1. WORK AND ENERGY
CHAPTER 5

2. WORK
A FORCE THAT CAUSES A DISPLACEMENT OF AN OBJECT DOES WORK ON THE OBJECT
IF THERE IS NO DISPLACEMENT THERE IS NO WORK DONE.
WORK= FORCE X DISTANCE

3. UNITS OF WORK WORK IS MEASURED IN JOULES.
NAMED FOR JAMES PRESCOTT JOULE WHO MADE MAJOR CONTRIBUTIONS TO UNDERSTNADING ENERGY, HEAT AND ELECTRICITY.

4. JOULE
NEWTON X METER = JOULE

5. WORK = FORCE X DISTANCE
WORK IS DONE ONLY WHEN COMPONENTS OF FORCE ARE PARALLEL TO THE DISPLACEMENT.

6. WORK = FORCE X DISTANCE
IF THE FORCE IS PERPENDICULAR TO THE DISPLACEMENT NO WORK HAS BEEN DONE.

7. EXAMPLE
IMAGINE PUSHING A CRATE ALONG THE GROUND

8. IF ALL FORCE IS HORIZONTAL, ALL YOUR EFFORT MOVES THE CRATE

9. WORK = FD(COS )
IF YOUR FORCE IS AT AN ANGLE ONLY THE HORIZONTAL COMPONENT MOVES THE CRATE.

10. NET WORK NET WORK CAN BE FOUND BY FIRST FINDING THE NET FORCE.
WORKnet = Fnet x D(COS )
Net work = net force x displacement x cosine of the angle between them

11. Practice
HOW MUCH WORK IS DONE ON A VACUUM CLEANER PULLED 3.0 M BY A FORCE OF 50.0 N AT AN ANGLE OF 30.O DEGREES ABOVE THE HORIZONTAL

12. SOLUTION F= 50.0 N  =30 D=3.0 m W=FD(COS  ) W = 50.0N X 3.0m(COS 30)
W = 130 J

13. PRACTICE
A 20.0 KG SUITCASE IS RAISED 3.0 m ABOVE A PLATFORM BY A CONVEYOR BELT. HOW MUCH WORK IN DONE ON THE SUITCASE?

14. SOLUTION M=20.0 KG D=3.0m W=FD F=MA F=20.0KG X 9.81m/s2 F= 196.2 N
W=196.2 N X 3.0m
W= J

15. THE SIGN OF WORK
WORK IS SCALAR
IT CAN BE POSITIVE
IT CAN BE NEGATIVE

16. POSITIVE WORK
WORK IS POSTIVE WHEN THE COMPONENT OF FORCE IS IN THE SAME DIRECTION AS THE DISPLACEMENT

17. NEGATIVE WORK
WORK IS NEGATIVE WHEN THE FORCE IS IN THE DIRECTIONOPPOSITE THE DISPLACEMENT
EX: WORK DONE BY THE FORCE OF FRICTION

18. WORK - SCILINKS HF2051
TAKE A LOOK AT WORK. READ THE LITERATURE AND VIEW THE ANIMATION.

19. ENERGY
KINETIC ENERGY
POTENTIAL ENERGY

20. KINETIC ENERGY ENERGY ASSOCIATED WITH MOTION Wnet=FX=(MA) X
X= DISPLACEMENT(D)

21. WORK TO ENERGY
Wnet=change in kinetic energy
KINETIC ENERGY =1/2MV2

22. PRACTICE
A 7.00 KG BOWLING BALL MOVES AT 3.00 M/S. HOW MUCH KINETIC ENERGY DOES THE BOWLING BALL HAVE? HOW FAST MUST A 2.45 G TABLE-TENNIS BALL MOVE IN ORDER TO HAVE THE SAME KINETIC ENERGY AS THE BOWLING BALL? IS THIS SPEED REASONABLE FOR THE TABLE-TENNIS BALL?

23. SOLUTION Mb-7.00 kg Mt=2.45 g Vb=3.00m/s KE= ? Vt=? KE=1/2MV2
KEb=1/2(7.00 kg)(3.00 m/s)2
KE= 31.5 J
31.5 J=1/2(.00245kg)V2
V=1.60 X 102 m/s

24. KINETIC ENERGY
SCALAR QUANTITY
SI UNIT IS JOULE

25. WORK-ENERGY THEOREM
THE NET WORK DONE BY A NET FORCE ACTING ON AN OBJECT IS EQUAL TO THE CHANGE IN THE KINETIC ENERGY OF THE OBJECT.
WNET=KE

26. PRACTICE
ON A FROZEN POND, APERSON KICKS A 10.0 kg SLED, GIVING IT AN INITIAL SPEED OF 2.2 m/s. How far does the sled move if the coefficient of kinetic friction between the sled and the ice is 0.10?

27. SOLUTION M=10.00 kg =0.10 Vf=0.00m/s D=? Vi=2.2m/s
KE=1/2MV2 Wnet=FxD(COS)
KE=1/2(10.00 kg)(2.2 m/s)2
KEi= 24.2 J KEf= 0 J
Wnet=KE
Wnet=0 J – 24.2 J
-24.2 J = 0.10(10.00kg)(9.81) D
2.4 m = D

28. POTENTIAL ENERGY POTENTIAL ENERGY IS STORED ENERGY ENERGY OF POSITION
PEg=MASS x GRAVITY x HEIGHT
PEg=GRAVITATIONAL POTENTIAL ENERGY
SI UNIT IS JOULES

29. ELASTIC POTENTIAL ENERGY
ENERGY STORED IN SPRINGS

30. RELAXED LENGTH OF THE SPRING
THE LENGTH OF A SPRING WHEN NO EXTERNAL FORCES ARE ACTING ON IT.

31. HOW TO DETERMINE THE ELASTIC POTENTIAL ENERGY
PEelastic=1/2kX2
ELASTIC POTENTIAL ENERGY = ½ X SPRING CONSTANT X (DISTANCE COMPRESSED OR STRETCHED)2
K IS THE SPRING CONSTANT

32. PROBLEM
A 70.O kg STUNTMAN IS ATTACHED TO A BUNGEE CORD WITH ANUNSTRETCHED LENGTH OF 15.0 m. HE JUMPS OFF A BRIDGE SPANNING A RIVER FROM A HEIGHT OF 50.0 m. WHEN HE FINALLY STOPS, THE CORD HAS STRETCHED LENGTH OF 44.0 m. TREAT THE STUNTMAN AS A POINT MASS, AND DISREGARD THE WEIGHT OF THE BUNGEE CORD.

33. PROBLEM CONTINUED
ASSUMING THE SPRING CONSTANT OF THE BUNGEE CORD IS 71.8 N/m, WHAT IS THE TOTAL POTENTIAL ENERGY RELATIVE TO THE WATER WHEN THE MAN STOPS FALLING?

34. SOLUTION M=70.0 kg k= 71.8 N/m H=50.0 m – 44.0 m = 6.0 m
X = 44.0 m – 15.0 m = 29.0 m
PE = 0 J AT RIVER LEVEL
PEtot=PEG + Peelastic
PEG=MGH
PEG=(70.0 kg)(9.81 m/s2)(6.0m)
PEG= 4.1 X 103J

35. SOLUTION CONTINUED Peelastic=1/2kX2 Peelastic=1/2(71.8 N/m)(29.0m)2
Peelastic=3.02 X 104 J

36. FINALLY THE ANSWER! PEtot=PEG + Peelastic
PEtot= 3.02 X 104 J X 103J
PEtot= 3.43 X 104 J

37. CONSERVATION OF ENERGY
CONSERVED MEANS CONSTANT
THE AMOUNT OF A QUANTITY STAYS THE SAME
THE QUANTITY CAN CHANGE FORM
EXAMPLE: MASS

38. MECHANICAL ENERGY THE SUM OF THE KINETIC AND POTENTIAL ENERGIES.
MECHANICAL ENERGY = KINETIC ENERGY + POTENTIAL ENERGY
ME = KE + PE

39. NONMECHANICAL ENERGY NOT ALL ENERGY IS MECHANICAL
EXAMPLES OF NONMECHANICAL ARE:
CHEMICAL
NUCLEAR
INTERNAL
ELECTRICAL

40. CONSERVATION OF MECHANICAL ENERGY
TOTAL MECHANICAL ENERGY REMAINS THE SAME IN THE ABSENCE OF FRICTION.
IF FRICTION IS PRESENT MECHANICAL ENERGY IS CONVERTED INTO HEAT, NOISE OR SOME OTHER FORM.

41. MECHANICAL FLOW CHART

42. CONSERVATION OF MECHANICAL ENERGY
INITIAL MECHANICAL ENERGY = FINAL MECHANICAL ENERGY (IN THE ABSENCE OF FRICTION)
KEi + PEi = KEf + PEf
½ MV2 + MGHi = ½ MV2f + MGHf

43. PRACTICE
STARTING FROM REST, A CHILD ZOOMS DOWN A FRICITONLESS SLIDE FROM AN INITIAL HEIGHT OF 3.00 M. WHAT IS HER SPEED AT THE BOTTOM OF THE SLIDE? ASSUME SHE HAS A MASS OF 25.0 KG.

44. SOLUTION Hi = 3.00M M=25.0 KG Vi= 0 Hf = 0 M Vf=?
Pei = 25KG X 9.81 X 3.00M
PE I = 736 J PE f = O J
KE I = 0 J
KE f = 1//2 25KGXV2

45. CONT. ME I = ME f PE I + KE I = PE f + KE f
736 J +0 J = 0 J + (0.500)(25.0KG)Vf2
Vf= 7.67M/S

46. POWER
POWER IS THE RATE AT WHICH WORK IS DONE.

47. UNITS
POWER IS MEASURED IN WATTS.
1 W (WATT) = JOULE/SECOND

48. POWER EQUATIONS POWER = WORK/ TIME WORK = FORCE X DISTANCE
P= F X D/T AND SINCE V=D/T
P= F X V

49. POWER RATING AND WORK
MACHINES WITH DIFFERENT POWER RATING DO THE SAME WORK IN DIFFERENT TIME INTERVALS.
THE MORE POWER THE FASTER THE RATE.

50. PRACTICE
A 193 KG CURTAIN NEEDS TO BE RAISED 7.5 M, AT A COSNTANT SPEED, IN A S CLOSE TO 5.0 S AS POSSIBLE. THE POWER RATINGS FOR THREE MOTORS ARE LSITED AS 1.0 KW, 3.5 KW, AND 5.5 KW. WHICH MOTOR IS BEST FOR THE JOB?

51. SOLUTION M= 193 KG T =5.0S D=7.5 M P=? P=W = FD = MGD T T T
P = 193KG X 9.81 M/S2 X 7.5M
5.0 S
P = 2800 WATTS OR 2.8 kW

52. Simple Machines Lecture
8-4

53. MACHINES
A DEVICE USED TO MULTIPLY FORCES OR CHANGE THE DIRECTION OF THE FORCES
WORK WE PUT INTO A MACHINE IS EQUAL TO THE WORK THE MACHINE PUTS OUT.
WORK IN = WORK OUT

54. TYPES OF SIMPLE MACHINES
A LEVER IS A SIMPLE MACHINE. AS WE DO WORK ON ONE END OF THE LEVER THE LEVER DOES WORK AT THE OTHER END ON THE LOAD.
INCLINED PLANE
WHEEL AND AXLE
WEDGE
PULLEYS
SCREW-INCLINE PLANE AROUND A CYLINDER

55. THE LEVER
f X D = WORK IN
WORK OUT =F X d

56. LEVERS THREE TYPES OF LEVERS
TYPE 1 - FULCRUM BETWEEN THE FORCE AND THE LOAD
TYPE 2 - LOAD BETWEEN THE INPUT FORCE AND THE FULCRUM
TYPE 3 - INPUT FORCE BETWEEN THE FULCRUM AND THE LOAD

57. TYPE 1
TYPE 1 – FULCRUM BETWEEN FORCE AND LOAD

58. TYPE 2
TYPE 2 – LOAD BETWEEN THE FULCRUM AND INPUT FORCE

59. TYPE 3
TYPE 3- FULCUM AT ONE END AND LOAD AT OTHER END WITH INPUT FORCE BETWEEN

60. PULLEY
A KIND OF LEVER THAT CAN BE USED TO CHANGE THE DIRECTION OF A FORCE.
PULLEYS MULTIPLY FORCES

61. SINGLE PULLEY BEHAVES LIKE A TYPE 1 LEVER
AXIS OF THE PULLEY ACTS AS THE FULCRUM
THE RADIUS OF THE PULLEY IS THE LEVER DISTANCE
BOTH LEVER DISTANCES ARE THE SAME SO THE FORCE IS NOT MULTIPLIED
A SINGLE PULLEY SIMPLY CHANGES DIRECTION.

62. tin = tout MECHANICAL ADVANTAGE
THE RATIO BETWEEN THE OUTPUT FORCE AND THE INPUT FORCE
M.A. = OUTPUT FORCE/INPUT FORCE
INPUT TORQUE = OUTPUT TORQUE
tin = tout
Fin X Din = Fout X Dout
M.A. = Fout = Din
Fin = Dout

63. SINGLE PULLEY diagram OUTPUT INPUT
MECHANICAL ADVANTAGE IS 1 BECAUSE IT ONLY CHANGES THE DIRECTION OF THE FORCE.

64. SINGLE PULLEY – TYPE 2 LEVER
OUTPUT
MECHANICAL ADVANTAGE = 2
THE LOAD IS SUSPENDED ½ WAY BETWEEN THE FULCRUM AND THE INPUT. EACH NEWTON OF INPUT PRODUCES 2 NEWTONS OF OUTPUT.
INPUT
THE LOAD IS NOW SUPPORTED BY 2 ROPES. EACH ROPE HOLDS ½ THE LOAD.

65. MECHANICAL ADVANTAGE
THE MECHANICAL ADVANTAGE FOR A SIMPLE PULLEY SYSTEM IS THE SAME AS THE NUMBER OF STRANDS OF ROPE THAT ACTUALLY SUPPORT THE LOAD.
1 STRAND : MA = 1
2 STRANDS : MA = 2

66. WHAT ABOUT THIS ONE?
M.A.=2
THERE ARE TWO ROPES SUPPORTING THE LOAD. THE THIRD ROPE CHANGES THE DIRECTION OF THE FORCE.

67. AND THIS ONE? OUTPUT = 500 N INPUT = 100 N M.A. = OUTPUT/INPUT
HOW MANY ROPES ARE THERE HOLDING THE LOAD?
500 N

68. MECHANICAL ADVANTAGE IDEALIZED PULLEY SYSTEM
APPLIED FORCE X INPUT DISTANCE = OUTPUT FORCE X OUTPUT DISTANCE
MECHANICAL ADVANTAGE CAN ALSO BE CALCULATED BY…
M.A.= INPUT DISTANCE/OUTPUT DISTANCE

69. EFFICIENCY A MEASURE OF HOW WELL A MACHINE WORKS.
NO MACHINE CAN PUT OUT MORE ENERGY THAN IS PUT INTO IT.
NO MACHINE CAN CREATE ENERGY.
MACHINES ONLY TRANSFER ENERGY.
ALL THE MACHINES WE HAVE DISCUSSED TO THIS POINT ARE IDEAL MACHINES. 100% EFFICIENT.

70. 100% EFFICIENCY NEVER HAPPENS
ENERGY IS LOST IN HEAT THROUGH BENDING, STRETCHING AND OTHER TYPES OF FRICTION.
THE LOWER THE EFFICIENCY THE GREATER AMOUNT OF ENERGY IS LOST TO HEAT.

71. CALCULATING EFFICIENCY
IT IS THE RATIO OF THE USEFUL WORK OUTPUT TO THE TOTAL WORK INPUT.
EFFICIENCY = USEFUL WORK OUTPUT
TOTAL WORK INPUT
EFFICIENCY = ACTUAL MECHANICAL ADVANTAGE
THEORETICAL MECHANICAL ADVANTAGE

72. INCLINE PLANE PUSHING A LOAD UP AN INCLINE. 5 METERS 1 METER
THEORETICAL M.A. OF 5. WHY IS THAT NOT TRUE IN REAL LIFE?

73. SPECIALIZED INCLINE PLANE
A SCREW IS AN INCLINE PLANE WRAPPED AROUND A CYLINDER. EACH TURN OF THE SCREW CAUSES THE LOAD TO BE RAISED ONE PITCH.
PITCH IS THE DISTANCE BETWEEN SCREWS.
CIRCULAR DISTANCE FOR ONE REVOLUTION DIVIDED BY THE PITCH EQUALS THE THEORETICAL MECHANICAL ADVANTAGE OF THE SCREW.
DUE TO THE AMOUNT OF FRICTION, EFFICIENCY IS ONLY ABOUT 20% SO THE M.A. WOULD BE APPROXIMATELY 1/5 OF THE THEORETICAL.

74. PRACTICE
A CHILD ON A SLED(TOTAL WEIGHT 500 N) IS PULLED UP A 10 M SLOPE THAT ELEVATES HER A VERTCAL DISTANCE OF 1 M.
A. WHAT IS THE THEORETCAL MECHANICAL ADVANTAGE OF THE SLOPE?
INPUT DISTANCE/OUTPUT DISTANCE
M.A. = 10 M/1 M = 10

75. PART B.
IF THE SLOPE IS WITHOUT FRICTION, AND SHE IS PULLED UP THE SLOPE AT CONSTANT SPEED, WHAT WILL BE THE TENSION IN THE ROPE?
SINCE M.A. = 10, IT WILL TAKE 1/10 THE AMOUNT BEING RAISED. SO 500-N/10 = 50 N

76. PART C
CONSIDERING THE PRACTICAL CASE WHERE FRICTION IS PRESENT, SUPPOSE THE TENSION IN THE ROPE WERE ACTUALLY 100 N. WHAT IS THE ACTUAL MECHANICAL ADVANTAGE OF THE SLOPE? WHAT WOULD THE EFFICIENCY BE?

77. SOLUTION M.A.= WEIGHT BEING RAISED/INPUT FORCE 500 N/100 N= 5
EFFICIENCY= ACTUAL M.A.
THEORETICAL M.A.
5/10 = .5 OR 50%

78. AUTOMOTIVE EFFICIENCY
EVEN THE BEST DESIGNED CAR ENGINES ONLY PRODUCE APPROXIMATELY 30-35% EFFICIENCY.
35% IS LOST IN WATER COOLING
35% IS LOST IN EXHAUST HEAT
30% BECOMES USEFUL MECHANICAL ENERGY