1. Work and Energy Work = Force X distance W = Fd Unit - Joules (N m), (kg m 2 /s 2 ) Force must be in the direction of the motion.

2. Work and Direction Lifting a box is work Fighting force of gravity. Lifting force in same direction box moves. Your lifting force and the boxes direction Fighting against the force of gravity

3. Work and Direction Pushing a box is work Applied force is in the direction of movement Working against force of friction Pushing on ice would be less work Frictional Force

4. Work and Direction Carrying a box is not work Your lifting force Fighting against the force of gravity But this is the direction of motion

5. Work and Direction Mowing the lawn is some work only do work in horizontal direction. The downward push, because of the angle of the handle, doesn’t do any work. This is the only part that does work This part is lost Your push

6. Vertical vs. Horizontal Work Lifting Gives an object Potential Energy of position. Pushing Gives an object Kinetic Energy of Motion.

7. Work or Not 1.A teacher pushes against a wall until he is exhausted. 2.A book falls off the table and falls freely to the ground. 3.A waiter carried a full try of meals across the room. 4.A rocket accelerates through space. 5.A rocket travels at a constant speed through space

8. Work: Example 1 How much work is done lifting a 5.0 kg backpack to a shelf 2.0 meters above the floor?

9. Superman does 36,750 J of work lifting this car 2.5m from the ground? What is the weight of the car? What is the mass of the car?

10. Work: Example 2 Mr. Fredericks pulls a box with 30 N of Force a distance of 5.0 m, at an angle of 50 o with the ground. How much work was done?  = 50 o Direction of motion

11. F x = (30N)(cos50 o ) = 19 N F y = (30N)(sin50 o ) = 23 N W = F x d = (19N)(5m) = 96J W = Fdcos   = 50 o FyFy FxFx

12. A sled is pulled with a force of 100.0 N at an angle of 65 o to the ground for 8.00 m. Calculate the work done. (338 J)

13. Work: Example 3 A 50-kg crate is pulled 40 m with a force of 100 N at an angle of 37 o. The floor is rough and exerts a frictional force of 50 N. Determine the work done on the crate by each force and the net work done on the crate. FpFp F fr FNFN mg 

14. Let’s deal with the vertical forces first W G = mgxcos90 o = 0 W N = F N xcos90 o = 0 No work is done in the vertical direction (the box is not lifted)

15. Now the horizontal forces W fr = F fr d W fr = (50 N)(40m) = -2000 J W p = F p dcos 37 o W p = (100N)(40 m)(cos 37 o ) = 3200 J W net = W G + W fr W net = 0 + 0 + 3200 J -2000 J = 1200 J

16. Example 4 A hiker walks up a hill at a constant speed. He is carrying a 15.0 kg backpack and the hill is 10.0 m high. How much work did he do? Surprisingly, he does no work in the horizontal direction.

17. The only forces are vertical  F = F H -mg ma = F H -mg (no acceleration) 0 =F H -mg F H = mg = (15.0 kg)(9.80 m/s 2 ) F H = 147 N W hiker = Fd = (147 N)(10.0 m) W hiker = 1470 J

18. How much work did gravity do? W G = -1470 J W NET = W hiker + W G W NET = 1470 J -1470 J = 0

19. Does the Earth Do Work on the Moon? W = Fdcos  W = Fd(cos 90 o ) W = Fd(0) W = 0 FRFR v

20. English Unit of Work Foot-pound – English unit of work. Pound – unit of Force W = Fd = (foot*pound)

21. Variable Force Work is really an area: W = ∫ Fdx (an integral tells you the area) WORK

22. Given the following graph, determine the work done by the following force in the first ten meters Distance (m)

23. Calculate the work: a)In the first 10 meters b)In the second 10 meters c)Overall Distance (m)

24. Calculate the work: a)In the first 20 meters b)From 20 to 40 meters c)From 40 to 80 metersd) Overall

25. Calculate the work: a)In the first 10 meters b)In the second 10 meters c)Overall

26. Calculate the work: a)In the first 10 meters b)In the second 10 meters c)Overall Distance (m)

27. A 35.0 kg suitcase is pulled 20.0 m with a force of 100.0 N at an angle of 45.0 o. The floor is has a  k of 0.200. a.Calculate the normal force on the suitcase (HINT: take the F py into account). (272 N) b.Calculate the work done by friction. (-1088 J) c.Calculate the work done by the pull. (1414 J) d.Calculate the net work on the suitcase. (326 J) FpFp F fr FNFN mg 

28. A 50.0 kg child and sled is pulled by a rope with a tension of 200.0 N and an angle of 35.0 o with the ground for 50.0 m. The sled and snow has a  k of 0.150. a.Calculate the normal force on the sled. (375 N) b.Calculate the work done by friction. (- 2813 J) c.Calculate the work done by the pull. (8191 J) d.Calculate the net work on the sled. (5379 J)

29. Energy Energy – The capacity to do work –1 Joule = 1 Newton-meter =1 kg-m 2 /s 2 –Two types Potential Energy Kinetic Energy

30. Definition - Energy of motion K = ½ mv 2 –m = mass (kg) –v = speed (m/s) –Unit = Joules Questions: a.If the mass of an object is doubled, what happens to KE? b.If the speed is doubled, what happens to KE?

31. A kitten is traveling with a kinetic energy of 10.0 J at 2.00 m/s a.Calculate his mass b.If Mr. Fredericks (56.0 kg) has the same kinetic energy, calculate his speed.

32. Kinetic Energy If Mr. Fredericks (56 kg), had the same kinetic energy as the Thing in the previous example, how fast would he move?

33. Kinetic Energy and Work W NET = KE f – KE I W NET =  KE

34. How much work is required to accelerate a 1000-kg car from 20-m/s to 30-m/s? (W = 250,000 J)

35. A 1000 kg car stops over 50.0 m. The coefficient of friction is 0.25. a)Calculate the force of friction. (-2450 N) b)Calculate the work done by friction. (-1.225 X 10 5 J) c) Calculate the initial speed of the car. (15.7 m/s)

36. A 400.0 g football travelling at 25.0 m/s is caught by a player. The players arms go back 75.0 cm while catching the ball. a)Calculate the initial kinetic energy of the ball. (125 J) b)Calculate the force the ball exerted on the player’s hand. (167 N)

37. A 140.0 g baseball is caught by a fielder. The glove moves 25.0 cm. The fielder experienced and average force of 204 N. a.Calculate the kinetic energy of the ball before being caught. (51 J) b.Calculate the initial speed of the ball. (27.0 m/s) c.Draw a free body diagram of the ball in the air. d.Draw a free body diagram of the ball while being caught by the glove.

38. Potential Energy Definition - Stored energy Two Types –Energy of position Gravitational PE - stored by placing something at a height above the ground Mechanical PE - compressing a spring or mechanical device –Energy stored in chemical bonds and atomic nuclei

39. Potential Energy of Position PE = mgy m = mass in kg g = acceleration of gravity (9.8 m/s 2 ) y = height from the ground in meters.

40. Potential Energy of Position. 15 J 45 J What would the potential energy of a ball be at the other steps?

41. Potential Energy and Work W =  PE W = mgy

42. How much work does a 50.0 kg woman do by climbing to the top of a 300 m hill?

43. Assume the roller coaster car has a mass of 1000 kg. Calculate the PE: a)At pts. A, B, and C b)Gained from A  B c)Lost B  A d)Lost B  C

44. Hooke’s Law F x = kx F=Force exerted on the spring k=spring constant x = distance compressed or stretched.

46. A force of 600 Newtons will compress a spring 0.5 meters. a)Calculate the spring constant of the spring. (1200 N/m) b)Calculate the force is necessary to stretch the spring by 2 meters. (2400 N) c)A force of 40 Newtons will stretch a different spring 0.1 meter. How far will a force of 80 Newtons stretch it? (0.2 m)

47. Springs PE = ½ kx 2 k = spring constant (measure of the stiffness of a spring) x = distance stretched from normal length

48. Spring Example 1 How much potential energy is stored in a spring if it is compressed 10.0 cm from its normal length. Assume it has a spring constant of 300 N/m PE = ½ kx 2 PE = ½ (300 N/m)(0.100 m) 2 PE = 1.5 N-m = 1.5 J

49. When 10 J is used to stretch a spring, the spring stretchs 20.0 cm. Calculate the spring constant.

50. Conservative and Nonconservative Forces Conservative Forces –Work is independent of the path taken –Gravity, elastic spring Nonconservative Forces –Work depends on the path taken –Friction, air resistance, tension in a cord –Also called dissipative forces

51. Conservative Forces Imagine carrying a box up 6 floors, then down 2. Overall, you only increased the PE by 4 floors (gravity is conservative). What if you went up 8 and down 4? 6 floors 4 floors

52. Nonconservative Forces Will it take more work to push the box from 1 to 2 on path A or path B? Or are they the same? B A

53. Law of Conservation of Mechanical Energy KE 1 + PE 1 = KE 2 + PE 2 ½ mv 2 1 + mgy 1 = ½ mv 2 2 + mgy 2 (assumes no friction)

55. Mr. Fredericks (56.0 kg) climbs up a 2.50 m slide. Calculate the speed at which he will leave the slide at the bottom. Neglect friction. 2.50 m

56. Suppose Mr. Fredericks wanted to leave the slide with a speed of 10.0 m/s. Calculate the height of the slide without using the mass. Neglect friction.

57. Law of Cons. Energy: Ex 1 Mr. Fredericks (100 kg) jumps from a 3 meter cliff. Calculate his velocity when he is: a)2 m above the ground b)1 m above the ground c)Just before he hits (0 m)

58. a) 2 m above the ground ½ mv 2 1 + mgy 1 = ½ mv 2 2 + mgy 2 0 + mgy 1 = ½ mv 2 2 + mgy 2 mgy 1 = m(½v 2 2 + gy 2 ) gy 1 = (½v 2 2 + gy 2 ) (9.8 m/s 2 )(3m) = ½v 2 2 + (9.8 m/s 2 )(2m) v 2 2 = 19.6 m 2 /s 2 v 2 = 4.4 m/s

59. b) 1 m above the ground ½ mv 2 1 + mgy 1 = ½ mv 2 2 + mgy 2 0 + mgy 1 = ½ mv 2 2 + mgy 2 mgy 1 = m(½v 2 2 + gy 2 ) gy 1 = (½v 2 2 + gy 2 ) (9.8 m/s 2 )(3m) = ½v 2 2 + (9.8 m/s 2 )(1m) v 2 2 = 39.4 m 2 /s 2 v 2 = 6.3 m/s

60. c) 0 m above the ground (just before he hits) ½ mv 2 1 + mgy 1 = ½ mv 2 2 + mgy 2 0 + mgy 1 = ½ mv 2 2 + 0 mgy 1 = ½ mv 2 2 (All PE gets turned to KE) gy 1 = ½v 2 2 v 2 2 = (9.8 m/s 2 )(3 m) v 2 = 7.7 m/s

61. Law of Cons. Energy: Ex 2 A rollercoaster starts at a height of 40 m. a)What is its speed at the bottom of the hill? (28.0 m/s) b)At what height will it have half of that speed? (30.0 m) 40 m

62. a)What is its speed at the bottom of the hill? ½ mv 2 1 + mgy 1 = ½ mv 2 2 + mgy 2 0 + mgy 1 = ½ mv 2 2 + 0 gy 1 = ½ v 2 2 v 2 = \/ 2gy 1 = \/ (2)(9.8 m/s 2 )(40 m) v 2 = 28 m/s

63. b) At what height will it have half of that speed? ½ mv 2 1 + mgy 1 = ½ mv 2 2 + mgy 2 0 + mgy 1 = ½ mv 2 2 + mgy 2 gy 1 = ½ v 2 2 + gy 2 (9.8 m/s 2 )(40 m) = ½ (14 m/s) 2 + (9.8 m/s 2 )y 2 y 2 = 30 m (above the low point)

64. Law of Cons. Energy: Ex 3 Superman is supposed to be able to leap tall buildings in a single bound. What initial velocity would he need to jump to the top of the Empire State Building (381 m)?

65. ½ mv 2 gr + mgy gr = ½ mv 2 top + mgy top ½ mv 2 gr + 0 = 0 + mgy top ½ mv 2 gr = mgy top ½ v 2 gr = gy top v gr = \/2gy top = (2 X 9.8m/s 2 X 381 m) 1/2 v gr = 86.4 m/s

66. So how fast is that in mph? Would an elephant have to jump faster?

67. We can ask if the reverse is true: if King Kong falls off the Empire State Building, will he hit at 193 mph?

68. Air drag, v max = 120-130 mph. Equivalent to jumping from about 150 meters (about 450 feet). Over 150 meters is academic.

69. Springs and Conservation of Energy Springs store potential energy ½ mv 2 1 + ½ kx 2 1 = ½ mv 2 2 + ½ kx 2 2 Use only for horizontal problems

70. Springs: Example 1 A 0.100 kg toy dart is compresses a spring 6.0 cm. The spring’s constant is 250 N/m. What speed will the dart leave the gun?

71. ½ mv 2 1 + ½ kx 2 1 = ½ mv 2 2 + ½ kx 2 2 0 + ½ kx 2 1 = ½ mv 2 2 + 0 ½ kx 2 1 = ½ mv 2 2 kx 2 1 = mv 2 2 v 2 2 = kx 2 1 /m =(250 N/m)(0.060m) 2 /(0.100 kg) v 2 2 = 9.0 m/s 2 v 2 = 3.0 m/s

72. A 1.50 kg block slides along a smooth table and collides with a spring (k = 1000 N/m). If block had an initial speed of 1.20 m/s, how far was the spring compressed? (4.65 cm)

73. Springs and Vertical Movement Car springs/struts Bungee Cord Make the zero the unstretched spring ½ mv 2 1 + mgy 1 + ½ ky 2 1 = ½ mv 2 2 + mgy 2 + ½ ky 2 2

74. A 2.60 kg ball is dropped. It falls 55.0 cm before hitting a spring. It compresses the spring 15.0 cm before coming to rest. Calculate the spring constant.

75. 55.0 cm -15.0 cm

76. ½ mv 2 1 + mgy 1 + ½ ky 2 1 = ½ mv 2 2 + mgy 2 + ½ ky 2 2 0 + mgy 1 + 0 = 0 + mgy 2 + ½ ky 2 2 mgy 1 = mgy 2 + ½ ky 2 2 (2.60 kg)(9.8m/s 2 )(0.550m) = (2.60 kg)(9.8m/s 2 )(-0.150m)+ ½ k(-0.150m) 2 14.0 J = -3.82 J + (0.01125m 2 )(k) 17.82 J = (0.01125m 2 )(k) k = 1580 N/m

77. A vertical spring has a spring constant of 450 N/m and is mounted on the floor. A 0.30 kg block is dropped from rest and compresses the spring by 2.50 cm. Calculate the height from which the block was dropped. Use the unstretched spring length as your zero point. (2.28 cm)

78. A 300 g block is dropped from rest 40.0 cm above a spring. It compresses the spring by 9.50 cm. Calculate the spring constant (322 N/m)

79. A vertical spring has a spring constant of 895 N/m is compressed by 15.0 cm. a) Calculate the upward speed can it give to a 0.360 kg ball when released. (7.28 m/s) b) Calculate how high above the top of the uncompressed spring the ball will fly. (2.70 m)

83. Law of Conservation of Mechanical Energy If nonconservative forces act, use: KE 1 + PE 1 = KE 2 + PE 2 + W fr ½ mv 2 1 + mgy 1 = ½ mv 2 2 + mgy 2 + F fr d Starting energySome of the starting energy lost to friction

84. Mr. Fredericks (100 kg) slides down a 3.5 m tall slide. He leaves the slide at a speed of 6.3 m/s. a)Calculate the Force of friction. (241 N) b)Calculate the coefficient of friction for the slide.(0.303) c)Would there be more or less friction if you raise the left side of the slide. 3.5 m6.0 m

85. ½ mv 2 1 + mgy 1 = ½ mv 2 2 + mgy 2 + F fr d 0 + mgy 1 = ½ mv 2 2 + 0 + F fr d mgy 1 = ½ mv 2 2 + F fr d F fr d = mgy 1 - ½ mv 2 2 F fr = (mgy 1 - ½ mv 2 2 )/d F fr = [(100 X 9.8 X 3.5) – (½ X 100 X 6.3 2 )]/6.0 F fr = 241 N

86. To calculate the coefficient of friction: F fr =  mgcos  (mgcos  is the Normal Force) tan  = 3.5/6.0  = 30.3 o  = F fr /mgcos   = 241 N/[(100 kg X 9.8 m/s 2 )(cos 30.3 o )]  = 0.28

87. Friction: Example 2 A 1000 kg roller coaster starts from a height of 40 m. On the second hill, it only rises to a height of 25 m. Calculate the force of friction. Assume a travel distance of 400 m.

88. ½ mv 2 1 + mgy 1 = ½ mv 2 2 + mgy 2 + F fr d 0 + mgy 1 = 0 + mgy 2 + F fr d mgy 1 = mgy 2 + F fr d F fr = (mgy 1 - mgy 2 )/d F fr = (1000 X 9.8 X 40 – 1000 X 9.8 X 25)/400 F fr =368 N

89. A delivery boy wishes to slide a 2.00 kg package up a 3.00 m long ramp. The ramp makes a 20 o angle with the ground, and has a coefficient of friction of 0.40. a)Calculate the height of the ramp (1.03 m) b)Calculate the normal force on the package (18.4 N) c)Calculate the minimum speed the package must have to go up the ramp. Remember to take Potential Energy and the energy lost to friction into account. (6.5 m/s) d)Draw free-body diagrams for the box at the bottom and in the middle of the ramp.

90. Mr. Fredericks (56.0 kg) slides down a 2.50 m tall slide. The slide has a coefficient of 0.250 and makes a 45.0 o angle with the ground. a)Calculate the length of the slide. (3.53 m) b)Calculate the normal force. (388 N) c)Calculate the force of friction. (97.0 N) d)Calculate how fast he will leave the slide. (6.06 m/s) 2.50 m

91. Power Definition – rate at which work is done –A powerful engine can do a lot of work quickly. –Running and walking up the steps require the same amount of work. –Running up steps requires more Power

92. Power Power = WorkP = W timet Metric Unit:Joules/s = Watt.

93. Power: Example 1 A donkey performs 15,000 J of work pulling a wagon for 20 s. What is the donkey’s power?

94. Power: Example 2 If the Angel snatches Juggernaut’s helmet (10 kg) and flies up 10 m with it in 0.75 seconds, what is the Angel’s power?

95. Suppose Mr. Fredericks (56.0 kg) can run up the Empire State Building in 2.00 seconds.The Empire State Building is 380 m tall. a.Calculate the work done (2.09 X 10 5 J) b.Calculate the power (1.05 X 10 5 W) c.How could the power be increased?

96. Example 4 A 60-kg runner and a 100-kg runner run 50 m up a 6.0 o slope. Calculate the work done by both runners. 50 m 6.0 o y

97. First we need to find the height they rise: sin 6.0 o = y 50 m y = 5.23 m Remember that work is a change in energy W =  PE W = PE f – PE i W = mgy – 0

98. W 60 = mgy = (60.0 kg)(9.8m/s 2 )(5.23 m) W 60 = 3075 J W 100 = mgy = (100.0 kg)(9.8m/s 2 )(5.23 m) W 100 = 5125 J

99. If it takes the 60 kg man 6 seconds, and the 100 kg man 10 seconds, who has more power? P = W/t P 60 = 3075 J/6 s = 513 W P 100 = 5125 J/10 s = 513 W

100. A 50.0 kg child/wagon is pushed from rest in a wagon. The wagon leaves with a speed of 6.00 m/s, and the person pushing has a power of 100.0 Watts. a)Calculate the work done on the child and wagon. (900. J) b)Calculate the time of the push. (9 s) c)If the coefficient of kinetic friction is 0.25, calculate how far the wagon will travel. (7.35 m)

101. Horsepower The English Unit of power is horsepower Foot-lb= Horsepower (hp) second 1 hp = 746 Watts 1 hp = ½ Columbus (who sailed in 1492)

102. Horsepower: Example 1 How much horsepower is required to power a 100 Watt lightbulb?

103. Horsepower Consider a 40 hp car engine that can go from 0 to 60 mi/hr in 20 seconds. A 160 hp car could go from zero to 60 mi/hr in 5 seconds. 4 times as powerful means it can do the same work in ¼ the time.

104. Horsepower: Example 2 Suppose you have a 1.34 hp lawnmower. How many Watts of power does it have?

105. Horsepower: Example 3 A crane lifts a 200 N box 5 meters in 3 seconds. What is the crane’s power in Watts and in horsepower?

106. You leave a 60 W lightbulb on outside your house for 12 hours a day for an entire month (30 days). How many kiloWatt hours will you be charged for?

107. Decide to turn your computer (75 W) off every night. Assume that it is off for 9 hours every night for 30 days. How many kilowatt-hours did you save?

108. Suppose you are charged for 1500 kWh for a given month. How many Joules of energy did you use that month? (Hint: Start from the power formula, and just use one hour)

109. Suppose you are charged for 3000 kWh for a given month. How many Joules of energy did you use that month? (Hint: Start from the power formula)

110. You push a 30.0 kg child up a slide that is 4.00 m long (hypotenuse) and has an angle of 33.0 o. The slide has a  k of 0.22. a.Calculate the height of the slide. (2.18 m) b.Calculate the initial speed that is needed to push the child to that height. Include friction and be sure to calculate the normal force. (7.56 m/s) c.What percentage of the initial energy was lost to friction? (25.3%) d.If you pushed for 1.3 s, what is your power in Watts? (659 W) e.What is his power in hp? (0.884 hp)

111. 2. a) 1.1 X 10 3 Jb) 5.4 X 10 3 J 4. 25 N 6. 7.8 J 8. 170 J 10.a) 420 Nb) -1800 Jc) -4100 J d) 5900 Je) 0 12. 5000 J 18. Square root 2, 4 20. -4.67 X 10 5 J 22. 44 m/s 24. 2.2526. -1.1 N

112. 26. -1.1 N 30. 71 J 32. a) 45.3 Jb) 12.3 Jc) 45.3 J 34. a) ½ k(x 2 -x o 2 )b) ½ kx o 2 (Same) 36. 49.5 m/s 38. 6.5 m/s 40. v b = 24 m/s, v c = 9.9 m/s, v d = 19 m/s 42. a) 100 N/mb) 22 m/s 2 50. 530 J 52. 12 m/s 54. 0.3174. 8.0 m/s

113. 58. 25.5 s 60. 0.134 hp 62. 360 W 74. 8.0 m/s