N - Purdue Physics

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Transcript N - Purdue Physics

This Week
• Forces on an object
• Newtons laws
Relating force to acceleration
• Riding in an elevator
What we feel going up and down
• Cars and Trains
• Reaction /action
What makes us walk or a car move
Sailing Up Wind
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Physics 214 Fall 2011
1
What causes motion
In our everyday life we observe that objects change their state of
motion. In fact everything that happens in the Universe results
from a change in motion. That is a static inert object does not
contribute to any of the things we consider to be useful. The
functioning of our body depends on continual change throughout
our bodies.
These changes are produced by forces and in our everyday life
there are just two forces.
Gravity acts on mass
F = Gm1m2/r2
Electric charge
F = kq1q2/r2
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Newtons Second and First Law
Second Law
The acceleration of an object is directly
proportional to the magnitude of the
imposed force and inversely
proportional to the mass. The
acceleration is in the same direction as
the force
F = ma F and a are vectors
unit is a Newton (or pound) 1lb = 4.448N
First Law
An object remains at rest or in uniform
motion in a straight line unless it is
acted on by an external force.
F = 0 a = 0 so v = constant
http://www.physics.purdue.edu/academic_programs/courses/phys214/movies.php (anim0003.mov) (anim0004.mov)
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Force at the earths surface
F = GMEm/rE2
rE
But F = ma = mg
so
g = GME/rE2
Famous experiment by Cavendish
Measured F = Gm1m2/r2 using two known masses in the
laboratory and so measured G
Then using
earth
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g = GME/rE2 he determined the mass of the
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Mass and weight
Newtons second law enables us to measure relative mass. If we
apply the same force to two objects and measure the
accelerations then.
F = m1a1 and F = m2a2 so m1/m2 = a2/a1
We then need to have one mass as a calibration and a kilogram is
the mass of a piece of platinum held in Paris.
Since gravity acts proportional to mass
then the force near the earths surface is
F = mg this is the weight of an object
so if we compare
F1 = m1g and F2 = m2g then
weight 1/weight 2 = m1/m2
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Inertia
Inertia = tendency of an object to resist changes in its
velocity.
Since F = ma
and a = Δv/t then Ft = mΔv
So if a force acts for a time t the change in velocity will
be smaller for larger masses so it is mass that
determines inertia.
In particular if t is very small and m is large then F can
also be large but Δv can still be very small.
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Friction
In our everyday world any object which is moving feels a force
opposing the motion --- this is friction.
 An object which is sliding
The air resistance on your car
These types of friction result in energy being lost and
minimizing friction is very important.
But
Friction is also useful and essential since with no friction
a car would not move but just spin it’s wheels
a car would not be able to turn a corner
we would not be able to walk
objects would slide off surfaces unless perfectly horizontal
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Force diagrams
When we are analyzing
a particular object we
have to take into account all the forces acting on the body both in
magnitude and direction.
The acceleration of the object is equal to F/m in the direction of F
where F is the net force acting.
As in the example above we know that if we want to move an
object that there is a force called friction which opposes what we
want to do.
In the case shown the 10N force is + and the 2N force - so the net
force is 8N and 8 = 5a
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Terminal velocity
As an object moves through the atmosphere the air
exerts a frictional retarding force which increases with
velocity. So an object that is dropped from a great
height first accelerates at 9.8m/s2 but this acceleration
decreases until the retarding force = mg and at that
point the acceleration is zero and the object has it’s
terminal velocity.
Similarly when a car is moving at constant speed on the
highway the forward force produced by the engine is
balanced by the frictional force of the air.
Fair
Ftire
Ff
mg
v
Mg – Ff =ma
Freaction – Fair = ma
Freaction
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Reaction action: Newtons Third Law
For every force that acts on a body there is an equal
and opposite reaction force.
Ff
F
F
Ff
You pull on the block the block pulls back on you
The floor exerts a frictional force holding the block back
The block exerts a frictional force on the floor trying to move the
floor to the right.
The block accelerates providing F > Ff
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Force analysis
To analyze the motion of an object we need to draw a diagram and
put in all the forces that are acting on the object.
We will only deal with problems that have an acceleration along a
single axis. The net force along an axis perpendicular to this axis
is zero
Ff
F
F
Ff
We normally choose + in the direction of the acceleration. We
now can use the equations
F = ma
v = v0 + at
d = v0t + ½ at2 v2 = v02 + 2ad
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Reaction action forces
N
For the book
Nt = mg
mg
For the table
Nearth = mg + mtableg
mtableg mg
Ne
For a stack of plates or a brick wall and even a mountain each
layer has to support the weight of everything higher. So for a
stack of 48 plates the force on the second plate from the top is
mg. For the bottom plate it is 47mg but of course each plate has
a net force of mg to balance it’s own weight.
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Forces in an elevator
W = mg = true weight with no acceleration
N = apparent weight
N – mg is the net force taking + to be up
N – mg = ma is the equation of motion
If N > mg a is positive and the apparent
weight is > than the true weight
If N < mg a is negative and the apparent
weight is less than the true weight
+
N
g
mg
IRRESPECTIVE OF THE DIRECTION OR MAGNITUDE
OF THE VELOCITY
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Connected objects
T
Both objects have the same acceleration
The coupling pulls back on the 4kg mass but accelerates the
2kg mass.
30 – 8 – T = 4a
T – 6 = 2a
Or 30 – 8 – 6 = 6a
a = 16/6 m/s2 T = 68/6N
If I have a freight train with 100 cars each of mass m then for the
coupling between the engine and the first car
T1 = 100ma
And between the last two cars T99 = ma or 100 times less
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Connected objects
6mg
7mg
In the example shown
each brick has a net
upward force = mg down.
But each brick has to
support all the bricks
above it. So the 7th brick
has a downward force on
it’s top surface of 6mg and
an upward force of 7mg
on it’s bottom surface.
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F
F3 = 3ma F = 4ma
4
In the example shown of 7 equal mass
objects being accelerated
F = 7ma. The tension in each coupling pulls
back on the mass ahead and accelerates the
mass behind.
So F4 =4ma but the net force on block 4 is
4ma – 3ma = ma. So the net force on each
block is ma but the tension in each coupling
reduces by ma as one goes down the chain.
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Acceleration of a car
Fair
Ftire
Freaction – Fair = ma
Freaction
Generally Fair is proportional to v2 so the difference
between 55mph and 80mph is a factor of 2.16. This
means you use 2.16 more gasoline to cover the same
distance
Travel from Indianapolis to Lafayette a distance of 60 miles
Car does 30miles/gallon at 55mph gas costs $4/gallon
At 55mph use 2 gallons
cost 8$
time = 65.45 minutes
At 80 mph use 4.32 gallons cost $17.28 time = 45 minutes
Cost of saving 20.45 minutes = $9.28
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Summary of Chapter 4
Forces are responsible for all physical phenomena
Gravitation and the electromagnetic force are responsible for all
the phenomena we normally observe in our everyday life.
Newton’s laws F = ma where F is net force
v = v0 + at
d = v0t + ½ at2 d = ½(v + v0)t v2 = v02 + 2ad
Every force produces an equal and opposite reaction
Weight = mg where g = 9.8m/s2 locally
Apparent weight in an elevator depends on the acceleration
a up weight is higher
a down weight is lower
If your weight becomes zero it’s time to worry because you are in
free fall!!
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Sailing Up Wind
The force on the sail
balances out the
force on the keel and
leaves a component
of force “against the
wind”. Then F = ma
and the boat will
increase speed until
the component of the
wind force equals the
drag force
The sail and keel forces are like lift forces on an airplane wing
An example is when the boat is moving perpendicular to the wind the
force of the winds on the sails remains constant. The sails are set at about
450 to the direction of motion and the wind. The boats equilibrium speed is
determined by the resistance to the motion. If the resistance can be made
small the boat speed can be very high. Sail iceboats have very little
resistance and reach speeds in excess of 90mph with wind speeds of 30mph
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1C-04 Bocce Ball Tracks
Two balls are released with the same initial velocity. One
travels a flat path the other goes down a slope increasing
it’s speed and then climbs a slope and slows down before
reaching the end. The balls roll to minimize friction.
V
V
V
Which ball will
reach the end
first ?
1
5
2
3
4
The time for the yellow ball is just d/v. The time for the blue ball
can be computed in 5 parts. Parts 1 and 5 are identical for both
balls but the blue ball is faster for part 3 and qualitatively one
would expect the blue ball to arrive first.
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1F-02 Stack of Washers
This is a demonstration of Inertia where a washer can be
removed from a stack if the blow is fast.
Why does it
work less
well as the
stack gets
shorter ?
Strike the stack quickly. So the friction will
be very short-lived and the stack will not
gain speed before the force is gone.
THIS TENDENCY TO RESIST CHANGES IN THEIR STATE OF
MOTION IS DESCRIBED AS INERTIA. MASS IS A MEASURE OF
THE AMOUNT OF INERTIA. SO FOR A FIXED FRICTIONAL
FORCE ACTING FOR A SHORT TIME. THE BIGGER THE MASS,
THE LESS IT WILL MOVE.
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1F-07 Table Cloth Jerk
Can the table cloth
be removed without
breaking any
dishes ?
THERE IS A FORCE ACTING ON THE DISHES, BUT IT LASTS
FOR A VERY SHORT TIME. COMBINED WITH THE
RELATIVELY LARGE MASS OF THE DISHES, THIS FORCE IS
OVER SO QUICKLY AND IS SO SMALL THAT THE DISHES
HARDLY MOVE.
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1F-03 Egg drop
Is it possible to
get the eggs in
the beakers
without
touching them ?
IF THE PAN IS HIT SHARPLY A FORCE WILL ACT ON THE
EGGS FOR A VERY SHORT TIME AND THEY WILL NOT
MOVE HORIZONTALLY. THE PAN HAS TO BE HIT HARD
ENOUGH SO THAT HAS MOVED OUT OF THE WAY BEFORE
THE EGGS DROP ANY APPRECIABLE DISTANCE
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1H-02 Fan Cart (Action-Reaction)
Can a fan attached to a cart propel the cart?
What if
the sail is
removed?
Reaction
Action
In which
direction
will it move ?
What if the
sail is
canted at an
angle ?
INTERNAL FORCES IN A SYSTEM CANCEL EACH OTHER WHEN THE SYSTEM AS A WHOLE
IS CONSIDERED. SO IF THE SAIL IS PERPENDICULAR THE FAN DOES NOT MOVE.
IF THE SAIL IS REMOVED THE FAN MOVES IN THE OPPOSITE DIRECTION TO WHICH IT
BLOWS AIR. THE FAN WOULD NOT MOVE IN OUTER SPACE.
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1H-03 CO2 Rocket
Imagine that you are sitting in a cart with a pile of bricks.
ROCKET PROPULSION !
How could you use the bricks to get yourself and the cart to
move ?
What would happen if you throw a brick out of the cart ?
Then you throw out another .
What if you throw smaller bricks faster and more frequently ?
Now, if the bricks were the size of molecules. . .
What happens when the fire extinguisher rapidly “throws” out
CO 2 molecules ?
THIS
IS DIFFERENT
THAN THE FAN CART. CO2 IS EXPELLED AT HIGH VELOCITY AND IN
Because
the repulse
TERMS OF FORCES THE REACTION FORCE CAUSES THE CART TO MOVE IN THE OPPOSITE
DIRECTION. THE QUANTITY CALLED MOMENTUM IS CONSERVED AND THE MASS OF THE
CO2 X AVERAGE SPEED = TOTAL MASS OF THE CART X AVERAGE VELOCITY. THIS IS HOW A
ROCKET IS ACCELERATED AND IT WORKS IN OUTER SPACE.
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1H-04 Hero's Engine
A glass bulb emits steam from small nozzles
What happens
when the Glass
Bulb begins to
emit steam ?
Reaction = Bulb Spins
Action = Ejects Steam
Same Principle
causes a Lawn
Sprinkler to Turn.
THE REACTION FORCE TO THE EJECTION OF MASS CAUSES THE OBJECT TO SPIN.
THIS IS THE SAME AS THE CO2 ROCKET IN THAT MATERIAL IS EXPELLED AT HIGH
VELOCITY
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1J-04 Scale Paradox 1
A Scale Measures the Force acting on it
NOW, What
is the
reading on
the scale ?
What is
the
reading on
the scale ?
WALL
mg
mg
mg
T
T
T
T
mg
T = mg
mg
T = mg
mg
mg
THE TENSION IN THE CORD IS THE SAME FOR BOTH CASES. THE SCALE
MEASURES THE TENSION IN THE CORD. FOR EXAMPLE THE TENSION IN A ROPE
IS THE SAME IF TWO PEOPLE PULL ON EACH END WITH FORCE F OR IF ONE
PERSON PULLS WITH FORCE F TO A ROPE TIED TO A WALL.
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Questions Chapter 4
Q8 A 3-kg block is observed to accelerate at a rate twice that of a
6-kg block. Is the net force acting on the 3-kg block therefore
twice as large as that acting on the 6-kg block? Explain.
The net force is the same
Q9 Two equal-magnitude horizontal forces act on a box as shown
in the diagram. Is the object accelerated horizontally? Explain.
-F
F
No the net force is zero
Q10 Is it possible that the object pictured in question 9 is moving,
given the fact that the two forces acting on it are equal in size but
opposite in direction? Explain.
Yes, constant velocity
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Q18 The acceleration due to gravity on the moon is approximately
one-sixth the gravitational acceleration near the earth’s surface. If
a rock is transported from the earth to the moon, will either its
mass or its weight change in the process? Explain.
It’s mass will not change but it’s weight wil be 6 times less
Q22 The engine of a car is part of the car and
Ftire
cannot push directly on the car in order to
accelerate it. What external force acting on the
car is responsible for the acceleration of the car
on a level road surface? Explain.
Fair
Freaction
It’s the reaction force between the tires and the road
Q23 It is difficult to stop a car on icy road surface. It is also
difficult to accelerate a car on this same icy road? Explain.
Because of a lack of friction the wheels will skid or spin
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Q25 When a magician performs the tablecloth trick, the objects on
the table do not move very far. Is there a horizontal force acting on
these objects while the tablecloth is being pulled off the table? Why
do the objects not move very far? Explain.
Yes but the force acts for a very short time and the objects
start to move, then when the cloth is gone friction stops them.
Q30 Two masses, m1 and m2, connected
by a string, are placed upon a fixed
frictionless pulley as shown in the
diagram. If m2 is larger than m1, will the
two masses accelerate? Explain.
Yes m2 will fall and m1 will rise
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m
m
2
1
29
Q31 Two blocks with the same mass are connected by a string
and are pulled across a frictionless surface by a constant force, F,
exerted by a string (see diagram).
A. Will the two blocks move with constant velocity? Explain.
B. Will the tension in the connecting string be greater than, less
than, or equal to the force F? Explain.
F
A. They will accelerate F = ma
B. The tension will be less
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Q33 If you get into an elevator on the top
floor of a large building and the elevator
begins to accelerate downward, will the
normal force pushing up on your feet be
greater than, equal to, or less than the force
of gravity pulling downward on you?
Explain.
+
N
g
N – mg = ma but a is negative so N is smaller than mg
The only force pulling you down is gravity so if you are
accelerating down the force due to gravity must be larger
than the reaction force N ( N is apparent weight)
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a
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Ch 4 E4
A 2.5kg block is pulled
with a force of 80N and
friction is 5N
a) What is the
acceleration?
5N
2.5 kg
Net force = 75 N
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80 N
a = 75/2.5 = 30 m/s2
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Ch 4 E6
A 6kg block is being
pushed with a force P
and has an
acceleration of
3.0m/s2
a) What is the net force?
b) If P is 20N what is Ff?
P
a = 3m/s2
6 kg
Ff
a) F = ma = 6 x 3 = 18 N
b) F = P – Ff
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Ff = 2N
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Ch 4 E14
A 4kg rock is dropped and experiences
air resistance of 15N
a) What is the acceleration?
4 kg
15 N
Mg
F = 4 x 9.8 – 15
F = ma = 24.2 N
a = 24.2/4 = 6.05m/s2
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Ch 4 E16
g
N
A vertical force of 6N presses on a book.
a) What is the gravitational force?
b) What is the normal force ?
6N
0.4 kg
a) Gravitational Force = mg = 3.92 N
b) Upward Force = 6 + 3.92 = 9.92 N
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Ch 4 E18
m = 60 KG
A 60kg person is in an elevator
With an upward acceleration of
1.2m/s2
a) What is the net force?
b) What is the gravitational force?
c) What is the normal force?
a) Net Force
a = 1.2 m/s2
mg
F = Ma = 60 x 1.2
= 72 N
b) mg = 60 x 9.8 = 588 N
c) N = 588 + 72 = 660 N
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Ch 4 CP4
A 60kg crate is lowered from a height of 1.4m
and the tension is 500N
a) Will the crate accelerate?
b) What is the acceleration?
c) How long to reach the floor?
d) How fast does the crate hit the floor?
a) Net Force
= 60 x 9.8 – 500
= 588 – 500
g
60 kg
= 88 N
b) Will accelerate down
a = 88/60 = 1.47 m/s2
c) d = 1/2 at2
t = 1.38s
d) v = v0 + at
v = 2.03 m/s
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500 N
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Ch 4 CP6
A 60kg person accelerating down at 1.4m/s2
a) What is the true weight?
b) What is the net force?
c) What is N?
d) What is the apparent weight?
e) a, b, c, d with 1.4m/s2 up?
a) True weight = mg
N
1.4 m/s2
Mg
= 60 x 9.8 = 588 N
b) Net Force = Ma = 84 N
c) N = 588 – 84 = 504 N
d) 504 N
e) ↑1.4 m/s2
Net Force = 84 N↑
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N = 588 + 84 = 672 N
Physics 214 Fall 2011
W = 672 N
38
Review Chapters 1 - 4
-
d
+ x
Units----Length, mass, time SI units m, kg, second
Coordinate systems
Average speed = distance/time = d/t
Instantaneous speed = d/Δt
Vector quantities---magnitude and direction
Magnitude is always positive
Velocity----magnitude is speed
Acceleration = change in velocity/time =Δv/Δt
Force = ma Newtons
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Conversions, prefixes and
scientific notation
giga
1,000,000,000
109
billion
1 in
2.54cm
mega
1,000,000
106
million
1cm
0.394in
kilo
1,000
103
thousand
1ft
30.5cm
centi
1/100
10-
hundredth
1m
39.4in
thousandth
1km
0.621mi
1mi
5280ft
1.609km
1lb
0.4536kg
g =9.8
1kg
2.205lbs
g=9.8
0.01
3.281ft
2
milli
micro
1/1000
1/1,000,000
0.00
1
1/106
103
10-
millionth
6
nano
1/1,000,000,000
1/109
109
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billionth
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Speed, velocity and acceleration
v = Δd/Δt
a = Δv/Δt
The magnitude of a is not
related to the magnitude of v
the direction of a is not
related to the direction of v
2 3
4
1
v = v0 + at constant acceleration
d = v0t + 1/2at2
d,v0 v,a can be + or d = 1/2(v + v0) t
v2 = v02 + 2ad
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One dimensional motion and gravity
v = v0 + at d = v0t + 1/2at2
v2 = v02 + 2ad
d = ½(v + v0)t
+
g = -9.8m/s2
+
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At the top v = 0 and t = v0/9.8
At the bottom t = 2v0/9.8
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Equations
v = v0 + at
d = v0t + 1/2at2
d = ½(v + v0)t
v2 = v02 + 2ad
Sometimes you have to use two equations.
v0 = 15m/s v = 50m/s What is h?
v = v0 + at
50 = 15 + 9.8t t = 3.57 s
v0
`
h = v0t + 1/2at2
g
h
v
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h = 15 x 3.57 + 1/2x9.8x3.572
= 116m
h = ½(15 + 50) x 3.57 = 116m
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Projectile Motion
axis 1
axis 2
v1 = constant and d1 = v1t
vv = v0v + at
and d = v0vt + 1/2at2
v1
g
9.8m/s2
h
v
R
Use + down so g is + and h is +
h = v0vt + 1/2at2
v0v = 0, t2 = 2h/a
R = v 1t
v = v0v + at
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Complete Projectile
v0v
v1
9.8m/s2
v1
v1
v0v
highest point the vertical velocity is zero
vv = v0v + at
so t = v0v/9.8 h = v0vt + 1/2at2
end t = 2v0v/9.8 and R = v1 x 2v0v/9.8
and the vertical velocity is minus v0v
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Newton’s Second and First Law
Second Law F = ma unit is a Newton (or pound)
First Law
F = 0 a = 0 so v = constant
Third law For every force there is an equal and opposite
reaction force
N
Weight = mg
mg
Ff
F
F
Ff
F = ma
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v = v0 + at
d = v0t + ½ at2 d = ½(v + v0)t v2 = v02 + 2ad
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Examples
+
T
N
g
30 – 8 – T = 4a
T – 6 = 2a
30 – 8 – 6 = 6a
mg
N – mg = ma
a + N > mg
a – N < mg
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