Energy Chapter 11 Chapter 11: Energy 1/1/2019.

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Presentation transcript:

Energy Chapter 11 Chapter 11: Energy 1/1/2019

Kinetic Energy and Work KE = 0 W = F·d KE = W KE = ½mv2 Chapter 11: Energy 1/1/2019

Example A baseball having a mass of 0.2kg is thrown by a pitcher. The pitcher applies a force of 75N on the ball for a distance of 2m. How fast will the ball be traveling when it leaves the pitcher’s hand? Chapter 11: Energy 1/1/2019

½mv2 = F·d m Given: m= 0.2kg F = 75N d = 2m Equation: K = W v = 2F·d Example Given: m= 0.2kg F = 75N d = 2m Equation: K = W ½mv2 = F·d v = 2F·d m Chapter 11: Energy 1/1/2019

(0.2kg) v = 2(75N)(2m) v = 39 m/s (88mph) Example Chapter 11: Energy 1/1/2019

Stored Energy: Potential Energy(PE) Gravitational (PEg) Chapter 11: Energy 1/1/2019

Gravitational Potential Energy PE = F · d m Ug = (m·g)·h h Chapter 11: Energy 1/1/2019

You lift a 2kg textbook from the floor to a shelf 2m above the floor. Example You lift a 2kg textbook from the floor to a shelf 2m above the floor. 1. What is the books Ug relative to the floor? 2. What is the books Ug relative to the top of your head if you are 1.5 m tall? Chapter 11: Energy 1/1/2019

Example 2m 1.5m Chapter 11: Energy 1/1/2019

Given: m= 2kg hs = 2m hh = 1.5m Equation:Ug = m·g·h Example Given: m= 2kg hs = 2m hh = 1.5m With respect to the floor. Equation:Ug = m·g·h Ug = (2kg)(9.8m/s2)(2m) Ug = 39.2 J Chapter 11: Energy 1/1/2019

Equation:Ug = m·g·h Ug = (2kg)(9.8m/s2)(2m-1.5m) Ug = 9.8 J Example With respect to the top of your head. Equation:Ug = m·g·h Ug = (2kg)(9.8m/s2)(2m-1.5m) Ug = 9.8 J Chapter 11: Energy 1/1/2019

Problems: 1-4 Page:251 Problems: 5-8 Page:254-255 Due: 2/11/03 Homework: 11-1 Problems: 1-4 Page:251 Problems: 5-8 Page:254-255 Due: 2/11/03 Chapter 11: Energy 1/1/2019

Conservation of Energy Total energy of a system remains unchanged!! E = PE + KE Chapter 11: Energy 1/1/2019

Energy Conversion Chapter 11: Energy 1/1/2019

Chapter 11: Energy 1/1/2019

Energy in a falling object. E =KE + PE Ug = mgh h = 16m K = ½mv² h = 12m v = 2gd h = 8m h = 4m Chapter 11: Energy 1/1/2019

16 157 12 9 118 39 8 13 78 79 4 15 18 height h (m) Speed v= 2gd (m/s) Potential Ug=mgh (J) Kinetic K=½mv² Total E=Ug+K 16 157 12 9 118 39 8 13 78 79 4 15 18 Chapter 11: Energy 1/1/2019

Analyzing Collisions v1a=.2m/s v1a=1.2m/s m2=1kg v2b=0m/s m1=1kg Chapter 11: Energy 1/1/2019

1 1 1 1 0.5 0.5 0.5 0.25 p=mv K=½mv² Before (J) Case I After Case II Case III p=mv K=½mv² 1 1 1 1 0.5 0.5 0.5 0.25 Chapter 11: Energy 1/1/2019

Problems: 9-12 Page:261-262 Due: 2/12/03 Homework: 11-2 Chapter 11: Energy 1/1/2019

Problems: 13-16 Page:265 Due: 2/25/03 Homework: 11-2 Chapter 11: Energy 1/1/2019

Problems: 36, 37, 39, 42, 46, 49 and 56 Pages:269-270 Due: 2/13/03 Homework: 11-3 Problems: 36, 37, 39, 42, 46, 49 and 56 Pages:269-270 Due: 2/13/03 Chapter 11: Energy 1/1/2019

Problems: 52, 53, 57, 58 and 62 Pages: 270-271 Due: 2/14/03 Homework: 11-4 Problems: 52, 53, 57, 58 and 62 Pages: 270-271 Due: 2/14/03 Chapter 11: Energy 1/1/2019