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This review covers kinematics, force, projectiles, momentum, and impulse.

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- 1. The slides used in this tutorial are color coded. If you are experiencing difficulty with one aspect of your understanding than another you might find this coding useful. Slides with red backgrounds involve word problems. Slides with tan backgrounds involve matching concepts. Slides with olive backgrounds involve reading data tables. Slides with green backgrounds involve graphing.
- 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. SOLUTION: K U 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. SOLUTION: K U 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. 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. SOLUTION: 49.21 km h Dimensional analysis. 49.21 km/h = 13.67 m/s So the Xf remains 100.07 m 1 h 60 min 1000 m 1 km X 1 min 60 s XX M K H D 0 d c m 1 x x x
- 7. Dr. Fiala jumps in his un- started car. He accelerates at a rate of 4 m/s2 for 8 seconds. How far did Doc travel?
- 8. vi = 0 m/s vf = ti = 0 s Xf = tf = 8 s a = 4 m/s2 Xi = 0 m Xf = 128 m Doc’s final position.
- 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. 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. Time (s) Object #1 Position (m) Object #2 Position (m) Object #3 Position (m) Object #4 Position (m) 1 16 4 2 4 3 48 24 6 4 16 32
- 12. Time (s) Object #1 Position (m) Object #2 Position (m) Object #3 Position (m) Object #4 Position (m) 1 16 4 8 2 2 32 8 16 4 3 48 12 24 6 4 64 16 32 8
- 13. Time (s) Object #1 Velocity (m/s) Object #2 Velocity (m/s) Object #3 Velocity (m/s) Object #4 Velocity (m/s) 1 16 8 2 2 9 4 3 13 8 4 10
- 14. Time (s) Object #1 Velocity (m/s) Object #2 Velocity (m/s) Object #3 Velocity (m/s) Object #4 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. #2 #4
- 16. 0 m/s 0 m/s
- 17. Vy = 0 m/s ty = 5 s Yf = Yi + Vi t + ½ gt2 Vi = 48.5 m/s Vf 2 = Vi 2 + 2g Δy Vi = 48.5 m/s
- 18. Position Graph Can you predict the slope shape and orientation of both the velocity and acceleration graphs? Graph Options
- 19. Position Graph Can you predict the slope shape and orientation of the velocity graph? Graph Options V = 0 m/s V = 0 m/s These two graphs begin with positive velocity that is decreasing over time.
- 20. Position Graph Can you predict the slope shape and orientation of the acceleration graph? Graph Options V = 0 m/s This graph both decreasing positive velocity and increasing negative velocity over time caused by constant negative acceleration (yellow arrow).
- 21. Position Graph Graph Options
- 22. Velocity Graph Position Graph Acceleration Graph Vy = 0 m/s Vy = 0 m/s -V +V g = -9.8 m/s2
- 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. SOLUTION: K U Calculate final velocity without knowing time. Vi = 0 m/s Vf Xi = 0 m tf Xf = 1530 m a = 2.25 m/s2 ti = 0 s Vf= 82.98 m/s
- 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. SOLUTION: K U Solve for time. Vi = 0 m/s tf Xi = 0 m Vf Xf = 1530 m a = 2.25 m/s2 ti = 0 s Vf = 82.98 m/s tf= 36.88 s
- 27. 0 m/s2 0 m/s2 30 m 15 m33.75 m 92 m
- 28. 0 50 100 150 200 0 2 4 6 8 10 12 Position (m) Time (s) 0 20 40 60 80 100 0 2 4 6 8 10 Position (m) Time (s) -3 -2 -1 0 1 2 0 2 4 6 8 10 12 Position (m) Time (s) -20 0 20 40 60 80 100 120 0 2 4 6 8 10 12 Position (m) Time (s)
- 29. 0 20 40 60 80 100 120 140 160 180 0 2 4 6 8 10 12 Position (m) Time (s) 0 2 4 6 8 10 12 14 16 0 2 4 6 8 10 12 14 16 Time (s) Velocity (m/s)
- 30. Motion Map Graph Options Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs based on this motion map?
- 31. Motion Map Graph Options Can you predict the slope shape and orientation of the position graph? These three graphs illustrate an object moving to the left over time. This graph can be eliminated because it illustrates an object that begins moving back to the right over time.
- 32. Motion Map Graph Options Can you predict the slope shape and orientation of the velocity graph now based on this position graph? This is the only graph that illustrates and object moving to the left with changing velocity (curved slope, fast to slow to fast) over time.
- 33. Motion Map Graph Options Can you predict the slope shape and orientation of the acceleration graph now based on this velocity graph? This is the only graph that illustrates negative velocity (moving to the left) the whole time. It is under the influence of constant positive and then constant negative (yellow arrows) acceleration. V = 0 m/s V = 0 m/s This graph can be eliminated because it illustrates an object that is moving slowly at the beginning.
- 34. Position Graph Motion Map Acceleration GraphVelocity Graph
- 35. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs based on this motion map?
- 36. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs?
- 37. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs? Position Graph Velocity Graph
- 38. Graph Options Motion Map Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs? Position Graph Velocity Graph
- 39. Position Graph Motion Map Acceleration GraphVelocity Graph
- 40. Motion Map Graph Options Can you predict the slope shape and orientation of the position, velocity, and acceleration graphs based on this motion map?
- 41. Motion Map Graph Options Can you predict the slope shape and orientation of the velocity, and acceleration graphs now based on this position graph? Position Graph
- 42. Motion Map Graph Options Can you predict the slope shape and orientation velocity graph now based on this position graph? Position Graph These four graphs illustrate positive velocity over time. The ones circled in orange can be eliminated because they indicate changing acceleration which we will not study in this class. The one circled in green can be eliminated because the velocity does not change.
- 43. Motion Map Graph Options Can you predict the slope shape and orientation of the acceleration graph now based on this velocity graph? Position Graph Velocity Graph
- 44. Position Graph Motion Map Acceleration GraphVelocity Graph
- 45. 18 m/s 0 m/s3
- 46. 18 m/s 2 m/s17.5 m/s 0 m/s3 0 m/s3
- 47. 2.25 + 3.25 = 2.25 + 3.25 = 2.25 + 3.25 =
- 48. SOLUTION: Adding vectors. +2.25 + +3.25 = +5.5 +2.25 + -3.25 = -1.00 +2.25 + +3.25 = +3.95
- 49. Determine the mass of a 153.08 N object. 153 N ? kg
- 50. SOLUTION: K U Calculate mass. W= 153.08 N m g = -9.8 m/s2 m = 15.62 kg
- 51. 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.
- 52. SOLUTION: K U Calculate force of friction. m = 15.62 kg Ff a = 0 m/s2 g = -9.8 m/s2 Fa = 6 N Ff = -6 N
- 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. 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?
- 55. Force Diagram Motion Map Position Graph Velocity Graph Acceleration Graph
- 56. 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.
- 57. SOLUTION: K U Find applied force. m = 1315 kg F a = 3.23 m/s2 F = 4247.45 N (kg)(m/s2)
- 58. 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?
- 59. SOLUTION: K U Find applied force. m = 1315 kg Ff F = 4247.45 N a = 3.00 m/s2 Ff = - 302.45 N
- 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. 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?
- 62. Force Diagram Motion Map Position Graph Velocity Graph Acceleration Graph
- 63. 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.
- 64. SOLUTION: K U Find force of support. m = .448 kg Fs g = -9.8 m/s2 Fs = 0 N Fg = -4.39 N
- 65. Force Diagram Motion Map Position (ΔY) Graph Velocity (Vy) Graph Acceleration Graph Can you predict the motion map, and kinematic graphs for this freefalling object?
- 66. Force Diagram Motion Map Velocity (Vy) Graph Acceleration Graph Position (ΔY) Graph
- 67. 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.
- 68. SOLUTION: K U Find force in the horizontal. m = 44.10 kg F a = 1.1 m/s2 F = 48.51 N
- 69. Mass (kg) Force (N) 0 0 42 46.2 47.56 44 48.4 48.65 49.25 49.51 45.45 50
- 70. 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
- 71. 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.
- 72. SOLUTION: K U 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
- 73. 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.
- 74. SOLUTION: K U Find force of support. m = 308.71 kg Fs Fg = -3025.36 N g = -9.8 m/s2 a = .75 m/s2 Fs= 3256.89 N Fg = 3025.36 N Fs = 3256.89 N
- 75. Force Diagram Motion Map Position Graph Velocity (Vy) Graph Acceleration Graph Can you predict the motion map, and kinematic graphs for this elevator?
- 76. Force Diagram Motion Map Position Graph Velocity (Vy) Graph Acceleration Graph
- 77. This same elevator accelerates at a rate of .50 m/s2 as it begins its stop for the second floor. Draw a Force diagram of this situation including the magnitude of Fg and Fs.
- 78. SOLUTION: K U Find force of support. m = 308.71 kg Fs Fg = -3025.36 N g = -9.8 m/s2 a = .-50 m/s2 Fs= 2871.01 N Fg = 3025.36 N Fs = 2871.01 N
- 79. Force Diagram Motion Map Position Graph Velocity (Vy) Graph Acceleration Graph Can you predict the motion map, and kinematic graphs for the ENTIRE TRIP?
- 80. Force Diagram Motion Map Position Graph Velocity (Vy) Graph Acceleration Graph
- 81. 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.
- 82. SOLUTION: K U 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
- 83. These cables will snap if the mass of the trafffic light exceeds 10.1 kg. Does the traffic light exceed 10.1 kg?
- 84. SOLUTION: K U The cable does not break. T1 = 375.4 N m g = -9.8 m/s2 T1y Θ = 7.5° m= 10 kg
- 85. 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.
- 86. SOLUTION: K U Resultant velocity magnitude. Vi = 30 m/s Vf a 2 + b 2 = c 2 Vi = 5 m/s Vf= 30.41 m/s Vy = 30 m/s Vx = 5 m/s
- 87. 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. Vx = 5 m/s Vy = 30 m/s
- 88. SOLUTION: Vy = 30 m/s Vx = 5 m/s tan Θ = x y Θ = 9.46° tan Φ = y x Φ = 80.54° Θ (from +x) = 279.46° Resultant velocity angle measured from positive x.
- 89. 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.
- 90. SOLUTION: K U 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
- 91. 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?
- 92. Force Diagram Motion Map Position Graph Velocity Graph Acceleration Graph
- 93. Force Diagram Motion Map Position Graph Velocity Graph Acceleration Graph Fa = 117.79 N Fg = 147 N Fs = 147 N Ff = 0 N 62.8 m 4 s 4 s 4 s 31.4 m/s 7.85 m/s2
- 94. 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?
- 95. SOLUTION: K U Find time package spends in the air. Yf = -15 m Vf Yi = 0 m tf m = 15 kg g = -9.8 m/s2 tf = 1.75 s Vi = 0 m/s ti = 0 m/s
- 96. Force Diagram Motion Map Acceleration Graph Can you predict what the force diagram, and vertical kinematic graphs for this freefalling object? Velocity (Vy) Graph Position (ΔY) Graph
- 97. Force Diagram Motion Map Position (ΔY) Graph Velocity (Vy) Graph Acceleration Graph
- 98. Time (s) Vertical Position (m) Horizontal Position (m) Vertical Velocity (m/s) Horizontal Velocity (m/s) 0.11 31.41 0.27 -0.36 8.48 -2.65 0.49 15.39 -4.80 0.63 -1.94 -6.17 1.07 -5.61 33.61 1.22 38.32 -11.96 1.35 -8.93 -13.23 1.46 -10.44 45.86 1.75 -
- 99. Time (s) Vertical Position (m) Horizontal Position (m) Vertical Velocity (m/s) Horizontal Velocity (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
- 100. 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.
- 101. SOLUTION: K U Resolve velocity vector into “x” and “y” components just like force or any other vector. V= 27 m/s Viy Θ= 27° Vix g = -9.8 m/s2 Viy = 12.26 m/s Vix = 24.06 m/s Vx = V Vy = V 27°
- 102. How much time will it take for the baseball to reach the same height from which it was thrown?
- 103. SOLUTION: K U Find time in the air. g = -9.8 m/s2 tf Θ= 27° Viy = 12.26 m/s Viy = 24.06 m/s ti = 0 s Yi = 0 m Yf = 0 m tf = 2.5 s
- 104. How far will the baseball travel in 2.5 seconds?
- 105. SOLUTION: K U 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
- 106. What is the maximum height the baseball attained during its flight?
- 107. SOLUTION: K U 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 Xf = 60.15 m
- 108. Vy = 0 m/s Vf = Vi + g Δt tf = 3.06 s Yf = 45.9 m Yf = Yi + Vi t + ½ at2 AREA = ½ Base x Height
- 109. Force Vector Arrows for this Projectile Acceleration 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? Velocity Position
- 110. Force Acceleration Velocity Position
- 111. If it was a .448 kg apple that was thrown into the air at 30 m/s what was the apple’s intial momentum?
- 112. SOLUTION: K U Find momentum of apple. m = .448 kg p Vi = 30 m/s g = -9.8 m/s2 p = 13.44 kgm/s
- 113. 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?
- 114. SOLUTION: K U 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
- 115. 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?
- 116. SOLUTION: K U Find the velocity of two apples stuck together. m1 = .448 kg Vf m2 = .577 kg p g = -9.8 m/s2 Vi1 = 10.4 m/s Vi2 = 0 m/s p = 4.66 kgm/s2 Vf = 4.55 m/s
- 117. Determine the force applied if the rolling apples strike a wall and a come to a stop in .311 seconds.
- 118. SOLUTION: K U Find force needed to stop apples. m1 = .448 kg F 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 = 14.98 N

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