1. Energy
Chapter 11
Chapter 11: Energy
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2. Kinetic Energy and Work
KE = 0
W = F·d
KE = W
KE = ½mv2
Chapter 11: Energy
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3. 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
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4. 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
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5. (0.2kg) v = 2(75N)(2m) v = 39 m/s (88mph) Example Chapter 11: Energy
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6. Stored Energy: Potential Energy(PE)
Gravitational (PEg)
Chapter 11: Energy
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7. Gravitational Potential Energy
PE = F · d m
Ug = (m·g)·h h
Chapter 11: Energy
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8. 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
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9. Example 2m
1.5m
Chapter 11: Energy
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10. 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
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11. 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
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12. 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:
Due: 2/11/03
Chapter 11: Energy
1/1/2019

13. Conservation of Energy
Total energy of a system
remains unchanged!!
E = PE + KE
Chapter 11: Energy
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14. Energy Conversion
Chapter 11: Energy
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15. Chapter 11: Energy
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16. 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
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17. 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
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18. Analyzing Collisions v1a=.2m/s v1a=1.2m/s m2=1kg v2b=0m/s m1=1kg
Chapter 11: Energy
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19. 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
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20. Problems: 9-12 Page:261-262 Due: 2/12/03 Homework: 11-2
Chapter 11: Energy
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21. Problems: 13-16 Page:265 Due: 2/25/03 Homework: 11-2
Chapter 11: Energy
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22. 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:
Due: 2/13/03
Chapter 11: Energy
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23. 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:
Due: 2/14/03
Chapter 11: Energy
1/1/2019