1. Work, Energy, Power, and Machines

2. Energy Energy: the currency of the universe. Just like money, it comes in many forms! Everything that is accomplished has to be “paid for” with some form of energy. Energy can’t be created or destroyed, but it can be transformed from one kind into another and it can be transferred from one object to another.

3. Doing WORK is one way to transfer energy from one object to another. Work = Force x displacement W = Fd Unit for work is Newton x meter. One Newton-meter is also called a Joule, J.

4. Work = Force x displacement Work is not done unless there is a displacement. If you hold an object a long time, you may get tired, but NO work was done. If you push against a solid wall for hours, there is still NO work done.

6. Energy and Work have no direction associated with them and are therefore scalar quantities, not vectors. YEAH!!

7. For work to be done, the displacement of the object must be along the same direction as the applied force. They must be parallel. If the force and the displacement are perpendicular to each other, NO work is done by the force.

8. For example, in lifting a book, the force exerted by your hands is upward and the displacement is upward- work is done. Similarly, in lowering a book, the force exerted by your hands is still upward, and the displacement is downward. The force and the displacement are STILL parallel, so work is still done. But since they are in opposite directions, now it is NEGATIVE work.

9. On the other hand, while carrying a book down the hallway, the force from your hands is vertical, and the displacement of the book is horizontal. Therefore, NO work is done by your hands. Since the book is obviously moving, what force IS doing work??? The static friction force between your hands and the book is acting parallel to the displacement and IS doing work!

10. Work = Force x distance

11. So,….while climbing stairs or walking up an incline, only the vertical component of the displacement is used to calculate the work done in moving the object from the bottom to the top. Horizontal component of d Vertical component of d Your Force

12. Example How much work is done to carry a 5 kg cat to the top of a ramp that is 7 meters long and 3 meters tall? W = Force x displacement Force = weight of the cat Which is parallel to the weight- the length of the ramp or the height? d = height NOT length W = mg x h W = 5 x 10 x 3 W = 150 J 7 m 3 m

13. How much work do you do to carry a 30 kg cat from one side of the room to the other if the room is 10 meters long? ZERO, because your Force is vertical, but the displacement is horizontal.

14. Example A boy pushes a lawnmower 20 meters across the yard. If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees, how much work did he do?  F Displacement = 20 m F cos  W = (F cos  )d W = (200 cos 50) 20 W = 2571 J

15. Watch for those “key words” NOTE: If while pushing an object, it is moving at a constant velocity, the NET force must be zero. So….. Your applied force must be exactly equal to any resistant forces like friction.

16. A 5.0 kg box is pulled 6m across a rough horizontal floor (  = 0.4) with a force of 80N at an angle of 35 degrees above the horizontal. What is the work done by EACH force exerted on it? What is the NET work done? Does the gravitational force do any work? NO! It is perpendicular to the displacement. Does the Normal force do any work? No! It is perpendicular to the displacement. Does the applied Force do any work? Yes, but ONLY the horizontal component! W F = Fcos  x d = 80cos 35 x 6 = 393.19J Does friction do any work? Yes, but first, what is the normal force? It’s NOT mg! Normal = mg – Fsin  W f = -f x d = -  Nd = -  (mg – Fsin  d = -90.53J What is the NET work done? 393.19 – 90.53 = 302.66 J mg Normal FAFA  f

17. Energy and Work have no direction associated with them and are therefore scalar quantities, not vectors. YEAH!!

19. Power is the rate at which work is done- how fast you do work. Power = work / time P = W / t You may be able to do a lot of work, but if it takes you a long time, you are not very powerful. The faster you can do work, the more powerful you are.

20. The unit for power is Joule / seconds which is also called a Watt, W (just like the rating for light bulbs) In the US, we usually measure power developed in motors in “horsepower” 1 hp = 746 W

21. Example A power lifter picks up a 80 kg barbell above his head a distance of 2 meters in 0.5 seconds. How powerful was he? P = W / t W = Fd W = mg x h W = 80 x 10 x 2 = 1600 J P = 1600 / 0.5 P = 3200 W

22. Another way of looking at Power: Power = Force x velocity

24. Kinds of Energy

25. Kinetic Energy the energy of motion K = ½ mv 2 Kinetic Energy the energy of motion K = ½ mv 2

26. Potential Energy Stored energy It is called potential energy because it has the potential to do work.

27. Example 1: Spring potential energy in the stretched string of a bow or spring or rubber band. SPE = ½ kx 2 Example 2: Chemical potential energy in fuels- gasoline, propane, batteries, food! Example 3: Gravitational potential energy- stored in an object due to its position from a chosen reference point.

28. Gravitational potential energy GPE = weight x height GPE = mgh Since you can measure height from more than one reference point, it is important to specify the location from which you are measuring.

29. The GPE may be negative. For example, if your reference point is the top of a cliff and the object is at its base, its “height” would be negative, so mgh would also be negative. The GPE only depends on the weight and the height, not on the path that it took to get to that height.

31. Work and Energy Often, some force must do work to give an object potential or kinetic energy. You push a wagon and it starts moving. You do work to stretch a spring and you transform your work energy into spring potential energy. Or, you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy. Work = Force x distance = change in energy

32. Example of Work = change in energy How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)? Fd =  ½ mv 2 The work done by the forces acting = the change in the kinetic energy With TWICE the speed, the car has FOUR times the kinetic energy. Therefore it takes FOUR times the stopping distance. (What FORCE is doing the work??)

33. The Work-Kinetic Energy Theorem NET Work done =  Kinetic Energy W net = ½ mv 2 f – ½ mv 2 o

34. Example A 500kg car moving at 15m/s skids 20m to a stop. How much kinetic energy did the car lose?  K = ½ mv f 2 – ½ mv o 2  K = -½ (500)15 2  K = -56250J What force was applied to stop the car? F·d =  K F =  K / d F = -56250 / 20 F = -2812.5N

35. Example A 500kg car moving at 15m/s slows to 10m/s. How much kinetic energy did the car lose?  K = ½ mv f 2 – ½ mv o 2  K = ½ (500)10 2 - ½ (500)15 2  K = -31250J What force was applied to stop the car if the distance moved was 12m? F·d =  K F =  K / d F = -31250 / 12 F = -2604N

36. Example A 500kg car moving on a flat road at 15m/s skids to a stop. How much kinetic energy did the car lose?  K = ½ mv f 2 – ½ mv o 2  K = -½ (500)15 2  K = -56250J How far did the car skid if the effective coefficient of friction was 0.6? Stopping force = friction =  N =  mg F·d =  K  mg)·d =  K d =  K /  mg d = 56250 / (0.6 · 500 · 10) = 18.75m

37. Back to Power… Since Power = Work / time and Net work =  K… Power =  K / time In fact, Power can be calculated in many ways since Power = Energy / time, and there are MANY forms of energy!

39. Graphs If you graph the applied force vs. the position, you can find how much work was done by the force. Work = Fd = “area under the curve”. Total Work = 2N x 2m + 3N x 4m = 16 J Area UNDER the x-axis is NEGATIVE work = - 1N x 2m Force, N Position, m F d Net work = 16J – 2J = 14J

40. Back to the Work-Kinetic Energy Theorem… According to that theorem, net work done = the change in the kinetic energy W net =  K But, if the work can be found by taking the “area under the curve”, then it is also true that Area under the curve =  K = ½ mv f 2 – ½ mv o 2 so that the area can be used to predict the final velocity of an object given its initial velocity and its mass.

41. For example… Suppose from the previous graph (Area = W net = 14J), the object upon which the forces were exerted had a mass of 3kg and an initial velocity of 4m/s. What would be its final velocity? Area under the curve = ½ mv f 2 – ½ mv o 2 14 = ½ 3v f 2 – ½ 3(4) 2 v f = 5.0 m/s

42. The Spring Force If you hang an object from a spring, the gravitational force pulls down on the object and the spring force pulls up.

43. The Spring Force The spring force is given by F spring = kx Where x is the amount that the spring stretched and k is the “spring constant” which describes how stiff the spring is

44. The Spring Force If the mass is hanging at rest, then F spring = mg Or kx = mg (this is called “Hooke’s Law) The easiest way to determine the spring constant k is to hang a known mass from the spring and measure how far the spring stretches! k = mg / x

45. Graphing the Spring Force Suppose a certain spring had a spring constant k = 30 N/m. Graphing spring force vs. displacement: On horizontal axis- the displacement of the spring: x On vertical axis- the spring force = kx = 30x What would the graph look like? F s = kx In “function” language: f(x) = 30x

46. x1x1 x2x2 Spring Force vs. Displacement x F s = 30x How could you use the graph To determine the work done by The spring from some x 1 to x 2 ? Take the AREA under the curve! FsFs

47. Analytically… The work done by the spring is given by W s = ½ kx f 2 – ½ kx o 2 where x is the distance the spring is stretched or compressed (Which would yield the same result as taking the area under the curve!)

48. I love mrs. BRown

49. Mechanical Energy Mechanical Energy = Kinetic Energy + Potential Energy E = ½ mv 2 + mgh

50. “Conservative” forces - mechanical energy is conserved if these are the only forces acting on an object. The two main conservative forces are: Gravity, spring forces “Non-conservative” forces - mechanical energy is NOT conserved if these forces are acting on an object. Forces like kinetic friction, air resistance

51. Conservation of Mechanical Energy If there is no kinetic friction or air resistance, then the total energy of an object remains the same. If the object loses kinetic energy, it gains potential energy. If it loses potential energy, it gains kinetic energy. For example: tossing a ball upward

52. Conservation of Mechanical Energy The ball starts with kinetic energy… Which changes to potential energy…. Which changes back to kinetic energy K = ½ mv 2 PE = mgh K = ½ mv 2 Energy bottom = Energy top ½ mv b 2 = mgh t What about the energy when it is not at the top or bottom? E = ½ mv 2 + mgh

53. Examples dropping an object box sliding down an incline tossing a ball upwards a pendulum swinging back and forth A block attached to a spring oscillating back and forth

54. If there is kinetic friction or air resistance, energy will not be conserved. Energy will be lost in the form of heat. The DIFFERENCE between the original energy and the final energy is the amount of energy lost due to heat. Original energy – final energy = heat loss

55. Sometimes, mechanical energy is actually INCREASED! For example: A bomb sitting on the floor explodes. Initially: ½ mv 2 = 0 mgh = 0 ½ kx 2 = 0 E = 0 After the explosion, there’s lots of kinetic and gravitational potential energy!! Did we break the laws of the universe and create energy??? Of course not! NO ONE, NO ONE, NO ONE can break the laws! The mechanical energy that now appears came from the chemical potential energy stored within the bomb itself!

56. According to the Law of Conservation of Energy, energy cannot be created or destroyed. The total amount of mechanical energy in a system remains constant when there are no NONCONSERVATIVE forces doing work, but one form of energy may be transformed into another as conditions change. E 1 = E 2 K 1 + GPE 1 = K 2 + GPE 2 ½ mv 1 2 + mgh 1 = ½ mv 2 2 + mgh 2

57. The starting point in using Conservation of Energy to solve for an unknown is to locate a position where the total mechanical energy IS known and use that value as E o. (i.e. you know all of these quantities: the height, the velocity, the distance the spring is stretched)

59. Simple Machines and Efficiency Machine: A device that HELPS do work. A machine cannot produce more WORK ENERGY than the energy you put into it, but it can make your work easier to do.

60. Some common “simple machines” include levers, pulleys, wheels and axles, and inclined planes Ideally, with no friction, the work energy you get out of a machine equals the work energy you put into it. Ideally: Work in = work out

61. Work = Force x distance The work you put into a machine is called EFFORT work. The work you get out of the machine- is called RESISTANCE work, so ideally Effort Work = Resistance Work F effort d effort = F resistance d resistance (if there’s no NON-conservative forces!)

62. Levers The RESISTANCE force is the weight of the load being lifted. Which arrangement will require the least EFFORT force? How do you “pay” for a small effort force? Effort force

63. Inclined Planes Which arrangement will require the least EFFORT force? How do you “pay” for a smaller effort force? Effort Force Effort Distance Height = Resistance Distance Weight = Resistance Force

64. Two pulleys with a belt A motor is attached to one of the pulleys so that as it turns, the belt causes the second pulley to turn. Which pulley should the motor be attached to so that it requires the least effort force from the motor? To have the least effort force, the effort distance must be the greatest. In this case the effort distance is the number of turns around. Which pulley will have to go around more times? This is the pulley that the motor should be attached to for the least effort force.

65. Pulleys pulley simulation

66. Efficiency No machine is perfect. That is reflected in the “efficiency” of the machine. In the real world, the efficiency will always be less that 100%. It is found by

67. A man pushes a 50 kg box up a 25 m long incline that is 8 meters high by applying a force of 200 N. What is the effort (input) work? W effort = F effort d effort W = 200N x 25 m = 5000J What is the resistance (output) work? W resistance = F r d r W = mg x h W = 50 x 10 x 8 = 4000J What is the efficiency of the incline? Efficiency = 4000 J / 5000 J =.80 = 80%