© 2007 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their.

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1 © 2007 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. Lecture Outlines Chapter 8 Physics, 3 rd Edition James S. Walker

2 Chapter 8 Potential Energy and Conservation of Energy

3 Units of Chapter 8 Conservative and Nonconservative Forces Potential Energy and the Work Done by Conservative Forces Conservation of Mechanical Energy Work Done by Nonconservative Forces Potential Energy Curves and Equipotentials

4 8-1 Conservative and Nonconservative Forces Conservative force: the work it does is stored in the form of energy that can be released at a later time Example of a conservative force: gravity Example of a nonconservative force: friction Also: the work done by a conservative force moving an object around a closed path is zero; this is not true for a nonconservative force

5 Figure 8-1 Work against gravity

6 Figure 8-2 Work against friction (hastigheten är konstant)

7 8-1 Conservative and Nonconservative Forces Work done by gravity on a closed path is zero: (Systemet sett från sidan)

8 8-1 Conservative and Nonconservative Forces Work done by friction on a closed path is not zero: (Systemet sett uppifrån = fågelperspektiv)

9 Figure 8-5 Gravity is a conservative force

10 8-1 Conservative and Nonconservative Forces The work done by a conservative force is zero on any closed path:

11 Table 8-1 Conservative and Nonconservative Forces

12 Example 8-1a (p.208) Different Paths, Different Forces (m= 4,57 kg) W 1 = -mg1,0 + 0 = -mg1,0 ; W 2 = 0 – mg2,0 + 0 + mg1,0 = -mg1,0 W 1 = W 2 (oberoende av vägen) (mg1,0 meter = 45 J)

13 Example 8-1b (p.208) Different Paths, Different Forces (m = 4,57 kg, μ = 0,63) W 1 = - mg0,634,0= -110 J ; W 2 = - mg0,6312,0 = - 330 J, dvs W 1 ≠ W 2 (beror av vägen)

14 8-2 The Work Done by Conservative Forces If we pick up a ball and put it on the shelf, we have done work on the ball. We can get that energy back if the ball falls back off the shelf; in the meantime, we say the energy is stored as potential energy. (8-1)

15 8-2 The Work Done by Conservative Forces Gravitational potential energy:

16 Exercise 8-1 Vad är potentiella energin för ett system som består av en person (m=65 kg) på en 3,0- meters svikt? Sätt U = 0 vid vattenytan. U = mgh = 65 kg 9,81 m/s 2 3,0 m = 1,9 kJ

17 Example 8-2 (p.211) Pikes Peak or Bust (m=82,0 kg)

18 Exercise 8-2 Pikes Peak or Bust Vad är ändringen i den potentiella energin för systemet som består av en person (m=82,0 kg) när han klättrat de sista 100,0 meterna? a) Sätt U = 0 vid havsytan (y f = 4301 m, y i = 4201 m) ΔU = U f – U i = mgy f - mgy i = 82 kg9,81 m/s 2 (4301m - 4201 m) = 80,4 kJ b) Sätt U = 0 på bergstoppen ΔU = U f – U i = mgy f - mgy i = 0 – (82 kg9,81 m/s 2 (-100,0 m)) = 80,4 kJ

19 Example 8-3 Converting Food Energy to Mechanical Energy

20 Exemple 8-3 Converting Food Energy to Mechanical Energy Vad skulle kunna vara ändringen i den potentiella energin för systemet som består av en person (m=81,0 kg) om han kunde omsätta hela sitt kaloriintag (212 Cal = 212 kcal) i en direkt “höjdvinst”? 212 kcal = 212 kcal 4,186 J/cal = 887 kJ ΔU = U f – U i = mgy f - mgy i = 81,0 kg9,81 m/s 2 Δ y = 887 kJ Δy = 1120 m

21 8-2 The Work Done by Conservative Forces Springs: (8-4)

22 Exemple 8-4 Compressing Energy (and the Jump of a Flea) När kraften 120,0 N drar i en fjäder orsakar den en förlängning på 2,25 cm. Vad blir den (lagrade) potentiella energin om fjädern trycks ihop 3,50 cm? k = F/x = 120,0N/(0,0225 m) = 5330 N/m U = kx 2 /2 = 5330 N/m ( - 0,035 m) 2 /2 = 3,26 J

23 Example 8-4b Compressed Energy and the Jump of a Flea

24 8-3 Conservation of Mechanical Energy Definition of mechanical energy: (8-6) Using this definition and considering only conservative forces, we find: Or equivalently:

25 Photo 8-4 Mormon Flat Dam

26 8-3 Conservation of Mechanical Energy Energy conservation can make kinematics problems much easier to solve:

27 8-3 Conservation of Mechanical Energy Eftersom vi är i ett konservativt kraftfält så gäller att energin är bevarad, det vill säga E i = E f vilket är detsamma som U i + K i = U f + K f som skrivs (i ett ”standard y/x-system”) mgy i + mv i 2 /2 = mgy f + mv f 2 /2 som lätt kan omformas till v f 2 = v i 2 + 2g(y i –y f )

28 Example 8-5 (p.216) Graduation Fling (m = 0,120 kg, v y0 = 7,85 m/s) a) Använd rörelseekvationerna för att bestämma v y då y = 1,18 m b) Visa att E(y=0) = E(y=1,18) Friktionskrafter försummas.

29 8-5 Graduative fling a) v y 2 = v i 2 + 2a y Δy dvs v y 2 = (7,85) 2 + 2 (-9,81) (1,18) = (6,20) 2 b) U i + K i = 0 + 0,120 (7,85) 2 /2 = 3,70 (J) U y + K y = 0,120 9,81 1,18 + 0,120 (6,20) 2 /2 = = 1,39 + 2,31 = 3,70 (J)

30 Figure 8-9 Speed is independent of path (Inga friktionskrafter)

31 Example 8-6a (p.218) Catching a Home Run (m=0,15 kg, v 0 = 36 m/s, h = 7,2 m inga friktionskrafter) Beräkna K f och v f

32 Example 8-6 Catching a Home Run a) U i + K i = 0 + 0,15(36) 2 /2 = 97 (J) U f + K f = 0,159,817,2 + K f K f = 97 – 11 = 86 (J) b) mv f 2 /2 = 86 (J) v f = 34 (m/s)

33 Example 8-6b Catching a Home Run

34 Conceptual Checkpoint 8–1 Compare the Final Speeds (Först ned?)

35 Example 8-7a Skateboard Exit Ramp (m = 55 kg, v i = 6,5 m/s, v f = 4,1 m/s, h = ?, no energy losses due to friction)

36 Example 8-7 Skateboard Exit Ramp E i = E f U i + K i = 0 + 55(6,5) 2 /2 ≈ 1160 J U f + K f = 559,81h + 55(4,1) 2 /2 (J) U f = 1160 – 460 ≈ 700 (J) h = 700/(559,81) = 1,3 (m) {Practical problem p.220 Maximal höjd? Då U f = 1200 J → h = 2,2 m}

37 Example 8-7b Skateboard Exit Ramp

38 Conceptual Checkpoint 8–2 What is the Final Speed? (Ingen friktion) a) 1 m/s b) 2 m/s c) 3 m/s

39 Example 8-8 (p.221) Spring Time (m=1,70 kg, k = 955 N/m, d = 4,60 cm Om friktionen kan försummas, vad hade blocket för begynnelsehastighet?

40 Example 8-8 Spring Time mv f 2 /2 = kx 2 /2 = 955(0,046) 2 /2 = 1,01 (J) v f = 21,01/1,70 = 1,09 (m/s)

41 Active Example 8-1 Find the Speed of the Block when x= x 0

42 Aktive Example 8-1 Find the speed of the block Energin lagrad i fjädern ger både kinetisk och potentiell energi. Då x = x 0 gäller att mv f 2 /2 + mgd = kd 2 /2 = 955 (0,046) 2 /2 mv f 2 /2 = 1,01 – 1,79,810,046 = 0,24 (J) v f = [2 0,24/1,70] 1/2 = 0,535 (m/s)

43 8-4 Work Done by Nonconservative Forces In the presence of nonconservative forces, the total mechanical energy is not conserved: Solving, (8-9)

44 Example 8-9a A Leaf Falls in the Forest: Find the Nonconservative Work

45 Example 8-9 A leaf Falls in the Forest Find the Nonconservative Work Energi finns lagrad som potentiell energi (h = 5,30 m, m = 17,010 -3 kg) där U i = mgh = 17,010 -3 9,815,30 = 0,884 J K i =0 U f = 0 K f = mv f 2 /2 = 17,0 10 -3 (1,3) 2 /2 = 0,0143 J W nc = ΔE = E f - E i = - 0,870 J

46 Example 8-9b A Leaf Falls in the Forest: Find the Nonconservative Work

47 8-4 Work Done by Nonconservative Forces In this example, the nonconservative force is water resistance:

48 8-4 Work Done by Nonconservative Forces Eftersom hastigheten är noll i de två tillstånden gäller att E i = mgh E f = mg(-d) W nc = ΔE = E f – E i = -mg(h+d) = -5120 (J) som då h = 3,00 m ger d = 5120/mg – h = 5,49 – 3,00 = 2,49 (m)

49 Conceptual Checkpoint 8–3 Judging a Putt Initial speed a) 2v 0 b) 3v 0 c) 4v 0 ? Friktionskraften antas konstant.

50 Conceptual Checkpoint 8-3 Judging a Putt Energi finns lagrad som kinetisk energi som går åt till arbetet som friktionskraften F utför på sträckan d. W nc = - Fd ΔE = E f - E i = 0 - mv 0 2 /2 W nc = ΔE ger v 0 ~ d så att om d = 4d måste v = 2v 0, rätt svar a)

51 Example 8-10 Landing with a Thud (m 1 = 2,40 kg, m 2 = 1,80 kg, d= 0,500 m, μ k = 0,450) Vad är hastigheten när blocket slår i golvet?

52 Example 8-10 Landing with a Thud Energi finns lagrad som potentiell energi som går åt till arbetet som friktionskraften F utför på sträckan d. W nc = - Fd F = μ k N = μ k m 1 g E i = m 1 gh + m 2 gd E f = m 1 gh + (m 1 + m 2 )v f 2 /2 ΔE = E f - E i = (m 1 +m 2 )v f 2 /2 – m 2 gd = W nc = -Fd v f = [2gd(m 2 – μ k m 1 )/(m 1 +m 2 )] 1/2 = 1,30 m/s

53 Active Example 8-3 Marathon Man: Find the Height of the Hill

54 Active Example 8-3 Marathon Man Joggaren (m= 80,0 kg) har utfört ett ickekonservativt (positivt) (muskel)arbete W nc1 = 18,0 kJ Luftmotståndet har utfört ett ickekonservativt (neg.) W nc2 = - 4,42 kJ Joggarens fart på backens krön är = 3,50 m/s Bestäm backens höjd h E i = 0 E f = mgh + mv f 2 /2 = mgh + 490 J ΔE = E f - E i = W nc = (18,0 – 4,42) kJ h = [(18,0 – 4,42 – 0,49)kJ/(80,0 kg9,81 m/s 2 )= 16,7 m

55 8-5 Potential Energy Curves and Equipotentials The curve of a hill or a roller coaster is itself essentially a plot of the gravitational potential energy:

56 8-5 Potential Energy Curves and Equipotentials The potential energy curve for a spring:

57 Example 8-11 A Potential Problem Partikeln rör sig i ett konservativt kraftfält längs banan enligt figur. Om hastigheten är 2,30 m/s för x = 0, vad är hastigheten i punkten x = 2,00 m?

58 Example 8-11 A potential problem E i = K i + U i = 1,60 (2,30) 2 /2 + 9,35 J = 13,58 J E f = K f + U f = 1,60 v f 2 /2 + 4,15 J v f = [2(13,58 – 4,15)/(1,6)] 1/2 = 3,43 m/s

59 8-5 Potential Energy Curves and Equipotentials Contour maps are also a form of potential energy curve:

60 Summary of Chapter 8 Conservative forces conserve mechanical energy Nonconservative forces convert mechanical energy into other forms Conservative force does zero work on any closed path Work done by a conservative force is independent of path Conservative forces: gravity, spring

61 Summary of Chapter 8 Work done by nonconservative force on closed path is not zero, and depends on the path Nonconservative forces: friction, air resistance, tension Energy in the form of potential energy can be converted to kinetic or other forms Work done by a conservative force is the negative of the change in the potential energy Gravity: U = mgy Spring: U = ½ kx 2

62 Summary of Chapter 8 Mechanical energy is the sum of the kinetic and potential energies; it is conserved only in systems with purely conservative forces Nonconservative forces change a system’s mechanical energy Work done by nonconservative forces equals change in a system’s mechanical energy Potential energy curve: U vs. position