chapter 6 conservation of energy the law of conservation ... · the law of conservation of energy...
TRANSCRIPT
6-1
Chapter 6 Conservation of Energy
Today we begin with a very useful concept – Energy.
We will encounter many familiar terms that now have very specific definitions in physics.
Conservation of energy
Work
Potential
Power
In some cases, it can be argued that these terms have a physics definition that is similar to its
everyday usage.
The Law of Conservation of Energy
The total energy in the universe is unchanged by any physical process:
total energy before = total energy after
From page 187: “In ordinary language, conserving energy means trying not to waste useful
energy resources. In the scientific meaning of conservation, energy is always conserved no
matter what happens.”
Conservation of energy is one of the few universal principles of physics. No exception has ever
been found. It applies to physical, chemical, and biological systems.
Also from page 187: “Some problems can be solved using either energy conservation or
Newton’s second law. Usually the energy method is easier.” Using Newton’s second law
involves vector methods since forces are vector quantities. Much of the time, energy involves
scalar quantities, which are much easier to deal with (and more familiar). “When deciding which
of these two approaches to use to solve a problem, try using energy conservation first.”
Kangaroos are mentioned at the beginning of this chapter.
http://www.youtube.com/watch?v=hijYSR2MFiY
Forms of Energy
At the most fundamental level there are three kinds of energy
1. Kinetic energy – energy due to motion
2. Potential energy – energy due to
interaction
a. Gravitational potential energy –
interaction between the Earth
and a mass
b. Elastic potential energy –
interaction between a spring and
a mass
3. Rest energy – internal energy to a body
Energy is measured in Joules (J).
6-2
Work
Suppose a force F
causes an object to move a distance x parallel to F
The work done by a constant force F
is defined as
xFW
DO NOT MEMORIZE!
Suppose a constant force F
causes an object to move along r
not parallel to F
x
6-3
cosrFW
where is the angle between F
and r
. MEMORIZE.
Work is a scalar quantity and can be positive, negative, or zero.
Positive: between 0o and 90o
Negative: between 90o and 180o
Zero: = 90o
o Usually tension and normal force do no work. The exception is (c) below.
The work done by several forces can be found from the net force
cos
21
rF
WWWW
net
Ntotal
Work and Kinetic Energy
Choosing the x axis along Fnet, (using x = r cos )
xma
xFW
x
nettotal
We had an equation from Chapter 2
6-4
)(
2
22
21
22
ixfxx
xixfx
vvxa
xavv
Substituting into Wtotal
)(22
21
ixfxtotal vvmW
Since the net force is in the x-direction, ay and az are both zero. Only the x-component of the
velocity changes
2222222222
)()( ixfxiziyixfzfyfxif vvvvvvvvvv
and
)(22
21
iftotal vvmW
The translational kinetic energy is defined as
2
21 mvK
The work-kinetic energy theorem is
KWtotal
While this expression is foundational to this chapter, do not memorize. We shall derive a more
useful form.
Gravitational Potential Energy (1)
The weight can do work. Toss a ball up and it slows down. In our new language, its kinetic
energy decreases. The kinetic energy is converted into another form of energy we call
gravitational potential energy.
The change in gravitational potential energy
gravgrav WU
In terms of position
ymgUgrav
This equation is true even if the object does not travel in a straight line.
6-5
The gravitational potential energy is
mgyUgrav
The final form of our relation is
UKWnc
This is it. You need to know it. We have another entry into our cause and effect table.
Another useful form is
ffncii
ifif
nc
UKWUK
UUKK
UKW
Wnc is the work done by nonconservative forces. Nonconservative forces do not have a potential
energy. A good example is friction.
The mechanical energy is
UKE
Conservation of Mechanical Energy
When nonconservative forces do no work, mechanical energy is conserved:
fi EE
6-6
The zero of potential energy is arbitrary. Choose whatever is convenient, usually the ground.
The work done by a conservative force is independent of the path taken.
Problem: A 0.1-kg ball is thrown at 5 m/s from a 10 m tower. What is its speed when it is 5 m
above the ground? What is its speed when it hits the ground?
Solution: Use the conservation of mechanical energy.
m/s1.11
)m5m10(m/s8.9(2)m/s5(
)(2
22
21
2
12
2
221
2
2
121
1
2
221
2
2
121
1
2211
21
yygvv
vgyvgy
mvmgymvmgy
KUKU
EE
At the bottom, y2 = 0, v2 = 14.9 m/s.
Notice the direction of the throw is not mentioned. No matter which way the ball is thrown, it
has the same speed at the same height! This is very hard to prove using Newton’s second law.
Gravitational Potential Energy (2)
For objects far from the Earth,
r
GmMU E
While this looks very different from mgy, the text (p. 203) shows they are equivalent.
Example 6.8 What is the escape velocity for the Earth?
Solution: Use conservation of energy. We cannot use U = mgy for an object far from the earth!
2
212
21
f
f
E
i
i
E
ffii
fi
mvr
Gmmmv
r
Gmm
KUKU
EE
When an object escapes from the Earth, Earth’s gravity is not acting on it and rf → ∞. If the
object barely escapes the Earth, vf = 0.
6-7
i
E
i
i
i
E
r
Gmv
mvr
Gmm
2
02
21
The starting position is on the Earth’s surface and ri = RE = 6.36×106 m. The mass of the Earth is
mE = 5.97×1024 kg. The escape velocity is
m/s200,11)kg1037.6(
)kg1097.5)(kg/Nm1067.6(226
242211
i
E
ir
Gmv
This is
s
mi0.7
m1609
mi1
s
m200,11
iv
To summarize, we developed what is usually called the work-energy theorem:
UKWnc or ffncii UKWUK
where K is the kinetic energy
2
21 mvK ,
U is the potential energy (usually gravitational)
mgyUgrav
and Wnc is the total work performed by all nonconservative forces. In many situations (but not
all), mechanical energy is conserved and Wnc = 0.
Work done by a variable force.
The advantage of energy methods is seen when dealing with a variable force. Suppose the force
changes with displacement. An example of such a device is shown on the next page. How do
we calculate the work done?
We divide the overall displacement into a series of small displacements, x. Over each
of the smaller displacements, the force is almost constant. The work done for a small
displacement is
xFW x
6-8
The total work is the area under the curve shown above. This procedure was used to find the
gravitational potential energy for an object far from the Earth
r
mGMU E
Hooke’s Law and Ideal Springs
The force exerted by the archer
increases as the bowstring is drawn
back. Robert Hooke proposed an ideal
spring where the force is proportional
to the displacement
kxFx
The displacement of the spring from the
relaxed position is x. The constant k, is
called the spring constant. It is
measured in N/m and it gives the
strength of the spring. The larger k is,
the stiffer the spring. The minus sign
indicates that if the spring is stretched
to the right, the spring pulls back to the
left and vice versa.
6-9
Elastic Potential Energy
As the spring is pulled (or pushed) from its relaxed position, work is done on it. The work done
is independent of the path taken and accordingly, a potential energy can be defined. The elastic
potential energy is found to be
2
21 kxUelastic
Note that U = 0 when x = 0.
Power
Often the rate of energy conversion is important. We use the term power to refer to the rate of
energy conversion. Over an extended time, the average power is
t
EPav
Power is measured in Joules/second or watts (W). Be careful with W for work and W for watts.
Remember that work changes the mechanical energy of the system.
cos
cos
Fv
t
rF
t
EP
Problem 82. A spring gun (k = 28 N/m) is used to shoot a 56-g ball horizontally. Initially the
spring is compressed by 18 cm. The ball loses contact with the spring and leaves the gun when
the spring is still compressed by 12 cm. What is the speed of the ball when it hits the ground 1.4
m below the gun?
Solution: This appears to be a projectile problem. It is an energy problem with two potential
energies. Take the initial position to be at the top and the final position just before it hits the
ground,
6-10
m/s04.6
)kg056.0/())m12.0()m18.0)((N/m28()m4.1)(m/s8.9(2
/)(2
00
222
2
2
2
112
2
2212
221
1
2
121
2211
mxxkgyv
kxmvmgykx
UKUK
Problem 91. A 1500-kg car coasts in neutral down a 2.0º hill. The car attains a terminal speed
of 20.0 m/s. (a) How much power must the engine deliver to drive the car on a level road at 20.0
m/s? (b) If the maximum useful power that can be delivered by the engine is 40.0 kW, what is
the steepest hill the car can climb at 20.0 m/s?
Solution: At the terminal speed, the x-component of the weight of the car is opposed by air
resistance. When coasting down the hill, the free body diagram is
For the x-component (along the incline)
N513
2sin)m/s8.9)(kg1500(
sin
0sin
2
mgF
Fmg
maF
air
air
xx
(a) Now the car is moving along a horizontal road at constant velocity. From the FBD found
at the top of the next page:
airmotor
airmotor
xx
FF
FF
maF
0
6-11
The power delivered by the engine when the car moves at 20 m/s (notice the angle below is not
2º!) is
W300,100cos)m/s20)(N513(0coscos vFvFP airmotormotor
(b) Again a free body diagram is helpful.
Climbing with constant speed, ax = 0,
sin
0sin
mgFF
FmgF
maF
airmotor
airmotor
xx
The maximum power supplied by the engine is 40.0 kW. The power is
sin
0cos)sin(
cos
mgvvF
vmgF
vFP
air
air
motor
6-12
101.0
)m/s20)(m/s8.9)(kg1500(
)m/s20)(N513(W100.40
sin
2
3
mgv
vFP air
This corresponds to a 5.8º angle.