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# 1 2012 ppt semester 1 review and tutorial 2

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### 1 2012 ppt semester 1 review and tutorial 2

1. 1. The slides used in this video are color coded. If you are experiencing difficulty with one aspect of your understanding than another you might find this coding… useful! Slides with Slides with tan red backgrounds backgrounds involve involve word matching Slides with green problems. concepts. backgrounds Slides with olive involve backgrounds graphing. involve reading data tables.
2. 2. Dr. Fiala is traveling on his Harley at a constant 13.67 m/s. What is the distance traveled by Doc in 7.32 seconds?
3. 3. SOLUTION: Find the distance traveled. Vi&f = 13.67 m/s Xf ti = 0 s tf = 7.32 s Xi = 0 m Xf = m a = 0 m/s2
4. 4. SOLUTION: Find the distance traveled. Vi&f = 13.67 m/s Xf ti = 0 s tf = 7.32 s Xi = 0 m Xf= 100.07 m a = 0 m/s2
5. 5. Dr. Fiala notices he is now traveling at a constant 49.21 km/h. What is the distance in meters traveled by Doc in 7.32 seconds?
6. 6. SOLUTION: Dimensional analysis. M K H D 1 x x X 0 d c x X X 49.21 km/h = 13.67 m/s So the Xf remains 100.07 m m
7. 7. Dr. Fiala jumps in his unstarted car. He accelerates at a rate of 4 m/s2 for 8 seconds. How far did Doc travel?
8. 8. Doc’s final position. vi = 0 m/s ti = 0 s tf = 8 s a = 4 m/s2 Xi = 0 m vf = Xf = Xf = 128 m
9. 9. • Displacement Velocity Acceleration Inertia Force Momentum • • • • • The change in the rate or direction of motion. The resistance to a change in an object’s current state of motion. A change in position. A push or a pull that tends to accelerate an object. The movement of an object in a specific direction over time. The product of mass times velocity.
10. 10. Displacement is a change in position. Velocity is the movement of an object in a specific direction over time. Acceleration is the change in the rate or direction of motion of an object. Inertia is the resistance to a change in an object’s current state of motion. Force is a push or a pull that tends to accelerate an object. Momentum is the product of mass times velocity.
11. 11. Time (s) 1 Object #1 Object #2 Object #3 Object #4 Position (m) Position (m) Position (m) Position (m) 16 4 2 3 4 4 48 24 16 32 6
12. 12. Time (s) Object #1 Object #2 Object #3 Object #4 Position (m) Position (m) Position (m) Position (m) 1 16 4 8 2 2 32 8 16 4 3 48 12 24 6 4 64 16 32 8
13. 13. Time (s) 1 Object #1 Object #2 Object #3 Object #4 Velocity (m/s) Velocity (m/s) Velocity (m/s) Velocity (m/s) 16 8 9 2 3 4 13 8 10 2 4
14. 14. Time (s) Object #1 Object #2 Object #3 Object #4 Velocity (m/s) Velocity (m/s) Velocity (m/s) Velocity (m/s) 1 16 4 8 2 2 14.5 6 9 4 3 13 8 10 6 4 11.5 10 11 8
15. 15. #1 #3 #2 #4
16. 16. 2m /s m/s 1 0 m/s 0 m/s
17. 17. Vy = 0 m/s Yf = Yi + Vi t + ½ gt2 Vi = 48.5 m/s ty = 5 s Vf2 = Vi2 + 2g Δy Vi = 48.5 m/s
18. 18. Graph Options Position Graph Can you predict the slope shape and orientation of both the velocity and acceleration graphs?
19. 19. Graph Options Position Graph Can you predict the slope shape and orientation of both the velocity and acceleration graphs? V = 0 m/s
20. 20. Graph Options Position Graph Can you predict the slope shape and orientation of both the velocity and acceleration graphs? V = 0 m/s
21. 21. Graph Options Position Graph Can you predict the slope shape and orientation of both the velocity and acceleration graphs?
22. 22. Position Graph Vy = 0 m/s Velocity Graph Acceleration Graph +V Vy = 0 m/s -V g = -9.8 m/s2
23. 23. If Dr. Fiala starts from a full stop and accelerates at 2.25 m/s2, how incredibly fast will he be traveling when he has traveled 1530 meters?
24. 24. SOLUTION: Calculate final velocity without knowing time. Vi = 0 m/s Xi = 0 m Xf = 1530 m a = 2.25 m/s2 ti = 0 s Vf tf Vf= 82.98 m/s
25. 25. If Dr. Fiala starts from a full stop and accelerates at 2.25 m/s2, how long will it take him to drive 1530 meters?
26. 26. SOLUTION: Solve for time. Vi = 0 m/s Xi = 0 m Xf = 1530 m a = 2.25 m/s2 ti = 0 s Vf = 82.98 m/s tf Vf t = 36.88 s
27. 27. 0 m/s2 2.3 3 m 2 /s 2 .8 m/s 75 0 m/s2 30 m 33.75 m 15 m 92 m
28. 28. Velocity (m/s) 16 14 12 10 8 6 4 2 0 0 2 4 6 8 10 12 14 16 Time (s)
29. 29. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
30. 30. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
31. 31. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
32. 32. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
33. 33. Position Graph Motion Map Velocity Graph Acceleration Graph
34. 34. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
35. 35. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
36. 36. Graph Options Motion Map Velocity Graph Position Graph Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
37. 37. Graph Options Motion Map Velocity Graph Position Graph Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
38. 38. Position Graph Motion Map Velocity Graph Acceleration Graph
39. 39. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
40. 40. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs? Position Graph
41. 41. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs? Position Graph
42. 42. Graph Options Motion Map Velocity Graph Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs? Position Graph
43. 43. Position Graph Motion Map Velocity Graph Acceleration Graph
44. 44. 0 m/s3 18 m/s
45. 45. 0 m/s3 -1 18 m/s m/ s3 17.5 m/s 0 m/s3 2 m/s
46. 46. 2.25 + 3.25 = 2.25 + 3.25 = 2.25 + 3.25 =
47. 47. SOLUTION: Adding vectors. + 2.25 + 3.25 = 5.5 + 2.25 + 3.25 = 1.00 + 2.25 + 3.25 = 3.95 + - + + - +
48. 48. ? kg 153 N Determine the mass of a 153.08 N object.
49. 49. SOLUTION: Calculate mass. W = 153.08 N g = -9.8 m/s2 m m = 15.62 kg
50. 50. Determine the force of friction on a 15.62 kg object traveling at a constant horizontal velocity of 3.62 m/s while experiencing an applied force of 6 N.
51. 51. SOLUTION: Calculate force of friction. m = 51.62 kg a = 0 m/s2 g = -9.8 m/s2 Fa = 6 N Ff Ff = -6 N
52. 52. Force Diagrams Motion Map Match the force diagram to the motion map. Can you also predict the slope shape and orientation of the position, velocity, and acceleration graphs?
53. 53. Force Diagrams Motion Map Match the force diagram to the motion map. Can you also predict the slope shape and orientation of the position, velocity, and acceleration graphs?
54. 54. Force Diagram Motion Map Position Graph Velocity Graph Acceleration Graph
55. 55. Determine the force needed to accelerate Dr. Fiala’s car and its occupants at a rate of 3.23 m/s2 if the total mass of car and occupants is 1315 kg and there is no friction force.
56. 56. SOLUTION: Find applied force. m = 1315 kg a = 3.23 m/s2 F F = 4247.45 N (kg)(m/s ) 2
57. 57. This time, when we apply that 4247.45 N force to Dr. Fiala’s car and its occupants, the resulting acceleration is actually lower. It registers at a rate of only 3.00 m/s2. What is the magnitude for the force of friction causing the acceleration to be decreased?
58. 58. SOLUTION: Find applied force. m = 1315 kg F = 4247.45 N a = 3.00 m/s2 Ff Ff = - 302.45 N
59. 59. Force Diagrams Motion Map Match the force diagram to the motion map. Can you also predict the slope shape and orientation of the position, velocity, and acceleration graphs?
60. 60. Force Diagrams Motion Map Match the force diagram to the motion map. Can you also predict the slope shape and orientation of the position, velocity, and acceleration graphs?
61. 61. Force Diagram Motion Map Position Graph Velocity Graph Acceleration Graph
62. 62. When an object is freefalling it is weightless. Prove mathematically that a .448 kg apple is weightless during its freefall from a tree. Draw a force diagram of the apple during its fall from the tree.
63. 63. SOLUTION: Find force of support. m = .448 kg g = -9.8 m/s2 Fs Fs = 0 N Fg = -4.39 N
64. 64. Motion Map Force Diagram Position (ΔY) Graph Velocity (Vy) Graph Can you predict the motion map, and kinematic graphs for this freefalling object? Acceleration Graph
65. 65. Motion Map Force Diagram Position (ΔY) Graph Velocity (Vy) Graph Acceleration Graph
66. 66. Assuming a perfectly frictionless surface, ideal for launching students in a game of faculty bowling, Dr. Fiala uses a brand new gizmo that automatically applies a force that results in an acceleration of 1.1 m/s2. Experimentation resulted in a student with a mass of 44.10 kg, accelerating at 1.1 m/s2. Find the force generated by the gizmo for that student.
67. 67. SOLUTION: Find force in the horizontal. m = 44.10 kg a = 1.1 m/s2 F F = 48.51 N
68. 68. Mass (kg) Force (N) 0 0 42 46.2 47.56 44 48.4 48.65 49.25 49.51 45.45 50
69. 69. Mass (kg) Force (N) 0 0 42 46.2 43.25 47.56 44 48.4 44.23 48.65 44.77 49.25 45.01 49.51 45.45 50
70. 70. All of the students from the previous problem (combined mass) step into an elevator at the same time. Draw a force diagram of this situation including the magnitude of Fg and Fs.
71. 71. SOLUTION: Find force of gravity and force of support. m1 = 42 kg Fg m2 = 43.25 kg Fs m3 = 44 kg m4 = 44.23 kg m5 = 44.77 kg m6 = 45.01 kg m2 = 45.45 kg Fg= -3025.36 N g = -9.8 m/s2 Fs= 3025.36 N
72. 72. This same elevator accelerates at a rate of .75 m/s2 towards the second floor. Draw a force diagram of this situation including the magnitude of Fg and Fs.
73. 73. SOLUTION: Find force of support. m = 308.71 kg Fg = -3025.36 N g = -9.8 m/s2 a = .75 m/s2 Fs Fs = 3256.89 N Fs= N Fg = 3025.36 N 3256.89
74. 74. Motion Map Force Diagram Position Graph Velocity (Vy) Graph Can you predict the motion map, and kinematic graphs for this elevator? Acceleration Graph
75. 75. Motion Map Force Diagram Position Graph Velocity (Vy) Graph Acceleration Graph
76. 76. This same elevator accelerates 2 at a rate of .50 m/s as it begins its stop for the second floor. Draw a Force diagram of this situation including the magnitude of Fg and Fs.
77. 77. SOLUTION: Find force of support. m = 308.71 kg Fg = -3025.36 N g = -9.8 m/s2 a = .-50 m/s2 N Fs = 2871.01 N Fg = 3025.36 N Fs Fs= 2871.01
78. 78. Motion Map Force Diagram Position Graph Velocity (Vy) Graph Can you predict the motion map, and kinematic graphs for the ENTIRE TRIP? Acceleration Graph
79. 79. Motion Map Force Diagram Position Graph Velocity (Vy) Graph Acceleration Graph
80. 80. According to Newton’s 3rd law, an action force causes an equal on opposite reaction force. It is no wonder a truck windshield squashes a bug and not vice versa. A 2000 kg truck and a .0002 kg bug hit with a 50 N force. Take a closer look at why the truck wins the collision by calculating the acceleration exerienced by the bug and by the truck.
81. 81. SOLUTION: Why the bug doesn’t survive. mt = 2000 kg at mb = .0002 kg ab g = -9.8 m/s2 F = -50 N at = -.025 m/s2 ab = -250,000 m/s2
82. 82. .40 N 375 These cables will snap if the mass of the trafffic light exceeds 10.1 kg. Does the traffic light exceed 10.1 kg?
83. 83. SOLUTION: The cable does not break. T1 = 375.4 N g = -9.8 m/s2 Θ = 7.5° m T1y m= 10 kg
84. 84. Dr. Fiala attempts to walk due east at 5 m/s at the same time as a 30 m/s cold, winter wind is blowing due south. What is the magnitude of Dr. Fiala’s velocity.
85. 85. SOLUTION: Resultant velocity magnitude. Vi = 30 m/s Vi = 5 m/s Vx = 5 m/s Vy = 30 m/s Vf Vf= 30.41 m/s 2 2 a +b =c 2
86. 86. Vx = 5 m/s V =3 0.4 1m /s Vy = 30 m/s If Dr. Fiala continues his velocity and the wind continues to blow steadily, at what angle, as measured from positive “X”, is Dr. Fiala’s velocity.
87. 87. SOLUTION: Resultant velocity angle measured from positive x. tan Θ = x y Θ = 9.46° Vx = 5 m/s Vy = 30 m/s tan Φ = y x Φ = 80.54° Θ (from +x) = 279.46°
88. 88. Because of this wind, a 15 kg package is blown from Dr. Fiala’s arms and onto the ground. The 15 kg package reaches a velocity of 30.41 m/s in a time of 4 seconds. Find the force acting on the box horizontally if there is no friction.
89. 89. SOLUTION: Find applied force. Yf = -15 m a Yi = 0 m F m = 15 kg g = -9.8 m/s2 Vi = 0 m/s Vf = 31.41 m/s ti = 0 s a = 7.60 m/s2 ti = 4 s F = 114 N
90. 90. Force Diagrams Motion Map Match the force diagram to the motion map. Can you also predict the slope shape and orientation of the position, velocity, and acceleration graphs?
91. 91. Force Diagram Motion Map Position Graph Velocity Graph Acceleration Graph
92. 92. Force Ff = 0 N Diagram Motion Map Fs = 147 N Fa = 117.79 N Fg = 147 N Position Graph Velocity Graph 31.4 m/s 62.8 m 4s Acceleration Graph 7.85 m/s2 4s 4s
93. 93. If the package is blow horizontally at 30.41 m/s off a ledge onto a parking lot that is 15 meters below how much time will it spend in the air before striking the ground? What does the motion map look like?
94. 94. SOLUTION: Find time package spends in the air. Yf = -15 m Yi = 0 m m = 15 kg g = -9.8 m/s2 Vi = 0 m/s ti = 0 m/s Vf tf tf = 1.75 s
95. 95. Force Diagram Motion Map Position (ΔY) Graph Velocity (Vy) Graph Can you predict what the force diagram, and vertical kinematic graphs for this freefalling object? Acceleration Graph
96. 96. Force Diagram Motion Map Position (ΔY) Graph Velocity (Vy) Graph Acceleration Graph
97. 97. Time (s) Vertical Position (m) Horizontal Position (m) Vertical Velocity Horizontal Velocity (m/s) (m/s) 0.11 0.27 31.41 -0.36 0.63 -5.61 -4.80 -1.94 1.07 -2.65 15.39 0.49 8.48 1.22 -6.17 33.61 38.32 1.35 -8.93 1.46 -10.44 1.75 - -11.96 -13.23 45.86
98. 98. Time (s) Vertical Position (m) Horizontal Position (m) Vertical Velocity Horizontal Velocity (m/s) (m/s) 0.11 -0.06 3.46 -1.08 31.41 0.27 -0.36 8.48 -2.65 31.41 0.49 -1.18 15.39 -4.80 31.41 0.63 -1.94 19.79 -6.17 31.41 1.07 -5.61 33.61 -10.49 31.41 1.22 -7.29 38.32 -11.96 31.41 1.35 -8.93 42.40 -13.23 31.41 1.46 -10.44 45.86 -14.31 31.41 1.75 -15.01 54.97 -17.15 31.41
99. 99. Dr. Fiala throws a baseball in the air with an initial velocity of 27 m/s at an angle of 27° to the horizon. Create a velocity vector diagram and show, by parallelogram method, the “X” and “Y” components of the baseball’s velocity.
100. 100. SOLUTION: Resolve velocity vector into “x” and “y” components just like force or any other vector. V = 27 m/s Θ= 27° g = -9.8 m/s2 m/s 27 27° Viy Vix Viy = 12.26 m/s Vix = 24.06 m/s Vx = V Vy = V
101. 101. How much time will it take for the baseball to reach the same height from which it was thrown?
102. 102. SOLUTION: Find time in the air. g = -9.8 m/s2 Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m tf tf = 2.5 s
103. 103. How far will the baseball travel in 2.5 seconds?
104. 104. SOLUTION: Find range. g = -9.8 m/s2 Xf Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m Xi = 0 m tf = 2.5 s Xf = 60.15 m
105. 105. What is the maximum height the baseball attained during its flight?
106. 106. SOLUTION: Find Δy. g = -9.8 m/s2 Δy Θ= 27° Viy = 12.26 m/s Vix = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m Xi = 0 m Δy = 7.67 m tf = 2.5 s
107. 107. Vy = 0 m/s a= 9.8 m Vf = Vi + 2g Δt tf = 3.06 s - /s 2 Yf = Yi + Vi t + ½ at2 Yf = 45.9 m AREA = ½ Base x Height
108. 108. Force Vector Arrows for this Projectile Position  Velocity Using these vector arrows can you predict what the position, force, velocity and acceleration vector arrows would look like for this projectile at the start and at the top? Acceleration
109. 109. Force Position Velocity Acceleration        
110. 110. If it was a .448 kg apple that was thrown into the air at 30 m/s what was the apple’s intial momentum?
111. 111. SOLUTION: Find momentum of apple. m = .448 kg Vi = 30 m/s g = -9.8 m/s2 p p = 13.44 kgm/s
112. 112. What constant force is needed to get a change in the apple’s momentum from 13.44 kgm/s to 0 In 3.06 seconds?
113. 113. SOLUTION: Find force necessary to change momentum. m = .448 kg F Vi = 30 m/s g = -9.8 m/s2 ti = 0 s tf = 3.06 s Δp = -13.44 kgm/s F = 4.39 N
114. 114. After falling to the ground the .448 kg apple rolled at a constant 10.4 m/s where collided with a stationary .577 kg apple. If the two apples stuck together, at what velocity would they roll?
115. 115. SOLUTION: Find the velocity of two apples stuck together. m1 = .448 kg m2 = .577 kg g = -9.8 m/s2 Vi1 = 10.4 m/s Vi2 = 0 m/s Vf p p = 4.66 kgm/s2 Vf = 4.55 m/s
116. 116. Determine the force applied if the rolling apples strike a wall and a come to a stop in .311 seconds.
117. 117. SOLUTION: Find force needed to stop apples. m1 = .448 kg m2 = .577 kg ti = 0 s tf = .311 s g = -9.8 m/s2 Vi1 = 4.55 m/s Vi2 = 0 m/s p = 4.66 kgm/s F F = 14.98 N