Download Chapter 10 - Energy and Work (Cont`d) w./ QuickCheck Questions

Chapter 10 - Energy and Work
(Cont’d)
w./ QuickCheck Questions
© 2015 Pearson Education, Inc.
Anastasia Ierides
Department of Physics and Astronomy
University of New Mexico
October 22, 2015
Review of Last Time
Impulse & Momentum; conservation of
momentum
Isolated system; internal & external forces;
collisions (elastic/inelastic); explosions
Energy - different forms (kinetic, work, etc.)
Energy transformations - from one kind of
energy to another
QuickCheck Question 10.1
A child is on a playground swing, motionless at the highest point of her arc. What energy transformation takes place as she swings back down to the lowest point of her motion?
A.
B.
C.
D.
E.
K → Ug
Ug → K
Eth → K
Ug → Eth
K → Eth
QuickCheck Question 10.1
A child is on a playground swing, motionless at the highest point of her arc. What energy transformation takes place as she swings back down to the lowest point of her motion?
A.
B.
C.
D.
E.
K → Ug
Ug → K
Eth → K
Ug → Eth
K → Eth
Hint: Going
from a high
point back
down to
ground where
she is moving
QuickCheck Question 10.2
A skier is gliding down a gentle slope at a constant speed.
What energy transformation is taking place?
A.
B.
C.
D.
E.
K → Ug
Ug → K
Eth → K
Ug → Eth
K → Eth
QuickCheck Question 10.2
A skier is gliding down a gentle slope at a constant speed.
What energy transformation is taking place?
A.
B.
C.
D.
E.
K → Ug
Ug → K
Eth → K
Ug → Eth
K → Eth
Since the skier glides at a constant
speed, the kinetic energy doesn’t
change; but he is going downward
meaning that his gravitational
potential energy is decreasing. This
energy must then be transformed into
thermal energy created by friction
between the skis and the slope.
QuickCheck 10.3
A tow rope pulls a skier up the slope at constant speed. What
energy transfer (or transfers) is taking place?
A.
B.
C.
D.
E.
W → Ug
W→K
W → Eth
Both A and B.
Both A and C.
QuickCheck 10.3
A tow rope pulls a skier up the slope at constant speed. What
energy transfer (or transfers) is taking place?
A.
B.
C.
D.
E.
W → Ug
W→K
W → Eth
Both A and B.
Both A and C.
Since the skier is traveling at a
constant speed, the kinetic energy
doesn’t change, which means that
this energy must then be transformed
into thermal energy created by
friction between the skis and the
slope. He is also going upward
meaning that he is gaining
gravitational potential energy.
The Law of Conservation of Energy
The total
energy of
an isolated
system
remains
constant!
QuickCheck Question 10.4
A crane lowers a girder into place at constant speed. Consider
the work Wg done by gravity and the work WT done by the
tension in the cable. Which is true?
A.
B.
C.
D.
E.
Wg > 0 and WT > 0
Wg > 0 and WT < 0
Wg < 0 and WT > 0
Wg < 0 and WT < 0
Wg = 0 and WT = 0
QuickCheck Question 10.4
A crane lowers a girder into place at constant speed. Consider
the work Wg done by gravity and the work WT done by the
tension in the cable. Which is true?
A.
B.
C.
D.
E.
Wg > 0 and WT > 0
Wg > 0 and WT < 0
Wg < 0 and WT > 0
Wg < 0 and WT < 0
Wg = 0 and WT = 0
The downward force of gravity is
in the direction of motion
⇒ positive work.
The upward tension is in the
direction opposite the motion
⇒ negative work.
QuickCheck Question 10.5
Robert pushes the box to the left at constant speed. In doing
so, Robert does ______ work on the box.
A. positive
B. negative
C. zero
QuickCheck Question 10.5
Robert pushes the box to the left at constant speed. In doing
so, Robert does ______ work on the box.
A. positive
B. negative
C. zero
Force is in the direction of displacement
⇒ positive work
QuickCheck Question 10.9
Ball A has half the mass and eight times the kinetic energy of
ball B. What is the speed ratio vA/vB?
A.
B.
C.
D.
E.
16
4
2
1/4
1/16
QuickCheck Question 10.9
Ball A has half the mass and eight times the kinetic energy of
ball B. What is the speed ratio vA/vB?
A.
B.
C.
D.
E.
16
4
2
1/4
1/16
Remember:
Rearranging to solve
vA = (2 KA/mA)1/2
for the velocity, v,
vB (2 KB/mB)1/2
v = (2 K/m)1/2
vA/vB = [(KA/KB)(mB/mA)]1/2
= [(8)(2)]1/2
= [16]1/2
=4
QuickCheck Question 10.10
A light plastic cart and a heavy steel cart are both pushed with the
same force for a distance of 1.0 m, starting from rest. After the
force is removed, the kinetic energy of the light plastic cart is
________ that of the heavy steel cart.
A.
B.
C.
D.
greater than
equal to
less than
Can’t say. It depends on how big the force is.
QuickCheck Question 10.10
A light plastic cart and a heavy steel cart are both pushed with the
same force for a distance of 1.0 m, starting from rest. After the
force is removed, the kinetic energy of the light plastic cart is
________ that of the heavy steel cart.
Same force, same distance
⇒ same work done
Same work
⇒ change of kinetic energy
A.
B.
C.
D.
greater than
equal to
less than
Can’t say. It depends on how big the force is.
QuickCheck Question 10.11
Each of the boxes shown is pulled for 10 m across a level,
frictionless floor by the force given. Which box experiences
the greatest change in its kinetic energy?
QuickCheck Question 10.11
Each of the boxes shown is pulled for 10 m across a level,
frictionless floor by the force given. Which box experiences
the greatest change in its kinetic energy?
D
Work-energy equation:
All have same d, so largest work (and hence
∆K = W = Fd
largest ∆K) corresponds to largest force.
QuickCheck Question 10.12
Each of the 1.0 kg boxes starts at rest and is then pulled for
2.0 m across a level, frictionless floor by a rope with the noted
force at the noted angle. Which box has the highest final
speed?
QuickCheck Question 10.12
Each of the 1.0 kg boxes starts at rest and is then pulled for
2.0 m across a level, frictionless floor by a rope with the noted
force at the noted angle. Which box has the highest final
F|| = 5√3 N
F|| = 10√3 N
F|| = 5√2 N
speed?
B
Same d; only need
to focus on
F|| = F cos θ
F|| = 10√2 N
Work-energy equation:
∆K = W = Fd cos θ
cos (30°) = √3 /2
cos (45°) = √2 /2
cos (60°) = 1/2
F|| = 5 N
QuickCheck Question 10.13
Rank in order, from largest to smallest, the gravitational potential
energies of the balls.
A.
B.
C.
D.
1>2=4>3
1>2>3>4
3>2>4>1
3>2=4>1
QuickCheck Question 10.13
Rank in order, from largest to smallest, the gravitational potential
energies of the balls.
A.
B.
C.
D.
1>2=4>3
1>2>3>4
3>2>4>1
3>2=4>1
QuickCheck Question 10.14
Starting from rest, a marble first rolls down a steeper hill, then
down a less steep hill of the same height. For which is it going
faster at the bottom?
A.
B.
C.
D.
Faster at the bottom of the steeper hill.
Faster at the bottom of the less steep hill.
Same speed at the bottom of both hills.
Can’t say without knowing the mass of the marble.
QuickCheck Question 10.14
Starting from rest, a marble first rolls down a steeper hill, then
down a less steep hill of the same height. For which is it going
faster at the bottom?
Think energy transformation!
The work required to bring
both balls down from the
same height is the same, that
is, Ug = mgy.
A.
B.
C.
D.
Work-energy equation:
∆K = W = Fd,
where F = w = mg and d = y.
Faster at the bottom of the steeper hill.
Faster at the bottom of the less steep hill.
Same speed at the bottom of both hills.
Can’t say without knowing the mass of the marble.
QuickCheck Question 10.15
A small child slides down the four frictionless slides A–D. Rank
in order, from largest to smallest, her speeds at the bottom.
A.
B.
C.
D.
vD > vA > vB > vC
vD > vA = vB > vC
vC > vA > vB > vD
vA = vB = vC = vD
QuickCheck Question 10.15
A small child slides down the four frictionless slides A–D. Rank
in order, from largest to smallest, her speeds at the bottom.
A.
B.
C.
D.
vD > vA > vB > vC
vD > vA = vB > vC
vC > vA > vB > vD
vA = vB = vC = vD
The work required to slide down
all the slides from the same height
is the same, that is, Ug = mgh.
Work-energy equation:
∆K = W = Fd,
where F = w = mg and d = h.
Same height ⇒ same work, W! ⇒ same change in kinetic energy
QuickCheck Question 10.16
Three balls are thrown from a cliff with the same speed but at
different angles. Which ball has the greatest speed just before it
hits the ground?
A.
B.
C.
D.
Ball A.
Ball B.
Ball C.
All balls have the same speed.
QuickCheck Question 10.16
Three balls are thrown from a cliff with the same speed but at
different angles. Which ball has the greatest speed just before it
hits the ground?
Work-energy equation:
∆K = W = Fd,
where F = w = mg and d = h.
h
Same height
⇒ same work, W!
A. Ball A.
⇒ same change in kinetic energy
B. Ball B.
C. Ball C.
D. All balls have the same speed.
Thermal Energy
The sum of the kinetic energy of a substance’s atoms
and molecules and the elastic potential energy stored
in their molecular bonds is the substance’s thermal
energy.
Thermal Energy
Friction on a moving
object does work
Thermal Energy
Friction on a moving
object does work
Work due to friction
creates thermal energy
Example 10.9: Creating thermal
energy by rubbing
A 0.30 kg block of wood is rubbed back and forth against a wood
table 30 times in each direction. The block is moved 8.0 cm during
each stroke and pressed against the table with a force of 22 N. How
much thermal energy is created in this process?
Example 10.9: Creating thermal
energy by rubbing
A 0.30 kg block of wood is rubbed back and forth against a wood
table 30 times in each direction. The block is moved 8.0 cm during
each stroke and pressed against the table with a force of 22 N. How
much thermal energy is created in this process?
PREPARE The
hand holding the block does work to push the block
back and forth. Work transfers energy into the block + table system,
where it appears as thermal energy according to Equation 10.16. The
force of friction can be found from the model of kinetic friction
introduced in Chapter 5, fk = µkn; from Table 5.2 the coefficient of
kinetic friction for wood sliding on wood is µk = 0.20.
Example 10.9: Creating thermal
energy by rubbing
PREPARE To
find the normal force n acting on the block, we draw the
free-body diagram of the figure, which shows only the vertical forces
acting on the block.
Example 10.9: Creating thermal
energy by rubbing
PREPARE To
find the normal force n acting on the block, we draw the
free-body diagram of the figure, which shows only the vertical forces
acting on the block.
Example 10.9: Creating thermal
energy by rubbing
PREPARE To
find the normal force n acting on the block, we draw the
free-body diagram of the figure, which shows only the vertical forces
acting on the block.
SOLVE We want to use ΔEth = fk Δx, where the
kinetic friction is fk = µkn.
Example 10.9: Creating thermal
energy by rubbing
PREPARE To
find the normal force n acting on the block, we draw the
free-body diagram of the figure, which shows only the vertical forces
acting on the block.
SOLVE We want to use ΔEth = fk Δx, where the
kinetic friction is fk = µkn. The block is not
accelerating in the y-direction, so from the
free-body diagram Newton’s second law gives
Example 10.9: Creating thermal
energy by rubbing
PREPARE To
find the normal force n acting on the block, we draw the
free-body diagram of the figure, which shows only the vertical forces
acting on the block.
SOLVE We want to use ΔEth = fk Δx, where the
kinetic friction is fk = µkn. The block is not
accelerating in the y-direction, so from the
free-body diagram Newton’s second law gives
Σ Fy = n − w − F = may = 0
or
n = w + F = mg + F
= (0.30 kg)(9.8 m/s2) + 22 N
= 24.9 N
Example 10.9: Creating thermal
energy by rubbing
SOLVE The
friction force is then
fk = µkn = (0.20)(24.9 N) = 4.98 N.
Example 10.9: Creating thermal
energy by rubbing
SOLVE The
friction force is then
fk = µkn = (0.20)(24.9 N) = 4.98 N.
The total displacement of the block is 2 × 30 × 8.0 cm = 4.8 m. Thus
the thermal energy created is
ΔEth = fk Δx = (4.98 N)(4.8 m) = 24 J
Example 10.9: Creating thermal
energy by rubbing
SOLVE The
friction force is then
fk = µkn = (0.20)(24.9 N) = 4.98 N.
The total displacement of the block is 2 × 30 × 8.0 cm = 4.8 m. Thus
the thermal energy created is
ΔEth = fk Δx = (4.98 N)(4.8 m) = 24 J
ASSESS This
modest amount of thermal energy seems reasonable for
a person to create by rubbing.
In Class Exercise
Vigorously rub the top of your leg with your
dominant hand’s palm for about 10 seconds. If
you then pass the fingers of your other hand over
the spot where you rubbed, you’ll feel a distinct
warm area.
In Class Exercise
Vigorously rub the top of your leg with your
dominant hand’s palm for about 10 seconds. If
you then pass the fingers of your other hand over
the spot where you rubbed, you’ll feel a distinct
warm area.
Congratulations! You’ve just set some
100,000,000,000,000,000,000,000 atoms into
motion!
Using the Law of Conservation of
Energy
Using the Law of Conservation of
Energy, W = ∆E
We use the law of conservation of energy to
develop a before-and-after perspective:
∆K + ∆Ug + ∆Us + ∆Eth = W
Using the Law of Conservation of
Energy, W = ∆E
We use the law of conservation of energy to
develop a before-and-after perspective:
∆K + ∆Ug + ∆Us + ∆Eth = W
Kf + (Ug)f + (Us)f + ∆Eth = Ki + (Ug)i + (Us)i + W
Using the Law of Conservation of
Energy, W = ∆E
We use the law of conservation of energy to
develop a before-and-after perspective:
∆K + ∆Ug + ∆Us + ∆Eth = W
Kf + (Ug)f + (Us)f + ∆Eth = Ki + (Ug)i + (Us)i + W
Analogous to before-and-after of the law of
conservation of momentum
Using the Law of Conservation of
Energy, W = ∆E
We use the law of conservation of energy to
develop a before-and-after perspective:
∆K + ∆Ug + ∆Us + ∆Eth = W
Kf + (Ug)f + (Us)f + ∆Eth = Ki + (Ug)i + (Us)i + W
Analogous to before-and-after of the law of
conservation of momentum
In an isolated system, W = 0, because ∆E = 0:
Kf + (Ug)f + (Us)f + ∆Eth = Ki + (Ug)i + (Us)i
Using the Law of Conservation of
Energy, W = ∆E
Using the Law of Conservation of
Energy, W = ∆E
How to Choose an Isolated System
QuickCheck Question 10.18
A spring-loaded gun shoots a plastic ball with a launch speed of
2.0 m/s. If the spring is compressed twice as far, the ball’s launch
speed will be
A.
B.
C.
D.
E.
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
QuickCheck Question 10.18
A spring-loaded gun shoots a plastic ball with a launch speed of
2.0 m/s. If the spring is compressed twice as far, the ball’s launch
speed will be
We initially have spring potential
energy,
Us = 1/2 k x2,
and produce kinetic energy,
K = 1/2 m v2.
A.
B.
C.
D.
E.
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
QuickCheck Question 10.18
A spring-loaded gun shoots a plastic ball with a launch speed of
2.0 m/s. If the spring is compressed twice as far, the ball’s launch
speed will be
We initially have spring potential
energy,
Us = 1/2 k x2,
and produce kinetic energy,
K = 1/2 m v2.
A.
B.
C.
D.
E.
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
Equating the two
we have,
x2 ∝ v2
QuickCheck Question 10.18
A spring-loaded gun shoots a plastic ball with a launch speed of
2.0 m/s. If the spring is compressed twice as far, the ball’s launch
speed will be
We initially have spring potential
energy,
Us = 1/2 k x2,
and produce kinetic energy,
K = 1/2 m v2.
A.
B.
C.
D.
E.
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
Equating the two
we have,
x2 ∝ v2
Double the displacement, x
⇒ double the speed, v
QuickCheck Question 10.18
A spring-loaded gun shoots a plastic ball with a launch speed of
2.0 m/s. If the spring is compressed twice as far, the ball’s launch
speed will be
We initially have spring potential
energy,
Us = 1/2 k x2,
and produce kinetic energy,
K = 1/2 m v2.
A.
B.
C.
D.
E.
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
Equating the two
we have,
x2 ∝ v2
Double the displacement, x
⇒ double the speed, v
QuickCheck Question 10.19
A spring-loaded gun shoots a plastic ball with a launch speed of
2.0 m/s. If the spring is replaced with a new spring having twice
the spring constant (but still compressed the same distance), the
ball’s launch speed will be
A.
B.
C.
D.
E.
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
QuickCheck Question 10.19
A spring-loaded gun shoots a plastic ball with a launch speed of
2.0 m/s. If the spring is replaced with a new spring having twice
the spring constant (but still compressed the same distance), the
ball’s launch speed will be
Again, we begin with Us = 1/2 k x2,
and produce K = 1/2 m v2.
A.
B.
C.
D.
E.
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
Equating the two
we have,
v ∝ k1/2
QuickCheck Question 10.19
A spring-loaded gun shoots a plastic ball with a launch speed of
2.0 m/s. If the spring is replaced with a new spring having twice
the spring constant (but still compressed the same distance), the
ball’s launch speed will be
Again, we begin with Us = 1/2 k x2,
and produce K = 1/2 m v2.
A.
B.
C.
D.
E.
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
Equating the two
we have,
v ∝ k1/2
Double the spring constant, k
⇒ multiply the speed by 21/2 ≈ 1.4
QuickCheck Question 10.19
A spring-loaded gun shoots a plastic ball with a launch speed of
2.0 m/s. If the spring is replaced with a new spring having twice
the spring constant (but still compressed the same distance), the
ball’s launch speed will be
Again, we begin with Us = 1/2 k x2,
and produce K = 1/2 m v2.
A.
B.
C.
D.
E.
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
Equating the two
we have,
v ∝ k1/2
Double the spring constant, k
⇒ multiply the speed by 21/2 ≈ 1.4
Example 10.11: Speed at the bottom
of a water slide
While at the county fair, Katie tries the water slide, whose shape is
shown in the figure. The starting point is 9.0 m above the ground. She
pushes off with an initial speed of 2.0 m/s. If the slide is frictionless,
how fast will Katie be traveling at the bottom?
Example 10.11: Speed at the bottom
of a water slide
PREPARE Table
10.2
showed that the system
consisting of Katie and the
earth is isolated because the
normal force of the slide is
perpendicular to Katie’s
motion and does no work. If
we assume the slide is
frictionless, we can use the
conservation of mechanical
energy equation.
Example 10.11: Speed at the bottom
of a water slide
PREPARE Table
10.2
showed that the system
consisting of Katie and the
earth is isolated because the
normal force of the slide is
perpendicular to Katie’s
motion and does no work. If
we assume the slide is
frictionless, we can use the
conservation of mechanical
energy equation.
Conservation of mechanical
energy gives
Kf + (Ug)f = Ki + (Ug)i
or
SOLVE
Example 10.11: Speed at the bottom
of a water slide
PREPARE Table
10.2
showed that the system
consisting of Katie and the
earth is isolated because the
normal force of the slide is
perpendicular to Katie’s
motion and does no work. If
we assume the slide is
frictionless, we can use the
conservation of mechanical
energy equation.
Conservation of mechanical
energy gives
Kf + (Ug)f = Ki + (Ug)i
or
SOLVE
Example 10.11: Speed at the bottom
of a water slide
Taking yf = 0 m, we have
Example 10.11: Speed at the bottom
of a water slide
Taking yf = 0 m, we have
which we can solve to get
Example 10.11: Speed at the bottom
of a water slide
Taking yf = 0 m, we have
which we can solve to get
Notice that the shape of the slide does not matter because
gravitational potential energy depends only on the height above a
reference level.
Example 10.13: Pulling a bike trailer
Monica pulls her daughter Jessie in a bike trailer. The trailer and
Jessie together have a mass of 25 kg. Monica starts up a 100-mlong slope that’s 4.0 m high. On the slope, Monica’s bike pulls on
the trailer with a constant force of 8.0 N. They start out at the
bottom of the slope with a speed of 5.3 m/s. What is their speed at
the top of the slope?
Example 10.13: Pulling a bike trailer
PREPARE Taking
Jessie and the trailer as the system, we see that
Monica’s bike is applying a force to the system as it moves
through a displacement; that is, Monica’s bike is doing work on
the system. Thus we’ll need to use the full version of the law of
energy conservation.
Example 10.13: Pulling a bike trailer
SOLVE
If we assume there’s no friction, so that ΔEth = 0, then
Kf + (Ug)f = Ki + (Ug)i + W
or
Example 10.13: Pulling a bike trailer
Taking yi = 0 m and writing W = Fd, we can solve for the final
speed:
Example 10.13: Pulling a bike trailer
from which we find that vf = 3.7 m/s. Note that we took the work
to be a positive quantity because the force is in the same direction
as the displacement.
Example 10.13: Pulling a bike trailer
ASSESS A speed
of 3.7 m/s—about 8 mph—seems reasonable for
a bicycle’s speed. Jessie’s final speed is less than her initial speed,
indicating that the uphill force of Monica’s bike on the trailer is
less than the downhill component of gravity.
Energy in Collisions
Two types of collisions: elastic and perfectly
inelastic
Energy in Collisions
Two types of collisions: elastic and perfectly
inelastic
Momentum is conserved in all collisions, but
mechanical energy is only conserved in a
perfectly elastic collision
Energy in Collisions
Two types of collisions: elastic and perfectly
inelastic
Momentum is conserved in all collisions, but
mechanical energy is only conserved in a
perfectly elastic collision
In an inelastic collision, some mechanical
energy is converted to thermal energy
Example 10.15: Energy transformations
in a perfectly inelastic collision
The figure shows two train
cars that move toward each
other, collide, and couple
together. In Example 9.8,
we used conservation of
momentum to find the final
velocity shown in the
figure from the given
initial velocities. How
much thermal energy is
created in this collision?
Example 10.15: Energy transformations
in a perfectly inelastic collision
PREPARE We’ll
choose
our system to be the two
cars. Because the track is
horizontal, there is no
change in potential
energy. Thus the law of
conservation of energy is
Kf + ΔEth = Ki.
Example 10.15: Energy transformations
in a perfectly inelastic collision
PREPARE We’ll
choose
our system to be the two
cars. Because the track is
horizontal, there is no
change in potential
energy. Thus the law of
conservation of energy is
Kf + ΔEth = Ki.
The total energy before the collision must equal the total
energy afterward, but the mechanical energies need not be
equal.
Example 10.15: Energy transformations
in a perfectly inelastic collision
SOLVE The
initial kinetic energy is
Example 10.15: Energy transformations
in a perfectly inelastic collision
SOLVE The
initial kinetic energy is
Because the cars stick together and move as a single object
with mass m1 + m2, the final kinetic energy is
Example 10.15: Energy transformations
in a perfectly inelastic collision
From the conservation of energy equation on the previous
slide, we find that the thermal energy increases by
ΔEth = Ki − Kf = 4.7 × 104 J − 1900 J = 4.5 × 104 J
Example 10.15: Energy transformations
in a perfectly inelastic collision
From the conservation of energy equation on the previous
slide, we find that the thermal energy increases by
ΔEth = Ki − Kf = 4.7 × 104 J − 1900 J = 4.5 × 104 J
This amount of the initial kinetic energy is transformed into
thermal energy during the impact of the collision.
Example 10.15: Energy transformations
in a perfectly inelastic collision
From the conservation of energy equation on the previous
slide, we find that the thermal energy increases by
ΔEth = Ki − Kf = 4.7 × 104 J − 1900 J = 4.5 × 104 J
This amount of the initial kinetic energy is transformed into
thermal energy during the impact of the collision.
ASSESS About
96% of the initial kinetic energy is transformed
into thermal energy. This is typical of many real-world
collisions.
Elastic Collisions
Elastic collisions
obey conservation of
momentum and
conservation of
mechanical energy
Elastic Collisions
Elastic collisions
obey conservation of
momentum and
conservation of
mechanical energy
Momentum: m1 (v1x )i = m1 (v1x )f + m2 (v2 x )f
Energy:
1
1
1
2
2
2
m1 (v1x )i = m1 (v1x )f + m2 (v2 x )f
2
2
2
Example 10.16: Velocities in an air
hockey collision
On an air hockey table, a moving puck, traveling to the right at
2.3 m/s, makes a head-on collision with an identical puck at rest.
What is the final velocity of each puck?
Example 10.16: Velocities in an air
hockey collision
PREPARE We’ve shown
the final
velocities in the picture, but we
don’t really know yet which way
the pucks will move. Because
one puck was initially at rest, we
can use Equations 10.20 to find
the final velocities of the pucks. The pucks are identical, so we
have m1 = m2 = m.
Example 10.16: Velocities in an air
hockey collision
SOLVE We
use Equations 10.20
with m1 = m2 = m to get
The incoming puck stops dead, and the initially stationary puck
goes off with the same velocity that the incoming one had.
Example 10.17: Protecting your
head
A bike helmet—basically a shell of hard, crushable foam—is
tested by being strapped onto a 5.0 kg headform and dropped from
a height of 2.0 m onto a hard anvil. What force is encountered by
the headform if the impact crushes the foam by 3.0 cm?
Example 10.17: Protecting your
head
A bike helmet—basically a shell of hard, crushable foam—is
tested by being strapped onto a 5.0 kg headform and dropped from
a height of 2.0 m onto a hard anvil. What force is encountered by
the headform if the impact crushes the foam by 3.0 cm?
PREPARE The
work-energy equation can
be used to calculate the force on the
headform. We’ll choose the headform
and the earth to be the system; the foam
in the helmet is part of the environment.
We make this choice so that the force on
the headform due to the foam is an
external force that does work W on the
headform.
Example 10.17: Protecting your
head
SOLVE The
headform starts
at rest, speeds up as it falls,
then returns to rest during the
impact. Overall, Kf = Ki.
Furthermore, ΔEth = 0
because there’s no friction to
increase the thermal energy.
Only the gravitational
potential energy changes, so
the work-energy equation is
(Ug)f − (Ug)i = W
Example 10.17: Protecting your
head
The upward force of the foam
on the headform is opposite
the downward displacement
of the headform. We see that
the work done is negative:
W = −Fd,
assuming that the force is
constant. Using this result in
the work-energy equation and
solving for F, we find
Example 10.17: Protecting your
head
Taking our reference height to be y = 0 m at the anvil, we have
(Ug)f = 0. We’re left with (Ug)i = mgyi, so
Example 10.17: Protecting your
head
Taking our reference height to be y = 0 m at the anvil, we have
(Ug)f = 0. We’re left with (Ug)i = mgyi, so
This is the force that acts on the head to bring it to a halt in 3.0
cm. More important from the perspective of possible brain injury
is the head’s acceleration:
Example 10.17: Protecting your
head
ASSESS The
accepted threshold for serious brain injury is around
300g, so this helmet would protect the rider in all but the most
serious accidents. Without the helmet, the rider’s head would
come to a stop in a much shorter distance and thus be subjected to
a much larger acceleration.
Example 10.17: Protecting your
head
ASSESS The
accepted threshold for serious brain injury is around
300g, so this helmet would protect the rider in all but the most
serious accidents. Without the helmet, the rider’s head would
come to a stop in a much shorter distance and thus be subjected to
a much larger acceleration.
Power
Power is the rate at which energy is
transformed or transferred
Power
Power is the rate at which energy is
transformed or transferred
Power
Power is the rate at which energy is
transformed or transferred
The unit of power is the watt:
1 watt = 1 W = 1 J/s
Output Power of a Force
A force doing work transfers energy
Output Power of a Force
A force doing work transfers energy
The rate that this force transfers energy is the
output power of that force:
Output Power of a Force
A force doing work transfers energy
The rate that this force transfers energy is the
output power of that force:
QuickCheck Question 10.20
Four students run up the stairs in the time shown. Which student
has the largest power output?
QuickCheck Question 10.20
Four students run up the stairs in the time shown. Which student
has the largest power output?
QuickCheck Question 10.20
Four students run up the stairs in the time shown. Which student
has the largest power output?
The work, W, done by the student is against gravity, where
each student gains gravitational potential energy, Ug = mgh.
QuickCheck Question 10.20
Four students run up the stairs in the time shown. Which student
has the largest power output?
Ug = 7840 J
Ug = 7840 J
Ug = 6272 J
Ug = 15680 J
The work, W, done by the student is against gravity, where
each student gains gravitational potential energy, Ug = mgh.
QuickCheck Question 10.20
Four students run up the stairs in the time shown. Which student
has the largest power output?
Ug = 7840 J
P = 784 W
Ug = 7840 J
P = 980 W
Ug = 6272 J
P = 784 W
Ug = 15680 J
P = 627.2 W
The work, W, done by the student is against gravity, where
each student gains gravitational potential energy, Ug = mgh.
QuickCheck Question 10.20
Four students run up the stairs in the time shown. Which student
has the largest power output?
B.
Ug = 7840 J
P = 784 W
Ug = 7840 J
P = 980 W
Ug = 6272 J
P = 784 W
Ug = 15680 J
P = 627.2 W
The work, W, done by the student is against gravity, where
each student gains gravitational potential energy, Ug = mgh.
QuickCheck Question 10.21
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?
QuickCheck Question 10.21
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?
QuickCheck Question 10.21
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?
Remember:
F = ma
= m Δv/Δt
QuickCheck Question 10.21
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?
Remember:
F = ma
= m Δv/Δt
QuickCheck Question 10.21
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?
Remember:
F = ma
= m Δv/Δt
= m v/Δt
QuickCheck Question 10.21
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?
Remember:
F = ma
= m Δv/Δt
= m v/Δt
= mv2/Δt
QuickCheck Question 10.21
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?
Remember:
F = ma
= m Δv/Δt
= m v/Δt
v2 = 9
v2 = 4
v2 = 4
v2 = 1
v2 = 4
= mv2/Δt
m/Δt = .05
m/Δt = .1
m/Δt = .1
m/Δt = .2
m/Δt = .1
QuickCheck Question 10.21
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?
Remember:
F = ma
= m Δv/Δt
= m v/Δt
v2 = 9
v2 = 4
v2 = 4
v2 = 1
v2 = 4
= mv2/Δt
m/Δt = .05
m/Δt = .1
m/Δt = .1
m/Δt = .2
m/Δt = .1
Example 10.18: Power to pass a
truck
Your 1500 kg car is behind a truck traveling at 60 mph (27 m/
s). To pass the truck, you speed up to 75 mph (34 m/s) in 6.0 s.
What engine power is required to do this?
Example 10.18: Power to pass a
truck
Your 1500 kg car is behind a truck traveling at 60 mph (27 m/
s). To pass the truck, you speed up to 75 mph (34 m/s) in 6.0 s.
What engine power is required to do this?
PREPARE Your
engine is transforming the chemical energy of
its fuel into the kinetic energy of the car. We can calculate the
rate of transformation by finding the change ΔK in the kinetic
energy and using the known time interval.
Example 10.18: Power to pass a
truck
Your 1500 kg car is behind a truck traveling at 60 mph (27 m/
s). To pass the truck, you speed up to 75 mph (34 m/s) in 6.0 s.
What engine power is required to do this?
PREPARE Your
engine is transforming the chemical energy of
its fuel into the kinetic energy of the car. We can calculate the
rate of transformation by finding the change ΔK in the kinetic
energy and using the known time interval.
SOLVE We have
Example 10.18: Power to pass a
truck
SOLVE
so that
ΔK = Kf − Ki
= (8.67 × 105 J) − (5.47 × 105 J) = 3.20 × 105 J
Example 10.18: Power to pass a
truck
SOLVE
so that
ΔK = Kf − Ki
= (8.67 × 105 J) − (5.47 × 105 J) = 3.20 × 105 J
To transform this amount of energy in 6 s, the power required
is
Example 10.18: Power to pass a
truck
SOLVE
so that
ΔK = Kf − Ki
= (8.67 × 105 J) − (5.47 × 105 J) = 3.20 × 105 J
To transform this amount of energy in 6 s, the power required
is
This is about 71 hp. This power is in addition to the power
needed to overcome drag and friction and cruise at 60 mph, so
the total power required from the engine will be even greater
than this.
Chapter 11 - Using Energy
w./ QuickCheck Questions
© 2015 Pearson Education, Inc.
Basic Energy Model
Work and heat are
energy transfers
that change the
system’s total
energy
If the system is
isolated, the total
energy is
conserved
Stop To Think: Review
Christina throws a javelin into the air. As she propels it
forward from rest, she does 270 J of work on it. At its highest
point, its gravitational potential energy has increased by 70 J.
What is the javelin’s kinetic energy at this point?
A.
B.
C.
D.
E.
270 J
340 J
200 J
–200 J
–340 J
Stop To Think: Review
Christina throws a javelin into the air. As she propels it
forward from rest, she does 270 J of work on it. At its highest
point, its gravitational potential energy has increased by 70 J.
What is the javelin’s kinetic energy at this point?
W = 270 J
ΔUg = 70 J
A. 270 J
B.
C.
D.
E.
340 J
200 J
–200 J
–340 J
Stop To Think: Review
Christina throws a javelin into the air. As she propels it
forward from rest, she does 270 J of work on it. At its highest
point, its gravitational potential energy has increased by 70 J.
What is the javelin’s kinetic energy at this point?
W = 270 J
ΔUg = 70 J
A. 270 J
B.
C.
D.
E.
340 J
200 J
–200 J
–340 J
Using the work energy theorem we
have
W = ∆K + ΔUg
Stop To Think: Review
Christina throws a javelin into the air. As she propels it
forward from rest, she does 270 J of work on it. At its highest
point, its gravitational potential energy has increased by 70 J.
What is the javelin’s kinetic energy at this point?
W = 270 J
ΔUg = 70 J
A. 270 J
B.
C.
D.
E.
340 J
200 J
–200 J
–340 J
Using the work energy theorem we
have
W = ∆K + ΔUg
or
∆K = W - ΔUg
Stop To Think: Review
Christina throws a javelin into the air. As she propels it
forward from rest, she does 270 J of work on it. At its highest
point, its gravitational potential energy has increased by 70 J.
What is the javelin’s kinetic energy at this point?
W = 270 J
ΔUg = 70 J
A. 270 J
B.
C.
D.
E.
340 J
200 J
–200 J
–340 J
Using the work energy theorem we
have
W = ∆K + ΔUg
or
∆K = W - ΔUg
Transforming Energy
Energy cannot be
created or destroyed,
but it can be transformed
Transforming Energy
Energy cannot be
created or destroyed,
but it can be transformed
The energy in these
transformations is not
lost, but converted to
other forms that are
typically less useful to
us
Transforming Energy
Energy cannot be
created or destroyed,
but it can be transformed
The energy in these
transformations is not
lost, but converted to
other forms that are
typically less useful to
us
Transforming Energy
Energy cannot be
created or destroyed,
but it can be transformed
The energy in these
transformations is not
lost, but converted to
other forms that are
typically less useful to
us
Transforming Energy
Energy cannot be
created or destroyed,
but it can be transformed
The energy in these
transformations is not
lost, but converted to
other forms that are
typically less useful to
us
Transforming Energy
The work-energy equation includes work, an energy
transfer. We now include electric, radiant, and chemical
energy in our definition of work
Transforming Energy
The work-energy equation includes work, an energy
transfer. We now include electric, radiant, and chemical
energy in our definition of work
Work is positive when energy is transferred into the
system and negative when energy is transferred out of
the system
Transforming Energy
The work-energy equation includes work, an energy
transfer. We now include electric, radiant, and chemical
energy in our definition of work
Work is positive when energy is transferred into the
system and negative when energy is transferred out of
the system
When other forms of energy are transformed into thermal
energy the change is irreversible; the energy isn’t lost, but
it is lost to our use
QuickCheck Question 11.1
When you walk at a constant speed on level ground, what
energy transformation is taking place?
A. Echem → Ug
B. Ug → Eth
C. Echem → K
D. Echem → Eth
E. K → Eth
QuickCheck Question 11.1
When you walk at a constant speed on level ground, what
energy transformation is taking place?
A. Echem → Ug
B. Ug → Eth
C. Echem → K
D. Echem → Eth
E. K → Eth
Efficiency
This woman climbs the
stairs, converting her
body’s chemical energy
into gravitational
potential energy
Efficiency
This woman climbs the
stairs, converting her
body’s chemical energy
into gravitational
potential energy
How efficiently will
this woman climb to
the top of the stairs?
Efficiency
The larger the energy
losses in a system, the
lower its efficiency
Efficiency
The larger the energy
losses in a system, the
lower its efficiency
Reductions in efficiency are caused by:
Efficiency
The larger the energy
losses in a system, the
lower its efficiency
Reductions in efficiency are caused by:
1. Process limitations: energy loss due to practical
details such as the design of the mechanism
Efficiency
The larger the energy
losses in a system, the
lower its efficiency
Reductions in efficiency are caused by:
1. Process limitations: energy loss due to practical
details such as the design of the mechanism
2. Fundamental limitations: energy loss due to
physical laws (transformation of energy into
thermal energy)
Efficiency
35% efficiency
is close to the
theoretical
maximum for
power plants
due to
fundamental
limitations
Efficiency
Example 1.1: Lightbulb Efficiency
A 15 W compact fluorescent
bulb and a 75 W incandescent
bulb each produce 3.0 W of
visible-light energy. What are
the efficiencies of these two
types of bulbs for converting
electric energy into light?
Example 1.1: Lightbulb Efficiency
PREPARE The
problem
statement gives us values for
power. But 15 W is 15 J/s, so
we will consider the value for
the power to be the energy in 1
second. For each of the bulbs,
“what you get” is the visible light output—3.0 J of light every
second for each bulb. “What you had to pay” is the electric
energy to run the bulb. This is how the bulbs are rated. A bulb
labeled “15 W” uses 15 J of electric energy each second.
A 75 W bulb uses 75 J each second.
Example 1.1: Lightbulb Efficiency
SOLVE The
efficiencies of the
two bulbs are computed using
the energy in 1 second:
Example 1.1: Lightbulb Efficiency
SOLVE The
efficiencies of the
two bulbs are computed using
the energy in 1 second:
Both bulbs produce the same visible-light output, but
the compact fluorescent bulb does so with a significantly
lower energy input, so it is more efficient. Compact
fluorescent bulbs are more efficient than incandescent bulbs,
but their efficiency is still relatively low—only 20%.
ASSESS
Getting Energy From Food: Energy
Inputs
The chemical
energy in food
provides the
necessary
energy input
for your body
to function
Getting Energy From Food: Energy
Inputs
Chemical energy in food is broken down into
simpler molecules such as glucose and glycogen
and are metabolized in the cells by combining
with oxygen
Getting Energy From Food: Energy
Inputs
Chemical energy in food is broken down into
simpler molecules such as glucose and glycogen
and are metabolized in the cells by combining
with oxygen
Oxidation reactions “burn” the fuel you obtain
by eating
Getting Energy From Food: Energy
Inputs
Burning food transforms all of its chemical
energy into thermal energy
Getting Energy From Food: Energy
Inputs
Burning food transforms all of its chemical
energy into thermal energy
Thermal energy is measured in units of calories
(cal)
1.00 calorie = 4.19 joules
Getting Energy From Food: Energy
Inputs
Getting Energy From Food: Energy
Inputs
Example 11.2: Energy in Food
A 12 oz can of soda contains approximately 40 g (or a bit less
than 1/4 cup) of sugar, a simple carbohydrate. What is the
chemical energy in joules? How many Calories is this?
Example 11.2: Energy in Food
A 12 oz can of soda contains approximately 40 g (or a bit less
than 1/4 cup) of sugar, a simple carbohydrate. What is the
chemical energy in joules? How many Calories is this?
SOLVE From Table 11.1, 1 g of sugar contains 17 kJ of energy,
so 40 g contains
Example 11.2: Energy in Food
A 12 oz can of soda contains approximately 40 g (or a bit less
than 1/4 cup) of sugar, a simple carbohydrate. What is the
chemical energy in joules? How many Calories is this?
SOLVE From Table 11.1, 1 g of sugar contains 17 kJ of energy,
so 40 g contains
Example 11.2: Energy in Food
A 12 oz can of soda contains approximately 40 g (or a bit less
than 1/4 cup) of sugar, a simple carbohydrate. What is the
chemical energy in joules? How many Calories is this?
SOLVE From Table 11.1, 1 g of sugar contains 17 kJ of energy,
so 40 g contains
Converting to Calories, we get
Example 11.2: Energy in Food
A 12 oz can of soda contains approximately 40 g (or a bit less
than 1/4 cup) of sugar, a simple carbohydrate. What is the
chemical energy in joules? How many Calories is this?
SOLVE From Table 11.1, 1 g of sugar contains 17 kJ of energy,
so 40 g contains
Converting to Calories, we get
Example 11.2: Energy in Food
160 Calories is a typical value
for the energy content of a 12 oz can
of soda (check the nutrition label on
one to see), so this result seems
reasonable.
ASSESS
Using Energy in the Body: Energy
Outputs
At rest, your body uses energy to build and
repair tissue, digest food and keep warm
Using Energy in the Body: Energy
Outputs
At rest, your body uses energy to build and
repair tissue, digest food and keep warm
Your body uses approximately 100 W of energy
at rest which is converted almost entirely into
thermal energy, which is transferred as heat to
the environment
Using Energy in the Body: Energy
Outputs
At rest, your body uses energy to build and
repair tissue, digest food and keep warm
Your body uses approximately 100 W of energy
at rest which is converted almost entirely into
thermal energy, which is transferred as heat to
the environment
When you feel cold, your body is releasing its
heat to the environment
Using Energy in the Body: Energy
Outputs
Using Energy in the Body: Energy
Outputs
When you’re active, your
cells require oxygen to
metabolize carbohydrates
Using Energy in the Body: Energy
Outputs
When you’re active, your
cells require oxygen to
metabolize carbohydrates
Your body’s energy use,
or total metabolic energy
use, can be measured by
how much oxygen it is
using
Using Energy in the Body: Energy
Outputs
Efficiency of the Human Body
As you climb stairs, you change your potential
energy.
Efficiency of the Human Body
As you climb stairs, you change your potential
energy. If you climb 2.7 m and your mass is 68
kg then your potential energy is
Efficiency of the Human Body
As you climb stairs, you change your potential
energy. If you climb 2.7 m and your mass is 68
kg then your potential energy is
On average, you use 7200 J
to climb the stairs:
Efficiency of the Human Body
We compute the efficiency of climbing the
stairs:
Efficiency of the Human Body
We compute the efficiency of climbing the
stairs:
We typically consider the
body’s efficiency to be 25%
Energy Storage
Energy from food that is not used will be stored
Energy Storage
Energy from food that is not used will be stored
If energy input from food continuously exceeds
the energy outputs of the body, the energy is
stored as fat under the skin and around the
organs
Energy Storage
Energy from food that is not used will be stored
If energy input from food continuously exceeds
the energy outputs of the body, the energy is
stored as fat under the skin and around the
organs
So get out there and move! Unless winter is
coming, then stay put and get ready for
hibernation
😜
Example 11.6: Running out of fuel
The body stores about 400 g of carbohydrates. Approximately
how far could a 68 kg runner travel on this stored energy?
Example 11.6: Running out of fuel
The body stores about 400 g of carbohydrates. Approximately
how far could a 68 kg runner travel on this stored energy?
PREPARE Table 11.1 gives a value of 17 kJ per g of
carbohydrate. The 400 g of carbohydrates in the body contain
an energy of
Example 11.6: Running out of fuel
The body stores about 400 g of carbohydrates. Approximately
how far could a 68 kg runner travel on this stored energy?
PREPARE Table 11.1 gives a value of 17 kJ per g of
carbohydrate. The 400 g of carbohydrates in the body contain
an energy of
SOLVE Table
11.4 gives the power used in running at 15 km/h
as 1150 W. The time that the stored chemical energy will last
at this rate is
Example 11.6: Running out of fuel
And the distance that can be covered during this time at 15
km/h is
to two significant figures.
Example 11.6: Running out of fuel
And the distance that can be covered during this time at 15
km/h is
to two significant figures.
ASSESS A marathon is longer than this—just over 42 km. Even
with “carbo loading” before the event (eating highcarbohydrate meals), many marathon runners “hit the wall”
before the end of the race as they reach the point where they
have exhausted their store of carbohydrates. The body has
other energy stores (in fats, for instance), but the rate that they
can be drawn on is much lower.
Energy and Locomotion
When you walk at a
constant speed on level
ground, your kinetic
and potential energy are
constant
Energy and Locomotion
When you walk at a
constant speed on level
ground, your kinetic
and potential energy are
constant
You need energy to walk because the kinetic
energy of your leg and foot is transformed into
thermal energy in your muscles and shoes,
which is lost
Summary
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Things that are due
Reading Quiz #11
Due October 27, 2015 by 4:59 pm
Homework #7
Due November 2, 2015 at 11:59 pm
Exam #3
October 29, 2015 at 5:00 pm
EXAM #3
Covers Chapters 8-11 & Lectures 14-18
Thursday, October 29, 2015
Bring a calculator and cheat sheet (turn in with
exam, one side of 8.5”x11” piece of paper)
Practice exam will be available on website
along with solutions
QUESTIONS?