Chapter 11 Lecture

Chapter 11 Work  Chapter Goal: To develop a more complete understanding of energy and its conservation. Slide 11-2

Chapter 11 Preview Slide 11-3

Chapter 11 Preview Slide 11-3

Chapter 11 Preview Slide 11-5

Chapter 11 Preview Slide 11-6

Chapter 11 Preview Slide 11-7

Chapter 11 Preview Slide 11-8

Chapter 11 Reading Quiz Slide 11-9

The statement K = W is called the Reading Question 11.1 The statement K = W is called the Law of conservation of energy. Work-kinetic energy theorem. Kinetic energy equation. Weight-kinetic energy theorem. Answer: B Slide 11-10 10

The statement K = W is called the Reading Question 11.1 The statement K = W is called the Law of conservation of energy. Work-kinetic energy theorem. Kinetic energy equation. Weight-kinetic energy theorem. Answer: B Slide 11-11 11

Reading Question 11.2 The transfer of energy to a system by the application of a force is called Dot product. Power. Work. Watt. Energy transformations. Answer: C Slide 11-12 12

Reading Question 11.2 The transfer of energy to a system by the application of a force is called Dot product. Power. Work. Watt. Energy transformations. Answer: C Slide 11-13 13

Reading Question 11.3 A vector has magnitude C. Then the dot product of the vector with itself, is 0. C/2. C. 2C. C2. Answer: C Slide 11-14 14

Reading Question 11.3 A vector has magnitude C. Then the dot product of the vector with itself, is 0. C/2. C. 2C. C2. Answer: C Slide 11-15 15

The work done by a dissipative force like friction, Wdiss, is Reading Question 11.4 The work done by a dissipative force like friction, Wdiss, is Always zero. Always positive. Always negative. Answer: B Slide 11-16 16

The work done by a dissipative force like friction, Wdiss, is Reading Question 11.4 The work done by a dissipative force like friction, Wdiss, is Always zero. Always positive. Always negative. Answer: B Slide 11-17 17

Reading Question 11.5 Light bulbs are typically rated in terms of their watts. The watt W is a measure of The change in the bulb’s potential energy. The energy consumed. The force exerted on the bulb holder. The power dissipated. Answer: C Slide 11-18 18

Reading Question 11.5 Light bulbs are typically rated in terms of their watts. The watt W is a measure of The change in the bulb’s potential energy. The energy consumed. The force exerted on the bulb holder. The power dissipated. Answer: C Slide 11-19 19

Chapter 11 Content, Examples, and QuickCheck Questions Slide 11-20

The Basic Energy Model W > 0: The environment does work on the system and the system’s energy increases. W < 0: The system does work on the environment and the system’s energy decreases. Slide 11-21

The Basic Energy Model The energy of a system is a sum of its kinetic energy K, its potential energy U, and its thermal energy Eth. The change in system energy is: Energy can be transferred to or from a system by doing work W on the system. This process changes the energy of the system: Esys = W. Energy can be transformed within the system among K, U, and Eth. These processes don’t change the energy of the system: Esys = 0. Slide 11-22

QuickCheck 11.1 A skier is gliding down a slope at a constant speed. What energy transformation is taking place? K  Ug Ug  K Eth  K Ug  Eth K  Eth Answer: B Slide 11-23 23

QuickCheck 11.1 A skier is gliding down a slope at a constant speed. What energy transformation is taking place? K  Ug Ug  K Eth  K Ug  Eth K  Eth Answer: B Slide 11-24 24

Work and Kinetic Energy The word “work” has a very specific meaning in physics. Work is energy transferred to or from a body or system by the application of force. This pitcher is increasing the ball’s kinetic energy by doing work on it. Slide 11-25

Work and Kinetic Energy Consider a force acting on a particle which moves along the s-axis. The force component Fs causes the particle to speed up or slow down, transferring energy to or from the particle. The force does work on the particle: The units of work are N m, where 1 N m = 1 kg m2/s2 = 1 J. Slide 11-26

The Work-Kinetic Energy Theorem The net force is the vector sum of all the forces acting on a particle . The net work is the sum Wnet = Wi, where Wi is the work done by each force . The net work done on a particle causes the particle’s kinetic energy to change. Slide 11-27

An Analogy with the Impulse-Momentum Theorem The impulse-momentum theorem is: The work-kinetic energy theorem is: Impulse and work are both the area under a force graph, but it’s very important to know what the horizontal axis is! Slide 11-28

QuickCheck 11.2 A tow rope pulls a skier up the slope at constant speed. What energy transfer (or transfers) is taking place? W  Ug W  K W  Eth Both A and B. Both A and C. Answer: B Slide 11-29 29

QuickCheck 11.2 A tow rope pulls a skier up the slope at constant speed. What energy transfer (or transfers) is taking place? W  Ug W  K W  Eth Both A and B. Both A and C. Answer: B Slide 11-30 30

Work Done by a Constant Force A force acts with a constant strength and in a constant direction as a particle moves along a straight line through a displacement . The work done by this force is: Here  is the angle makes relative to . Slide 11-31

Example 11.1 Pulling a Suitcase Slide 11-32

Example 11.1 Pulling a Suitcase Slide 11-33

QuickCheck 11.3 A crane lowers a girder into place at constant speed. Consider the work Wg done by gravity and the work WT done by the tension in the cable. Which is true? Wg > 0 and WT > 0 Wg > 0 and WT < 0 Wg < 0 and WT > 0 Wg < 0 and WT < 0 Wg = 0 and WT = 0 Answer: B Slide 11-34 34

QuickCheck 11.3 A crane lowers a girder into place at constant speed. Consider the work Wg done by gravity and the work WT done by the tension in the cable. Which is true? Wg > 0 and WT > 0 Wg > 0 and WT < 0 Wg < 0 and WT > 0 Wg < 0 and WT < 0 Wg = 0 and WT = 0 The downward force of gravity is in the direction of motion  positive work. The upward tension is in the direction opposite the motion  negative work. Answer: B Slide 11-35 35

Tactics: Calculating the Work Done by a Constant Force Slide 11-36

Tactics: Calculating the Work Done by a Constant Force Slide 11-37

Tactics: Calculating the Work Done by a Constant Force Slide 11-38

QuickCheck 11.4 Robert pushes the box to the left at constant speed. In doing so, Robert does ______ work on the box. positive negative zero Answer: B Slide 11-39 39

QuickCheck 11.4 Robert pushes the box to the left at constant speed. In doing so, Robert does ______ work on the box. positive negative zero Answer: B Force is in the direction of displacement  positive work Slide 11-40 40

QuickCheck 11.5 A constant force pushes a particle through a displacement . In which of these three cases does the force do negative work? Both A and B. Both A and C. Answer: B Slide 11-41 41

QuickCheck 11.5 A constant force pushes a particle through a displacement . In which of these three cases does the force do negative work? Both A and B. Both A and C. Answer: B Slide 11-42 42

Example 11.2 Work During a Rocket Launch Slide 11-43

Example 11.2 Work During a Rocket Launch Slide 11-44

Example 11.2 Work During a Rocket Launch Slide 11-45

Example 11.2 Work During a Rocket Launch Slide 11-46

QuickCheck 11.6 Which force below does the most work? All three displacements are the same. The 10 N force. The 8 N force The 6 N force. They all do the same work. sin60 = 0.87 cos60 = 0.50 Answer: B Slide 11-47 47

QuickCheck 11.6 Which force below does the most work? All three displacements are the same. The 10 N force. The 8 N force The 6 N force. They all do the same work. sin60 = 0.87 cos60 = 0.50 Answer: B Slide 11-48 48

QuickCheck 11.7 A light plastic cart and a heavy steel cart are both pushed with the same force for a distance of 1.0 m, starting from rest. After the force is removed, the kinetic energy of the light plastic cart is ________ that of the heavy steel cart. greater than equal to less than Can’t say. It depends on how big the force is. Slide 11-49 49

QuickCheck 11.7 A light plastic cart and a heavy steel cart are both pushed with the same force for a distance of 1.0 m, starting from rest. After the force is removed, the kinetic energy of the light plastic cart is ________ that of the heavy steel cart. greater than equal to less than Can’t say. It depends on how big the force is. Same force, same distance  same work done Same work  change of kinetic energy Slide 11-50 50

Force Perpendicular to the Direction of Motion The figure shows a particle moving in uniform circular motion. At every point in the motion, Fs, the component of the force parallel to the instantaneous displacement, is zero. The particle’s speed, and hence its kinetic energy, doesn’t change, so W = K = 0. A force everywhere perpendicular to the motion does no work. Slide 11-51

QuickCheck 11.8 A car on a level road turns a quarter circle ccw. You learned in Chapter 8 that static friction causes the centripetal acceleration. The work done by static friction is _____. positive negative zero Answer: B Slide 11-52 52

QuickCheck 11.8 A car on a level road turns a quarter circle ccw. You learned in Chapter 8 that static friction causes the centripetal acceleration. The work done by static friction is _____. positive negative zero Answer: B Slide 11-53 53

The Dot Product of Two Vectors The figure shows two vectors, and , with angle  between them. The dot product of and is defined as: The dot product is also called the scalar product, because the value is a scalar. Slide 11-54

The Dot Product of Two Vectors The dot product as  ranges from 0 to 180. Slide 11-55

Example 11.3 Calculating a Dot Product Slide 11-56

The Dot Product Using Components If and , the dot product is the sum of the products of the components: Slide 11-57

Example 11.4 Calculating a Dot Product Using Components Slide 11-58

Work Done by a Constant Force A force acts with a constant strength and in a constant direction as a particle moves along a straight line through a displacement . The work done by this force is: Slide 11-59

Example 11.5 Calculating Work Using the Dot Product Slide 11-60

Example 11.5 Calculating Work Using the Dot Product Slide 11-61

The Work Done by a Variable Force To calculate the work done on an object by a force that either changes in magnitude or direction as the object moves, we use the following: We must evaluate the integral either geometrically, by finding the area under the curve, or by actually doing the integration. Slide 11-62

Example 11.6 Using Work to Find the Speed of a Car Slide 11-63

Example 11.6 Using Work to Find the Speed of a Car Slide 11-64

Example 11.6 Using Work to Find the Speed of a Car Slide 11-65

Example 11.6 Using Work to Find the Speed of a Car Slide 11-66

Conservative Forces The figure shows a particle that can move from A to B along either path 1 or path 2 while a force is exerted on it. If there is a potential energy associated with the force, this is a conservative force. The work done by as the particle moves from A to B is independent of the path followed. Slide 11-67

Nonconservative Forces The figure is a bird’s-eye view of two particles sliding across a surface. The friction does negative work: Wfric = kmgs. The work done by friction depends on s, the distance traveled. This is not independent of the path followed. A force for which the work is not independent of the path is called a nonconservative force. Slide 11-68

Mechanical Energy Consider a system of objects interacting via both conservative forces and nonconservative forces. The change in mechanical energy of the system is equal to the work done by the nonconservative forces: Mechanical energy isn’t always conserved. As the space shuttle lands, mechanical energy is being transformed into thermal energy. Slide 11-69

Example 11.8 Using Work and Potential Energy Slide 11-70

Example 11.8 Using Work and Potential Energy Slide 11-71

Example 11.8 Using Work and Potential Energy Slide 11-72

Finding Force from Potential Energy The figure shows an object moving through a small displacement s while being acted on by a conservative force . The work done over this displacement is: Because is a conservative force, the object’s potential energy changes by U = −W = −FsΔs over this displacement, so that: Slide 11-73

Finding Force from Potential Energy In the limit s  0, we find that the force at position s is: The force on the object is the negative of the derivative of the potential energy with respect to position. Slide 11-74

Finding Force from Potential Energy Figure (a) shows the potential-energy diagram for an object at height y. The force on the object is (FG)y = mg. Figure (b) shows the corresponding F-versus-y graph. At each point, the value of F is equal to the negative of the slope of the U-versus-y graph. Slide 11-75

Finding Force from Potential Energy Figure (a) is a more general potential energy diagram. Figure (b) is the corresponding F-versus-x graph. Where the slope of U is negative, the force is positive. Where the slope of U is positive, the force is negative. At the equilibrium points, the force is zero. Slide 11-76

QuickCheck 11.9 A particle moves along the x-axis with the potential energy shown. At x = 4 m, the x-component of the force on the particle is –4 N. –2 N. 0 N. 2 N. 4 N Answer: B Slide 11-77 77

QuickCheck 11.9 A particle moves along the x-axis with the potential energy shown. At x = 4 m, the x-component of the force on the particle is –4 N. –2 N. 0 N. 2 N. 4 N. Answer: B Slide 11-78 78

Thermal Energy Figure (a) shows a mass M moving with velocity with macroscopic kinetic energy Kmacro = ½ Mvobj2. Figure (b) is a microphysics view of the same object. The total kinetic energy of all the atoms is Kmicro. The total potential energy of all the atoms is Umicro. The thermal energy of the system is: Slide 11-79

Dissipative Forces As two objects slide against each other, atomic interactions at the boundary transform the kinetic energy Kmacro into thermal energy in both objects. K  Eth Kinetic friction is a dissipative force. Slide 11-80

Dissipative Forces The figure shows a box being pulled at a constant speed across a horizontal surface with friction. Both the surface and the box are getting warmer as it slides. Dissipative forces always increase the thermal energy; they never decrease it. Slide 11-81

Example 11.9 Calculating the Increase in Thermal Energy Slide 11-82

Conservation of Energy For a system with both internal interaction forces and external forces, the energy equation is: Slide 11-83

Conservation of Energy Slide 11-84

The Basic Energy Model For a system with both internal interaction forces and external forces, Esys, the total energy of the system, changes only if external forces transfer energy in or out of the system. Slide 11-85

The Basic Energy Model When a system is isolated, Esys, the total energy of the system, is constant. Slide 11-86

Energy Bar Charts We may express the conservation of energy concept as an energy equation. We may also represent this equation graphically with an energy bar chart. Slide 11-87

QuickCheck 11.10 How much work is done by the environment in the process represented by the energy bar chart? –2 J –1 J 0 J 1 J 2 J Answer: B Slide 11-88 88

QuickCheck 11.10 How much work is done by the environment in the process represented by the energy bar chart? –2 J –1 J 0 J 1 J 2 J Answer: B The system started with 5 J but ends with 4 J. 1 J must have been transferred from the system to the environment as work. Slide 11-89 89

Problem-Solving Strategy: Solving Energy Problems Slide 11-90

Problem-Solving Strategy: Solving Energy Problems Slide 11-91

Example 11.10 Energy Bar Chart I Slide 11-92

Example 11.10 Energy Bar Chart I Slide 11-93

Example 11.11 Energy Bar Chart Il Slide 11-94

Example 11.11 Energy Bar Chart Il Slide 11-95

Example 11.12 Energy Bar Chart Ill Slide 11-96

Example 11.12 Energy Bar Chart Ill Slide 11-97

Power The rate at which energy is transferred or transformed is called the power P. Highly trained athletes have a tremendous power output. The SI unit of power is the watt, which is defined as: 1 watt = 1 W = 1 J/s The English unit of power is the horsepower, hp. 1 hp = 746 W Slide 11-98

Example 11.13 Choosing a Motor Slide 11-99

Example 11.13 Choosing a Motor Slide 11-100

Examples of Power Slide 11-101

Power When energy is transferred by a force doing work, power is the rate of doing work: P = dW/dt. If the particle moves at velocity while acted on by force , the power delivered to the particle is: Slide 11-102

QuickCheck 11.11 Four students run up the stairs in the time shown. Which student has the largest power output? Answer: B Slide 11-103 103

QuickCheck 11.11 Four students run up the stairs in the time shown. Which student has the largest power output? Answer: B Slide 11-104 104

Example 11.14 Power Output of a Motor Slide 11-105

Example 11.14 Power Output of a Motor Slide 11-106

Chapter 11 Summary Slides

General Principles Slide 11-108

General Principles Slide 11-109

General Principles Slide 11-110

Important Concepts Slide 11-111

Important Concepts Slide 11-112

Important Concepts Slide 11-113

Important Concepts Slide 11-114