Chapter 10. Energy

1. Chapter 10. Energy Chapter Goal: To introduce the ideas of kinetic and potential energy and to learn a new problem-solving strategy based on conservation of energy.

2. Ch. 10 – Student Learning Objectives • To begin developing a concept of energy—what it is, how it is transformed, and how it is transferred. • To introduce the concepts of kinetic and potential energy. • To learn Hooke’s law for springs and the new idea of a restoring force.

3. Conservation of Mechanical Energy – The Big Picture • Two basic types of mechanical energy: • Kinetic energy (K) is an energy of motion. • Gravitational potential energy (Ug) is an energy of position. • Under some circumstances (e.g. freefall) these two kinds of energy can be transformed back and forth without any loss from the system.

4. Conservation of Mechanical Energy For some systems (e.g. systems in freefall), mechanical energy is conserved:

5. ½ mv2f + mgyf = ½ mv20 + mgy0 where y is the height above an arbitrary zero, (not the displacement). This result is true for motion along any frictionless surface, regardless of the shape. This is a generalization of a relationship we already use for free fall: v2f = v20 + 2(-g)(yf – y0)

6. Conservation of Kinetic and Potential Energy

7. A box slides along the frictionless surface shown in the figure. It is released from rest at the position shown. Is the highest point the box reaches on the other side at level a, at level b, or level c? • At level a • At level b • At level c

8. Analysis of energy conservation problems: The before and after picture Bob uses a slingshot to launch a 20.0 g pebble up with an initial speed of 25.0 m/s. How high above his hand does the pebble go?

9. Energy Bar charts – a visual aid

10. Analysis of energy conservation problems: bar charts Bob uses a slingshot to launch a 20.0 g pebble up with an initial speed of 25.0 m/s. How high above his hand does the pebble go? Now solve using the appropriate parts of the conservation of mechanical energy equation.

11. Analysis of energy conservation problems: bar charts Bob uses a slingshot to launch a 20.0 g pebble up with an initial speed of 25.0 m/s. How high above his hand does the pebble go? Now solve using the appropriate parts of the conservation of mechanical energy equation: 32 meters

12. Workbook Problems 5-7 • A car runs out of gas and coasts up a hill until finally stopping • A pendulum is held out at 45 degrees and released from rest. A short time later, it swings through the lowest point on the arc. • A ball starts from rest on on the top of one hill, rolls without friction through a valley and just barely makes it to the top of an adjacent hill before stopping

14. EOC #4 • What is the kinetic energy of a 1500 kg car traveling at 30 m/s? • From what height would the car have to be dropped (!) to have the same amount of energy upon impact?

15. 6.75 x 105 Joules Solve for the amount of initial potential energy needed to give the car a final kinetic energy equal to part a to get y = 46 m EOC#4 K0 + Ug0 = Kf + Ugf

16. The Zero of Potential Energy • You can place the origin of your coordinate system, and thus the “zero of potential energy,” wherever you choose  and be assured of getting the correct answer to a problem. • The reason is that only ΔU has physical significance, not   Ug itself.

17. Workbook Problem #2 We see a 1 kg object that is initially 1 m above the ground and rises to a height of 2 m. Anjay, Brittany and Carlos each measure its position, but each of them uses a different coordinate system. Determine Ui, Ufand ∆U for each coordinate system. Which of these, if any remain constant for all 3 coordinate systems?

18. Workbook Problem #2 - Answer ∆ U remains the same for all.

19. Conservation of Mechanical Energy The sum of the kinetic energy and the potential energy of a system is called the mechanical energy. the kinetic energy and the potential energy can change, as they are transformed back and forth into each other, but their sum remains constant.

20. Conservation of Mechanical Energy • Under what conditions is Emech conserved? • What happens to the energy when Emech is not conserved? • Are there potential energies other than gravitational, and how do you calculate them?

21. Hooke’s Law If you stretch a rubber band, a force appears that tries to pull the rubber band back to its equilibrium, or unstretched, length. A force that restores a system to an equilibrium position is called a restoring force. If s is the position of the end of a spring, and se is the equilibrium position, we define Δs = s – se. If (Fsp)s is the s-component of the restoring force, and k is the spring constant of the spring, then Hooke’s Law states that The minus sign is the mathematical indication of a restoring force.

22. Hooke’s Law

23. Workbook #10 A spring has an unstretched length of 10 cm. It exerts a restoring force F when stretched to a legth of 11 cm. For what length is the restoring force 3F? • 33 cm • 13 cm • 2 cm • 14 cm

24. The graph shows force versus displacement for three springs. Rank in order, from largest to smallest, the spring constants k1, k2, and k3. • k1, k2, k3 • k3, k2, k1 • k1 =k2 = k3

25. Elastic Potential Energy

26. Elastic Potential Energy • Us = ½ k (∆s)2 • ∆s = (s-se) where se is the equilibrium position of the end of the spring • ∆Us = ½ k [(sf - se)2 - (s0 - se)2]. Often sf or s0 = se as the spring begins or ends in equilibrium position. at s=se, the ball will lose contact with the spring

27. EOC #21 A student places his 500 g physics book on a frictionless table and pushes it against a spring, compressing it by 4.0 cm. The spring constant is 1250 N/m. He then releases the book. What is the speed as it slides away from the spring? Draw a before/after sketch and an energy bar chart for this situation.

28. EOC #21 Knowns m=.5 kg k = 1250 N/m x1 = -.04 m (origin placed at equilibrium of spring) x2 = 0 m v1 = 0m/s Find: v2, the speed of the book when it is released Draw an energy bar chart to determine what energy transformation takes place

29. EOC #21 • The elastic potential energy of the spring is converted into the kinetic energy of the book. • Solve by replacing each energy term in the bar chart with a value:

30. v2 = 2.0 m/s

31. Problem with Ug and Us (EOC #44) A 1000 kg safe is 2.0 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 50 cm. What is the spring constant of the spring? Draw a before and after picture and decide on a zero. Draw an energy bar chart.

32. Note that mgy1 is negative for the zero chosen. A 1000 kg safe is 2.0 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 50 cm. What is the spring constant of the spring?

33. Using the non-zero terms from the energy bar chart, we get: k= 196,000 N/m Note that mgy1 is negative for the zero chosen. A 1000 kg safe is 2.0 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 50 cm. What is the spring constant of the spring?

34. Problem with Ug and Us A 20-cm-tall spring with spring constant of 5000 N/m is placed on the ground as shown. A 100-N block, is released from rest, 15 cm above the top of the spring, compressing it. What is the height H of the spring at the point of maximum compression? (Hint: quadratic equation).

35. Problem with Ug and Us A 20-cm-tall spring with spring constant of 5000 N/m is placed on the ground as shown. A 100-N block, is released from rest, 15 cm above the top of the spring, compressing it. What is the height H of the spring at the point of maximum compression? y1 = -10 cm in the coordinate system chosen; and therefore H = 10 cm

36. Collisions revisited • Kinetic energy of the colliding objects are changed into microscopic elastic potential energy and then back again to kinetic energy. • This very seldom happens without energy being transferred from the 2-object system to the environment (as noise or heat).

37. Inelastic collision • During an inelastic collision, momentum is conserved, but much of the kinetic energy transferred out of the system. • During an elastic collision, kinetic energy and momentum are conserved. • Most real collisions fall somewhere between the two extremes, but we will use this simplified model. Elastic collision

38. Elastic Collisions Consider a head-on, perfectly elastic collision of a ball of mass m1 having initial velocity (vix)1, with a ball of mass m2 that is initially at rest. The solution, worked out in the text, is

39. Elastic Collisions What is vf1 if the two balls have the same mass? What is vf2 if the two balls have the same mass? When will vf1 be in the opposite direction of vi1? When will it be in the same direction? How about vf2 ? Can it ever be in the opposite direction? Explain

40. Elastic Collisions What happens to the speed of each as m1 >> m2 ? What happens to the speed of each as m2 >> m1 ? Final speeds (and directions) depend on the relationship between the masses.

41. EOC - 28 Ball 1 with a mass of 100 g and traveling at 10 m/s collides head-on with ball 2, which has a mass of 300 g, initially at rest. a. What is the final velocity of each ball if the collision is perfectly elastic? b. What is the final velocity of each ball if the collision is perfectly inelastic? c. What percentage of the system’s original kinetic energy is lost in the inelastic collision?

42. EOC - 28 Ball 1 with a mass of 100 g and traveling at 10 m/s collides head-on with ball 2, which has a mass of 300 g, initially at rest. a. What is the final velocity of each ball if the collision is perfectly elastic? vf1 = -5m/s, vf2 = 5 m/s b. What is the final velocity of each ball if the collision is perfectly inelastic? vf1 = vf2 = 2.5 m/s c. What percentage of the system’s original energy is lost in the inelastic collision? 75% Where did it go?

43. Energy Diagrams – For systems with no energy loss • A graph showing a system’s potential energy and total energy as a function of position is called an energy diagram. • The object’s kinetic energy is TE – Ug. The object’s potential energy is read directly off the PE graph. • The particle can’t have more energy than TE. Even though the PE curve goes above the TE line, the particle cannot reach any position to the right of the intersection of TE and PE. This is the turning point.

44. Energy Diagram of a mass on a horizontal spring You can only change the height of the TE line. The PE curve is determined by the spring. The object cannot reach a position where the PE curve is above the TE line. The intersection of the two is the turning point.

45. An oscillating mass

46. Interpreting an energy diagram of an object that has several kinds of potential energies • If the particle starts at the intersection of the TE and PE curves, it has no K. • As PE decrease, K must increase. • At the turning point, its speed K are both zero.

47. Equilibrium: ΣF = 0 • Minimums and maximums on the PE curve are points of equilibrium. • A minimum is a point of stable equilibrium (eg the bottom of the hill). • A maximum is a point of unstable equilibrium. • We’ll investigate the relationship between force and potential energy in Ch. 11.

48. A particle with the potential energy shown in the graph is moving to the right. It has 1 J of kinetic energy at x = 1 m. Where is the particle’s turning point? • 2m • 5m • 6m • Not on graph

49. EOC #29 - This is the PE curve for a 20-g particle that is released from rest at x = 1 m. • Will the particle move to the left or right. How do you know? • What is the maximum speed of the particle? Where does this happen? • What is/are the turning point/s of the motion?

50. EOC #29 - This is the PE curve for a 20-g particle that is released from rest at x = 1 m. • Will the particle move to the left or right. How do you know? – right, particle cannot move left since TE = 4 J • What is the maximum speed of the particle? Where does this happen? 17.3 m/s at x = 4 m. • What is/are the turning point/s of the motion? x=1m and x=6m.