Lecture Outlines Chapter 8 Physics, 3

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Lecture Outlines Chapter 8

Physics, 3^rd Edition James S. Walker

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) W₁ = -mg•1,0 + 0 = -mg•1,0 ; W₂ = 0 – mg•2,0 + 0 + mg•1,0 = -mg•1,0 W₁ = W₂ (oberoende av vägen) (mg•1,0 meter = 45 J)

Example 8-1b (p.208) Different Paths, Different Forces (m = 4,57 kg, μ = 0,63) W₁= - mg•0,63•4,0= -110 J ;

W₂ = - mg•0,63•12,0 = - 330 J, dvs W₁ ≠ W₂ (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

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 kg•9,81 m/s²•(4301m - 4201 m) = 80,4 kJ

b) Sätt U = 0 på bergstoppen

ΔU = U_f – U_i = mgy_f - mgy_i = 0 – (82 kg•9,81 m/s²•(-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 kg•9,81 m/s²• Δy ^{= 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 = kx²/2 = 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_i²/2 = mgy_f + mv_f²/2

som lätt kan omformas till v_f² = v_i² + 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_y² = v_i² + 2a_yΔy dvs

v_y² = (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)

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

Example 8-6a (p.218) Catching a Home Run (m=0,15 kg, v₀ = 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 = 0,15•9,81•7,2 + K_f

K_f = 97 – 11 = 86 (J) b)

mv_f²/2 = 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 = 0 + 55•(6,5)²/2 ≈ 1160 J

U_f + K_f = 55•9,81•h + 55•(4,1)²/2 (J) U_f = 1160 – 460 ≈ 700 (J)

h = 700/(55•9,81) = 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_f²/2 = kx²/2 = 955•(0,046)²/2 = 1,01 (J) v_f = 2•1,01/1,70 = 1,09 (m/s)

Active Example 8-1 Find the Speed of the Block when x= x₀

Aktive Example 8-1 Find the speed of the block Energin lagrad i fjädern ger både kinetisk och potentiell energi. Då x = x₀ gäller att

mv_f²/2 + mgd = kd²/2 = 955 • (0,046)²/2 mv_f²/2 = 1,01 – 1,7•9,81•0,046 = 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 = 17,0•10^-3 kg) där

U_i = mgh = 17,0•10^-3 •9,81•5,30 = 0,884 J K_i =0

U_f = 0

K_f = mv_f²/2 = 17,0 10^-3 • (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) = -5120 (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) 2v₀ b) 3v₀ c) 4v₀? 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 = - F•d

ΔE = E_f - E_i = 0 - mv₀²/2 W_nc = ΔE ger

v₀ ~ d så att om d = 4d måste v = 2v₀, rätt svar a)

Example 8-10 Landing with a Thud

(m₁ = 2,40 kg, m₂ = 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 = - F•d

F = μ_kN = μ_km₁g E_i = m₁gh + m₂gd

E_f = m₁gh + (m₁+ m₂)v_f²/2

ΔE = E_f - E_i = (m₁+m₂)v_f²/2 – m₂gd = W_nc = -F•d v_f = [2gd(m₂ – μ_km₁)/(m₁+m₂)]^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_f²/2 = 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 kg•9,81 m/s²)= 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_f²/2 + 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 kx²

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