ICT role in 21st century education and it's challenges.
Forces & Changes in Motion
1. Forces & Changes in Motion
Introduction to Dynamics
Canadian Academy, Kobe
2. Felix Baumgartner: 14 October 2012
Broke world records set by Joe Kittinger in 1960
Quick stats:
• Jump height: 39,045m (Joe K: 31,300m)
• Max speed: 1,342.8 km/h (Joe K: 988km/h)
• Freefall time: 260s (Joe K: 276s)
Joe Kittinger, 1960: Felix Baumgartner, 2012: 8-min news video: http://gu.com/p/3b59k
http://en.wikipedia.org/wiki/Joseph_Kittinger Highlight reel: http://www.youtube.com/watch?v=FHtvDA0W34I
3. Why did Felix have to jump from so high?
Karman line “space”
The Karman line is the ‘edge of
space’ by political definition. There is
no true edge of space, though.
What is terminal velocity?
http://hypertextbook.com/facts/Ji
anHuang.shtml
Felix’s Jump, 2012 (39,045m)
Joe’s Jump, 1960 (31,300m)
At lower altitudes, air pressure (and
density) increase – more air particles
give greater air resistance, so a slower
terminal velocity (not enough to
break the speed of sound).
http://en.wikipedia.org/wiki/Atmosphere_of_Earth
5. Beeeeeeeeefcaaaaaaaaaaaaake!
Explain what is happening in terms of forces.
Can you describe the forces in terms of vectors?
MrT
http://www.free-clipart-pictures.net/wrestling_clipart.html
6. Beeeeeeeeefcaaaaaaaaaaaaake!
Explain what is happening in terms of forces.
Can you describe the forces in terms of vectors?
MrT
Force is a vector – it has magnitude and direction.
Therefore, we can create vector diagrams for force!
http://www.free-clipart-pictures.net/wrestling_clipart.html
7. Forces and Changes in Motion
Unit Question: “How can interactions cause change?”
Areas of interaction:.
• Health and Social Education
Making informed decisions on road and sports safety.
Criterion Assessment Tasks
A: One World “Acceleration Kills”
B: Communication in Science “Acceleration Kills”
C: Knowledge & Understanding Unit test, Quia Quizzes, class whiteboarding
D: Scientific Inquiry (last unit)
E: Processing Data Can a spring be used to accurately measure force?
F: Attitudes in Science (last unit)
8. Forces and Changes in Motion
Assessment Statements
Describe the different common forces:
Frictional, Normal, Tension, Spring, Air Resistance, Applied,
Gravitational (Weight), Electrical, Magnetic
Identify the forces, their agents and directions acting on a single object
Explain how the magnitude of a force can be measured
State Newton’s first law of motion: Inertia
Draw free body diagrams
State Newton’s second law of motion: Acceleration and define net force
Distinguish between balanced forces (equilibrium) and unbalanced forces
on an object
Explain the effect of balanced or unbalanced forces on an object
Explain the effect of mass on the acceleration of an object with a constant
net force
Calculate the weight of an object on Earth from its mass.
Calculate forces and accelerations using net force statements
State Newton’s third law of motion: Interaction
Identify action and reaction force pairs
9. Little tricks to impress little kids
Flick the card away with your fingernail.
Explain what happens!
Catch the coin balanced on your elbow.
Explain what happens!
http://www.inklesstales.com/science/coin.shtml
10. Explain what is going on
As a group explain what happens.
Use whiteboard and markers.
Don’t use any sources.
Nominate a speaker.
Present.
Think about:
- forces acting ‘before flick’
- forces acting ‘after flick’
After hearing the other groups’ explanations, think
about how you might need to modify your own.
11. Newton’s First Law of Motion: Inertia
An object will not change its motion unless acted on by an unbalanced force.
(Like a boxer’s brain)
Why might boxing with
gloves be more dangerous
than boxing without?
http://sportige.com/top-10-boxing-photos/
12. What do you think?
The ball needs a force to
keep it moving
The ball needs a force to
stop it moving.
Clipart people from: http://www.clker.com/search/krug/1
13. Newton’s First Law of Motion: Inertia
An object will not change its motion
unless acted on by an unbalanced force.
• if it is at rest, it will stay at rest
• if it is in motion, it will remain at the same velocity
The ball needs a force to
keep it moving
The ball needs a force to
stop it moving.
Clipart people from: http://www.clker.com/search/krug/1
14. Newton’s First Law of Motion: Inertia
An object will not change its motion
unless acted on by an unbalanced force.
• if it is at rest, it will stay at rest
• if it is in motion, it will remain at the same velocity
Objects with a greater mass have more inertia.
It takes more force to change their motion.
300kg
30kg
http://www.clker.com/clipart-man-push.html
15. Buckle up in the back…
Explain, using the term inertia why
wearing a seatbelt is important!
http://www.youtube.com/watch?v=e6Qhmdk4VNs&NR=1
16. But what are FORCES? Forces exert a push or a pull on an object.
Use the text or the PhysicsClassroom (http://goo.gl/umXYp) to describe different
types of forces.
• Give examples: object, agent and direction
• Can you draw free body diagrams of the examples?
At-a-distance forces:
Fnorm
Gravitational (weight) (Fgravity)
Electrical (Felectric)
Magnetic (Fmag) Fapp Ffriction
Why is this arrow longer?
Contact forces: Fgrav
Normal (Fnorm)
Frictional (Ffriction) e.g. A book is pushed across a tabletop.
- static
- kinetic Force Agent Direction
Tension (Ftension) Weight (gravity) Earth Down
Spring (Fspring)
Normal Table Up
Applied (Fapp)
Elastic (Felastic) Applied (finished) MrT Left
Buoyant (Fbuoyant)
Air resistance (Fair) Friction Tabletop Right
17. Mass, Weight & Gravity Force
Mass is a measure of the amount of matter in an
object. We measure it in kg.
Weight is the force of the object being pulled to
the Earth (or moon, Neptune, etc). It is measured
in Newtons (N).
On Earth, the force of gravity (g) is 9.8 N/kg.
Therefore the weight of an object is its mass
(m)multiplied by 9.8.
Fgravity = m.g
“The” kilogram. From:
http://sicktek.com/the-standard-kilogram-is-losing-mass/
Fgrav In all of our examples and questions,
we work under the assumption that
m g 1kg = 10N
18. Mass vs Weight In all of our examples and
questions, we work under the
Calculate the weights of these masses (on Earth): assumption that
1. 50kg
2. 500g 1kg = 10N
3. 7g
4. 1,000kg
Calculate the masses of these weights (on Earth):
1. 100N
2. 8.7N
3. 0.034N
4. 1.02 x 104N
19. Check out my awesome
follow-through!
It will make the ball
fly faster
It will make no
difference to the
flight of the ball
It will allow for
better control of the
ball in flight
Golfer from: http://www.clker.com/clipart-2404.html Clipart people from: http://www.clker.com/search/krug/1
20. Spring Force Describe the forces acting on the mass
mass Balanced forces
Equilibrium
mass
Stretched
Compressed
mass
21. Spring Force Describe the forces acting on the mass
mass Balanced forces
Equilibrium
The spring is
mass
pulling on the mass
Stretched
Compressed
The spring is
mass
pushing on the mass
22. Using springs to measure force
Can the change in shape of a spring be used
to measure the magnitude of a force?
Criterion E: Processing Data
Attach 0 – 6 weights (each set is 0.5N (± 0.1N)
Record the extension of the spring.
Repeat 3 times.
Calculate and plot the means with error bars
as half the range.
Using the equation of the line, find the
weight of the mystery object (#1, #2 or #3)
mass
Compare your answer to the true value.
Equilibrium Calculate % error. Show your working.
mass
Stretched
23. The bigger guy is being pulled
down more by gravity
The smaller guy is being
pushed up less by the water
They are both affected by water
and gravity equally.
Clipart people from: http://www.clker.com/search/krug/1 http://www.youtube.com/watch?v=IJWlI0Q8s-I
24. Measuring common forces
Gravity:
- Masses with different Newton-meters
Bouyancy Force How do the range and
- Use the Logger Pro setup uncertainties of each of our
Newton-meters compare?
Tension Force
- String and Newton-meter setup
Normal Force
- Bathroom scales
- LoggerPro force plate Set up the buoyancy-meter
like this. Do not get it wet.
Frictional Forces: Submerge them most of the
- Static way and be consistent.
- Kinetic
- Different surfaces ⌘ 0 will zero the sensor.
- Effect of weight
25. What happens when the 500g mass
is attached to two Newton-meters?
Both meters will read 5N
They will both read 2.5N
The top one will read 5N and the
bottom one will read zero.
Clipart people from: http://www.clker.com/search/krug/1
26. Pull
The second Newton
What do you think….
meter will read higher
than the first. … and WHY?
Both Newton meters
will read the same
The first newton
meter will read higher
than the second
27. Water pushes
them all equally
because they have
the same volume
Water pushes them
all equally because
they have the same
density
Water pushes them all
equally because they
have the same buoyancy
28. What would happen to
the readings if the water
were extremely salty?
What would happen to
the air and water
readings if the aluminium
block was re-shaped into
a bigger volume?
29. 87kg 66kg
100N 100N
Who weighs more on the bathroom scales?
Image source unknown.
30. 87kg 66kg
100N 100N
Who weighs more on the bathroom scales?
77kg Fgrav 76kg
m g
Free body diagrams are vector diagrams.
They show the direction and magnitude of
each force.
Draw free body diagrams for the forces
acting on the men.
Image source unknown.
31. 87kg 66kg
870N FN (scale) 100N 100N 660N FN (scale)
100N FN (sink) Who weighs more on the bathroom scales?
77kg Fgrav 76kg 100N FN (sink)
870N Fg
m g
Free body diagrams are vector diagrams.
660N Fg
They show the direction and magnitude of
each force.
Draw free body diagrams for the forces
acting on the men.
Image source unknown.
32. Free Body Diagrams
Free body diagrams are vector diagrams. They show the direction and magnitude of the
forces acting on an object. Forces are drawn as arrows acting from the centre of a box
representing the object.
Can you identify:
the slowing car, the accelerating bike, FN
the launching angry bird and the sinking stone?
Fapplied Fapplied Ffriction
Fbuoyant
Fg
FN
Fgravity Fgravity Ffriction
Fg
33. Free Body Diagrams Free body art-attack
All free clipart images from: http://www.clker.com
34. Free Body Diagrams Free body art-attack
Accelerating
All free clipart images from: http://www.clker.com
35. Free Body Diagrams Free body art-attack
Accelerating
All free clipart images from: http://www.clker.com
36. Free Body Diagrams Free body art-attack
Accelerating
All free clipart images from: http://www.clker.com
37. Free Body Diagrams Free body art-attack
All free clipart images from: http://www.clker.com
38. Free Body Diagrams Free body art-attack
All free clipart images from: http://www.clker.com
39. Free Body Diagrams Free body art-attack
All free clipart images from: http://www.clker.com
40. Free Body Diagrams Free body art-attack
Accelerating
Accelerating
All free clipart images from: http://www.clker.com
41. The Earth has a larger mass, so
pulls the moon harder than the
moon pulls the Earth
The Earth and the moon pull on
each other equally.
The moon pulls more on the Earth
than the Earth does on the moon.
http://wallpaperart.altervista.org/Immagini/luna-terra-sfondo-1280x800.jpg
42. The Earth has a larger mass, so
pulls the moon harder than the
moon pulls the Earth
The Earth and the moon pull on
each other equally.
The moon pulls more on the Earth
than the Earth does on the moon.
Check out the answer here:
it might surprise you!
http://wallpaperart.altervista.org/Immagini/luna-terra-sfondo-1280x800.jpg
43. Newton’s Third Law of Motion: Interaction
For every action, there is an equal and opposite reaction.
• In all interactions
• Regardless of mass or size
http://photo-dict.faqs.org/phrase/630/Newton's-Cradle.html
44. Newton’s Third Law of Motion: Interaction
For every action, there is an equal and opposite reaction.
• In all interactions
• Regardless of mass or size
FBD for the man
FBD for the block
http://www.clker.com/clipart-man-push.html
45. Newton’s Third Law of Motion: Interaction
For every action, there is an equal and opposite reaction.
• In all interactions
• Regardless of mass or size
FNormal
FNormal
FNormal FNormal
(block)
Fg Fg (man)
FBD for the man
FBD for the block
http://www.clker.com/clipart-man-push.html
46. Newton’s Third Law of Motion: Interaction
For every action, there is an equal and opposite reaction.
• In all interactions
• Regardless of mass or size
FNormal FNormal
FNormal FNormal
(block (man)
) Fg Fg
The normal forces of the man acting on the block and
the block acting on the man are equal and opposite.
They are a reaction force pair.
http://www.clker.com/clipart-man-push.html What other reaction force pairs can you see?
47. Newton’s Third Law of Motion: Interaction
For every action, there is an equal and opposite reaction.
• In all interactions
• Regardless of mass or size This means forces are in
reaction force pairs.
http://www.clker.com/clipart-man-push.html
48. Newton’s Third Law of Motion: Interaction
For every action, there is an equal and opposite reaction.
• In all interactions
• Regardless of mass or size This means forces are in
Identify the reaction force pairs in
reaction force pairs.
these situations and draw force pair diagrams:
Extension – the information in green will help you label the magnitude of the forces.
1. A (70kg) person standing on a desk
2. A (100g) ball (slows from 20ms-1 0ms-1 in 0.1s when it) hits a wall.
3. A bow pushes an arrow forward with a force of 50N.
4. A bike tyre pushes down on the ground. (The bike weighs 20kg).
5. A rocket pushes hot air down and out, so that it takes off.
1. How many reaction force pairs can you identify here?
http://www.canstockphoto.com/illustration/dribbling.html
49. The harder he pushes, the
faster it moves.
The harder he
pushes, the greater the
acceleration.
He needs to keep pushing
harder to maintain
constant velocity.
http://www.clker.com/clipart-man-push.html
50. Free body diagrams, force and acceleration
http://phet.colorado.edu/en/simulation/forces-and-motion
Set up the investigation like this:
51. Free body diagrams, force and acceleration
http://phet.colorado.edu/en/simulation/forces-and-motion
Draw free body diagrams
(include labels, forces and Net force if it is present):
- Pushing less than the force of static friction
- Pushing just more than static friction
- Letting go immediately after the crate moves 100kg crate, wood floor
Identify: Static Friction Crate ____ N Fridge ____ N
Kinetic Friction Crate ____ N Fridge ____ N
Explain what happens to the velocity and acceleration of the objects
when the following are applied:
- A constant force
- A force which is let go 200kg fridge, wood floor
Compare changes in velocity and acceleration between the 100kg
crate and 200kg fridge.
Set up the experiment again with ice and bouncy walls.
What happens to the graphs and free body diagrams now?
100kg crate, ice floor, bouncy walls
52. Free body diagrams, force and acceleration
http://phet.colorado.edu/en/simulation/forces-and-motion
Draw free body diagrams
(include labels, forces and Net force (Sum) if it is present):
- Pushing less than the force of static friction
- Pushing just more than static friction
- Letting go immediately after the crate moves 100kg crate, wood floor
FN FN FNet FNet FN
F fr static Fapplied Ffr kinetic Fapplied Ffr kinetic
Fg Fg Fg
What happens to the What is the What is the
applied force and the significance of the significance of the
static friction force as the magnitude of the magnitude of the
applied force increases Net force vector? Net force vector?
(but the crate stays still)?
53. Net Force and Free Body Diagrams
Net force is a the sum of the forces acting on an object. It is a vector – direction and
magnitude of each force are important.
y We consider the forces acting in the same plane
(not perpendicular to each other).
x
Fapplied
Fbuoyant 200N
4.5N
Fgravity
∑Fy = 4.5N – 9N ∑Fy = 200N – 80N
Fgravity 80N
9.0N = -4.5N = 120N
∑Fx = 0 ∑Fx = 0
54. Net Force and Free Body Diagrams
Net force is a the sum of the forces acting on an object. It is a vector – direction and
magnitude of each force are important.
y We consider the forces acting in the same plane
(not perpendicular to each other).
x
Fapplied
Fbuoyant 200N
4.5N
∑Fy = 120N
∑Fy = -4.5N
Fgravity
∑Fy = 4.5N – 9N ∑Fy = 200N – 80N
Fgravity 80N
9.0N = -4.5N = 120N
∑Fx = 0 ∑Fx = 0
55. Net Force and Free Body Diagrams
Net force is a the sum of the forces acting on an object. It is a vector – direction and
magnitude of each force are important.
y We consider the forces acting in the same plane
(not perpendicular to each other).
x
FN FN
80N 80N
160N 160N
Ffriction Fapplied Ffriction
Fg Fg 80N
80N 80N
∑Fy =
∑Fy =
∑Fx =
∑Fx =
56. Net Force and Free Body Diagrams
Net force is a the sum of the forces acting on an object. It is a vector – direction and
magnitude of each force are important.
y We consider the forces acting in the same plane
(not perpendicular to each other).
x
FN FN
80N 80N
160N 160N
Ffriction Fapplied Ffriction
Fg Fg 80N
80N 80N
∑Fx = -160N ∑Fx = -80N
∑Fy = 80N – 80N
∑Fy = 80N – 80N
= 0N
= 0N
∑Fx = 160 – 80N
∑Fx = -160N (left)
= -80N (left)
57. Free body diagrams, force and acceleration 200kg
http://phet.colorado.edu/en/simulation/forces-and-motion 100kg
Which set of graphs represents the 100kg crate and which
one is the 200kg fridge? How do you know?
Force (N)
Acceleration
(ms-2)
Velocity
(ms-1)
Time (s) Time (s)
58. Free body diagrams, force and acceleration
http://phet.colorado.edu/en/simulation/forces-and-motion
Which set of graphs represents the 100kg crate and which
one is the 200kg fridge? How do you know?
Force (N)
Same force
Acceleration
(ms-2)
Velocity
(ms-1)
Time (s) Time (s)
59. Newton’s Second Law of Motion: Acceleration
Unbalanced forces produce motion. Acceleration is
directly proportional to the net force and inversely
proportional to the mass of the object.
• Bigger net forces, greater acceleration*
• Larger masses, smaller acceleration#
F net
m a
60. Newton’s Second Law of Motion: Acceleration
Unbalanced forces produce motion. Acceleration is
directly proportional to the net force and inversely
proportional to the mass of the object.
• Bigger net forces, greater acceleration*
• Larger masses, smaller acceleration#
*If the forces are unbalanced, Balanced forces give inertia
- constant velocity
F
there is a net force – and so there
- rest
must be acceleration!
net
m a
#The heavier it is, the less Acceleration due to
it will accelerate (more gravity is 9.8ms-2
mass = more inertia)
62. Explain this:
5kg 1kg
Why do they hit the ground at the same time?
Are the forces acting on them the same?
http://www.youtube.com/watch?v=_mCC-68LyZM
63. Explain this:
5kg 1kg
Why do they hit the ground at the same time?
Are the forces acting on them the same?
http://www.youtube.com/watch?v=_mCC-68LyZM
F = m.a
F = m.a
Acceleration due to
gravity is 9.8ms-2
64. It takes a greater force to accelerate a larger mass.
5kg 1kg
49N 9.8N
Why do they hit the ground at the same time?
Are the forces acting on them the same?
http://www.youtube.com/watch?v=_mCC-68LyZM
F = m.a
More inertia = 5kg x 9.8ms-2
= 49N Different forces.
Same acceleration.
Less inertia F = m.a
= 1kg x 9.8ms-2
Acceleration due to
gravity is 9.8ms-2 = 9.8N
65. Now can you explain this?
Why do they hit the ground at the same time?
Are the forces acting on them the same?
Would it be the same on Earth? Why?
http://www.youtube.com/watch?v=-4_rceVPVSY
66. Newton’s Second Law of Motion: Acceleration
Try these calculations.
In each case, assume no friction or air resistance.
1. A 5kg object accelerates at 2ms-2.
Calculate the net force.
Remember:
Force is in Newtons.
2. A 5N net force causes an object to accelerate at 3ms-2. Mass is in kg.
Calculate the mass of the object. Acceleration is in ms-2
So you might need to
adjust units sometimes.
3. A 250N net force acts on a 10kg object .
Calculate the acceleration of the object.
Practice on this Quia Quiz: Simple Second Law (http://www.quia.com/quiz/3326680.html)
67. Newton’s Second Law of Motion: Acceleration
Try these calculations.
In each case, assume no friction or air resistance.
1. A 5kg object accelerates at 2ms-2.
Calculate the net force.
F = m.a so F = 5 x 2 = 10N
Remember:
Force is in Newtons.
2. A 5N net force causes an object to accelerate at 3ms-2. Mass is in kg.
Calculate the mass of the object. Acceleration is in ms-2
So you might need to
adjust units sometimes.
3. A 250N net force acts on a 10kg object .
Calculate the acceleration of the object.
Practice on this Quia Quiz: Simple Second Law (http://www.quia.com/quiz/3326680.html)
68. Newton’s Second Law of Motion: Acceleration
Try these calculations.
In each case, assume no friction or air resistance.
1. A 5kg object accelerates at 2ms-2.
Calculate the net force.
F = m.a so F = 5 x 2 = 10N
Remember:
Force is in Newtons.
2. A 5N net force causes an object to accelerate at 3ms-2. Mass is in kg.
Calculate the mass of the object. Acceleration is in ms-2
So you might need to
m = F/a so m = 5/3 = 1.67kg adjust units sometimes.
3. A 250N net force acts on a 10kg object .
Calculate the acceleration of the object.
Practice on this Quia Quiz: Simple Second Law (http://www.quia.com/quiz/3326680.html)
69. Newton’s Second Law of Motion: Acceleration
Try these calculations.
In each case, assume no friction or air resistance.
1. A 5kg object accelerates at 2ms-2.
Calculate the net force.
F = m.a so F = 5 x 2 = 10N
Remember:
Force is in Newtons.
2. A 5N net force causes an object to accelerate at 3ms-2. Mass is in kg.
Calculate the mass of the object. Acceleration is in ms-2
So you might need to
m = F/a so m = 5/3 = 1.67kg adjust units sometimes.
3. A 250N net force acts on a 10kg object .
Calculate the acceleration of the object.
a = F/m so a = 250/10 = 25ms-2
Practice on this Quia Quiz: Simple Second Law (http://www.quia.com/quiz/3326680.html)
70. Newton’s Second Law of Motion: Acceleration
Try these calculations.
In each case, assume no friction or air resistance.
1. A 5kg object accelerates from rest to 5ms-1 in 2s.
Calculate the net force.
2. A 5N net force causes an object to accelerate from 3ms-1 to resting in 0.5s.
Calculate the mass of the object.
3. A 250N net force acts on a 10kg object at rest.
Calculate the velocity of the object after 7 seconds.
Practice: http://www.quia.com/quiz/3327710.html
71. Newton’s Second Law of Motion:
Try these calculations.
In each case, assume no friction or air resistance.
1. A 5kg object accelerates from rest to 5ms-1 in 2s.
Calculate the net force.
2. A 5N net force causes an object to accelerate from 3ms-1 to resting in 0.5s.
∆v
Calculate the mass of the object. ∆t
3. A 250N net force acts on a 10kg object at rest.
Calculate the velocity of the object after 7 seconds.
Practice: http://www.quia.com/quiz/3327710.html
72. Newton’s Second Law of Motion:
Try these calculations.
In each case, assume no friction or air resistance.
1. A 5kg object accelerates from rest to 5ms-1 in 2s.
Calculate the net force.
F = 5.(5/2) so F = 5 x 2.5 = 12.5N
2. A 5N net force causes an object to accelerate from 3ms-1 to resting in 0.5s.
∆v
Calculate the mass of the object. ∆t
3. A 250N net force acts on a 10kg object at rest.
Calculate the velocity of the object after 7 seconds.
Practice: http://www.quia.com/quiz/3327710.html
73. Newton’s Second Law of Motion:
Try these calculations.
In each case, assume no friction or air resistance.
1. A 5kg object accelerates from rest to 5ms-1 in 2s.
Calculate the net force.
F = 5.(5/2) so F = 5 x 2.5 = 12.5N
2. A 5N net force causes an object to accelerate from 3ms-1 to resting in 0.5s.
∆v
Calculate the mass of the object. ∆t
m = 5/(-3/0.5) so F = 5/-6 = 0.83kg
You can’t have negative mass!
3. A 250N net force acts on a 10kg object at rest.
Calculate the velocity of the object after 7 seconds.
Practice: http://www.quia.com/quiz/3327710.html
74. Newton’s Second Law of Motion:
Try these calculations.
In each case, assume no friction or air resistance.
1. A 5kg object accelerates from rest to 5ms-1 in 2s.
Calculate the net force.
F = 5.(5/2) so F = 5 x 2.5 = 12.5N
2. A 5N net force causes an object to accelerate from 3ms-1 to resting in 0.5s.
∆v
Calculate the mass of the object. ∆t
m = 5/(-3/0.5) so F = 5/-6 = 0.83kg
You can’t have negative mass!
3. A 250N net force acts on a 10kg object at rest.
Calculate the velocity of the object after 7 seconds.
a = F/m ∆v/∆t = F/m ∆v/7s = 250N/10kg
v = 25Nkg-1 x 7
v = 175ms-1
Practice: http://www.quia.com/quiz/3327710.html
75. Newton’s Second Law of Motion:
A more complex example. Try using triangles and don’t forget the units.
A friction force of 17N slows a rolling bike from 30kmh-1 to 22.8kmh-1 in
8 seconds. Calculate the mass of the bike. Assume no air resistance.
Practice: http://www.quia.com/quiz/3327710.html
76. Newton’s Second Law of Motion:
A more complex example. Try using triangles and don’t forget the units.
A friction force of 17N slows a rolling bike from 30kmh-1 to 22.8kmh-1 in
8 seconds. Calculate the mass of the bike. Assume no air resistance.
17N
m a
Practice: http://www.quia.com/quiz/3327710.html
77. Newton’s Second Law of Motion:
A more complex example. Try using triangles and don’t forget the units.
A friction force of 17N slows a rolling bike from 30kmh-1 to 22.8kmh-1 in
8 seconds. Calculate the mass of the bike. Assume no air resistance.
17N
∆v
m a 8s
Practice: http://www.quia.com/quiz/3327710.html
78. Newton’s Second Law of Motion:
A more complex example. Try using triangles and don’t forget the units.
A friction force of 17N slows a rolling bike from 30kmh-1 to 22.8kmh-1 in
8 seconds. Calculate the mass of the bike. Assume no air resistance.
17N 30kmh-1 - 22.8kmh-1
∆v
m a 8s Units!
Practice: http://www.quia.com/quiz/3327710.html
79. Newton’s Second Law of Motion:
A more complex example. Try using triangles and don’t forget the units.
A friction force of 17N slows a rolling bike from 30kmh-1 to 22.8kmh-1 in
8 seconds. Calculate the mass of the bike. Assume no air resistance.
17N – 7.2kmh-1
= -2ms-1
∆v 3.6
m a 8s 1ms-1 = 3.6kmh-1
Practice: http://www.quia.com/quiz/3327710.html
80. Newton’s Second Law of Motion:
A more complex example. Try using triangles and don’t forget the units.
A friction force of 17N slows a rolling bike from 30kmh-1 to 22.8kmh-1 in
8 seconds. Calculate the mass of the bike. Assume no air resistance.
17N
-2ms-1
m a 8s
Practice: http://www.quia.com/quiz/3327710.html
81. Newton’s Second Law of Motion:
A more complex example. Try using triangles and don’t forget the units.
A friction force of 17N slows a rolling bike from 30kmh-1 to 22.8kmh-1 in
8 seconds. Calculate the mass of the bike. Assume no air resistance.
17N
m -0.25ms-2 m = 17/0.25 = 68kg
You can’t have negative mass!
Practice: http://www.quia.com/quiz/3327710.html
82. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 1:
A 2.50kg book sits on a desk. Determine the magnitude of the
force the desk must apply on the book to keep it from falling.
2.5kg
83. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 1:
A 2.50kg book sits on a desk. Determine the magnitude of the
force the desk must apply on the book to keep it from falling.
∑Fy = Normal – gravity = 0
FN ∑Fx = 0
___N
2.5kg
Fgravity
___N
84. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 1:
A 2.50kg book sits on a desk. Determine the magnitude of the
force the desk must apply on the book to keep it from falling.
∑Fy = Normal – (2.5kg x 10N/kg) = 0
FN ∑Fx = 0
___N
2.5kg
Fgravity
___N
85. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 1:
A 2.50kg book sits on a desk. Determine the magnitude of the
force the desk must apply on the book to keep it from falling.
∑Fy = Normal – 25N = 0
FN ∑Fx = 0
___N
2.5kg
Normal – 25N = 0
Normal = 25N
Fgravity
___N
86. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 2:
A 100kg fridge is pushed to the right along a floor with an
acceleration of 2ms-2. It is opposed by a friction force of 75N.
Calculate the applied force.
87. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 2:
A 100kg fridge is pushed to the right along a floor with an
acceleration of 2ms-2. It is opposed by a friction force of 75N.
Calculate the applied force.
∑Fx ∑Fx = m.a
FN =
=
Fapp = ∑Fx + Ffr
Ffr Fapp
=
Fgravity
∑Fx = ?
∑Fy = ?
88. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 2:
A 100kg fridge is pushed to the right along a floor with an
acceleration of 2ms-2. It is opposed by a friction force of 75N.
Calculate the applied force.
∑Fx = ?N ∑Fx = m.a
FN =
1000N =
Fapp = ∑Fx + Ffr
Ffr Fapp
75N ?N =
Fgravity
1000N ∑Fx = ?
∑Fy = 0
89. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 2:
A 100kg fridge is pushed to the right along a floor with an
acceleration of 2ms-2. It is opposed by a friction force of 75N.
Calculate the applied force.
∑Fx = ?N ∑Fx = m.a
FN = 100 x 2
1000N =
Fapp = ∑Fx + Ffr
Ffr Fapp
75N ?N =
Fgravity
1000N ∑Fx = ?
∑Fy = 0
90. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 2:
A 100kg fridge is pushed to the right along a floor with an
acceleration of 2ms-2. It is opposed by a friction force of 75N.
Calculate the applied force.
∑Fx = 200N ∑Fx = m.a
FN = 100 x 2
1000N = 200N
Fapp = ∑Fx + Ffr
Ffr Fapp
75N ?N = ?N (right)
Fgravity
1000N ∑Fx = 200N
∑Fy = 0
91. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 2:
A 100kg fridge is pushed to the right along a floor with an
acceleration of 2ms-2. It is opposed by a friction force of 75N.
Calculate the applied force.
∑Fx = 200N ∑Fx = m.a
FN = 100 x 2
1000N = 200N
Fapp = ∑Fx + Ffr
Ffr Fapp
75N 275N = 275N (right)
Fgravity
1000N ∑Fx = 200N
∑Fy = 0
92. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 3:
An 85kg rocket has a thrust force of 1800N. It accelerates
upwards at 2ms-2. Determine the magnitude of air
resistance against the rocket.
FThrust
___N
∑Fy
Fg 85kg
Fair ___N
93. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 3:
An 85kg rocket has a thrust force of 1800N. It accelerates
upwards at 2ms-2. Determine the magnitude of air
resistance against the rocket.
∑Fy =
FThrust ∑Fx = 0
___N
∑Fy
Fg 85kg
Fair ___N
94. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 3:
An 85kg rocket has a thrust force of 1800N. It accelerates
upwards at 2ms-2. Determine the magnitude of air
resistance against the rocket.
∑Fy = FThrust – Fg – Fair
FThrust ∑Fx = 0
___N
∑Fy
Fg ____N 85kg
Fair ___N
95. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 3:
An 85kg rocket has a thrust force of 1800N. It accelerates
upwards at 2ms-2. Determine the magnitude of air
resistance against the rocket.
∑Fy = 1800 – 850 – Fair
FThrust ∑Fx = 0
1800N
∑Fy ∑Fy =
=
Fg Fair = 1800 – 850 – ∑Fy
850N 85kg
=
Fair ?N =
96. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 3:
An 85kg rocket has a thrust force of 1800N. It accelerates
upwards at 2ms-2. Determine the magnitude of air
resistance against the rocket.
∑Fy = 1800 – 850 – Fair
FThrust ∑Fx = 0
1800N
∑Fy ∑Fy = 85kg x 2ms-2
= 170N
Fg Fair = 1800 – 850 – ∑Fy
850N 85kg
= 1800 – 850 – 170
Fair ?N =
97. More complex problems Assume gravity on
Earth is 10N per kg.
In these problems, you will need to:
- Draw a free body diagram (with all relevant forces)
y
- Write Net force statements
- Use the Net force statements to answer the question x
Example 3:
An 85kg rocket has a thrust force of 1800N. It accelerates
upwards at 2ms-2. Determine the magnitude of air
resistance against the rocket.
∑Fy = 1800 – 850 – Fair
FThrust ∑Fx = 0 85kg
1800N
∑Fy ∑Fy = 85kg x 2ms-2
= 170N
Fg Fair = 1800 – 850 – ∑Fy
850N = 1800 – 850 – 170
Fair ?N = 780N (down)
98. Make your own problems
In a drag race, the drivers apply the accelerator as hard as possible
until they cross the finish line. A parachute deploys to stop them.
Sometimes the parachute fails and they go into the sand.
Parachute-less drag racer:
http://www.youtube.com/watch?v=VxY7zXE0Hjc
¼ mile = 400m y
Fair + Ffriction = 2000N x
1250kg, 250kmh-1 1300kg, 257kmh-1 Mass, top speed
99. Can you fix these incorrect laws of motion?
http://www.veritasium.com/
101. …or does it?
The ISS orbits the Earth at a
constant velocity of around
7.7-7.6 kms-1 (~27 500 kmh-1)
So why are the astronauts OK?
http://science.nationalgeographic.com/science/space/space-exploration/international-space-station-article/
102. Fatal Accelerations
"The NHTSA standard for a sudden impact
acceleration on a human that would cause
severe injury or death is 75 g's for a "50th
percentile male", 65 g's for a "50th percentile
female", and 50 g's for a "50th percentile child".
These figures assume the human is taking the impact on the chest/stomach, the
back, sides or the head. The average value is about 65 g's, so I used that for the
fatal impact acceleration on a human being.”
Reed, Kevin. Re: In a vacuum, can an ant survive a fall that would kill a human? MadSci Network. 4 November 2003.
104. Criterion A: One World
Level Level descriptor
0 The student does not reach a standard described by any of the descriptors below.
1–2 The student states how science is applied and how it may be used to address a specific problem or issue in a local or global context.
The student states the effectiveness of science and its application in solving the problem or issue.
3–4 The student describes how science is applied and how it may be used to address a specific problem or issue in a local or global context.
The student describes the effectiveness of science and its application in solving the problem or issue.
The student describes the implications of the use and application of science interacting with at least one of the following factors: moral,
ethical, social, economic, political, cultural and environmental.
5–6 The student explains how science is applied and how it may be used to address a specific problem or issue in a local or global context.
The student discusses the effectiveness of science and its application in solving the problem or issue.
The student discusses and evaluates the implications of the use and application of science interacting with at least two of the following
factors: moral, ethical, social, economic, political, cultural and environmental.
Students should be able to:
• explain the ways in which science is applied and used to address a specific problem or issue
• discuss the effectiveness of science and its application in solving the problem or issue
• discuss and evaluate the moral, ethical, social, economic, political, cultural and environmental
implications of the use of science and its application in solving specific problems or issues.
Describe: to give a detailed account.
Discuss: to give an account including, where possible, a range of arguments for and against the
relative importance of various factors and comparisons of alternative hypotheses.
Evaluate: to assess the implications and limitations.
Explain : to give a clear account, including causes and reasons or mechanisms.
State: to give a specific name, value or other brief answer without explanation or calculation.
105. Criterion B: Communication in Science
Level Level descriptor
0 The student does not reach a standard described by any of the descriptors below.
1–2 The student uses a limited range of scientific language correctly.
The student communicates scientific information with limited effectiveness.
When appropriate to the task, the student makes little attempt to document sources of information.
3–4 The student uses some scientific language correctly.
The student communicates scientific information with some effectiveness.
When appropriate to the task, the student partially documents sources of information.
5–6 The student uses sufficient scientific language correctly.
The student communicates scientific information effectively.
When appropriate to the task, the student fully documents sources of information correctly.
Students should be able to:
• use scientific language correctly
• use appropriate communication modes and formats
• acknowledge the work of others and the sources of information used by appropriately documenting
them using a recognized referencing system.
• Suitable assessment tasks for criterion B include scientific investigation reports, research essays,
case studies, written responses, debates and multimedia presentations among others.
Students should be able to use different communication modes, including verbal (oral, written) and visual (graphic,
symbolic), as well as appropriate communication formats (laboratory reports, essays, and multimedia presentations)
to effectively communicate scientific ideas, theories, findings and arguments in science
Document: to credit fully all sources of information used by referencing (or citing), following one recognized
referencing system. References should be included in the text and also at the end of the piece of work in a reference
list or bibliography.
106. More free body diagrams…
y ∑Fx = N FN
FN
x Ffriction
870N (scale)
Fapplied Ffriction
Fbuoyant Ftension Fg
Fg
FN Ftension
FN Fapplied
Fg
Ffriction Fapplied FN Fspring Fspring
Fg
Fair
Fg FN F
g Ftension
Fg Fg Fbuoyant
Fapplied Ffriction Felastic Felastic
Fg Fapplied
Fair
Fg
107. What do you think?
Ideas based on
Concept Cartoons:
http://www.conceptcartoons.com
Clipart people from: http://www.clker.com/search/krug/1
108. @IBiologyStephen Please consider a donation to charity via Biology4Good.
Click here for more information about Biology4Good charity donations.
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