
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By what factor does the
kinetic energy of a car
change when its speed
is tripled?
a) no change at all
b) factor of 3
c) factor of 6
d) factor of 9
e) factor of 12
Because the kinetic energy is mv2, if the speed
increases by a factor of 3, then the KE will increase by a
factor of 9.
Kinetic Energy I
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Slowing Down
a) 20 m
b) 30 m
c) 40 m
d) 60 m
e) 80 m
If a car traveling 60 km/hr can
brake to a stop within 20 m,
what is its stopping distance if
it is traveling 120 km/hr?
Assume that the braking force
is the same in both cases.
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F d = Wnet = KE = 0 – mv2,
and thus, |F| d = mv2.
Therefore, if the speed doubles,
the stopping distance gets four
times larger.
Slowing Down
a) 20 m
b) 30 m
c) 40 m
d) 60 m
e) 80 m
If a car traveling 60 km/hr can
brake to a stop within 20 m,
what is its stopping distance if
it is traveling 120 km/hr?
Assume that the braking force
is the same in both cases.
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Work Done by a Variable Force
We can interpret the work done graphically:
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Work Done by a Variable Force
If the force takes on several successive constant values:
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Work Done by a Variable Force
We can then approximate a continuously varying force by a succession of constant values.
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Work Done by a Variable Force
The force needed to stretch a spring an amount x is F = kx.
Therefore, the work done in stretching the spring is