1. I. Force & Acceleration Chapter 3 Forces Newton’s Second Law Friction*
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Chapter 3 Forces
I. Force & Acceleration
Newton’s Second Law
Friction
Air Resistance
Calculations *

2. F = ma A. Newton’s Second Law Newton’s Second Law of Motion*
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A. Newton’s Second Law
Newton’s Second Law of Motion
The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
F = ma *

3. F = ma F m a F m F: force (N) m: mass (kg) a: accel (m/s2)*
07/16/96 m F a F m
F = ma
F: force (N)
m: mass (kg)
a: accel (m/s2)
1 N = 1 kg ·m/s2 *

4. a F m B. Calculations F = ? F = ma m = 40 kg F = (40 kg)(4 m/s2)*
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B. Calculations
What force would be required to accelerate a 40 kg mass by 4 m/s2?
GIVEN:
F = ?
m = 40 kg
a = 4 m/s2
WORK:
F = ma
F = (40 kg)(4 m/s2)
F = 160 N m F a *

5. a F m m = 4.0 kg a = F ÷ m F = 30 N a = (30 N) ÷ (4.0 kg) a = ?*
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A 4.0 kg shotput is thrown with 30 N of force. What is its acceleration?
GIVEN:
m = 4.0 kg
F = 30 N
a = ?
WORK:
a = F ÷ m
a = (30 N) ÷ (4.0 kg)
a = 7.5 m/s2 m F a *

6. C. Friction Suppose you give a skateboard a push with your hand.
According to Newton’s first law of motion, if the net force acting on a moving object is zero, it will continue to move in a straight line with constant speed.
Does the skateboard keep moving with constant speed after it leaves your hand?

7. Recall that when an object slows down it is accelerating.
By Newton’s second law, if the skateboard is accelerating, there must be a net force acting on it.

8. The force that slows the skateboard and brings it to a stop is friction.
Friction is the force that opposes the sliding motion of two surfaces that are touching each other.
The amount of friction between two surfaces depends on two factorsthe kinds of surfaces and the force pressing the surfaces together.

9. If two surfaces are in contact, welding or sticking occurs where the bumps touch each other.
These microwelds are the source of friction.

10. Sticking Together
The larger the force pushing the two surfaces together is, the stronger these microwelds will be, because more of the surface bumps will come into contact.
To move one surface over the other, a force must be applied to break the microwelds.

11. Static Friction
Suppose you have filled a cardboard box with books and want to move it.
It’s too heavy to lift, so you start pushing on it, but it doesn’t budge.
If the box doesn’t move, then it has zero acceleration.

12. According to Newton’s second law, if the acceleration is zero, then the net force on the box is zero.
Another force that cancels your push must be acting on the box.

13. That force is the friction due to the microwelds that have formed between the bottom of the box and the floor.
Static friction is the frictional force that prevents two surfaces from moving past each other.

14. Sliding Friction You ask a friend to help you move the box.
Pushing together, the box moves. Together you and your friend have exerted enough force to break the microwelds between the floor and the bottom of the box.

15. If you stop pushing, the box quickly comes to a stop.
This is because as the box slides across the floor, another forcesliding frictionopposes the motion of the box.
Sliding friction is the force that opposes the motion of two surfaces sliding past each other.

16. Rolling Friction
As a wheel rolls over a surface, the wheel digs into the surface, causing both the wheel and the surface to be deformed.

17. Static friction acts over the deformed area where the wheel and surface are in contact, producing a frictional force called rolling fiction.
Rolling friction is the frictional force between a rolling object and the surface it rolls on.

18. D. Air Resistance Air Resistance a.k.a. “fluid friction” or “drag”*
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D. Air Resistance
Air Resistance
a.k.a. “fluid friction” or “drag”
force that air exerts on a moving object to oppose its motion
depends on:
speed
surface area
shape
density of fluid *

19. Fair Fgrav Terminal Velocity*
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Terminal Velocity
maximum velocity reached by a falling object
reached when… Fgrav = Fair
Fair
no net force
 no acceleration
 constant velocity
Fgrav *

20. Fgrav = Fair Falling with air resistance*
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Falling with air resistance
heavier objects fall faster because they accelerate to higher speeds before reaching terminal velocity
Fgrav = Fair
larger Fgrav
 need larger Fair
 need higher speed *

21. *
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MythBusters! *

22. II. Gravity Chapter 3 Forces Gravitational Force Projectile Motion*
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Chapter 3 Forces
II. Gravity
Gravitational Force
Projectile Motion
Free fall and Weightlessness
Calculations *

23. *
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A. Gravity
Gravity
Any two objects in the universe exert an attractive force on each other
increases as...
mass increases
distance between them decreases
Force and mass are directly related
All objects exert a pull on each other *

24. Who experiences more gravity - the astronaut or the politician?*
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Who experiences more gravity - the astronaut or the politician?
Which exerts more gravity -
the Earth or the moon?
All matter accelerates toward Earth at 9.8m/s2
more mass
less distance *

25. Would you weigh more on Earth or Jupiter?*
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Would you weigh more on Earth or Jupiter?
Jupiter because...
greater mass
greater gravity
greater weight *

26. Accel. due to gravity (g)*
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Accel. due to gravity (g)
In the absence of air resistance, all falling objects have the same acceleration!
On Earth: g = 9.8 m/s2
Bowling ball
Feather *

27. W = mg Weight The measure of the force of gravity on an object WEIGHT*
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Weight
The measure of the force of gravity on an object
W = mg
W: weight (N)
m: mass (kg)
g: acceleration due to gravity (m/s2)
MASS
always the same (kg)
The amount of matter an object possesses (constant)
WEIGHT
depends on gravity (N)
On the moon: you’d weigh 1/6 of what you weigh on Earth *

28. g W m W = 557 N m = W ÷ g m = ? m = (557 N) ÷ (9.8 m/s2) g = 9.8 m/s2*
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Mrs. J. weighs 557 N. What is her mass?
GIVEN:
W = 557 N
m = ?
g = 9.8 m/s2
WORK:
m = W ÷ g
m = (557 N) ÷ (9.8 m/s2)
m = 56.8 kg m W g *

29. B. Projectile Motion Projectile any object thrown in the air*
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B. Projectile Motion
Projectile
any object thrown in the air
acted upon only by gravity
follows a parabolic path called a trajectory
has horizontal and vertical velocities
PROJECTILE MINI-LAB *

30. Gravity is an unbalanced force*
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Horizontal Velocity
depends on inertia
remains constant
Vertical Velocity
depends on gravity
accelerates downward at 9.8 m/s2
Gravity is an unbalanced force *

31. C. Circular Motion Centripetal Acceleration*
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C. Circular Motion
Centripetal Acceleration
acceleration toward the center of a circular path
caused by centripetal force
B-BALL DEMO
PLATE DEMO *

32. friction provides centripetal force*
On the ground...
friction provides centripetal force
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Centripetal Force: friction between the tires and the road surface
Car wants to keep moving in a straight line
Ice *

33. gravity provides centripetal force*
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In orbit...
gravity provides centripetal force
The Earth and the moon in orbit
ROUND LAB *

34. *
07/16/96 *

35. *
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D. Free-Fall
Free-Fall
when an object is influenced only by the force of gravity
Weightlessness
sensation produced when an object and its surroundings are in free-fall
object is not weightless!
CUP DEMO *

36. *
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Weightlessness
surroundings are falling at the same rate so they don’t exert a force on the object *

37. III. The Third Law of Motion*
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Chapter 3 Forces
III. The Third Law of Motion
Newton’s Third Law
Momentum
Conservation of Momentum *

38. A. Newton’s Third Law Newton’s Third Law of Motion*
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A. Newton’s Third Law
Newton’s Third Law of Motion
When one object exerts a force on a second object, the second object exerts an equal but opposite force on the first. *

39. Aren’t these “balanced forces” resulting in no acceleration?*
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Problem:
How can a horse pull a cart if the cart is pulling back on the horse with an equal but opposite force?
Aren’t these “balanced forces” resulting in no acceleration?
NO!!! *

40. forces are equal and opposite but act on different objects *
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Explanation:
forces are equal and opposite but act on different objects
they are not “balanced forces”
the movement of the horse depends on the forces acting on the horse *

41. Action-reaction pairs
Trampoline
Row boat
Sitting in a chair
Billiards
Jumping off a dock

42. Action-Reaction Pairs*
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Action-Reaction Pairs
The hammer exerts a force on the nail to the right.
The nail exerts an equal but opposite force on the hammer to the left. *

43. Action-Reaction Pairs*
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Action-Reaction Pairs
The rocket exerts a downward force on the exhaust gases.
The gases exert an equal but opposite upward force on the rocket. FG FR *

44. F m Action-Reaction Pairs Both objects accelerate.*
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Action-Reaction Pairs
Both objects accelerate.
The amount of acceleration depends on the mass of the object. F m
Small mass  more acceleration
Large mass  less acceleration *

45. p = mv v p m B. Momentum Momentum*
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B. Momentum
Momentum
A property that is due to mass and velocity (how much force is needed to change motion)
p = mv m p v
p: momentum (kg ·m/s)
m: mass (kg)
v: velocity (m/s) *

46. v p m p = ? p = mv m = 280 kg p = (280 kg)(3.2 m/s) v = 3.2 m/s*
07/16/96
Find the momentum of a bumper car if it has a total mass of 280 kg and a velocity of 3.2 m/s.
GIVEN:
p = ?
m = 280 kg
v = 3.2 m/s
WORK:
p = mv
p = (280 kg)(3.2 m/s)
p = 896 kg·m/s m p v *

47. v p m p = 675 kg·m/s v = p ÷ m m = 300 kg v = (675 kg·m/s)÷(300 kg)*
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The momentum of a second bumper car is 675 kg·m/s. What is its velocity if its total mass is 300 kg?
GIVEN:
p = 675 kg·m/s
m = 300 kg
v = ?
WORK:
v = p ÷ m
v = (675 kg·m/s)÷(300 kg)
v = 2.25 m/s m p v *

48. C. Conservation of Momentum*
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C. Conservation of Momentum
Law of Conservation of Momentum
The total momentum in a group of objects doesn’t change unless outside forces act on the objects.
pbefore = pafter *

49. Elastic Collision KE is conserved Inelastic Collision*
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Elastic Collision
KE is conserved
Inelastic Collision
KE is not conserved *

50. Momentum doesn’t change unless the object’s mass, velocity or both change
When objects collide, momentum is transferred
Billiards, bowling

51. *
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Some math to prove the Law of Conservation of Momentum
A 5-kg cart traveling at 1.2 m/s strikes a stationary 2-kg cart and they connect. Find their speed after the collision.
BEFORE
Cart 1:
m = 5 kg
v = 4.2 m/s
Cart 2 :
m = 2 kg
v = 0 m/s
AFTER
Cart 1 + 2:
m = 7 kg
v = ?
p = 21 kg·m/s m p v
p = 0
v = p ÷ m
v = (21 kg·m/s) ÷ (7 kg)
v = 3 m/s
pbefore = 21 kg·m/s
pafter = 21 kg·m/s *

52. *
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A 50-kg clown is shot out of a 250-kg cannon at a speed of 20 m/s. What is the recoil speed of the cannon?
BEFORE
Clown:
m = 50 kg
v = 0 m/s
Cannon:
m = 250 kg
AFTER
Clown:
m = 50 kg
v = 20 m/s
Cannon:
m = 250 kg
v = ? m/s
p = 0
p = 1000 kg·m/s
p = 0
p = kg·m/s
pbefore = 0
pafter = 0 *