1. PHYS 20 LESSONS Unit 4: Energetics Lesson 4: Conservation of Mechanical Energy

2. Reading Segment: Conservation of Mechanical Energy To prepare for this section, please read: Unit 4: p. 14

3. D. Conservation of Energy Recall: There are two major advantages to the energetics perspective:

4. D. Conservation of Energy Recall: There are two major advantages to the energetics perspective: 1. Energy is a scalar - does not have direction - no need for vector math 2. It is conserved We will now analyze situations where total mechanical energy is conserved.

5. Conservation of Mechanical Energy - total mechanical energy stays constant (conserved) when: * the system is closed - no objects are added / lost * no friction * only mechanical forms of energy change - only F g and F s do work on the object - mechanical energy can convert only to another form of mechanical energy

6. Equations: If mechanical energy is conserved, then you can use two equations: 1. Em T remains constant Em Ti = Em Tf Epg i + Ek i = Epg f + Ek f

7. 2. No overall change in energy  Ep g +  Ek = 0  Ep g = -  Ek That is, if Ep g increases by 100 J, then Ek decreases by 100 J (no overall change) So, one form of mechanical energy transforms into another form of mechanical energy

8. e.g. Consider an object thrown downward (No air resistance) Epg i = 600 J Ek i = 200 J Epg f Ref h (h = 0) Ek f What is the total mechanical energy at the start?

9. Em Ti Epg i = 600 J 800 JEk i = 200 J Epg f Ref h (h = 0) Ek f What is the total mechanical energy at the end?

10. Em Ti Epg i = 600 J 800 JEk i = 200 J Em Ti = Em Tf Em Tf Epg f Ref h (h = 0) 800 JEk f So, what is its kinetic energy at the end?

11. Em Ti Epg i = 600 J 800 JEk i = 200 J Em Tf Epg f = 0 Ref h (h = 0) 800 JEk f = 800 J

12. Epg i = 600 J Ek i = 200 J Calculate and interpret: - the change in Ep g - the change in Ek Epg f = 0 Ek f = 800 J

13. Epg i = 600 J Ek i = 200 J  Epg = Epg f - Epg i = 0 - 600 J = - 600 J object loses 600 J of Epg Epg f = 0 (since it loses height) Ek f = 800 J

14. Epg i = 600 J Ek i = 200 J  Ek = Ek f - Ek i = 800 J - 200 J = + 600 J object gains 600 J of Ek Epg f = 0 (since it gains speed) Ek f = 800 J

15. Epg i = 600 J Ek i = 200 J The object loses 600 J of Epg, but it gains 600 J of Ek We say that the Ep g has transformed into Ek. Epg f = 0 Ek f = 800 J

16. Animations: 1. Dropped object: http://departments.weber.edu/physics/amiri/director/dcrfiles/en ergy/FallingBallS.dcr Can you explain what happens in the animation?

17. You should have noticed: The Ep g goes down by 1078 J, but at the same time, the Ek goes up by 1078 J That is, the Ep g has transformed entirely into Ek Thus, the total mechanical energy will remain the same throughout the entire motion

18. Other animations showing conservation of energy: 2. Pendulum: http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=27 3. Rollercoaster: http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/qt/energy/ coastwin.html

19. Ex. A ball is dropped from a height H and it lands with a speed of 7.0 m/s. If the system is conservative, a) find H b) sketch a Ep g vs t, Ek vs t, and a Em T vs t graph

20. a) Em Ti = Em Tf Rest Epg i + Ek i = Epg f + Ek f Ref h (h = 0) Be certain to state7.0 m/s where the height is zero

21. a) Em Ti = Em Tf Rest Epg i + Ek i = Epg f + Ek f Ref h (h = 0) 7.0 m/s Since it starts at rest, Ek i = 0 Since it has no height at the end, Epg f = 0

22. Em Ti = Em Tf Epg i + Ek i = Epg f + Ek f mgh i = 0.5 mv f 2 gh i = 0.5 v f 2 If you divide both sides of the equation by m, the mass cancels Thus, the answer does not depend on mass. i.e. It is true for any mass

23. Em Ti = Em Tf Epg i + Ek i = Epg f + Ek f mgh i = 0.5 mv f 2 gh i = 0.5 v f 2 h i = 0.5 v f 2 = 0.5 (7.0 m/s) 2 g9.81 m/s 2 H = 2.5 m

24. b) Ep g Ek tt Em T t

25. b) Ep g Ek height speed decreases increases tt Em T constant t

26. Ex. 3 A 1.20 kg car travels the following path. No friction. A 6.70 m/s v ? C 8.70 m B 4.20 m 1.90 m D Find its speed at C.

27. A 6.70 m/s v ? C 8.70 m B 4.20 m 1.90 m D (h = 0) The reference height (h = 0) is at D, the lowest location in the diagram

28. A 6.70 m/s v ? C 8.70 m B 4.20 m 1.90 m D (h = 0) Total mechanical energy remains the same. So, Em TA = Em TC Epg A + Ek A = Epg C + Ek C

29. Ref h = D Epg A = mgh A = (1.20 kg) (9.81 N/kg) (8.70 m) = 102.42 J Ek A = 0.5mv A 2 = 0.5 (1.20 kg) (6.70 m/s) 2 = 26.93 J Epg C = mgh C = (1.20 kg) (9.81 N/kg) (4.20 m) = 49.44 J

30. Find Ek C : Em TA = Em TB Epg A + Ek A = Epg C + Ek C 102.42 J + 26.93 J = 49.44 J + Ek C Ek C = 79.908 J

31. Find speed at C: Ek C = 0.5 mv 2 v 2 = Ek C 0.5 m v = Ek C = 79.908 J = 11.5 m/s 0.5 m0.5 (1.20 kg)

32. Practice Problems Try these problems in the Physics 20 Workbook: Unit 4 p. 15 #1 - 3

33. Ex. 4 A 14.0 kg object is currently moving at a height of 75.0 cm. If it then loses 92.0 J of Ek, find the object's new height. Assume a conservative system.

34. Solution The total mechanical energy must remain the same So, if the object loses 92.0 J of Ek, it must gain 92.0 J of Ep g That is, the Ek converts (transforms) into Ep g

35. Find Epg i : Epg i = mgh i = (14.0 kg) (9.81 N/kg) (0.750 m) = 103.0 J

36. Find Epg f : Since Ep g must increase by 92.0 J, Epg f = Epg i + 92.0 J = 103.0 J + 92.0 J = 195.0 J

37. Find the final height: Epg f = mgh f h f = Epg f = 195.0 J = 1.42 m mg (14.0 kg) (9.81 N/kg)

38. Practice Problems Try these problems in the Physics 20 Workbook: Unit 4 p. 15 #4 - 8