1. CH 6

2. Work - Work is calculated by multiplying the force applied by the distance the object moves while the force is being applied. W = Fs

3. Ex. 4 - A flatbed truck accelerating at a = +1.5 m/s 2 is carrying a 120 kg crate. The crate does not slip as the truck moves s = 65 m. What is the total work done on the crate by all the forces acting on it?

4. The Work-Energy Theorem. W = KE f - KE 0 = 1/2 mv f 2 - 1/2 mv 0 2

5. Ex. 6 - A 54 kg skier is coasting down a 25° slope. A kinetic frictional force of f k = 70 N opposes her motion. Her initial speed is v 0 = 3.6 m/s. Ignoring air resistance, determine the speed v f at a displacement 57 m downhill.

6. Gravitational Potential Energy is energy due to the distance an object is able to fall. PE = mgh PE is also measured in joules.

7. The work done by the gravitational force on an object does not depend on the path taken by the object. This makes gravitational force a conservative force.

8. Conservation of Mechanical Energy The total mechanical energy (E = KE + PE) of an object remains constant as the object moves, provided that the net work done by external nonconservative forces is zero.

9. Ex. 9 - A motorcyclist drives horizontally off a cliff to leap across a canyon. When he drives off, he has a speed of 38.0 m/s. Find the speed with which the cycle strikes the ground on the other side if he is 35 m below his starting point when he strikes the ground.

11. Ex. 10 - A 6.00-m rope is tied to a tree limb and used as a swing. A person starts from rest with the rope held in a horizontal orientation. Determine how fast the person is moving at the lowest point on the circular arc of the swing.

12. Power is the rate at which work is done. P = W/t The unit is the joule/second, which is called the watt. 1 horsepower = 746 watts

13. W/t = Fs/t W/t is power, and s/t is average speed v, so P = Fv

14. Ex. 15 - A 1.10 x 10 3 kg car, starting from rest, accelerates for 5.00 s. The magnitude of the acceleration is a = 4.60 m/s 2. Determine the average power generated by the net force that accelerates the vehicle.

15. Energy of all types can be converted from one form to another. The Principle of Conservation of Energy: Energy can be neither created nor destroyed, but can only be converted from one form to another.

16. CH 7

17. The impulse of a force is the product of the average force and the time interval during which the force acts. Impulse = F ave Δt The unit is the newtonsecond (Ns)

18. The linear momentum p of an object is the product of the object’s mass m and the velocity v. p = mv The unit is the kilogrammeter/second (kgm/s)

19. The impulse-momentum theorem, the impulse is equal to the change in momentum. F Δt = mv f - mv 0 impulse finalinitial momentum momentum

20. Ex. 1 - A baseball (m = 0.14 kg) has an initial velocity of v 0 = -38 m/s as it approaches a bat. The ball leaves the bat with a velocity of v f = +58 m/s. (a) Determine the impulse applied to the ball by the bat. (b) If the time of contact is Δt = 1.6 x 10 -3 s, find the average force exerted on the ball by the bat.

21. This is the principle of conservation of linear momentum. The total linear momentum of an isolated system remains constant. (mv f1 + mv f2 ) = (mv 01 + mv 02 ) or: P f = P 0

22. Ex. 5 - A freight train is being assembled in a switching yard. Car 1 has a mass of m 1 = 65 x10 3 kg and moves with a velocity of v 01 = +0.80 m/s. Car 2, with a mass of m 2 = 92 x 10 3 kg and a velocity of v 02 = +1.2 m/s, overtakes car 1 and couples to it. Find the common velocity v f of the two cars after they become coupled.

23. Ex. 7 - When a gun fires a blank, is the recoil greater than, the same as, or less than when the gun fires a standard bullet?

24. An elastic collision is one in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision. K.E. is conserved in the collision.

25. An inelastic collision is one in which the total kinetic energy of the system is not the same before and after the collision; if the objects stick together after colliding, the collision is said to be completely inelastic. Kinetic energy is not conserved. The coupling boxcars is an example of an inelastic collision.

26. Ex. 9 - A ballistic pendulum consists of a block of wood (mass m 2 = 2.50 kg) suspended by a wire. A bullet (mass m 1 = 0.0100 kg) is fired with a speed v 01. Just after the bullet collides with it, the block (now containing the bullet) has a speed v f and then swings to a maximum height of 0.650 m above the initial position. Find the speed of the bullet.

27. In an isolated system momentum is conserved, P f = P 0. Remember that momentum is a vector quantity; when a collision in two dimensions occurs the x and y components are conserved separately.

28. CH 8

29. The angle through which a rigid body rotates about a fixed axis is called the angular displacement.

31. Angular velocity is the angular displacement divided by elapsed time.  =  /  t

32. The unit is radians per second. rad/s

33. Example 3. A gymnast on a high bar swings through two revolutions in time of 1.90 s. Find the average angular velocity (in rad/s) of the gymnast.

34. Angular acceleration  is the rate of change of angular velocity.  =  /  t

35. Example 4. A jet engine’s turbine fan blades are rotating with an angular velocity of -110 rad/s. As the plane takes off, the angular velocity of the blades reaches -330 rad/s in a time of 14 s. Find the angular acceleration.

36. The equations for rotational dynamics are similar to those for linear motion.  =  0 +  t

37.  =  0 t + ½  t 2  2 =  0 2 + 2  

38. The tangential velocity v T is the speed in m/s around the arc. The magnitude is called the tangential speed. v T = r 

39. The centripetal acceleration formula is a c = v T 2 /r. This can be expressed in terms of angular speed since v T = r .

40. a c = v T 2 /r becomes a c = (r  ) 2 /r a c = r  2 (  is rad/s)

41. When objects roll there is a relationship between the angular speed of the object and the linear speed at which the object moves forward.

42. Linear speed is equal to tangential speed. v = r  It follows that linear acceleration is equal to tangential acceleration. a = r 

43. Right-Hand Rule. When the fingers of your right hand encircle the axis of rotation, and your fingers point in the direction of the rotation, your extended thumb points in the direction of the angular velocity vector.

45. The direction of the angular acceleration vector is found the same way. The direction is determined by the change in angular velocity.

46. If the angular velocity is increasing, the angular acceleration vector points in the same direction as the angular velocity.

47. If the angular velocity is decreasing, the angular acceleration vector points in the opposite direction as the angular velocity.

48. CH 9

49. Torque Ƭ is the magnitude of the force multiplied by the lever arm. Ƭ = Fl

50. ∑F x = 0 and ∑F y = 0 ∑Ƭ = 0 The above must be true for all equilibrium conditions.

51. Ƭ = mr 2  The net external torque Ƭ is directly proportional to the angular acceleration . The constant of proportionality is I = mr 2, the moment of inertia. (Si unit kgm 2.)

52. The moment of inertia depends on the location and orientation of the axis relative to the particles that make up the object.

53. There are different formulas for moment of inertia. The moment of inertia depends on the shape of the object, the distribution of the mass in the object, and the location of the pivot point.

54. Newton’s second law for a rigid body rotating about a fixed axis. Torque = moment of inertia X angular acceleration Ƭ = I  (  must be in rad/s 2 )

55. Work is equal to force times displacement, W = Fx. Angular displacement  is equal to linear displacement/radius, x/r. So x = r . Thus W = Fx becomes W = Fr . Since Fr is equal to torque Ƭ, rotational work is equal to torque multiplied by angular displacement. W R = Ƭ  (  must be in radians and the work unit is the joule J.)

56. Rotational KE = ½ I  2 (  = must be in rad/s and the unit is the joule J.)

57. Total kinetic energy is not just ½ mv 2, but ½ mv 2 + ½ I  2.

58. The rotational analog to displacement is angular displacement , the rotational analog to velocity is angular velocity . For acceleration it is angular acceleration . For work it is rotational work Ƭ . The rotational analog to kinetic energy is rotational kinetic energy ½ I  2 and

59. the rotational analog to momentum is angular momentum. The formula for momentum is p = mv. In angular momentum m is replaced with moment of inertia I, and velocity is replaced with angular velocity . Angular momentum L = I .

60. L = I   must be in rad/s and I must be in kgm 2. L is in kgm 2 /s

61. If the net force on an object is zero, the momentum remains constant. Law of Conservation of Momentum.

62. Similarly, If the net torque on an object is zero, the angular momentum remains constant. Law of Conservation of Angular Momentum.

63. CH 10

64. A force must be applied to a spring to stretch or compress it. By Newton’s third law, the spring must apply an equal force to whatever is applying the force to the spring.

65. This reaction force is often called the “restoring force” and is represented by the equation F = -kx.

66. When the restoring force has the mathematical form given by F = -kx, the type of motion resulting is called “simple harmonic motion”.

67. PE elastic = 1/2 kx 2 where k is the spring constant, and x is the distance the spring is compressed or stretched beyond its unstrained length. The unit is the joule (J).

68. A simple pendulum is a mass m suspended by a pivot P. When the object is pulled to one side and released, it will swing back and forth in a motion approximating simple harmonic motion.

69. A series of substitutions finds that, for small angles, 2πf = √g/L f is frequency, g is 9.80, and L is length.

70. Formulas for frequency and period of an oscillating spring: 2πf = √ k/m 2π/ T = √ k/m

71. Formulas for period of an oscillating spring and pendulum: T P = 2π √ L/g T S = 2π √ m/k