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- 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/3b59khttp://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 particlesgive greater air resistance, so a slower terminal velocity (not enough to break the speed of sound).http://en.wikipedia.org/wiki/Atmosphere_of_Earth
- 4. Class arm-wrestling competition Pool A Pool B Pool C Pool Dhttp://www.free-clipart-pictures.net/wrestling_clipart.html
- 5. Beeeeeeeeefcaaaaaaaaaaaaake! Explain what is happening in terms of forces. Can you describe the forces in terms of vectors? MrThttp://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 TasksA: One World “Acceleration Kills”B: Communication in Science “Acceleration Kills”C: Knowledge & Understanding Unit test, Quia Quizzes, class whiteboardingD: 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 30kghttp://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 differenttypes 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 theCalculate the weights of these masses (on Earth): assumption that1. 50kg2. 500g 1kg = 10N3. 7g4. 1,000kgCalculate the masses of these weights (on Earth):1. 100N2. 8.7N3. 0.034N4. 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 flightGolfer 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 forcesGravity:- Masses with different Newton-metersBouyancy 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 setupNormal 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 equallybecause they havethe same volumeWater pushes themall equally becausethey have the same density Water pushes them all equally because theyhave the same buoyancy
- 28. What would happen tothe readings if the water were extremely salty? What would happen to the air and waterreadings if the aluminiumblock 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 DiagramsFree body diagrams are vector diagrams. They show the direction and magnitude of theforces acting on an object. Forces are drawn as arrows acting from the centre of a boxrepresenting 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-attackAll 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-attackAll free clipart images from: http://www.clker.com
- 38. Free Body Diagrams Free body art-attackAll free clipart images from: http://www.clker.com
- 39. Free Body Diagrams Free body art-attackAll 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/Newtons-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 blockhttp://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 blockhttp://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 accelerationhttp://phet.colorado.edu/en/simulation/forces-and-motionSet up the investigation like this:
- 51. Free body diagrams, force and accelerationhttp://phet.colorado.edu/en/simulation/forces-and-motionDraw 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 floorIdentify: Static Friction Crate ____ N Fridge ____ N Kinetic Friction Crate ____ N Fridge ____ NExplain what happens to the velocity and acceleration of the objectswhen the following are applied:- A constant force- A force which is let go 200kg fridge, wood floorCompare changes in velocity and acceleration between the 100kgcrate 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 accelerationhttp://phet.colorado.edu/en/simulation/forces-and-motionDraw 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 DiagramsNet force is a the sum of the forces acting on an object. It is a vector – direction andmagnitude 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 DiagramsNet force is a the sum of the forces acting on an object. It is a vector – direction andmagnitude 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 DiagramsNet force is a the sum of the forces acting on an object. It is a vector – direction andmagnitude 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 DiagramsNet force is a the sum of the forces acting on an object. It is a vector – direction andmagnitude 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 200kghttp://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 accelerationhttp://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)
- 61. Huh? Acceleration due to gravity is 9.8ms-2
- 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-2Practice 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-1Practice: 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 aPractice: 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 8sPractice: 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-1Practice: 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 8sPractice: 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 xExample 1:A 2.50kg book sits on a desk. Determine the magnitude of theforce 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 xExample 1:A 2.50kg book sits on a desk. Determine the magnitude of theforce 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 xExample 1:A 2.50kg book sits on a desk. Determine the magnitude of theforce 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 xExample 1:A 2.50kg book sits on a desk. Determine the magnitude of theforce 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 xExample 2:A 100kg fridge is pushed to the right along a floor with anacceleration 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 xExample 2:A 100kg fridge is pushed to the right along a floor with anacceleration 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 xExample 2:A 100kg fridge is pushed to the right along a floor with anacceleration 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 xExample 2:A 100kg fridge is pushed to the right along a floor with anacceleration 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 xExample 2:A 100kg fridge is pushed to the right along a floor with anacceleration 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 xExample 2:A 100kg fridge is pushed to the right along a floor with anacceleration 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 xExample 3:An 85kg rocket has a thrust force of 1800N. It acceleratesupwards at 2ms-2. Determine the magnitude of airresistance 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 xExample 3:An 85kg rocket has a thrust force of 1800N. It acceleratesupwards at 2ms-2. Determine the magnitude of airresistance 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 xExample 3:An 85kg rocket has a thrust force of 1800N. It acceleratesupwards at 2ms-2. Determine the magnitude of airresistance 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 xExample 3:An 85kg rocket has a thrust force of 1800N. It acceleratesupwards at 2ms-2. Determine the magnitude of airresistance 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 xExample 3:An 85kg rocket has a thrust force of 1800N. It acceleratesupwards at 2ms-2. Determine the magnitude of airresistance 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 xExample 3:An 85kg rocket has a thrust force of 1800N. It acceleratesupwards at 2ms-2. Determine the magnitude of airresistance 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 problemsIn a drag race, the drivers apply the accelerator as hard as possibleuntil 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/
- 100. Speed kills…http://www.bikeroute.com/Recumbents/News/Archives/000054.html http://www.hurst-pc.org.uk/news/vas.html
- 101. …or does it?The ISS orbits the Earth at aconstant velocity of around7.7-7.6 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 gs for a "50th percentile male", 65 gs for a "50th percentile female", and 50 gs 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 gs, 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.
- 103. Follow the GoogleDoc link from MrT
- 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.comClipart 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. This is a Creative Commons presentation. It may be linked and embedded but not sold or re-hosted.

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