Download Chapter 8C -- Conservation of Energy

Chapter 8C - Conservation of Energy
AA PowerPoint
PowerPoint Presentation
Presentation by
by
Paul
Paul E.
E. Tippens,
Tippens, Professor
Professor of
of Physics
Physics
Southern
Southern Polytechnic
Polytechnic State
State University
University
©
2007
A waterfall in
Yellowstone Park
provides an
example of
energy in
nature. The
potential energy
of the water at
the top is
converted into
kinetic energy at
the bottom.
Objectives: After completing this
module, you should be able to:
• Define and give examples of conservative
and nonconservative forces.
• Define and apply the concept of
conservation of mechanical energy for
conservative forces.
• Define and apply the concept of
conservation of mechanical energy
accounting for friction losses.
Potential Energy
Potential
Potential Energy
Energy isis the
the ability
ability to
to do
do work
work
by
by virtue
virtue of
of position
position or
or condition.
condition.
h
m
Example: A mass held a
distance h above the earth.
mg
If released, the earth can
do work on the mass:
Earth
Work = mgh
Is this Positive!
work + or - ?
Gravitational Potential Energy
Gravitational
Gravitational Potential
Potential Energy
Energy UU isis equal
equal to
to
the
the work
work that
that can
can be
be done
done BY
BY gravity
gravity due
due to
to
height
height above
above aa specified
specified point.
point.
UU =
= mgh
mgh
Gravitational P. E.
Example: What is the potential energy when
a 10 kg block is held 20 m above the street?
U = mgh = (10 kg)(9.8 m/s2)(20 m)
UU == 1960
1960 JJ
The Origin of Potential Energy
Potential
Potential energy
energy isis aa property
property of
of the
the Earth
Earthbody
body system.
system. Neither
Neither has
has potential
potential energy
energy
without
without the
the other.
other.
Work
done
by
F
lifting force F
h
provides
positive
mg
potential energy,
mgh, for earthbody system.
Only external forces can add or remove energy.
energy
Conservative Forces
AA conservative
conservative force
force isis one
one that
that does
does
zero
zero work
work during
during aa round
round trip.
trip.
Weight is conservative.
F
Work done by earth
h
on the way up is
mg
negative, - mgh
Work on return is
positive, +mgh
Net
Net Work
Work =
= -- mgh
mgh +
+ mgh
mgh =
= 00
The Spring Force
The force exerted by a spring
is also conservative.
F
x
m
When stretched, the spring
does negative work, - ½kx2.
On release, the spring does
positive work, + ½kx2
Net
Net work
work == 00 (conservative)
(conservative)
x
m
F
Independence of Path
A
Work
Work done
done by
by conservative
conservative forces
forces isis
independent
independent of
of the
the path.
path.
C
C
Force
due to
gravity
mg
B
B
A
Because
vertical
of
Work
(A only
C) =the
Work
(A Bcomponent
C)
Why?
the weight does work against gravity.
Nonconservative Forces
Work
Work done
done by
by nonconservative
nonconservative forces
forces
cannot
cannot be
be restored.
restored. Energy
Energy isis lost
lost and
and
cannot
-dependent!
cannot be
be regained.
regained. ItIt isis path
path-dependent!
B
A
f
A
B
m
f
Friction forces are nonconservative forces.
Work of Conservative Forces
is Independent of Path:
For gravitational force:
B
C
(Work)AB= -(Work)BCA
Zero net work
For friction force:
A
(Work)AB  -(Work)BCA
The
The work
work done
done against
against friction
friction isis greater
greater
for
for the
the longer
longer path
path (BCD).
(BCD).
Stored Potential Energy
Work done by a conservative force is stored in
the system as potential energy.
m
x
xo
The
The potential
potential energy
energy isis
equal
equal to
to the
the work
work done
done in
in
compressing
compressing the
the spring:
spring:
F(x) = kx to compress
Potential energy of
compressed spring:
Displacement is x
U  Work  kx
1
2
2
Conservation of Energy
(Conservative forces)
In
In the
the absence
absence of
of friction,
friction, the
the sum
sum of
of the
the
potential
potential and
and kinetic
kinetic energies
energies isis aa constant,
constant,
provided
provided no
no energy
energy isis added
added to
to system.
system.
h
y
v=0
v
mg
At top: Uo = mgh; Ko = 0
At y: Uo = mgy; Ko = ½mv2
At y=0: Uo = 0; Ko = ½mvf 2
0
vf
EE == UU ++ KK == Constant
Constant
Constant Total Energy
for a Falling Body
TOP: E = U + K = mgh
h
K=0
y
At any y: E = mgh + ½mv2
v
Bottom: E = ½mv2
mgh = mgy + ½mv2 = ½mvf2
Total E is same at any point.
(Neglecting Air Friction)
0
U=0
vf
Example 1: A 2-kg ball is released from
a height of 20 m. What is its velocity
when its height has decreased to 5 m?
20m
Total
= Total E at 5 m
Total EEtop
top = Total E at 5 m
v=0
mgh = mgy + ½mv2
2gh = 2gy +
v2
5m
v2
= 2g(h - y) = 2(9.8)(20 - 5)
v=
(2)(9.8)(15)
0
v
vv == 17.1
17.1 m/s
m/s
Example 2: A roller coaster boasts a
maximum height of 100 ft. What is the
speed when it reaches its lowest point?
Assume zero friction:
At top: U + K = mgh + 0
Bottom: U + K = 0 + ½mv2
Total energy is conserved
mgh = ½mv2
v=
(2)(32 ft/s2)(100 ft)
v = 2gh
vv =
= 80
80 ft/s
ft/s
Conservation of Energy
in Absence of Friction Forces
The
The total
total energy
energy isis constant
constant for
for aa conservative
conservative
system,
system, such
such as
as with
with gravity
gravity or
or aa spring.
spring.
Begin: (U + K)o = End: (U + K)f
Height?
mgho
Spring?
½kxo2
Velocity? ½mvo2
=
mghf
Height?
½kxf2
Spring?
½mvf2
Velocity?
Example 3. Water at the bottom of a falls has
a velocity of 30 m/s after falling 35 ft.
What is the water speed
at the top of the falls?
ho = 35 m; vf = 30 m/s2
First look at beginning point—top of falls.
Assume y = 0 at bottom for reference point.
Height? Yes (35 m)
mgho
Spring? No
½kxo2
Velocity? Yes (vo)
½mvo2
Example 3 (Cont.) Water at the bottom of falls
has a velocity of 30 m/s after falling 35 ft.
What is the water speed
at the top of the falls?
ho = 35 m; vf = 30 m/s2
Next choose END point at bottom of falls:
Height? No (0 m)
mghf
Spring? No
½kxf2
Velocity? Yes (vf)
½mvf2
Example 3 (Cont.) Water at the bottom of falls
has a velocity of 30 m/s after falling 35 ft.
ho = 35 m; vf = 30 m/s2
What is the water speed
at the top of the falls?
Total energy at top = Total energy at bottom
mgh  mv  0  mv
1
2
2
0
1
2
2
f
2 gh  v  v
2
0
2
f
v02  v 2f  2 gh  (25.8 m/s) 2  2(9.8 m/s 2 )(33.2 m)
v0  14.9 m /s
2
2
vvoo== 3.86
3.86 m/s
m/s
Example 4. A bicycle with initial velocity 10
m/s coasts to a net height of 4 m. What is
the velocity at the top, neglecting friction?
E(Top) = E(Bottom)
vf = ?
4m
1
2
Etop = mgh + ½mv2
vo = 10 m/s
EBot = 0 + ½mvo2
mv  mgh  mv
2
f
1
2
2
0
1
2
v  v  gh
2
f
1
2
2
0
v  v  2 gh  (10 m/s)  2(9.8 m/s )(4 m)
2
f
2
0
v f  21.6 m 2 /s 2
2
2
vvff == 4.65
4.65 m/s
m/s
Example 5: How far up the 30o-incline
will the 2-kg block move after release?
The spring constant is 2000 N/m and it
is compressed by 8 cm.
mgho
½kxo2
½mvo2
=
mghf
End
Begin
½kxf2
s
30o
h
½mvf2
Conservation of Energy:
kx02
(2000 N/m)(0.08m) 2
h

2
2mg
2(2 kg)(9.8 m/s )
½kxo2 = mghf
hh == == 0.327
0.327 m
m
Example (Cont.): How far up the 30oincline will the 2-kg block move after
release? The spring constant is 2000
N/m and it is compressed by 8 cm.
Continued:
h = 0.327 m = 32.7 cm
h
sin 30o =
s
h
32.7 cm
s=
=
sin 30o Sin 30o
End
Begin
s
30o
ss == 65.3
65.3 cm
cm
h
Energy Conservation and
Nonconservative Forces.
f
Work
Work against
against friction
friction
forces
forces must
must be
be accounted
accounted
for.
for. Energy
Energy isis still
still
conserved,
conserved, but
but not
not
reversible.
reversible.
Conservation of Mechanical Energy
(U
= (U
(U +
+ K)
K)ff +
+ Losses
Losses
(U +
+ K)
K)oo =
Problem Solving Strategies
1. Read the problem; draw and label a sketch.
2. Determine the reference points for gravitational and/or spring potential energies.
3. Select a beginning point and an ending
point and ask three questions at each point:
a. Do I have height?
UU == mgh
mgh
b. Do I have velocity?
22
KK == ½mv
½mv
c. Do I have a spring?
22
UU == ½kx
½kx
Problem Solving (Continued)
4. Apply the rule for Conservation of Energy.
mgho
½kxo2
½mvo2
=
mghf
½kxf2
½mvf2
+
Work
against
friction:
fk x
5. Remember to use the absolute (+) value
of the work of friction. (Loss of energy)
Example 6: A mass m is connected to a cord
of length L and held horizontally as shown.
What will be the velocity at point B? (d = 12 m,
L = 20 m)
A
1. Draw & label.
L v
B
c
2. Begin A and end B.
d
r
3. Reference U = 0.
0
(U + K)o =(U + K)f + loss
mgL + 0 = mg(2r) + ½mvc2
2gL - 4gr = vc2
U=0
(Multiply by 2, simplify)
Next find r from figure.
Example (Cont.): A mass m is connected to a
cord of length L and held horizontally as
shown. What will be the velocity at point B?
(d = 12 m, L = 20 m)
A
2gL - 4gr = vc2
L
vc
r=L-d
r = 20 m - 12 m = 8 m
vc2 =2gL - 4gr = 2g(L - 2r)
B
r
d
U=0
vc2 = 2(9.8 m/s2)[20 m - (2)(8 m)]
vc =
2(9.8 m/s2)(4 m)
vvcc == 8.85
8.85 m/s
m/s
Example 7: A 2-kg mass m located 10 m above
the ground compresses a spring 6 cm. The
spring constant is 40,000 N/m and k = 0.4.
What is the speed when it reaches the bottom?
2 kg
h
s
30o
n
f
Begin
mg Cos
End
mg Sin 30o
30o
30o
mg
Conservation: mgh + ½kx2 = ½mv2 + fkx
(Work)f = (kn) x = (mg Cos 30o) x
Continued . . .
Example (Cont.): A 2-kg mass m located 10 m
above the ground compresses a spring 6 cm.
The spring constant is 40,000 N/m and k = 0.4.
What is the speed when it reaches the bottom?
mgh + ½kx2 = ½mv2 + fkx
2 kg
h
10 m
x
30o
x=
10 m
= 20 m
Sin 30o
fkx = (mg Cos 30o) x
fkx = (0.4)(2 kg)(9.8 m/s2)(0.866)(20 m) = 136 J
mgh = (2 kg)(9.8 m/s2)(10 m) = 196 J
½kx2 = ½(40,000 N/m)(0.06 m)2 = 72.0 J
Example (Cont.): A 2-kg mass m located 10 m
above the ground compresses a spring 6 cm.
The spring constant is 40,000 N/m and k = 0.4.
What is the speed when it reaches the bottom?
mgh + ½kx2 = ½mv2 + fkx
2 kg
h
10 m
x
30o
mgh = 196 J
½kx2 = 72.0 J
fkx = 136 J
½mv2 = mgh + ½kx2 - fkx
½(2 kg) v2 = 196 J + 72 J - 136 J = 132 J
vv =11.4
=11.4 m/s
m/s
Summary:
Energy Gains or Losses:
Gravitational
Gravitational Potential
Potential Energy
Energy
UU == mgh
mgh
Spring
Spring Potential
Potential Energy
Energy
U  kx
Kinetic
Kinetic Energy
Energy
K  mv
Work
Work Against
Against Friction
Friction
1
2
1
2
2
2
Work
Work =
= fx
fx
Summary:
Conservation of Energy
The basic rule for conservation of energy:
mgho
½kxo2
½mvo2
=
mghf
½kxf2
½mvf2
+
Work
against
friction:
fk x
Remember to use the absolute (+) value of
the work of friction. (Loss of energy)
CONCLUSION: Chapter 8C
Conservation of Energy