Chapter 10. Energy - GSU P&Aphysics.gsu.edu/dhamala/Physics2211/Chapter10.pdfChapter 10. Energy This...
Transcript of Chapter 10. Energy - GSU P&Aphysics.gsu.edu/dhamala/Physics2211/Chapter10.pdfChapter 10. Energy This...
Chapter 10. EnergyThis pole vaulter can lift
herself nearly 6 m (20 ft) off
the ground by transforming
the kinetic energy of her
run into gravitational
potential energy.potential energy.
Chapter Goal: To introduce
the ideas of kinetic and
potential energy and to
learn a new problem-
solving strategy based on
conservation of energy.
Topics:
• A “Natural Money” Called Energy
• Kinetic Energy and Gravitational
Potential Energy
• A Closer Look at Gravitational Potential
Chapter 10. Energy
• A Closer Look at Gravitational Potential
Energy
• Restoring Forces and Hooke’s Law
• Elastic Potential Energy
• Elastic Collisions
• Energy Diagrams
Kinetic and Potential EnergyThere are two basic forms of energy. Kinetic energy is an energy of motion
Gravitational potential energy is an energy of positionposition
The sum K + Ug is not changed when an object is in freefall. Its initial and final values are equal
This is the conservation law for free fall
motion: the quantity
Free-Fall motion
2 2 2 ( )fy iy i f
v v g y y− = −− = −− = −− = −
2 21 1
2 2fy f iy i
v gy v gy+ = ++ = ++ = ++ = +
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motion: the quantity
has the same value before and after the
motion.
21
2y
v gy++++
Free-Fall Motion
Then
ivrrrr
2 2 2 ( )fy iy i f
v v g y y− = −− = −− = −− = −
fx ixv v====
2 2 2 21 1 1 1
2 2 2 2fx fy f ix iy i
v v gy v v gy+ + = + ++ + = + ++ + = + ++ + = + +
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fvrrrr
2 2 2 2
2 21 1
2 2f f i i
v gy v gy+ = ++ = ++ = ++ = +
Conservation law: the quantity
has the same value before and after the
motion.
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2v gy++++
Frictionless surface: acceleration y
iy
fy
αααα
sina g αααα====arrrr
s 2 2 2f i
v v as− =− =− =− =
Motion with constant acceleration:
sin
i fy y
sαααα
−−−−====
y y−−−−
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2 2 2 sin 2 2sin
i f
f i i
y yv v g gy gyαααα
αααα
−−−−− = = −− = = −− = = −− = = −
2 21 1
2 2f f i i
v gy v gy+ = ++ = ++ = ++ = +Conservation law: the quantity
has the same value before and after the
motion.
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2v gy++++
y
iy
fy
Conservation law: the quantity
has the same value at the initial and the
final points
21
2v gy++++
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In all cases the velocity at the final point is the same (FRICTIONLESS MOTION)
21
2v gy++++
21
2mv mgy++++Conservation law: or
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2K mv==== Kinetic Energy – energy of motion
gU mgy==== Gravitational Potential Energy – energy of position
11The units of energy is Joule:
2
2
mJ kg
s====
mech gE K U= += += += + Mechanical Energy
Conservation law of mechanical energy (without friction):
constantmech
E K U= + == + == + == + =
Zero of potential energy
y y
0
,1iy
,2iy
y
, ,1 ,1 , ,1i g i f g fK U K U+ = ++ = ++ = ++ = +
, ,2 ,2 , ,2i g i f g fK U K U+ = ++ = ++ = ++ = +
,1 , ,1 , ,1( )
f i g i g fK K U U= + −= + −= + −= + −
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0
,2fy,1f
y
,2 , ,2 , ,2( )
f i g i g fK K U U= + −= + −= + −= + −
,1 , ,2 , ,2 ,2 ,2
,1 ,1 , ,1 , ,1 ,2
( ) ( )
( ) ( )
g g i g f i f
i f g i g f g
U U U mg y y
mg y y U U U
∆ = − = − =∆ = − = − =∆ = − = − =∆ = − = − =
= − = − = ∆= − = − = ∆= − = − = ∆= − = − = ∆
Only the change of potential energy has the physical meaning
, ,i g i f g fK U K U+ = ++ = ++ = ++ = +
2
0
1
2i
K mv====
, 0g iU mgy====
, 10
g fU mgy= == == == =
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, 10
g fU mgy= == == == =
2
1
1
2f
K mv====
2 2
1 0 0
1 1
2 2mv mv mgy= += += += +
2
1 0 02 4 100 10.2 /v v gy m s= + = + ≈= + = + ≈= + = + ≈= + = + ≈
Restoring Force: Hooke’s Law
spF k s= − ∆= − ∆= − ∆= − ∆
The sign of a restoring force is always opposite to the sign of displacement
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Restoring Force: Elastic Potential Energy
spF k s= − ∆= − ∆= − ∆= − ∆
21( )
2s
U k s= ∆= ∆= ∆= ∆
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The elastic potential energy is always positive
Restoring Force: Elastic Potential Energy
21( )
2s
U k s= ∆= ∆= ∆= ∆
2 21 1( ) constant
2 2s
K U mv k s+ = + ∆ =+ = + ∆ =+ = + ∆ =+ = + ∆ =
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Restoring Force: Elastic Potential Energy 21( )
2s
U k s= ∆= ∆= ∆= ∆
202 21 1( ) constant
2 2s
K U mv k s+ = + ∆ =+ = + ∆ =+ = + ∆ =+ = + ∆ =
Law of Conservation of Mechanical Energy
2 21 1( ) constant
2 2g s
K U U mv mgy k s+ + = + + ∆ =+ + = + + ∆ =+ + = + + ∆ =+ + = + + ∆ =
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(((( ))))2 ,2 2 ,2 1 ,1 1 ,1fx ix fx ixm v m v m v m v− = − −− = − −− = − −− = − −
1 ,1 2 ,2 1 ,1 2 ,2ix ix fx fxm v m v m v m v+ = ++ = ++ = ++ = +
,1 ,2 ,1 ,2ix ix fx fxp p p p+ = ++ = ++ = ++ = +
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,1 ,2 ,1 ,2ix ix fx fxp p p p+ = ++ = ++ = ++ = +
, ,ix total fx totalp p====
The law of conservation of momentum
Perfectly Elastic Collisions
A A
,A ivrrrr
vrrrr
,A fvrrrr
During the collision the kinetic energy will be transformed into elastic energy and then elastic energy will transformed back into kinetic energy
Perfectly Elastic Collision: Mechanical Energy is Conserved (no internal friction)
2 2
, ,
1 1
2 2
1 1
A A i B B im v m v+ =+ =+ =+ =
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B
B
,B ivrrrr
,B fvrrrr
2 2
, ,
1 1
2 2A A f B B f
m v m v= += += += +
Perfectly inelastic collision:
A collision in which the two objects stick together and move with a
common final velocity – NO CONSERVATION OF MECHANICAL
ENERGY
Perfectly Elastic Collisions
A
B
A
,A ivrrrr
,B ivrrrr
,A fvrrrr
vrrrr
Conservation of Momentum and Conservation of Energy
isolated system
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B
,B fv
2 2 2 2
, , , ,
1 1 1 1
2 2 2 2A A i B B i A A f B B f
m v m v m v m v+ = ++ = ++ = ++ = +
, , , ,A A i B B i A A f B B fm v m v m v m v+ = ++ = ++ = ++ = +
r r r rr r r rr r r rr r r r
Now we have enough equations to find the final velocities.
Example:
,A iv ,
0B i
v ====
,A fv
,B fv
2 2 2
, , ,
1 1 1
2 2 2A A i A A f B B f
m v m v m v= += += += +
, , ,A A i A A f B B fm v m v m v= += += += +
, ,
A B
A f A i
A B
m mv v
m m
−−−−====
++++
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, ,
2A
B f A i
A B
mv v
m m====
++++
is always positive, can be positive (if ) or
negative (if )
,B fv
,A fv
A Bm m>>>>
A Bm m<<<<
,0
A fv ====
, ,B f A iv v====
ifA B
m m====
Chapter 10. Summary SlidesChapter 10. Summary SlidesChapter 10. Summary SlidesChapter 10. Summary Slides
Chapter 10. Clicker QuestionsChapter 10. Clicker QuestionsChapter 10. Clicker QuestionsChapter 10. Clicker Questions
Rank in order,
from largest to
smallest, the
gravitational
potential energies
of balls 1 to 4.
A. (Ug)1 > (Ug)2 = (Ug)4 > (Ug)3
B. (Ug)4 > (Ug)3 > (Ug)2 > (Ug)1
C. (Ug)1 > (Ug)2 > (Ug)3 > (Ug)4
D. (Ug)4 = (Ug)2 > (Ug)3 > (Ug)1
E. (Ug)3 > (Ug)2 = (Ug)4 > (Ug)1
Rank in order,
from largest to
smallest, the
gravitational
potential energies
of balls 1 to 4.
A. (Ug)1 > (Ug)2 = (Ug)4 > (Ug)3
B. (Ug)4 > (Ug)3 > (Ug)2 > (Ug)1
C. (Ug)1 > (Ug)2 > (Ug)3 > (Ug)4
D. (Ug)4 = (Ug)2 > (Ug)3 > (Ug)1
E. (Ug)3 > (Ug)2 = (Ug)4 > (Ug)1
A small child slides down the four frictionless slides A–D. Each has the
same height. Rank in order, from largest to smallest, her speeds vA to vD at
the bottom.
A. vC > vA = vB > vD
B. vC > vB > vA > vD
C. vD > vA > vB > vC
D. vA = vB = vC = vD
E. vD > vA = vB > vC
A small child slides down the four frictionless slides A–D. Each has the
same height. Rank in order, from largest to smallest, her speeds vA to vD at
the bottom.
A. vC > vA = vB > vD
B. vC > vB > vA > vD
C. vD > vA > vB > vC
D. vA = vB = vC = vD
E. vD > vA = vB > vC
A box slides along the frictionless surface shown in the figure. It is
released from rest at the position shown. Is the highest point the box
reaches on the other side at level a, at level b, or level c?reaches on the other side at level a, at level b, or level c?
A. At level a
B. At level b
C. At level c
A box slides along the frictionless surface shown in the figure. It is
released from rest at the position shown. Is the highest point the box
reaches on the other side at level a, at level b, or level c?reaches on the other side at level a, at level b, or level c?
A. At level a
B. At level b
C. At level c
The graph shows force versus
displacement for three
springs. Rank in order, from
largest to smallest, the spring
constants k1, k2, and k3.
A. k1 > k3 > k2
B. k3 > k2 > k1
C. k1 = k3 > k2
D. k2 > k1 = k3
E. k1 > k2 > k3
The graph shows force versus
displacement for three
springs. Rank in order, from
largest to smallest, the spring
constants k1, k2, and k3.
A. k1 > k3 > k2
B. k3 > k2 > k1
C. k1 = k3 > k2
D. k2 > k1 = k3
E. k1 > k2 > k3
A spring-loaded gun shoots a plastic ball with a speed of 4 m/s. If the spring is compressed twice as far, the ball’s speed will be
A. 1 m/s.
B. 2 m/s.
C. 4 m/s.
D. 8 m/s.
E.16 m/s.
A spring-loaded gun shoots a plastic ball with a speed of 4 m/s. If the spring is compressed twice as far, the ball’s speed will be
A. 1 m/s.
B. 2 m/s.
C. 4 m/s.
D. 8 m/s.
E.16 m/s.
A particle with the potential energy shown in the graph is moving to the
right. It has 1 J of kinetic energy at x = 1 m. Where is the particle’s
turning point?
A. x = 1 m
B. x = 2 mB. x = 2 m
C. x = 5 m
D. x = 6 m
E. x = 7.5 m