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Energy
Topics:
• Important forms of energy
• How energy can be transformed and transferred
• Definition of work
Definition of work
• Concepts of kinetic, potential, and thermal energy
• The law of conservation of energy
• Elastic collisions
Sample question:
When flexible poles became available for pole vaulting, athletes
were able to clear much higher bars. How can we explain this
using energy concepts?
A “Natural Money” Called Energy
Income
System
Liquid Asset:
Cash
Saved Asset:
Stocks
Transfers into and out of system
Transformations Transformations
within system
Key concepts:
• Definition of the system.
Expenses
• Transformations within the system.
y
• Transfers between the system and the environment.
Forms of Energy
Mechanical Energy
gy
Ug
K
Thermal
E
Energy
Us
Other forms include
Eth
Echem
Enuclear
The Basic Energy Model
The unit of work and energy is the joule (J). 1 J = 1 Nm = 1 kg m2/s2.
Checking Understanding
A skier is moving down a slope at a constant speed
speed. What
energy transformation is taking place?
A. K → U g
B. U g → Eth
C. U s → U g
D. Ug → K
E. K → Eth
Answer
A skier is moving down a slope at a constant speed
speed. What
energy transformation is taking place?
B. U g → Eth
Slide 10-13
Checking Understanding
A child is on a playground swing
swing, motionless at the highest
point of his arc. As he swings back down to the lowest point of
his motion, what energy transformation is taking place?
A.
B.
C
C.
D.
E.
K → Ug
U g → Eth
Us → Ug
Ug → K
K → Eth
Slide 10-14
Answer
A child is on a playground swing
swing, motionless at the highest
point of his arc. As he swings back down to the lowest point of
his motion, what energy transformation is taking place?
D. U g → K
Slide 10-15
Kinetic Energy
Kinetic Energy
1 2
K = mv
2
is an object’s translational kinetic energy.
This is the energy an object has because of its state of motion.
It can be shown that, in general
Wnet = ΔK .
Work-Kinetic Energy Theorem
We have seen earlier that K f − K i = Wnet .
We define the change in kinetic energy as
ΔK = K f − K i . The equation above becomes
the work-kinetic
k ki i energy theorem
h
:
ΔK = K f − K i = Wnet
⎡Change in the kinetic ⎤ ⎡ net work done on ⎤
⎢energy of a particle ⎥ = ⎢ the particle
⎥
⎣
⎦ ⎣
⎦
The work-kinetic energy theorem holds for both positive and negative values of Wnet .
If
Wnet > 0 → K f − K i > 0 → K f > K i
If
Wnet < 0 → K f − K i < 0 → K f < K i
Example: The extinction of the dinosaurs and the majority of species on Earth in the Cretaceous Period (65 Myr ago) is thought to have been caused by an asteroid striking the Earth near the Yucatan Peninsula The resulting ejecta caused widespread global climate Earth near the Yucatan Peninsula. The resulting ejecta
caused widespread global climate
change. If the mass of the asteroid was 1016 kg (diameter in the range of 4‐9 miles) and had a speed of 30.0 km/sec, what was the asteroid’s kinetic energy?
(
)(
1 2 1 16
K = mv = 10 kg
k 30 ×103 m/s
/
2
2
= 4.5 × 10 24 J
)
2
This is equivalent to ~109 Megatons of TNT.
Work by a Constant Force
Work by a Constant Force
Work is an energy transfer by the application of a force. For work to be done there must be a nonzero displacement.
The unit of work and energy is the joule (J). 1 J = 1 Nm = 1 kg m2/s2.
12
m
m
Finding an Expression for Work :
Consider a bead of mass m that can move
without friction along a straight wire along
G
the x-axis. A constant force F applied at an
angle φ to the wire is acting on the bead.
W = Fd cos ϕ
G G
W = F ⋅d
Example: A ball is tossed straight up. What is the work done by the force of gravity on the ball as it rises?
y
Δy
FBD for rising ball:
x
w
Wg = wΔy cos180°
= −mgΔy
Example: A box of mass m is towed up a frictionless incline at constant speed. The applied force F is parallel to the incline. What is the net work done on the box?
y
F
An FBD for the box:
N
F
x
θ
θ
w
Apply Newton’s 2nd Law:
∑F
∑F
x
= F − w sin θ = 0
y
= N − w cos θ = 0
Work by a Variable Force
Work by a Variable Force
Work can be calculated by finding the area underneath a plot of the applied force in the Work
can be calculated by finding the area underneath a plot of the applied force in the
direction of the displacement versus the displacement.
16
Example: What is the work done by the variable force shown below?
Fx (N)
F3
F2
F1
x1
x2
x3
x (m)
x (m)
The work done by F1 is
W1 = F1 ( x1 − 0 )
The work done by F2 is
W2 = F2 ( x2 − x1 )
The work done by F3 is
W3 = F3 ( x3 − x2 )
The net work is then W1+W2+W3.
17
Gravitational Potential Energy
Gravitational Potential Energy
Objects have potential energy because of their location (or configuration). There are potential energies associated with different (but not all!) forces. Such a force is called a conservative force.
In general
Wcons = − ΔU
The change in gravitational potential energy (only near the surface of the Earth) is ΔU g = mgΔy
where Δy is the change in the object’s vertical position with respect to some reference point that you are free to choose.
19
Example: What is the change in gravitational potential energy of the box if it is placed on the table? The table is 1.0 m tall and the mass of the box is 1.0 kg. First: Choose the reference level at the floor. U = 0 here.
ΔU g = mgΔy = mg ( y f − yi )
(
)
= (1.0 kg ) 9.8 m/s 2 (1.0 m − 0 m ) = +9.8 J
20
Example continued:
Now take the reference level (U = 0) to be on top of the table so that yi = −1.0 m and yf = 0.0 m.
ΔU g = mgΔy = mg ( y f − yi )
(
)
= (1 kg ) 9.8 m/s 2 (0.0m − (− 1.0 m )) = +9.8 J
The results do not depend on the location of The
results do not depend on the location of
U = 0.
Mechanical energy is E = K +U
Whenever nonconservative forces do no work, the mechanical energy of a system is conserved. That is Ei = Ef or ΔK = −ΔU.
22
Example: A cart starts from position 4 with v = 15.0 m/s to the left. Find the speed of the cart at positions 1, 2, and 3. Ignore friction. p
, ,
g
E4 = E3
U 4 + K 4 = U 3 + K3
1 2
1 2
mgy4 + mv4 = mgy3 + mv3
2
2
v3 = v42 + 2 g ( y4 − y3 ) = 20.5 m/s
/
Example continued:
E4 = E2
Or use E3=E2
U 4 + K4 = U 2 + K2
1
1
mgy4 + mv42 = mgy2 + mv22
2
2
v2 = v42 + 2 g ( y4 − y2 ) = 18.0 m/s
E4 = E1
U 4 + K 4 = U 1 + K1
1 2
1 2
mgy4 + mv4 = mgy1 + mv1
2
2
v1 = v42 + 2 g ( y4 − y1 ) = 24.8 m/s
/
Or use E3=E1
E2=E1
Elastic potential energy / Spring force
By hanging masses on a spring we find that stretch ∝
applied force. This is Hooke’s law.
li d f
Thi i H k ’ l
For an ideal spring: F
d l
k
x = ‐kx
Fx is the magnitude of the force exerted by the free end of the is the magnitude of the force exerted by the free end of the
spring, x is the measured stretch of the spring, and k is the spring constant (a constant of proportionality; its units are N/m).
A larger value of k implies a stiffer spring
A larger value of k implies a stiffer spring.
25
Example: (a) If forces of 5.0 N applied to each end of a spring cause the spring to stretch 3.5 cm from its relaxed length, how far does a force of 7.0 N cause the same spring to
3.5 cm from its relaxed length, how far does a force of 7.0 N cause the same spring to stretch?
For springs F∝x This allows us to write
For springs F∝x. This allows us to write Solving for x2:
F1 x1
= .
F2 x2
F2
⎛ 7.0 N ⎞
x1 = ⎜
x2 =
⎟(3.5 cm ) = 4.9 cm.
F1
⎝ 5.0 N ⎠
Example continued:
(b) What is the spring constant of this spring?
F1 5.0 N
k= =
= 1.43 N/cm.
x1 3.5 cm
Or
F2 7.0 N
k=
=
= 1.43 N/cm.
x2 4.9 cm
Example: An ideal spring has k = 20.0 N/m. What is the amount of work done (by an external agent) to stretch the spring 0.40 m from its relaxed length? g )
p g
g
Fx (N)
kx1
x1=0.4 m
=0 4 m
x (m)
W = Area under curve
1
1 2 1
2
= (kx1 )( x1 ) = kx1 = (20.0 N/m )(0.4 m ) = 1.6 J
2
2
2
28
Elastic potential energy
Elastic potential energy
The work done in stretching/compressing a spring transfers energy to the spring.
1 2
U s = kx
2
29
Example: A box of mass 0.25 kg slides along a horizontal, frictionless surface with a speed of 3.0 m/s. The box encounters a spring with k = 200 N/m. How far is the spring speed of 3.0 m/s. The box encounters a spring with k 200 N/m. How far is the spring
compressed when the box is brought to rest?
Ei = E f
U i + Ki = U f + K f
1 2 1 2
0 + mv = kx + 0
2
2
⎛ m⎞
⎟v = 0.11 m
x = ⎜⎜
⎟
k
⎠
⎝
30
Conceptual Example
A car sits at rest at the top of a hill
hill. A small push sends it
rolling down a hill. After its height has dropped by 5.0 m, it is
moving at a good clip. Write down the equation for
conservation
ti off energy, noting
ti the
th choice
h i off system,
t
the
th initial
i iti l
and final states, and what energy transformation has taken
place.
Slide 10-17
Checking Understanding
Each of the boxes,
boxes with masses noted
noted, is pulled for
10 m across a level, frictionless floor by the noted force. Which
box experiences the largest change in kinetic energy?
Slide 10-18
Answer
Each of the boxes,
boxes with masses noted
noted, is pulled for
10 m across a level, frictionless floor by the noted force. Which
box experiences the largest change in kinetic energy?
Slide 10-19
Checking Understanding
Each of the boxes,
boxes with masses noted
noted, is pulled for
10 m across a level, frictionless floor by the noted force. Which
box experiences the smallest change in kinetic energy?
Slide 10-20
Answer
Each of the boxes,
boxes with masses noted
noted, is pulled for
10 m across a level, frictionless floor by the noted force. Which
box experiences the smallest change in kinetic energy?
Slide 10-21
Example
A 200 g block on a frictionless surface is pushed against a
spring with spring constant 500 N/m, compressing the spring
by 2.0 cm. When the block is released, at what speed does it
shoot
h t away from
f
the
th spring?
i ?
Slide 10-23
Example
A 2.0
2 0 g desert locust can achieve a takeoff
speed of 3.6 m/s (comparable to the best
human jumpers) by using energy stored in an
internal “spring” near the knee joint
joint.
A. When the locust jumps, what energy
transformation takes place?
B What
B.
Wh t is
i the
th minimum
i i
amountt off energy
stored in the internal spring?
C. If the locust were to make a vertical leap,
h
how
hi
high
h could
ld it jjump?
? IIgnore air
i
resistance and use conservation of energy
concepts to solve this problem.
D. If 50% of the initial kinetic energy is
transformed to thermal energy because of
air resistance, how high will the locust
jump?
Slide 10-24
Power
•
•
Same mass...
mass
Both reach 60 mph...
Same final kinetic
energy, but
different times mean
different powers
powers.
Average Power
Average Power
ΔE
Pav =
Δt
Instantaneous Power
P = Fv cosθ
Slide 10-25
Example : A race car with a mass of 500.0 kg completes a quarter‐mile (402 m) race in a time of 4 2 s starting from rest The car’ss final speed is 125 m/s. What is the engine
time of 4.2 s starting from rest. The car
final speed is 125 m/s What is the engine’ss average power output? Neglect friction and air resistance.
ΔE ΔU + ΔK
Pav =
=
Δt
Δt
1 2
mv f
ΔK 2
=
=
= 9.3 × 105 watts
Δt
Δt
Four toy cars accelerate from rest to their top speed in a certain
amount of time. The masses of the cars, the final speeds, and the
time to reach this speed are noted in the table. Which car has the
greatest power?
Car
Mass (g) Speed (m/s) Time (s)
A
100
3
2
B
200
2
2
C
200
2
3
D
300
2
3
E
300
1
4
Slide 10-26
Answer
Four toy cars accelerate from rest to their top speed in a certain
amount of time. The masses of the cars, the final speeds, and the
time to reach this speed are noted in the table. Which car has the
greatest power?
Car
A
Mass (g) Speed (m/s) Time (s)
100
3
2
Slide 10-27
Four toy cars accelerate from rest to their top speed in a certain
amount of time. The masses of the cars, the final speeds, and the
time to reach this speed are noted in the table. Which car has the
smallest power?
Car
Mass (g) Speed (m/s) Time (s)
A
100
3
2
B
200
2
2
C
200
2
3
D
300
2
3
E
300
1
4
Slide 10-28
Answer
Four toy cars accelerate from rest to their top speed in a certain
amount of time. The masses of the cars, the final speeds, and the
time to reach this speed are noted in the table. Which car has the
smallest power?
Car
E
Mass (g) Speed (m/s) Time (s)
300
1
4
Slide 10-29
A typical human head has a mass of 5
5.0
0 kg
kg. If the head is moving
at some speed and strikes a fixed surface, it will come to rest. A
helmet can help protect against injury; the foam in the helmet
allows the head to come to rest over a longer distance
distance, reducing
the force on the head. The foam in helmets is generally designed
to fail at a certain large force below the threshold of damage to
th h
the
head.
d If thi
this fforce is
i exceeded,
d d th
the ffoam b
begins
i tto compress.
If the foam in a helmet compresses by 1.5 cm under a force of
2500
500 N (be
(below
o tthe
e tthreshold
es o d for
o da
damage
age to tthe
e head),
ead), what
at is
s tthe
e
maximum speed the head could have on impact?
Use energy concepts to solve this problem.
Slide 10-31
Data
D
t for
f one stage
t
off the
th 2004 Tour
T
de
d France
F
show
h
that
th t Lance
L
Armstrong’s average speed was 15 m/s, and that keeping Lance
and his bike moving at this zippy pace required a power of 450
W.
A. What was the average forward force keeping Lance and his
bike moving forward?
B. To put this in perspective, compute what mass would have
this weight.
Slide 10-32
Question
Each of the 1
1.0
0 kg boxes starts at rest and is then pushed for 2
2.0
0
m across a level, frictionless floor by a rope with the noted force.
Which box has the highest final speed?
Slide 10-33
Answer
Each of the 1
1.0
0 kg boxes starts at rest and is then pushed for 2
2.0
0
m across a level, frictionless floor by a rope with the noted force.
Which box has the highest final speed?
Slide 10-34
Additional Clicker Questions
Three balls are thrown off a cliff with the same speed
speed, but in
different directions. Which ball has the greatest speed just
before it hits the ground?
A. Ball A
B Ball
B.
B ll B
C. Ball C
D. All balls have
the same speed
Slide 10-41
Answer
Three balls are thrown off a cliff with the same speed
speed, but in
different directions. Which ball has the greatest speed just
before it hits the ground?
D. All balls have
the same speed
Slide 10-42
Questions
A 20-cm-long
20 cm long spring is attached to a wall
wall. When pulled
horizontally with a force of 100 N, the spring stretches to a
length of 22 cm. What is the value of the spring constant?
A. 5000 N/m
B. 500 N/m
C. 454 N/m
Slide 10-43
Answer
A 20-cm-long
20 cm long spring is attached to a wall
wall. When pulled
horizontally with a force of 100 N, the spring stretches to a
length of 22 cm. What is the value of the spring constant?
C. 454 N/m
Slide 10-44