This document discusses the conservation of mechanical energy and different types of potential and kinetic energy. It provides examples of how to calculate changes in speed using the principle of conservation of energy for situations involving gravitational potential energy, elastic potential energy, and kinetic energy such as a rollercoaster, pendulum, and hot dog cart passing over two hills. The key steps shown are identifying the relevant energies, setting the initial and final energies equal to each other, and solving the resulting equation to find the unknown speed or height.

1. Conservation Of…
Mechanical
Energy

2. Types

3. Types

6. Kinetic Energy

7. Kinetic Energy
a.k.a.
KE

8. Energy of
Motion

11. Gravitational Potential Energy

12. Gravitational Potential Energy
a.k.a
g

13. Energy object has because work
was done to put an object in its
present position in gravitational field

14. Energy object has because work
was done to put an object in its
t=
present position in gravitational field
gh
ei
h
th
wi
ct
je
b g
O

17. Elastic Potential Energy

18. Elastic Potential Energy
a.k.a.
s

19. Energy
object/spring
has because of
work done to
stretch or
compress the
spring

20. Energy
object/spring
ng s
ri
Sp
hasdbecause of
se
eswork done to
pr s
om d=
C lvestretch or
vo
In
compress the
spring

23. In a closed system…

24. In a closed system…
What's
that??

25. Nothing leaves…

26. Nothing Enters…

27. In a closed system…

29. What???

32. Simply…
IF:

33. Then:
i (before) f (after)

38. A Rollercoaster

39. A Rollercoaster
PEg KE
PEg KE
PEg KE
PEg KE

40. A Rollercoaster
PEg KE
PEg KE
PEg KE
PEg KE

41. A Rollercoaster
PEg KE
PEg KE
PEg KE
PEg KE

42. A Rollercoaster
PEg KE
PEg KE
PEg KE
PEg KE

43. A Rollercoaster
PEg KE
PEg KE
PEg KE
PEg KE

45. 1
4
2
3

46. A Pendulum
1
4
2
3

47. A Pendulum
1
4
PE=8J 2
3
KE=0J

48. A Pendulum
1
4
PE=8J 2
3
KE=0J
PE=4J
KE=4J

49. A Pendulum
1
4
PE=8J 2
3
KE=0J
PE=4J
PE=0J
KE=4J
KE=8J

50. A Pendulum
1
4
PE=8J 2
PE=8J
3
KE=0J
PE=4J KE=0J
PE=0J
KE=4J KE=8J

51. A Pendulum
1
4
PE=8J 2
PE=8J
3
KE=0J
PE=4J KE=0J
PE=0J
KE=4J KE=8J

54. Vertical Spring With An
Attached Mass

55. Vertical Spring With An
Attached Mass
PEs

56. Vertical Spring With An
Attached Mass
PE g+ KE
PEs

57. Vertical Spring With An
Attached Mass
PEg + PEs
PEg+ KE
PEs

58. Vertical Spring With An
Attached Mass
PEg + PEs
PEg+ KE
PEs

61. Horizontal Spring With An Attached
Mass

62. Horizontal Spring With An Attached
Mass
Equilibrium

63. Horizontal Spring With An Attached
Mass
Equilibrium
PEs

64. Horizontal Spring With An Attached
Mass
Equilibrium
PEs
KE

65. Horizontal Spring With An Attached
Mass
Equilibrium
PEs
KE
PEs

66. Horizontal Spring With An Attached
Mass
Equilibrium
PEs
KE
PEs

67. How To Solve…

68. Example Problem:
A 300 kg Hotdog cart rolls up and
down a street. It passes the top of
hill A. Which is 50m high at a
speed of 8 meters per second.
How fast is the hot dog cart going
to pass the next hill (B) which is
30m high?

69. Example Problem:
1: cart rolls up and
ep
A 300 kg Hotdog
t
S
down a street. It passes the top of
hill A. Which is 50m high at a
speed of 8 meters per second.
How fast is the hot dog cart going
to pass the next hill (B) which is
30m high?

70. Example Problem:
A 300 kg Hotdog cart rolls up and
down a street. It passes the top of
hill A. Which is 50m high at a
speed of 8 meters per second.
How fast is the hot dog cart going
to pass the next hill (B) which is
30m high?

71. 2:
ep
St

72. Hill A
Hill B
m=300kg
V=8m/s
V=?
h=50m
h=30m

73. Knowns
Hill A
Hill B
m=300kg
V=8m/s
V=?
h=50m
H=30m

74. Knowns
Hill A
Hill B
3:
M=300kg
p gy
te er
S
En
h e rs
y t sf e
V=8m/s
t if n
en ra V=?
Id T
h=50m
H=30m

75. Hill A
Hill B
m=300kg
PE i + KE i
V=8m/s
KEf
h=50m
V=?
H=30m

76. 4:
ep
St

77. Ei = Ef

78. Ei = Ef
te
i tu
st l as
ub u
S rm
Fo
gy
er
En

79. Ei = Ef
mgh+.5mv2 = .5mv2

80. Ei = Ef
mgh+.5mv2 =e
.5mv2
th
t y
u e
o
s ; th
os e s
rs l
Cs e
a anc
m c

81. Ei = Ef
mgh+.5mv2 = .5mv2

82. Ei = Ef
mgh+.5mv2 = .5mv2

83. Ei = Ef
mgh+.5mv2 = .5mv2
(9.81m/s2)(20m)+(.5)(8m)=.5v2

84. Ei = Ef
mgh+.5mv^2 = .5mv^2
(9.81m/s^2)(20)+(.5)(8)=.5v^2

85. Ei = Ef
mgh+.5mv2 = .5mv2
(9.81m/s2)(20)+(.5)(8)=.5v2
.5
.5
V2= (9.81m/s2)(20m)+(.5)(8m/s)
V=20.01m/s