Force powerpoint

1. CHAPTER 11 FORCE

2. NEWTON’S FIRST LAW
An object at rest will remain at rest and
an object in motion will continue in
motion at a constant velocity unless
acted upon by a net force

3. NEWTON’S FIRST LAW
Newton’s First Law of Motion
 “Law of Inertia”
Inertia
 tendency of an object to resist any
change in its motion
 increases as mass increases

4. INERTIA
 Inertia is the tendency to resist change in
motion
 Change in motion is always caused by a
force
 An object will keep doing what it’s doing
until a force acts on it

5. MASS AND INERTIA
 Massive objects resist change in motion
 Objects with more mass need larger forces to
change their motion

6. NEWTON’S 1ST LAW
 An object at rest (still) will
stay at rest until a force acts
on it.
 An object in motion will stay
in motion until a force acts
on it.

7. NEWTON’S 1ST LAW
 You aren’t wearing a seatbelt while driving
(NOT a good idea). You run your car into a
wall. Your car stops, but you remain in
motion, so you fly through the windshield.
 If you’re wearing your seatbelt, you slow
down with the car. Airbags can reduce the
force of the impact.

8. FORCE, MASS, AND ACCELERATION
Exerting a larger
force on an
object causes a
greater change
in velocity.
 A greater change in velocity means there is greater
acceleration.

9. 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.
 AKA-When mass goes up,
acceleration goes down, and when
force goes up, acceleration goes up
F = ma

10. FORCE, MASS, AND ACCELERATION
If the same force is exerted on a more
massive and a less massive object, the
more massive object can’t accelerate as
much.
The more massive object will have a
lower velocity.

11. NEWTON’S SECOND LAW
F = ma
F: force (N)
m: mass (kg)
a: accel (m/s2)
1 N = 1 kg ·m/s2

a
m
F
a
F
m

12. GRAVITY
Gravity
 force of attraction between any two
objects in the universe
 increases as...
mass increases (one or both
objects)
distance decreases

13. GRAVITY
Who experiences more gravity - the
astronaut or the politician?
less
distance
more
mass
 Which exerts more gravity -
the Earth or the moon?

14. GRAVITY
Weight
 the force of gravity on an object
MASS
always the same
(kg)
WEIGHT
depends on gravity
(N)
W = mg
W: weight (N)
m: mass (kg)
g: acceleration
due to gravity
(m/s2)

15. GRAVITY
Would you weigh more on Earth
or Jupiter?
greater gravity
greater weight
greater mass
 Jupiter because...

16. GRAVITY
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
m
W
g 
elephant
m
W
g 
feather
Animation from “Multimedia Physics Studios.”

17. 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

18. CALCULATIONS
 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

19. CALCULATIONS
 Mrs. Ellis weighs 572 N. What is her
mass?
GIVEN:
F(W) = 572 N
m = ?
a(g) = 9.8 m/s2
WORK:
m = F ÷ a
m = (572 N) ÷ (9.8 m/s2)
m = 58.4 kg
m
F
a

20. CONCEPTEST
Is the following statement true or false?
 An astronaut has less mass on the moon
since the moon exerts a weaker
gravitational force.
 False! Mass does not depend on
gravity, weight does. The astronaut has
less weight on the moon.

21. 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

22. FREE-FALL
Weightlessness
 surroundings are falling at the same
rate so they don’t exert a force on the
object

23. FREE-FALL
NASA’s KC-135 - “The Vomit Comet”
Go
to
NASA.
Go
to
CNN.com.
Go
to
Space
Settlement
Video
Library.
Space Shuttle Missions

24. CONCEPTEST 1
TRUE or FALSE:
An astronaut on the Space Shuttle
feels weightless because there is no
gravity in space.
FALSE!
There is gravity which is causing the
Shuttle to free-fall towards the Earth.
She feels weightless because she’s
free-falling at the same rate.

25. 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.

26. NEWTON’S THIRD LAW
 One object exerts a force on the second object
 The second exerts a force back that is equal in
strength, but opposite in direction

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

28. NEWTON’S THIRD LAW
 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
 Explanation:

29. NEWTON’S THIRD LAW
 Action and Reaction
forces do not cancel –
they can act on different
objects or spread in
different directions

30. NEWTON’S THIRD LAW
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.

31. NEWTON’S THIRD LAW
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

32. NEWTON’S THIRD LAW
Action-Reaction Pairs
 Both objects accelerate.
 The amount of acceleration
depends on the mass of the object.

a F
m
 Small mass  more acceleration
 Large mass  less acceleration

33. MOMENTUM
Momentum
 quantity of motion
p = mv
p: momentum (kg ·m/s)
m: mass (kg)
v: velocity (m/s)
m
p
v

34. MOMENTUM:
 inertia in motion
 when you’re moving,
momentum keeps you
moving
 an object with lots of
momentum is hard to stop

35. MOMENTUM:
 p = mv
 p = momentum (in kg m/s)
 m = mass (in kg)
 v = velocity (in m/s)
 momentum is a vector quantity, so a complete
answer will include magnitude and direction
m
p
v

36. IS IT POSSIBLE FOR A MOTORCYCLE AND
A LARGE TRUCK TO HAVE THE SAME
MOMENTUM?
m = 40,000 kg
m = 400 kg

37. MOMENTUM
 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

38. MOMENTUM
 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

39. EXAMPLE:
WHAT IS THE MOMENTUM OF A 150 KG
DEFENSIVE TACKLE AT 5 M/S NORTH
TOWARDS THE END ZONE?

40. EXAMPLE:
WHAT IS THE MOMENTUM OF A 325
POUND DEFENSIVE TACKLE AT 5 M/S
NORTH TOWARDS THE END ZONE?
m = 150 kg)
v = 5 m/s north
p = mv
momentum
p = (150 kg) (5 m/s)
p = 750 kg m/s north

41. TO CHANGE THE MOMENTUM OF AN
OBJECT, YOU MUST CHANGE…
 mass and/or velocity
 usually velocity changes
 change in velocity is acceleration
 to make an object accelerate, a force is required
 So a force causes a change in the momentum of an
object.
 Size of force and the time the force acts both affect
change in momentum.

42. THE IMPORTANCE OF “FOLLOWING
THROUGH” IN SPORTS
 The follow-through
increases the time of
collision and thus
increases the change in
momentum, making the
ball have a higher
velocity!

43. IMPULSE:
 what changes the momentum
 Depends on force and time of collision
 Impulse = Ft
 Impulse = change of momentum
 units for impulse are same as units for momentum
 Impulse = Δp
OR
 Ft = m(Δv)
F
I
t
m
I
Δv

44. EXAMPLE:
A HOCKEY PLAYER APPLIES AN AVERAGE
FORCE OF 80.0 N TO A 0.25 KG HOCKEY
PUCK FOR A TIME OF 0.10 SECONDS.
DETERMINE THE IMPULSE EXPERIENCED
BY THE HOCKEY PUCK.

45. EXAMPLE:
A HOCKEY PLAYER APPLIES AN AVERAGE
FORCE OF 80.0 N TO A 0.25 KG HOCKEY
PUCK FOR A TIME OF 0.10 SECONDS.
DETERMINE THE IMPULSE EXPERIENCED
BY THE HOCKEY PUCK.
F = 80.0 N
t = 0.10 s
Impulse = Ft
impulse
Impulse = (80.0 N) (0.1 s)
Impulse = 8 N s
Impulse = 8 kg m/s
kg m/s2
N
F
I
t

46. EXAMPLE #2
 What is the change in momentum (impulse) of a
950 kg car that travels from 40 m/s to 31 m/s?
GIVEN:
I = ?
m = 950 kg
Δv = 40-31 m/s
= 9 m/s
WORK:
I = m Δ v
I = (950 kg)(9m/s)
I = 8550 kg·m/s
m
I
Δv

47. EXAMPLE #3
 If the from the last problem had an impulse of 8550
kg*m/s, and it took 30 seconds to change speeds,
then what force caused the change?
GIVEN:
I = 8550 kg* m/s
F = ? N
t= 30 s
WORK:
F= I÷ t
F = (8550)÷ (30)
F = 285 N
F
I
t

48. THE EFFECT OF COLLISION TIME UPON THE
FORCE
Impulse (kg m/s) Time (s) Force (N)
100 1 100
100 2 50
100 4 25
100 10 10
100 25 4
100 50 2
100 100 1
100 1000 0.1
F
I
t

49. TIME AND FORCE ARE ________
PROPORTIONAL.
 For an object in a collision,
 to decrease the effect of the force, the time
must be increased
 To Increase the effect of the force, the time
must be decreased.
inversely

50. AIR BAGS
 Air bags increase the
time allowed to stop
the momentum of the
driver and passenger,
decreasing the force
of impact.
 Time increases 100
times!
 Force decreases
100 times!

51. USE OF MATS IN SPORTS
 Mats increase the time
allowed to stop the
momentum of the
athlete, decreasing the
force of impact.

52. “RIDING THE PUNCH”
 When a boxer recognizes
that he will be hit in the head
by his opponent, the boxer
often relaxes his neck and
allows his head to move
backwards upon impact.
 “Riding the punch” increases
the time the force is applied,
decreasing the force of
impact.

53. 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

54. CONSERVATION OF MOMENTUM
Elastic Collision
 KE is conserved
 Inelastic Collision
 KE is not conserved

55. 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
p = 0
pbefore = 21 kg·m/s pafter = 21 kg·m/s
m
p
v
v = p ÷ m
v = (21 kg·m/s) ÷ (7 kg)
v = 3 m/s

56. CONSERVATION OF MOMENTUM
 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
v = 0 m/s
AFTER
Clown:
m = 50 kg
v = 20 m/s
Cannon:
m = 250 kg
v = ? m/s
p = 0
p = 0
pbefore = 0
p = 1000 kg·m/s
pafter = 0
p = -1000 kg·m/s

57. CONSERVATION OF MOMENTUM
 So…now we can solve for velocity.
GIVEN:
p = -1000 kg·m/s
m = 250 kg
v = ?
WORK:
v = p ÷ m
v = (-1000kg·m/s)÷(250kg)
v = - 4 m/s
(4 m/s backwards)
m
p
v