Download Chapter 10. Energy

Chapter 10. Energy
This 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.
Chapter Goal: To introduce
the ideas of kinetic and
potential energy and to
learn a new problemsolving strategy based on
conservation of energy.
Chapter 10. Energy
Topics:
• A “Natural Money” Called Energy
• Kinetic Energy and Gravitational
Potential Energy
• A Closer Look at Gravitational Potential
Energy
• Restoring Forces and Hooke’s Law
• Elastic Potential Energy
• Elastic Collisions
• Energy Diagrams
Money-Energy Analogy
From the Parable of the Lost Penny
Money—Energy Analogy
From the law of conservation of energy
Kinetic and Potential Energy
There are two basic forms of energy. Kinetic energy
is an energy of motion
Gravitational potential energy is an energy of
position
The sum K + Ug is not changed when an object is
in freefall. Its initial and final values are equal
Kinetic and Potential Energy
Free-Fall motion
2
fy
2
iy
v − v = 2 g ( yi − y f )
1 2
1 2
v fy + gy f = viy + gyi
2
2
This is the conservation law for free fall
motion: the quantity
1 2
v y + gy
2
has the same value before and after the
motion.
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Free-Fall Motion
2
fy
2
iy
v − v = 2 g ( yi − y f )
v fx = vix
r
vi
Then
r
vf
1 2 1 2
1 2 1 2
v fx + v fy + gy f = vix + viy + gyi
2
2
2
2
1 2
1 2
v f + gy f = vi + gyi
2
2
Conservation law: the quantity
1 2
v + gy
2
has the same value before and after the
motion.
8
y
yi
Frictionless surface: acceleration
r
a
a = g sin α
Motion with constant acceleration:
s
v 2f − vi2 = 2as
yi − y f
s=
sin α
yf
α
2
f
2
i
v − v = 2 g sin α
Conservation law: the quantity
1 2
v + gy
2
yi − y f
sin α
= 2 gyi − 2 gy
1 2
1 2
v f + gy f = vi + gyi
2
2
has the same value before and after the
motion.
9
y
Conservation law: the quantity
yi
1 2
v + gy
2
has the same value at the initial and the
final points
yf
In all cases the velocity at the
final point is the same
(FRICTIONLESS MOTION)
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1 2
v + gy
2
Conservation law:
or
1
mv 2 + mgy
2
1
K = mv 2
2
Kinetic Energy – energy of motion
U g = mgy
Gravitational Potential Energy – energy of position
E mech = K + U g
Mechanical Energy
Conservation law of mechanical energy (without
friction):
Emech = K + U = constant
The units of energy is Joule:
m2
J = kg 2
s
11
12
13
Zero of potential energy
y
y
yi ,1
yi ,2
K i + U g ,i ,1 = K f ,1 + U g , f ,1
K i + U g ,i ,2 = K f ,2 + U g , f ,2
0
K f ,1 = K i + (U g ,i ,1 − U g , f ,1 )
y f ,1
0
y f ,2
K f ,2 = K i + (U g ,i ,2 − U g , f ,2 )
∆U g ,1 = (U g ,i ,2 − U g , f ,2 ) = mg ( yi ,2 − y f ,2 ) =
= mg ( yi ,1 − y f ,1 ) = (U g ,i ,1 − U g , f ,1 ) = ∆U g ,2
Only the change of potential energy has the physical meaning
14
K i + U g ,i = K f + U g , f
1
K i = mv02
2
U g , i = mgy0
U g , f = mgy1 = 0
1
K f = mv12
2
1
1
2
mv1 = mv02 + mgy0
2
2
v1 = v02 + 2 gy0 = 4 + 100 ≈ 10.2m / s
15
Restoring Force: Hooke’s Law
Hooke’s Law:
Fsp = − k ∆s
spring constant
16
Restoring Force: Hooke’s Law
Fsp = − k ∆s
The sign of a restoring force is
always opposite to the sign of
displacement
17
Restoring Force: Elastic Potential Energy
Fsp = − k ∆s
1
U s = k ( ∆s ) 2
2
The elastic potential energy is
always positive
18
Restoring Force: Elastic Potential Energy
1
U s = k ( ∆s ) 2
2
1
1
2
K + U s = mv + k ( ∆s )2 = constant
2
2
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Restoring Force: Elastic Potential Energy
1
1
2
K + U s = mv + k ( ∆s )2 = constant
2
2
1
U s = k ( ∆s ) 2
2
20
Law of Conservation of Mechanical Energy
1
1
2
K + U g + U s = mv + mgy + k ( ∆s )2 = constant
2
2
21
(
m2 v fx ,2 − m2 vix ,2 = − m1v fx ,1 − m1vix ,1
m1vix ,1 + m2 vix ,2 = m1v fx ,1 + m2 v fx ,2
pix ,1 + pix ,2 = p fx ,1 + p fx ,2
pix ,total = p fx ,total
The law of conservation of momentum
22
)
Elastic Collisions
Perfectly Elastic Collisions
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)
A
B
r
v A,i
r
v A, f
A
r
v B ,i
r
vB, f
1
1
2
m Av A,i + m B v B2 ,i =
2
2
1
1
2
= m Av A , f + m B v B2 , f
2
2
B
Perfectly inelastic collision:
A collision in which the two objects stick together and move with a
common final velocity – NO CONSERVATION OF MECHANICAL
24
ENERGY
Perfectly Elastic Collisions
Conservation of Momentum and Conservation of Energy
isolated system
A
B
r
v A,i
r
v A, f
A
r
v B ,i
r
vB, f
B
r
r
r
r
m Av A,i + m B v B ,i = m Av A, f + m B v B , f
1
1
1
1
2
2
2
m Av A,i + m B v B , i = m Av A, f + m B v B2 , f
2
2
2
2
Now we have enough equations to find the final velocities.
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Example:
v A,i
v B ,i = 0
vB, f
v A, f
m Av A,i = m Av A, f + m B v B , f
1
1
1
m Av A2 ,i = m A v A2 , f + m B v B2 , f
2
2
2
v A, f
m A − mB
v A,i
=
m A + mB
vB , f =
2m A
v A,i
m A + mB
v B , f is always positive, v A , f can be positive (if m A > m B ) or
negative (if
v A, f = 0
v B , f = v A,i
m A < mB )
if
m A = mB
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Chapter 10. Summary Slides
General Principles
General Principles
Important Concepts
Important Concepts
Important Concepts
Applications
Applications
Chapter 10. Clicker Questions
Rank in order,
from largest to
smallest, the
gravitational
potential energies
of balls 1 to 4.
A.
B.
C.
D.
E.
(Ug)1 > (Ug)2 = (Ug)4 > (Ug)3
(Ug)4 > (Ug)3 > (Ug)2 > (Ug)1
(Ug)1 > (Ug)2 > (Ug)3 > (Ug)4
(Ug)4 = (Ug)2 > (Ug)3 > (Ug)1
(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.
B.
C.
D.
E.
(Ug)1 > (Ug)2 = (Ug)4 > (Ug)3
(Ug)4 > (Ug)3 > (Ug)2 > (Ug)1
(Ug)1 > (Ug)2 > (Ug)3 > (Ug)4
(Ug)4 = (Ug)2 > (Ug)3 > (Ug)1
(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.
B.
C.
D.
E.
vC > vA = vB > vD
vC > vB > vA > vD
vD > vA > vB > vC
vA = vB = vC = vD
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.
B.
C.
D.
E.
vC > vA = vB > vD
vC > vB > vA > vD
vD > vA > vB > vC
vA = vB = vC = vD
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?
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?
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.
B.
C.
D.
E.
k1 > k3 > k2
k3 > k2 > k1
k1 = k3 > k2
k2 > k1 = k3
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.
B.
C.
D.
E.
k1 > k3 > k2
k3 > k2 > k1
k1 = k3 > k2
k2 > k1 = k3
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 m
C. x = 5 m
D. x = 6 m
E. x = 7.5 m
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 m
C. x = 5 m
D. x = 6 m
E. x = 7.5 m