1. “It is important that students bring a certain ragamuffin, barefoot irreverence to their studies; they are not here to worship what is known, but to question it.” Jacob Chanowski Reading: Chapter 6. Raw grades are on Blackboard. Homework 3 is on-line now and due on Friday at 5pm.

2. Chapter 6 Work and Energy Work W = Fs*cos q

3. A B C What happens between points A and B?

4. A B C Work is done by gravity.

5. A B C What happens between points A and B? What happens to the speed of the person?

6. A B C What happens between points A and B? What happens to the speed of the person? It increases. The work done by gravity goes into changing the speed of the person.

7. Back to our skier We found that gravity did work on this skier. Where did that work go?

8. Back to our skier We found that gravity did work on this skier. Where did that work go? It went into making the skier go faster!

9. Work – Energy equivalency principle. The total work done on an object is equal to the change in its energy. W total = ∆E

10. Work – Energy equivalency principle. The total work done on an object is equal to the change in its energy. In our slide and skiing examples, work done by gravity is converted into motion (kinetic energy) of the person.

11. So now we need to define some energies.

12. Kinetic Energy (KE) is the energy of motion. KE is measured in Joules (just like work). KE = ½ mv 2 NOTE that because we have v 2, we are directionally impaired and answers will NOT give the direction of the velocity.

13. Work-Energy Theorem The total work done on an object is equal to the change in its energy – usually kinetic but it can be other forms. W total = ∆KE = KE f -KE o Fs*cos q = ½ m(v f 2 -v o 2 )

14. Kinetic Energy Example What is the kinetic energy of a 0.1kg hockey puck traveling at 25m/s? KE = ½ mv 2 A) 1.25J B) 27.1J C) 31.3J D) 37.4J E) 0J

15. Kinetic Energy Example What is the kinetic energy of a 0.1kg hockey puck traveling at 25m/s? KE = ½ mv 2 KE= ½ (0.1)(25 2 ) = 31.3J

16. Kinetic Energy Example A football is thrown across the field and caught by a person standing still. Just before it's caught, it's moving at 28m/s. If the mass of the football is 0.6kg, how much work was done on the football in catching it? Where do we start?

17. Kinetic Energy Example A football is thrown across the field and caught by a person standing still. Just before it's caught, it's moving at 28m/s. If the mass of the football is 0.6kg, how much work was done on the football in catching it? Where do we start? W=F. s cosq We don't know how much force was applied or over what distance.

18. Kinetic Energy Example A football is thrown across the field and caught by a person standing still. Just before it's caught, it's moving at 28m/s. If the mass of the football is 0.6kg, how much work was done on the football in catching it? Where do we start? We don't know how much force was applied or over what distance. But we do know that its final motion is 0, so ALL of its kinetic energy has been converted into work.  KE → W

19. Kinetic Energy Example A football is thrown across the field and caught by a person standing still. Just before it's caught, it's moving at 28m/s. If the mass of the football is 0.6kg, how much work was done on the football in catching it? KE → W  KE = ½ mv 2

20. Kinetic Energy Example A football is thrown across the field and caught by a person standing still. Just before it's caught, it's moving at 28m/s. If the mass of the football is 0.6kg, how much work was done on the football in catching it? KE → W KE = ½ mv 2 =1/2 (0.6)(28 2 ) = 235.2J All this energy was lost to the work of catching the ball.

21. Two women push a car (m=870kg) with a force of 235N for 23m. How fast is the car going when they're done? (Assuming no friction.)

22. Two women push a car (m=870kg) with a force of 235N for 23m. How fast is the car going when they're done? The work of the women is converted into kinetic energy: W →  KE

23. Two women push a car (m=870kg) with a force of 235N for 23m. How fast is the car going when they're done? W=F. scosq → KE=1/2 mv 2 235(23)=1/2 (870) v 2

24. Another type of energy: Gravitational Potential Energy. The skydiver's kinetic energy increases after he leaves the plane. That energy must come from somewhere.

25. Another type of energy: Gravitational Potential Energy: PE. The skydiver is converting PE into kinetic energy. PE = mg(h f -h o ) Most often PE = mgh. Caution: g is positive for energy!!!!!

26. Potential energy is the stored ability to do work: stored energy.

27. EXTREMELY IMPORTANT: ENERGY IS NEVER LOST. IT IS ONLY CONVERTED FROM ONE FORM TO ANOTHER. (It may be lost to “a system” like a cup of coffee loses heat to a room, but it is still there.)

28. The top of this platform is 10m above the water. How fast will a diver be moving when he/she hits the water? (Note that this problem can be solved without knowing the diver's mass!!!) Where to start?

29. The top of this platform is 10m above the water. How fast will a diver be moving when he/she hits the water? (Note that this problem can be solved without knowing the diver's mass!!!) Where to start? We could use velocity and acceleration from Chapter 3: v f 2 =v o 2 +2gy with v o =0 and y=10m. This gives an answer v f = -14m/s

30. The top of this platform is 10m above the water. How fast will a diver be moving when he/she hits the water? (Note that this problem can be solved without knowing the diver's mass!!!) Where to start? We could use v f 2 =v o 2 +2gy from Chapter 3: v f = -14m/s Or we could convert the change in PE (use the surface of the water as h=0) to a change in KE (the diver begins at rest): PE o = KE f mgh = ½ mv f 2 → v f 2 = 2gh (look familiar?)

31. Converting forms of energy As the ball rolls down the incline, it is converting gravitational potential energy into kinetic energy. Neglecting wind and friction, no energy is lost, only converted.

32. Converting forms of energy If I know the energy at one point, I know it at ALL points. E A =E B =E C Only its form could change.

33. Converting forms of energy If I know the energy at one point, I know it at ALL points. E A =E B =E C Only its form could change. mgh A =mgh B + 1/2 mv B 2 =mgh C + 1/2 mv C 2

34. Converting forms of energy E A =E B =E C I can define the axes to make it even simpler. mgh A = 1/2 mv B 2 =mgh C + 1/2 mv C 2

35. Converting forms of energy E A =E B =E C If the ball is 2kg, how much energy does it have? Where to begin? mgh A = 1/2 mv B 2 =mgh C + 1/2 mv C 2

36. Converting forms of energy E A =E B =E C If the ball is 2kg, how much energy does it have? Where to begin? We only have enough information to solve the energy for Point A. mgh A = 1/2 mv B 2 =mgh C + 1/2 mv C 2

37. Converting forms of energy E A =E B =E C If the ball is 2kg, how much energy does it have? We only have enough information to solve the energy for Point A. E A =mgh=2(9.8)7 mgh A = 1/2 mv B 2 =mgh C + 1/2 mv C 2

38. Converting forms of energy E A =E B =E C If the ball is 2kg, how much energy does it have? We only have enough information to solve the energy for Point A. E A =mgh=2(9.8)7 = 137.2 J mgh A = 1/2 mv B 2 =mgh C + 1/2 mv C 2

39. Converting forms of energy E A =E B =E C If the ball is 2kg, how fast is it moving at points B and C? E A = 137.2 J mgh A = 1/2 mv B 2 =mgh C + 1/2 mv C 2 At B: 137.2J = ½(2)v B 2 At C: 137.2J=2(9.8)3+½(2)v C 2

40. Converting forms of energy E A =E B =E C If the ball is 2kg, how fast is it moving at points B and C? E A = 137.2 J mgh A = 1/2 mv B 2 =mgh C + 1/2 mv C 2 At B: 137.2J = v B 2 At C: 78.4 = v C 2

41. Converting forms of energy E A =E B =E C If the ball is 2kg, how fast is it moving at points B and C? E A = 137.2 J mgh A = 1/2 mv B 2 =mgh C + 1/2 mv C 2 At B: v B = 11.7 m/s At C: v C = 8.85 m/s

42. Conservation of Energy Would the speed of the ball at points B and/or C be different if the ball took the blue or orange path? (still neglecting friction and other forces)

43. Conservation of Energy Would the speed of the ball at points B and/or C be different if the ball took the blue or orange path? (still neglecting friction and other forces) No! The speed will be the same. Energy is not lost, only converted. So the path doesn't matter!

44. Law of Conservation of Energy – energy can neither be created nor destroyed – it can only change forms.

45. Mechanical => Kinetic (motion) & Potential (spring) Other forms of Energy => Chemical, Heat, Nuclear, Solar, Geothermal, Sound, etc. Electromagnetic Energy

46. Nonconservative interactions: Where energy is lost from the system. How could our ball example be a nonconservative example?

47. Nonconservative interactions: Where energy is lost from the system. How could our ball example be a nonconservative example? If we allow friction or wind resistance.

48. Nonconservative interactions: Where energy is lost from the system. How could our ball example be a nonconservative example? If we allow friction or wind resistance. What's the speed of the ball at points B and C if m K =0.2?