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Chapter 8 Potential Energy and Conservation of Energy

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

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

Figure 8-1 Work against gravity

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

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

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

Figure 8-5 Gravity is a conservative force

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

Table 8-1 Conservative and Nonconservative Forces

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

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 = J, dvs W 1 ≠ W 2 (beror av vägen)

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)

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

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

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

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 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

Example 8-3 Converting Food Energy to Mechanical Energy

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

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

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

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

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

Photo 8-4 Mormon Flat Dam

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

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 )

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.

8-5 Graduative fling a) v y 2 = v i 2 + 2a y Δy dvs v y 2 = (7,85) (-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)

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

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

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)

Example 8-6b Catching a Home Run

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

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)

Example 8-7 Skateboard Exit Ramp E i = E f U i + K i = (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}

Example 8-7b Skateboard Exit Ramp

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

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?

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)

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

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)

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

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

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, kg) där U i = mgh = 17, ,815,30 = 0,884 J K i =0 U f = 0 K f = mv f 2 /2 = 17, (1,3) 2 /2 = 0,0143 J W nc = ΔE = E f - E i = - 0,870 J

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

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

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) = (J) som då h = 3,00 m ger d = 5120/mg – h = 5,49 – 3,00 = 2,49 (m)

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

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)

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?

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

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

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 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

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:

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

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?

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

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

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

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

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