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Chapter 7
Potential energy and conservation of energy
In this chapter we will explore one of the most
fundamental concepts in physics, energy , and one of
the most important principles, conservation of
energy.
When a conservative force does work, this work is
stored in the form of potential energy (symbol: U)
and can be retrieved. This is not the case with nonconservative forces.
(7-1)
y
Potential Energy U
1
U is defined for conservative
forces only
B
A
2
O
Point A (xA , yA)
Point B (xB , yB)
x
WAB = work done by a conservative force F from A to B
WAB does not depend on the path taken. This can be
expressed mathematically as follows: WAB = W(A,B)
Also it can be written as: WAB = U(A) - U(B)
U is the potential energy. Units: J
(7-2)
.
O
m
.
A
F
(7-3)
.
B
xo
path
x1
x-axis
We assume that F depends on position i.e. F = F(x)
WAB  U ( x0 )  U ( x1 )
x1
Also WAB   F ( x)dx

x0
x1
U ( x1 )  U ( x0 )    F ( x)dx
x0
Point A at xo is called the “reference point” . The choice of
xo and U(xo) does not affect the result of our calculations.
Thus we will choose xo and U(xo) for maximum convenience
Example: Potential energy of the
gravitational force
y
path
h
.A
m
dy
mg
.O
floor
Reference point: Point O at yo = 0
Point A at y1 = h
h
h
h
0
0
0
U ( y1 )  U ( y0 )    Fy dy    (mg )dy  mg  dy
U ( y1 )  U ( y0 )  mg  y 0  mgh
h
We choose: U(y0) = 0
Thus we get :
U  mgh
(7-4)
Summary
If we know F(x) along the x-axis we can determine U at any
point P using the following equation:
.
O
m
.
F(x)
R
xo
.
P
path
x
x-axis
x
U ( x)  U ( xo )    F ( x)dx 
xo
The equation above gives U(x) for every point P on the
x-axis with coordinate x.
F(x)  U(x)
(7-5)
What about the reverse problem? i.e. if we know U(x) can we
get the force F(x)? Our starting point is the definition of U:
x
U ( x)  U ( xo )    F ( x)dx 
xo
g(x)
Theorem from calculus:
x
If f ( x)   g ( x )dx 
a
O
a
x
Thus:
x’
dU
 F
dx

df
 g ( x)
dx
In our case U plays the role of
function f and -F plays the role of
function g

F 
dU
dx
(7-6)
Conservation of mechanical energy
vi
.
vf
m
.
O
A
xi
F
(7-7)
.
B
path
xf
x-axis
WAB  U ( x f )  U ( xi ) (eqs.1)
WAB is also given by the work-energy theorem:
WAB
mv 2f
mvi2


2
2
(eqs.2)
mv 2f
mvi2
Combine eqs.1 and eqs.2  U ( x f )  U ( xi ) 

2
2
2
i
mv 2f
mv
U ( xi ) 
 U (x f ) 
2
2

vi
.
vf
m
.
O
A
.
B
xi
2
i
F
(7-8)
path
xf
x-axis
mv 2f
mv
U ( xi ) 
 U (x f ) 
or
U i  Ki  U f  K f
2
2
We define as total mechanical energy E
the sum of potential energy U and kinetic energy K
The conservation of mechanical energy is expressed
by one of the following equations:
Ei  E f
E  0
y
Example: An object of mass m is shot
straight up with an initial velocity vo .
B
.
m
mg
vo
A
.
h
Determine the maximum height h of the
object
The only force acting on the object is the
gravitational force. The gravitational
force is conservative.
floor U = 0
We will use conservation of mechanical energy : E A  EB
mvo2
mvo2
EA 
, EB  mgh  mgh 
2
2
vo2
h
2g

(7-9)
Potential energy of a spring (spring constant k)
x
U ( x)  U ( xo )    F ( x )dx 
F ( x )  kx 
xo
We choose the reference point x0  0
x
x
U ( x)    kx dx   k  x dx 
0
0
x
 x 
kx 2
U ( x)  k   
2
 2 0
2
2
kx
U ( x) 
2
(7-10)

and
U ( xo )  0
a “before”
b “after”
Example (7-3) page 178
In the Atwood machine shown
in the figure m1 = 1.37 kg,
m2 = 1.51 kg. The system is
released from rest with h2 = 0.84
m. Find the speed v of m2 just
before it hits the floor
m1v 2 m2 v 2
Ea  m1 gh1  m2 gh2 , Eb  m1 g (h1  h2 ) 
+
2
2
m1v 2 m2 v 2
Ea  Eb  m1 g (h1  h2 ) 
+
 m1 gh1  m2 gh2 
2
2
(m1  m2 )
v  2 gh2
 0.89 m / s
(7-11)
(m1  m2 )
Energy diagram
(7-12)
U
.
.
A
xA
B
motion allowed xB
E
An energy diagram is a
plot of the potential
energy U(x) versus x
x
The blue horizontal line
is the total energy
E=U+K
The points A and B for which E = U are called turning points .
Motion is restricted for some parts of the x-axis only.
mv 2
This comes from the fact that K 
 0  E U  0 
2
U  E Motion is allowed for: x A  x  xB
Motion is forbidden
for: x  xA and x  xB
f
.
A (7-13)
.
C
.
Maxima and minima of a
function f(x)
B
O
xA
xB
x
The function f(x) plotted in
the figure has one minimum
at point B and two maxima
at points A and C
xC
Minima and and maxima of a function occur at points for which
df
the first derivative
 0 The sign of the second derivative
dx
can be used to differentiate between minima and maxima as follows:
d2 f
>0 for minima
2
dx
d2 f
<0 for maxima
2
dx
U
Equilibrium
.
B
C
.
A
O
xA
.
xB
x
Consider an object that
moves along the x-axis
under the action of a force
whose potential energy U
is plotted in the figure.
xC
A point on the x-axis is a position of equilibrium if F = 0
dU
dU
F 

 0 at positions of equilibrium
dx
dx
Conclusion: Equilibrium positions coincide with the minima
or maxima of U(x). Points at x A , x B , and x C in this example
are positions of equilibrium.
(7-14)
(7-15)
U
.
C
FC
FB
.
xC
xA
A
FA = 0
Stable Equilibrium
An object is said to be in a position of
stable equilibrium if: when the object
is moved by a small distance from that
point, the force direction is such
as to bring it back
.
B
xB
x
Point A in the figure is an equilibrium position because FA = 0
If we move the object to point B,
dU
dU
 0. Thus F  
 0
dx
dx
FB points towards A
dU
If we move the object to point C,
 0
dx
FC points towards A.
dU
Thus F  
0
dx
Conclusion: point A is a position of stable equilibrium
U
.
FC C
xC
.
A
(7-16)
FA = 0
.
xA
B
FB
xB
x
Unstable Equilibrium
An object is said to be in a position of
unstable equilibrium if: when the
object is moved by a small distance
from that point, the force direction is
such as to take the object further away
Point A in the figure is an equilibrium position because FA = 0.
If we move the object to point B,
dU
dU
 0 F 
0
dx
dx
FB points away from A
If we move the object to point C,
dU
dU
 0  F 
 0
dx
dx
FC points away A.
Conclusion: point A is a position of unstable equilibrium
E = U+K
Summary
1. Motion is allowed for
parts of the x-axis:
UE
2. At turning points
(points A and B)
U = E At the turning
points v = 0 and K = 0
3. At any point on the plot:
2K
K  E  U and v 
m
4. Minima of the energy diagram correspond to positions of
stable equilibrium. Maxima of the energy diagram
correspond to positions of unstable equilibrium
(7-17)
Motion in two dimensions
The potential energy U is given by the integral:
rB
U(rB - rA )    F  d r
rA  reference point
rA
2
m(vx2  v y2 )
mv
Kinetic Energy K 

2
2
Mechanical Energy E  K  U
(7-18)
The reverse problem: If we know U(x,y,z) can we
determine F ?
The recipe:
F  Fx i  Fy j  Fz k
U
Fx  
x
U
Fy  
y
U
Fz  
z
(7-19)
We will encounter problems that fall in the following two
categories:
1. All forces acting on the object(s) under study are
conservative. In this case we use the expression:
Ei  E f
or simply:
E  0
2. Some forces acting on the object(s) under study are
conservative and some are non-conservative. Fnet = Fc + Fnc
Here Fc is the vector sum of all conservative forces, and
Fnc is the vector sum of all non-conservative forces. In this
case we use the equation:
E  Wnc
Where Wnc is the work done by Fnc
(7-20)
Example (7-6) page 188
A ball of mass m = 10 kg is attached
to a wire of length L = 5 m that can
swing freely from a support. The
ball is pulled aside so that the wire
makes an angle 1 = 31  from the
vertical.
After 10 swings the maximum angle that the ball reaches is
2 = 25  Calculate the work Wf done by air resistance
(a non-conservative force) on the ball during these 10 swings
(7-21)
Calculation of the potential
energy U for the ball
O
U = mgh
C
B
h = OA - OB
m
OA = L
h
A
From triangle OBC we have; OB = OCcos
Thus OB = Lcos

OC = L
and h = L(1 - cos)
U = mgL(1 - cos)
(7-22)
1. “Before”
2. “After”
1
E = Wf
2
E = E2 - E1
E1 = mgL(1 - cos1)
E2 = mgL(1 - cos2)  E = mgL(cos1 - cos2)
Wf = mgL(cos1 - cos2) = 109.85 [cos(31) -cos(25)]
Wf = -24 J
(7-23)