Transcript 141S13-NotesCh5c-June03

5.5 Conservation of Energy
Much of physics is concerned with conservation laws – we’ll
learn the 3 most important of these in PC141. A conservation
law dictates that a particular physical quantity is conserved, i.e. it
is constant in time.
Conservation of energy is perhaps the most important of these
laws. It dictates that the total energy of the universe is
conserved. That doesn’t mean that the total kinetic or potential
energy is conserved, but it does mean that we can relate changes
in one to changes in the other. By keeping track of these
changes, we can solve many problems in classical mechanics
without having to resort to free-body diagrams and Newton’s 2nd
law (and just when you were becoming comfortable with
those…). Another way of stating this conservation law is that
energy can never be created or destroyed.
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 1
5.5 Conservation of Energy
Of course, keeping track of energy changes in the whole universe
when you just want to analyze a simple apparatus on the table in
front of you isn’t a pleasant prospect. Therefore, it helps to
define systems. A system is defined as a quantity of matter,
enclosed by boundaries (either real or imaginary). For the first
few problems of this chapter, we will start by defining the system
under study, just to get used to the concept.
A closed or isolated system (by far the most common in PC141)
is one for which there is no interaction of any kind across the
boundary. Since no energy can pass through this boundary, we
can restate the law of conservation of total energy as:
the total energy of a closed (i.e. isolated)
system is always conserved
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 2
5.5 Conservation of Energy
On the other hand, an open system is one for which energy or
matter can interact with the outside world. An example is the
earth. If we imagine a fictitious boundary enclosing the earth
and its atmosphere, it’s still an open system, since solar energy
can enter the system, while thermal radiation can leave it.
A somewhat less precise conservation law exists for open
systems:
the total energy of an open system changes by
exactly the amount of net work done on the system
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 3
5.5 Conservation of Energy
Conservative and Nonconservative Forces
One thing that we skipped over in chapter 4 is that forces can be
classified as conservative or nonconservative:
a force is conservative if the work done by it in moving an
object is independent of the object’s path; i.e. it depends
only on the initial and final positions of the object.
As a corollary of this statement, one can also prove that
a force is conservative if the work done by it in moving an
object through a round trip is zero
Another point (not mentioned in the text, but important at
higher levels of physics) is that a conservative force can only
depend on the position or configuration of a system, and not any
other parameters (such as velocity or acceleration).
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 4
5.5 Conservation of Energy
Conservative and Nonconservative Forces cont’
On the other hand,
a force is nonconservative if the work done by it in moving
an object depends on the object’s path
Friction, for example, is a nonconservative force. Taking a longer
path from an initial point to a final point produces more work
done by friction – this is manifested as an increase heat (thermal
energy).
In general, a conservative force allows you to “store” energy as
potential energy, while a nonconservative force does not. In fact,
the concept of potential energy is only meaningful with
conservative forces.
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 5
5.5 Conservation of Energy
Conservation of Total Mechanical Energy
Having learned about kinetic energy (K) and potential energy (U),
we can finally sum them together to produce total mechanical
energy (E):
𝐸 =𝐾+𝑈
For a conservative system, the total mechanical energy is
constant. If a system evolves in some way, we can relate the
initial and final kinetic and potential energies as
𝐾𝑓 + 𝑈𝑓 = 𝐾𝑖 + 𝑈𝑖
And, since 𝐾 =
1
𝑚𝑣 2 ,
2
1
1
2
𝑚𝑣𝑓 + 𝑈𝑓 = 𝑚𝑣𝑖2 + 𝑈𝑖
2
2
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 6
5.5 Conservation of Energy
Conservation of Total Mechanical Energy cont’
Hence, we see that while kinetic and potential energies in a
conservative system may change, their sum is always constant.
For example, an object projected upward from rest will lose KE
and gain PE on the way to the top of its trajectory, then lose PE
and gain KE on the way back down. At any point in the
trajectory, we can use the object’s height to find its PE, then use
conservation of total mechanical energy to find its KE, from
which we can find its speed. The results will match those found
in the kinematic equations of chapter 2. An example is provided
on p. 161 of the text.
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 7
5.5 Conservation of Energy
Conservation of Total Mechanical Energy cont’
Another way of expressing the conservation of total mechanical
energy is shown here. From 2 slides ago,
𝐾𝑓 + 𝑈𝑓 = 𝐾𝑖 + 𝑈𝑖
Rearranging, we get 𝐾𝑓 − 𝐾𝑖 + 𝑈𝑓 − 𝑈𝑖 = 0, or
∆𝐾 + ∆𝑈 = 0
This equation requires a bit of an explanation, which we’ll cover
in class.
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 8
5.5 Conservation of Energy
This figure
from the text
(p. 163)
illustrates the
conservation
of E for the
case of a
mass
dropping
onto a
spring.
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 9
5.5 Conservation of Energy
The short video shown here illustrates concepts of kinetic and
potential energy as they relate to half-pipe snowboarding.
Videos are not
embedded into the PPT
file. You need an
internet connection to
view them.
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 10
5.5 Conservation of Energy
Total Energy and Nonconservative Forces
The presence of nonconservative forces such as friction does not
mean that we can’t solve problems purely using energy
considerations. We just need to know how to deal with the work
done by the nonconservative force.
Starting with the work-KE theorem, 𝑊 = ∆𝐾 = 𝐾𝑓 − 𝐾𝑖 . Then,
we subdivide the work into a portion done by conservative
forces (𝑊𝑐 ) and a portion done by nonconservative forces (𝑊𝑛𝑐 ).
But we know that the work done by conservative forces is 𝑊𝑐
= −∆𝑈 = − 𝑈𝑓 − 𝑈𝑖 . Therefore,
𝑊𝑛𝑐 = 𝐾𝑓 − 𝐾𝑖 − (𝑈𝑓 −𝑈𝑖 ) = (𝐾𝑓 + 𝑈𝑓 ) − (𝐾𝑖 + 𝑈𝑖 )
Then, 𝑊 = 𝐸𝑓 − 𝐸𝑖 = ∆𝐸. The work done by nonconservative
forces is equal to the change in mechanical energy.
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 11
Problem #1: Work Done by a Nonconservative Force
WBL LP 5.15
If a nonconservative force acts on an object, and does work, then…
A
…the object’s kinetic energy is conserved
B
…the object’s potential energy is conserved
C
…the mechanical energy is conserved
D
…the mechanical energy is not conserved
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 12
Problem #2: Two Identical Stones
WBL LP 5.19
Two identical stones are thrown from the top of a tall building.
Stone 1 is thrown vertically downward with an initial speed v, and
stone 2 is thrown vertically upward with the same initial speed.
Neglecting air resistance, which stone hits the ground with a
greater speed?
A
Stone 1
B
Stone 2
C
Both will have the same speed
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 13
Problem #3: How Far Does it Go?
WBL Ex 5.49
A 1.00-kg block (M) is on a flat frictionless surface. It is
attached to a spring initially at its relaxed length (k =
50.0 N/m). A string is attached to the block, and runs
over a pulley to a 450-g dangling mass (m). If the
dangling mass is released from rest, how far does it fall
before stopping?
Solution: In class
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 14
Problem #4: Bouncing Ball
WBL Ex 5.51
A 0.20-kg rubber ball is dropped from a height of 1.0 m above the floor and it
bounces back to a height of 0.70 m.
a) What is the ball’s speed just before hitting the ground?
b) What is the speed of the ball just as it leaves the ground?
c) How must energy was lost and where did it go?
Solution: In class
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 15
5.6 Power
Ever since we started chapter 4, we’ve neglected one very
important kinematic parameter – time. Quite often, we are
interested not only in what quantity of work is done, but in how
long it took to do the work. The time rate of doing work is called
power.
The average power 𝑷 is the total work done divided by the time
required to do the work:
𝑊
𝑃=
𝑡
From its definition, we see that power has SI units of Joules per
second. 1 J/s is defined as 1 Watt (W)
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 16
5.6 Power
If the work is done by a constant force of magnitude F that acts
while an object moves through a displacement of d, then
𝑊 𝐹𝑑
𝑑
𝑃=
=
=𝐹
= 𝐹𝑣
𝑡
𝑡
𝑡
(that is, it’s the product of force and the magnitude of average
velocity). If the force and displacement are at an angle 𝜃 to each
other, then
𝐹 (cos 𝜃) 𝑑
𝑑
𝑃=
= 𝐹 (cos 𝜃)
= 𝐹 𝑣 cos 𝜃 = F ∙ v
𝑡
𝑡
Since power expresses the time rate of work, a process that is
“twice as powerful” as another can do twice the work in a given
time, or the same amount of work in half the time.
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 17
5.6 Power
Efficiency
In many branches of science, we are concerned with how well
energy is converted from one form to another. The mechanical
efficiency (ε) denotes the useful work output compared with the
energy input:
work output
𝜀=
× 100 %
energy input
It can also be expressed in terms of power:
𝑃𝑜𝑢𝑡
𝜀=
× 100 %
𝑃𝑖𝑛
Since this isn’t an engineering course, we won’t spend too much
time thinking about efficiency. However, it may pop up from
time to time in assignments and exams.
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 18
Problem #5: Race Car
WBL Ex 5.67
A race car is driven at a constant velocity of 200 km/h on a straight, level track.
The power delivered to the wheels is 150 kW. What is the total resistive force on
the car?
Solution: In class
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 19
Problem #6: Construction Hoist
WBL Ex 5.73
A construction hoist exerts an upward force of 500 N on an object with a mass of
50 kg. If the hoist started from rest, determine the power it expended to lift the
object vertically for 10 s under these circumstances.
Solution: In class
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 20
Problem #7: Vertical Spring
WBL Ex 5.77
You hang an ideal spring of force constant k is vertically from the ceiling, and
attach an object of mass m to the loose end. You then lower the mass slowly to
its equilibrium position using your hand.
a) Show that the spring’s change in length is given by 𝒅 =
b) Show that the work done by the spring is 𝑾𝒔𝒑 =
c) Show that the work done by gravity is 𝑾𝒈 =
works do not sum to zero
𝒎𝒈
𝒌
𝒎𝟐 𝒈𝟐
− 𝟐𝒌
𝒎𝟐 𝒈𝟐
,
𝒌
and explain why these two
𝒎𝟐 𝒈𝟐
− 𝟐𝒌 ,
d) Show that the work done by your hand is 𝑾𝒉𝒂𝒏𝒅 =
hand exerted an average force of half the object’s weight.
and that the
Solution: In class
PC141 Intersession 2013
Day 11 – June 3 – WBL 5.5-5.6
Slide 21