Chapter 8
Conservation of
Energy
Definitions of conservative
Work done by a conservative force depends only
on the initial and final positions → path independent
 F  dl

a 1b


F  dl 
 F  dl   F  dl  0
a1b
a 2b
a 2b
 F  dl   F  dl  0
a1b
 F  dl  0
b 2a
b
1 2
Loop integral
a
Definition2: Work done by the force on an object
moving around any closed path is 0.
2
y
Conservative forces
a
Gravity G  m g
W 

hb
ha
 mgdy  mgha  mghb
Elastic force F   kx
W 
xb
xa
ha
1
1
2
 kxdx  kx a  kxb 2
2
2
Mm
Gravitational force F  G 2
r
rb
W   Fdl cos    r
b
a
1 1
  GMm (  )
ra rb
a
Mm
G 2 dr
r
 L
mg
C
hb
o
b
x
b
dr
dl

rb
r
M
m
ra
dlcos( - )  dr
a
3
Potential energy
Work done by a conservative force:
W 

b
a
F  dl  f  ra   f  rb   U a  U b
Potential energy: associated with the position or
configuration of an object (or objects)
U  mgh
1 2
U  kx
2
Mm
U  G
r
Work done by a conservative force is the negative
of the change in corresponding potential energy.
4
Moving with spring
Example1: Mass m is released from rest when the
spring (k) is unstretched, what is the lowest height?
Solution: Two conservative forces:
W G    U G  m gh
1 2
W k   U k   kh
2
Work-energy principle:
WG  Wk  Ek  0
2mg
h
k
m
h
m
5
Properties of potential energy
The capacity or “potential” to do work
Energy stored in the form of potential energy
1. Always associated with a conservative force
2. The choice of where U=0 is arbitrary, usually
chosen wherever it’s most convenient
Ua 

Z
a
F  dl , (U Z  0)
3. Potential energy is not somebody has by itself,
it is associated with the interaction system
6
Potential energy and force
One-dimensional case:
U  x 
 F  x  dx    F  x  dx
Z
x
x
Z
So the conservative force is
dU
F  x  
dx
In 3-dimension case:
U
U
U
F  x, y, z   
i
j
k
x
y
z
Partial derivative:
U U U
,
,
x y z
7
Find out the force
Example2: U(x,y)=2xy2+x, what is the force?
Solution: How to take partial derivative?
U
2
Fx  
  2 y  1 (treat y as constant)
x
U
(treat x as constant)
Fy  
  4 xy
y
F  Fx i  F y j    2 y 2  1  i  4 xyj
8
System and work
Consider a system of particles
o
r21
Internal forces & external forces
r1 m
1
f
Work done by a pair of forces
dW  f12  dr1  f 21  dr2
 f 21  ( dr2  dr1 )
 fdr
r: distance between m1 and m2 Fn
r21
21
f1i
m1
fi1
f1n
fn1
mn

Fi
F1
  f 21  dr1  f 21  dr2
 f 21  dr21
m2
r2
mi
fni
fin
9
Work done by a pair of forces
dW  fdr
Work done by a pair of forces depends on the
change of distance between two particles.
1. It is independent of reference frames.
2. The total work is 0 if the distance between two
particles remains constant.
L
10
Mechanical energy
Consider a conservative system
kinetic energy
potential energy
 E k  W net    U total
  E k   U total   E k 2  E k 1    U 2  U 1   0
  E k 2  U 2    E k1  U 1   0
Define total mechanical energy: E  E k  U
E 2  E 1  C onstant
Total mechanical energy is a conserved quantity
11
Conservation of mechanical energy
If only conservative forces are doing work,
the total mechanical energy of a system stays
constant.
Principle of conservation of mechanical energy
1. Only nonconservative forces can change the
total mechanical energy of system.
2. Don’t count twice about the work done by
conservative forces.
3. It can only be applied in inertial frames.
12
Moving along a ring
Example3: A spring (k=50N/m, L=0.10m) is fixed to
a horizontal frictionless ring (R=L=0.10m) at A and
connected with a small object m=1.0kg, if m starts
from rest at B, determine its speed v at C.
Solution: conservation of mechanical energy
1 2
A
mv   U  0
2
1
U  k
2

2R  L

2
1
2
 k 2R  L 
2
  0.207 J  v  0 .6 4 m / s
C
m
R
What about a vertical ring?
B
13
Homework
A spring (k=500N/m, original length L=0.10m) is
fixed to a vertical frictionless ring (R=L=0.10m) at A
and connected with a small object m=1.0kg, if m
starts from rest at B, determine (a) its speed v at C,
(b) the minimum length of spring.
A
m
R
C
B
14
Skating on a sphere
Example4: A skier of mass m slides from rest at
the top of a frictionless solid sphere of radius R.
a) At what angle  will he leave the sphere?
b) If friction were present, how would  change?
Solution: a) m g cos   N  m v 2 / R
N
Leave the sphere: N = 0
Conservation principle:
1 2
mgR (1  cos  )  mv
2
 co s   2 / 3
R 

v
mg
b) with friction, v↘,  ↗
15
The law of conservation of energy
Consider dissipative forces such as friction
mechanical energy ↘
thermal energy ↗
Taking into account all forms of energy, we have
the law of conservation of energy:
The total energy is neither increased nor decreased in any process.
Energy can be transformed from one form to
another, and transferred from one body to another,
but the total amount remains constant.
16
Energy with dissipative forces
Consider nonconservative (dissipative) forces
 E k  W net  W C  W N C    U  W N C
 E   E k   U  W NC
Example5: A mass m with v0 compresses a massless
unstretched spring (k) maximally L, determine µ.
Solution:
 E  W NC
1 2 1 2
kL  mv 0    mgL
2
2
v02
kL


2 gL 2 mg
v0
m
17
Falling chain
Example6: A chain with length h hanging out of the
desk falls from rest, known m, L, µ. Determine:
a) Work done by friction during the separation
b) Separating speed of the chain
Solution: a) Friction is varying!
Lhx
 dx
W    mg
L
0


Lh
 mg
2L
o
If the chain already moved x
and is going to move dx
L-h
x
h
(L  h ) 2
18
 mg
2
W 
(L  h )
2L
b) Separating speed of the chain
1 2
W   E k   U  mv   U
2
Potential energy of a continuous object:
equivalent to a particle in its center of mass
g 2 2
2
v
[( L  h )   ( L  h) ]
L
L-h
o
L
h
h
 U   mg  (  mg )
2
L
2
x
h
19
Gravitational potential energy
The gravitational force is
mM
F  G 2 rˆ
r
where r is the distance between two objects
“-” means it is an attraction force
The gravitational force is conservative, the
corresponding potential energy is
mM
U  G
r
Usually we choose U = 0 at r = ∞
20
Fly me to the Moon
Example7: A spaceship (m=10000kg) starts from the Earth
and lands on the Moon. Determine the energy required if
no air friction. ( Re=6.4106m, me=6.01024kg,
Rm=1.7106m, mm=7.41022kg, L= 3.8108m, G=6.710-11 )
Rm
Solution: energy required: change in potential energy
U 0  G
U  G
me m
m m
G m
Re
Rm  L
  6.2  10 11 J
me m
m m
G m
Re  L
Rm
 3.8 1010 J
L
  E  U  U 0  5.8  10 11 J
Re
Is this enough in reality?
21
Launching a satellite
Example8: A satellite is launched from the ground
into a circular orbit of radius r, determine the initial
speed v0. (known Re, me, no air friction)
Solution: Uniform circular motion in orbit
Gme
me m
v2
(orbit speed)
v
G 2 m
r
r
r
Conservation of mechanical energy
me m 1 2
1
me m
2
mv0  G
 mv  G
r
2
Re 2
Re
Gme
Re
 v0  2
(1  )  2 gRe (1  )
2r
Re
2r
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Cosmic velocity
Gme
Re
Re
v0  2
(1  )  2 gRe (1  )
Re
2r
2r
a) Let r = Re, v1  gR e  7.9km /s
This is called the first cosmic velocity
Satellite moves in a circle nearby the ground
b) Let r = ∞, v 2  2 gR e  11.2km /s
This is called the second cosmic velocity or
the escape velocity
Satellite can escape from the Earth
23
Escape velocity
v 2  2 Gm e / R e
With escape velocity, mechanical energy E = 0
me m
1 2
E  mv2  G
0
2
Re
E < 0, bound state
E > 0, unbound state
Escape velocity v≧c
black hole
Schwarzschild radius
2GM
R 2
c
24
Challenging question
To escape the solar system, a spaceship must
overcome the gravitational force of both the Earth
and the Sun. Determine the 3rd cosmic velocity for
this escape. (Ignore other bodies in the solar system)
v  v  (vS  v0 )
2
E
 16.7km / s
2
Power
The rate at which work is done:
dW F  dl

 F v
P
dt
dt
In SI units, unit of power is Watt (W)
Example9: A 70-kg man runs up stairs in 4.0s. The
vertical height of stairs is 4.5m. a) How much
energy is required? b) The average power.
Solution: a) E   U  m g h  3.1  10 3 J
b) P  E / t  m gh / t  770 W
Human power in real life
26
Potential energy diagrams
Potential energy in a diatomic molecule can be
written:
U  r    ar  6  br  12
where r is the distance between two atoms, a and b
are positive constants. Shown in a diagram:
a) F(r) between two atoms
dU
6 a 12b
  7  13
dr
r
r
1/6
b) Where F=0? r0   2b 
 a 
F r   
Equilibrium point
U r 
F is negative of the
slope of U-r curve
r0
r
27
Potential energy diagrams
c) if r>r0, F<0, attraction force
if r<r0, F>0, repulsion force
d) r0 relates to a minimum U(r)
U r 
F is negative of the
slope of U-r curve
r0
r
stable equilibrium
returns toward E-point when displaced slightly
e) Point of maximum U(r)
U(r)
unstable equilibrium
moves farther away when
displaced slightly
r
28