Chapter 10 Lecture

Chapter 10 Preview Slide 10-5

Chapter 10 Preview Slide 10-6

Chapter 10 Preview Slide 10-7

Reading Question 10.3 Mechanical energy is The energy due to internal moving parts. The energy of motion. The energy of position. The sum of kinetic energy plus potential energy. The sum of kinetic, potential, thermal, and elastic energy. Answer: B Slide 10-13 5

Reading Question 10.3 Mechanical energy is The energy due to internal moving parts. The energy of motion. The energy of position. The sum of kinetic energy plus potential energy. The sum of kinetic, potential, thermal, and elastic energy. Answer: B Slide 10-14 6

Reading Question 10.5 A perfectly elastic collision is a collision Between two springs. That conserves thermal energy. That conserves kinetic energy. That conserves potential energy. That conserves mechanical energy. Answer: E Slide 10-17 7

Reading Question 10.5 A perfectly elastic collision is a collision Between two springs. That conserves thermal energy. That conserves kinetic energy. That conserves potential energy. That conserves mechanical energy. Answer: E Slide 10-18 8

The Basic Energy Model Within a system, energy can be transformed from one type to another. The total energy of the system is not changed by these transformations. This is the law of conservation of energy. Energy can also be transferred from one system to another. The mechanical transfer of energy to a system via forces is called work. Slide 10-23

Kinetic Energy and Gravitational Potential Energy Define kinetic energy as an energy of motion: Define gravitational potential energy as an energy of position: The sum K + Ug is not changed when an object is in free fall. Its initial and final values are equal: Slide 10-25

Example 10.1 Launching a Pebble Slide 10-31

Example 10.1 Launching a Pebble Slide 10-32

Example 10.1 Launching a Pebble Slide 10-33

Energy Bar Charts A pebble is tossed up into the air. The simple bar charts below show how the sum of K + Ug remains constant as the pebble rises and then falls. Slide 10-34

QuickCheck 10.4 Rank in order, from largest to smallest, the gravitational potential energies of the balls. 1 > 2 = 4 > 3 1 > 2 > 3 > 4 3 > 2 > 4 > 1 3 > 2 = 4 > 1 Answer: D Slide 10-38 15

QuickCheck 10.4 Rank in order, from largest to smallest, the gravitational potential energies of the balls. 1 > 2 = 4 > 3 1 > 2 > 3 > 4 3 > 2 > 4 > 1 3 > 2 = 4 > 1 Answer: D Slide 10-39 16

Gravitational Potential Energy on a Frictionless Surface – Slide 1 of 4 Figure (a) shows an object of mass m sliding along a frictionless surface. Figure (b) shows a magnified segment of the surface that, over some small distance, is a straight line. Define an s-axis parallel to the direction of motion Newton’s second law along the axis is: Slide 10-45

QuickCheck 10.6 A small child slides down the four frictionless slides A–D. Rank in order, from largest to smallest, her speeds at the bottom. vD > vA > vB > vC vD > vA = vB > vC vC > vA > vB > vD vA = vB = vC = vD Answer: D Slide 10-51 18

QuickCheck 10.6 A small child slides down the four frictionless slides A–D. Rank in order, from largest to smallest, her speeds at the bottom. vD > vA > vB > vC vD > vA = vB > vC vC > vA > vB > vD vA = vB = vC = vD Answer: D Slide 10-52 19

Example 10.3 The Speed of a Sled Slide 10-53

Example 10.3 The Speed of a Sled Slide 10-54

Example 10.3 The Speed of a Sled Slide 10-55

Problem-Solving Strategy: Conservation of Mechanical Energy Slide 10-56

QuickCheck 10.7 Three balls are thrown from a cliff with the same speed but at different angles. Which ball has the greatest speed just before it hits the ground? Ball A. Ball B. Ball C. All balls have the same speed. Answer: D Slide 10-57 24

QuickCheck 10.7 Three balls are thrown from a cliff with the same speed but at different angles. Which ball has the greatest speed just before it hits the ground? Ball A. Ball B. Ball C. All balls have the same speed. Answer: D Slide 10-58 25

QuickCheck 10.8 A hockey puck sliding on smooth ice at 4 m/s comes to a 1-m-high hill. Will it make it to the top of the hill? Yes. No. Can’t answer without knowing the mass of the puck. Can’t say without knowing the angle of the hill. Answer: A Slide 10-59 26

QuickCheck 10.8 A hockey puck sliding on smooth ice at 4 m/s comes to a 1-m-high hill. Will it make it to the top of the hill? Yes. No. Can’t answer without knowing the mass of the puck. Can’t say without knowing the angle of the hill. Answer: A Slide 10-60 27

Restoring Forces and Hooke’s Law The figure shows how a hanging mass stretches a spring of equilibrium length L0 to a new length L. The mass hangs in static equilibrium, so the upward spring force balances the downward gravity force. Slide 10-61

Restoring Forces and Hooke’s Law The figure shows measured data for the restoring force of a real spring. s is the displacement from equilibrium. The data fall along the straight line: The proportionality constant k is called the spring constant. The units of k are N/m. Slide 10-62

Hooke’s Law One end of a spring is attached to a fixed wall. (Fsp)s is the force produced by the free end of the spring. s = s – se is the displacement from equilibrium. The negative sign is the mathematical indication of a restoring force. Slide 10-63

QuickCheck 10.9 The restoring force of three springs is measured as they are stretched. Which spring has the largest spring constant? Answer: A Slide 10-64 31

QuickCheck 10.9 The restoring force of three springs is measured as they are stretched. Which spring has the largest spring constant? Steepest slope. Takes lots of force for a small displacement. Answer: A Slide 10-65 32

Example 10.5 Pull Until It Slips Slide 10-66

Example 10.5 Pull Until It Slips Slide 10-67

Example 10.5 Pull Until It Slips Slide 10-70

Example 10.5 Pull Until It Slips Slide 10-71

Elastic Potential Energy The figure shows a before-and-after situation in which a spring launches a ball. Integrating the net force from the spring, as given by Hooke’s Law, shows that: Here K = ½ mv2 is the kinetic energy. We define a new quantity: Slide 10-74

QuickCheck 10.10 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be 1.0 m/s. 2.0 m/s. 2.8 m/s 4.0 m/s. 16.0 m/s. Slide 10-76 38

QuickCheck 10.10 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be 1.0 m/s. 2.0 m/s. 2.8 m/s 4.0 m/s. 16.0 m/s. Conservation of energy: Double x double v Slide 10-77 39

QuickCheck 10.11 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twice the spring constant (but still compressed the same distance), the ball’s launch speed will be 1.0 m/s. 2.0 m/s. 2.8 m/s. 4.0 m/s. 16.0 m/s. Slide 10-78 40

QuickCheck 10.11 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twice the spring constant (but still compressed the same distance), the ball’s launch speed will be 1.0 m/s. 2.0 m/s. 2.8 m/s. 4.0 m/s. 16.0 m/s. Conservation of energy: Double k  increase v by square root of 2 Slide 10-79 41

Example 10.6 A Spring-Launched Plastic Ball Slide 10-80

Example 10.6 A Spring-Launched Plastic Ball Slide 10-83

Energy Diagrams Shown is a more general energy diagram. The particle is released from rest at position x1. Since K at x1 is zero, the total energy TE = U at that point. The particle speeds up from x1 to x2. Then it slows down from x2 to x3. The particle reaches maximum speed as it passes x4. When the particle reaches x5, it turns around and reverses the motion. Slide 10-89

Equilibrium Positions: Stable Consider a particle with the total energy E2 shown in the figure. The particle can be at rest at x2, but it cannot move away from x2: This is static equilibrium. If you disturb the particle, giving it a total energy slightly larger than E2, it will oscillate very close to x2. An equilibrium for which small disturbances cause small oscillations is called a point of stable equilibrium. Slide 10-90

QuickCheck 10.12 A particle with the potential energy shown is moving to the right. It has 1.0 J of kinetic energy at x = 1.0 m. In the region 1.0 m < x < 2.0 m, the particle is Speeding up. Slowing down. Moving at constant speed. I have no idea. Slide 10-99 46

QuickCheck 10.12 A particle with the potential energy shown is moving to the right. It has 1.0 J of kinetic energy at x = 1.0 m. In the region 1.0 m < x < 2.0 m, the particle is Speeding up. Slowing down. Moving at constant speed. I have no idea. Slide 10-100 47

QuickCheck 10.13 A particle with the potential energy shown is moving to the right. It has 1.0 J of kinetic energy at x = 1.0 m. Where is the particle’s turning point? 1.0 m. 2.0 m. 5.0 m. 6.0 m. It doesn’t have a turning point. Slide 10-101 48

QuickCheck 10.13 A particle with the potential energy shown is moving to the right. It has 1.0 J of kinetic energy at x = 1.0 m. Where is the particle’s turning point? 1.0 m. 2.0 m. 5.0 m. 6.0 m. It doesn’t have a turning point. Slide 10-102 49

QuickCheck 10.14 A particle with this potential energy could be in stable equilibrium at x = 0.0 m. 1.0 m. 2.0 m. Either A or C. Either B or C. Slide 10-103 50

QuickCheck 10.14 A particle with this potential energy could be in stable equilibrium at x = 0.0 m. 1.0 m. 2.0 m. Either A or C. Either B or C. Slide 10-104 51

Elastic Collisions During an inelastic collision of two objects, some of the mechanical energy is dissipated inside the objects as thermal energy. A collision in which mechanical energy is conserved is called a perfectly elastic collision. Collisions between two very hard objects, such as two billiard balls or two steel balls, come close to being perfectly elastic. Slide 10-106

Using Reference Frames: Quick Example A 200 g ball moves to the right at 2.0 m/s. It has a head-on, perfectly elastic collision with a 100 g ball that is moving toward it at 3.0 m/s. What are the final velocities of both balls? Slide 10-113

Using Reference Frames: Quick Example Figure (a) shows the situation just before the collision in the lab frame L. Figure (b) shows the situation just before the collision in the frame M that is moving along with ball 2. Slide 10-114

Using Reference Frames: Quick Example We can use Equations 10.42 to find the post-collision velocities in the moving frame M: Transforming back to the lab frame L: Slide 10-115

Chapter 11 Preview Slide 11-3

Chapter 11 Preview Slide 11-3

Chapter 11 Preview Slide 11-8

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

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 11.3 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 61

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 62

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 63

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 64

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 65

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 66

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 67

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 68

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 69

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 70

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 11.8 Using Work and Potential Energy Slide 11-70

Example 11.8 Using Work and Potential Energy Slide 11-71

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

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 78

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 79

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

Example 11.14 Power Output of a Motor Slide 11-105

Example 11.14 Power Output of a Motor Slide 11-106

Important Concepts Slide 11-112

Important Concepts Slide 11-113