2. Physics 1 SSS 1 Kinematics Dynamics Work/Energy/Power

3. Kinematics Constant Velocity Constant Acceleration Projectile Motion

7. 1D Motion Graphs

8. Consider the position vs time graph for objects A and B below. Describe how the motion of object A is different from that of object B. -Car A is traveling with a constant velocity -At t=0 car B is traveling faster than car A but slowing down. -At t=5sec car A catches up to car B

11. Answer: C = D > A > B. The displacements of C and D during the first three seconds was 6 m, next was A at 4 m and least is B at 2 m.

12. A bulldozer moves according to the velocity-time graph above. The positive direction is East. When is the bulldozer speeding up? What can the area under the curve tell you? What can the slope tell you? Velocity Time Graph Review

13. Independence of x and y Projectile Motion

15. Angled Launch

16. Screen clipping taken: 9/22/2014, 10:01 PM Answer: (1) A; (2) C; (3) I; and (4) B.

17. Answer: D > C = B > A.

18. Answer: A = E > C = D =F > B.

19. Dynamics Forces Newton’s Laws of Motion

21. Forces

22. Newton’s 1 st Law “Inertia”

23. Newton’s 2 nd Law F=ma

25. Frictional Forces

33. Static Equilibrium

34. Tension

35. Stationary

36. Accelerating

37. Work, Energy, Power, And The Conservation of Energy

38. WORK In order for work to be done, three things are necessary: There must be an applied force. The force must act through a certain distance, called the displacement. The force must have a component along the displacement.

39. Work is a scalar quantity equal to the product of the magnitudes of the displacement and the component of the force in the direction of the displacement. W = F. x or W = F cos  x UNITS: N.m this unit is called a Joule (J)

40. As long as this person does not lift or lower the bag of groceries, he is doing no work on it. The force he exerts has no component in the direction of motion. Work done by forces that oppose the direction of motion, such as friction, will be negative.

41. Centripetal forces do no work, as they are always perpendicular to the direction of motion.

42. If the force acting on an object varies in magnitude and/or direction during the object’s displacement, graphical analysis can be used to determine the work done. The Force is plotted on the y-axis and the distance through which the object moves is plotted on the x-axis. The work done is represented by the area under the curve.

44. In mechanics we are concerned with two kinds of energy: KINETIC ENERGY: K, energy possessed by a body by virtue of its motion. Units: Joules (J) POTENTIAL ENERGY: PE, energy possessed by a system by virtue of position or condition. PE = m g h Units: Joules (J)

45. WORK-ENERGY PRINCIPLE: The work of a resultant external force on a body is equal to the change in kinetic energy of the body. W =  KUnits: Joules (J)

46. W =  PE

47. 5.2 What average force F is necessary to stop a 16 g bullet traveling at 260 m/s as it penetrates into wood at a distance of 12 cm? v f = 0 m/s m = 0.016 kg v o = 260 m/s x = 0.12 m W = ΔK = - 4506.7 N WE

48. Pendulums and Energy Conservation Energy goes back and forth between K and U. At highest point, all energy is U. As it drops, U goes to K. At the bottom, energy is all K.

49. h Pendulum Energy ½mv max 2 = mgh For minimum and maximum points of swing K 1 + U 1 = K 2 + U 2 For any points 1 and 2.

50. Springs and Energy Conservation Transforms energy back and forth between K and U. When fully stretched or extended, all energy is U. When passing through equilibrium, all its energy is K. At other points in its cycle, the energy is a mixture of U and K.

51. Spring Energy mm -x m x 0 ½kx max 2 = ½mv max 2 For maximum and minimum displacements from equilibrium K 1 + U 1 = K 2 + U 2 = E For any two points 1 and 2 All U All K

52. Law of Conservation of Energy (with dissipative forces) E = U + K + E int = C  U +  K +  E int = 0 E int is thermal energy. Mechanical energy may be converted to and from heat.

53. Work done by non-conservative forces W net = W c + W nc  U = -W c  K = W net  E int = W nc  K = -  U +  E int  E int =  U +  K

54. CONSERVATIVE AND NON-CONSERVATIVE FORCES The work done by a conservative force depends only on the initial and final position of the object acted upon. An example of a conservative force is gravity. The work done equals the change in potential energy and depends only on the initial and final positions above the ground and NOT on the path taken.

55. Friction is a non-conservative force and the work done in moving an object against a non-conservative force depends on the path. For example, the work done in sliding a box of books against friction from one end of a room to the other depends on the path taken.

56. For mechanical systems involving conservative forces, the total mechanical energy equals the sum of the kinetic and potential energies of the objects that make up the system and is always conserved.

58. A roller-coaster car moving without friction illustrates the conservation of mechanical energy.

60. In real life applications, some of the mechanical energy is lost due to friction. The work due to non-conservative forces is given by: W NC = ΔPE + ΔK or W NC = E f - E o

62. 5.3 A ballistic pendulum apparatus has a 40-g ball that is caught by a 500-g suspended mass. After impact, the two masses rise a vertical distance of 45 mm. Find the velocity of the combined masses just after impact. m 1 = 0.04 kg m 2 = 0.500 kg h = 0.045 m K 0 = PE f COE

63. PE f = (m 1 +m 2 ) gh f = (0.04+0.500)(9.8)(0.045) = 0.24 J K 0 = PE f = 0.24 J K 0 = PE f m 1 = 0.04 kg m 2 = 0.500 kg h = 0.045 m = 0.94 m/s

64. 5.4 The tallest and fastest roller coaster in the world is the Steel Dragon in Japan. The ride includes a vertical drop of 93.5 m. The coaster has a speed of 3 m/s at the top of the drop. a. Neglect friction and find the speed of the riders at the bottom.? v A = 3 m/s h A = 93.5 m h B = 0 m At point A: PE A + K A At point B: K B PE A + K A = K B A B = 42.9 m/s (about 96 mi/h) COE

65. b. Find the work done by non-conservative forces on a 55 kg rider during the descent if the actual velocity at the bottom is 41 m/s. W NC = E f - E 0 = K B - (PE A + K A ) v A = 3 m/s v B = 41 m/s h A = 93.5 m h B = 0 m m = 55 kg = - 4416.5 J

66. 5.5 A 20-kg sled rests at the top of a 30˚ slope 80 m in length. If μ k = 0.2, what is the velocity at the bottom of the incline? m = 20 kg θ = 30° r = 80 m μ k = 0.2 W NC = E f - E o = K f - PE 0 COE

67. m = 20 kg θ = 30° x = 80 m μ k = 0.2 x h h = x sin θ h = 80 sin 30° = 40 m PE 0 = mgh 0 = 20(9.8)(40) = 7840 J F f = μ k F N = μ k F gy = μ k F g cos30° = (0.2)(20)(9.8)cos30° = 34 N W NC = - F f r = - 34 (80) = - 2720 J W NC = K f - PE 0 K f = PE 0 + W NC = 7840 - 2720 = 5120 J

68. = 22.6 m/s

69. ELASTIC FORCE The force F s applied to a spring to stretch it or to compress it an amount x is directly proportional to x. F s = - k x Units: Newtons (N) Where k is a constant called the spring constant and is a measure of the stiffness of the particular spring. The spring itself exerts a force in the opposite direction:

70. This force is sometimes called restoring force because the spring exerts its force in the direction opposite to the displacement. This equation is known as the spring equation or Hooke’s Law.

74. The elastic potential energy is given by: PE s = ½ kx 2 Units: Joules (J)

75. 5.7 A dart of mass 0.100 kg is pressed against the spring of a toy dart gun. The spring (k = 250 N/m) is compressed 6.0 cm and released. If the dart detaches from the spring when the spring reaches its normal length, what speed does the dart acquire? m = 0.1 kg k = 250 N/m x = 0.06 m PE s = K ½ kx 2 = ½ mv 2 = 3 m/s

77. L L h

78. Power The rate of which work is done. When we run upstairs, t is small so P is big. When we walk upstairs, t is large so P is small.

79. Power in Equation Form  P = W/t  work/time  P = F V  (force )(velocity)

80. Unit of Power SI unit for Power is the Watt. 1 Watt = 1 Joule/s Named after the Scottish engineer James Watt (1776-1819) who perfected the steam engine.

81. POWER Is the rate at which work is performed.P = work/time The difference between walking and running up these stairs is power. The change in gravitational potential energy is the same. UNITS: = Watt

82. 5.6 A 1100-kg car starting from rest, accelerates for 5.0 s. The magnitude of the acceleration is 4.6 m/s 2. What power must the motor produce to cause this acceleration? m = 1100 kg v o = 0 m/s t = 5 s a = 4.6 m/s 2 F = ma = (1100)(4.6) = 5060 N v f = v o + at = 0 + 4.6 (5) = 23 m/sThe average velocity is: 23/2 = 11.5 m/s P = Fv = (5060)(11.5) = 5.82x10 4 W

83. How We Buy Energy… The kilowatt-hour is a commonly used unit by the electrical power company. Power companies charge you by the kilowatt- hour (kWh), but this not power, it is really energy consumed. 1 kW = 1000 W 1 h = 3600 s 1 kWh = 1000J/s 3600s = 3.6 x 10 6 J