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Transcript
```7. Conservation of Energy
1.
2.
3.
4.
Conservative & Non-conservative Forces
Potential Energy
Conservation of Mechanical Energy
Potential Energy Curves
How many different energy conversions
take place as the Yellowstone River
plunges over Yellowstone Falls?
P.E.  K.E., sound, & heat
Work against conservative force (path independent),
e.g., gravity:
Work can be stored  potential energy.
Work against non-conservative force (path dependent),
e.g., friction:
Work is dissipated  heat & entropy.
7.1. Conservative & Non-conservative Forces
F is conservative if

C
F  dr  0

WAB =  WBA = WAB
WAB   F  d r
is path-independent
WBA + WAB = 0
WAB
i.e.,
for every closed path C.
B
A
WAB
BA
F is non-conservative if there is a closed path C such that

Mathematica

C
F  dr  0
B
WAB   F  d r
A
is path-dependent
Example: Work done on climber by gravity
Going up:
W1 = ( m g ) h
=mgh
F   m g ˆj
Going down:
r  h ˆj
r  h ˆj
W2 = ( m g ) ( h)
=mgh
F   m g ˆj
Round trip:
W = W 1 + W2 = 0
Horizontal displacement requires no work.
Gravity is conservative.
Example: Work done on trunk by friction
Ff    m g ˆi
r  L ˆi
Going right:
W1 = (  m g ) L =   m g L
Going left:
W2 = (  m g ) ( L) =   m g L
r   L ˆi
Ff   m g ˆi
Round trip: W = W1 + W2 =  2  m g L  0
Friction is non-conservative.
GOT IT? 7.1.
If it takes the same amount of work to push a trunk across a rough floor as it
does to lift a weight to the same distance straight upward.
How do the amounts of work compare if the trunk & weight are moved along
curved paths between the same starting & end points?
Ans. Work is greater for the trunk.
7.2. Potential Energy
Conservative force:
Potential energy = stored work
=  ( work done by force )
B
U AB  WAB   F  d r
A
Note: only difference of potential energy matters.
U    F  x  d x
x2
1-D case:
Constant F:
x1
U  F  x2  x1 
Gravitational Potential Energy
Horizontal component of
path does not contribute.
mg
Vertical lift:
U  mg y
U    m g  h  mgh
Example 7.1. Riding the Elevator
A 55 kg engineer takes an elevator from her office on the 33rd floor to the 59th floor.
Later, she descends to street level.
If she takes her office as the zero of potential energy,
59
and if the distance between floors is 3.5 m,
what’s her potential energy
33
(a) in her office.
(b) on the 59th floor.
(c) at street level ?
1
(a)
U 33  0
(b)
U59  m g  h59  h33    55 kg   9.8 m / s 2   3.5 m  59  33   49 kJ
(c)
U1  m g  h1  h33    55 kg   9.8 m / s 2   3.5 m 1  33   60 kJ
Application: Pumped Storage
Excess electric energy
stored by pumping
water to higher ground.
(see Prob. 29)
Northfield Mountain Pumped Storage Project, MA, USA
Elastic Potential Energy
Ideal spring:
F  k x  k  x  x0 
x0 = equilibrium position
x
2
1


2
U    F  x  d x     k  x  x0   d x  k
x1
x1
 2 x  x0 x 
x1
1

 k   x22  x12   x0  x2  x1    1 k   x  x  2   x  x  2 
2
0
1
0

2

2 
x2
x2
Let

U  0 at x  x0
1
2
U  x    k  x  x0  d x  k  x  x0 
x0
2
1
2
 k  x 
parabolic
2
x
U is always positive
Setting x0 = 0 :
x x0
x = x0
x x0
U  x 
1
k x2
2
Example 7.2. Springs vs Gasoline
A car’s suspension consists of springs with overall effective k = 120 kN/m.
How much these springs need be compressed to store the same amount
of energy as in 1 gram of gasoline?
From Appendix C: Energy contents of gasoline = 44 MJ / kg.
103 kg  44 103 kJ / kg  
x2 
11 2
m
15
1
120 kN / m  x 2
2
x  0.86 m  86 cm
Springs can’t compete with gasoline as energy source.
Example 7.3. Climbing Rope
Climbing ropes are springy to cushion falls.
Consider rope with F =  k x + b x2 , where k = 223 N/m, b = 4.10 N/m2.
Find the potential energy when it’s stretched 2.62 m, taking U = 0 at x = 0.
U     k x  b x 2  d x
x
0

1
 1

    k x 2  b x3 
3
 2

x

0
1
1
2
3
 223 N / m   2.62 m    4.10 N / m2   2.62 m 
2
3
 741 J
which is  3% less than U = ½ k x2 = 765 J.
1
1
k x 2  b x3
2
3
7.3. Conservation of Mechanical Energy
K  Wnet  Wc  Wnc  U  Wnc
Mechanical energy:
E  K U
E  K  U  Wnc
Law of Conservation of Mechanical Energy:
E  K  U  0
E  K  U  constant
if
Wnc  0
( no non-conservative forces )
Example 7.4. Tranquilizing an Elephant
A biologist uses a spring-loaded gun to shoot tranquilizer darts into an elephant.
The gun’s spring has k = 940 N/m, & is compressed x = 25 cm before firing a 38-g dart.
Assuming the gun points horizontally, at what speed does the dart leave the gun?
Initial state:
Final state:
E  E0
v
E0  U 0 
EK

v
1
2
k  x 
2
1
m v2
2
k
x
m
940 N / m
2
25

10
m   39 m / s

3
38 10 kg
Problem is harder to solve by 2nd law.
Example 7.5. Spring & Gravity
A 50-g block is placed against a spring at the bottom of a frictionless slope.
The spring has k = 140 N/m and is compressed 11 cm.
When the block is released, how high up the slope does it rise?
E0  U 0 
Initial state:
1
k x02
2
Final state:
E U  m g h
E  E0
k x02
h
2mg
h

140 N / m  11 102 m 
2  50 10 kg  9.8 m / s
3
2
2

 1.7 m
GOT IT? 7.3. Bowling Ball
A bowling ball is tied to the end of a long rope and suspended from the ceiling.
A student holds the ball to her nose, then releases it from rest.
Should she duck as it swings back?
Example 7.6. Sliding Block
A block of mass m is launched from a spring of constant k that is compressed a distance x0.
The block then slides on a horizontal surface of frictional coefficient .
How far does the block slide before coming to rest?
Initial state:
Launch:
1
k x02
2
1
E1  K1  m v 2
2
E0  U 0 
Work done against friction:
Wnc   f f x    m g x
Final state:
Conservation of energy :
E1  E0  0
0  E1  Wnc
E 0

k x02
x 
2mg
7.4. Potential Energy Curves
Frictionless roller-coaster track
How fast must a car be coasting at
point A if it’s to reach point D?
Criterion:
turning
points
EA  U C
1
m v A2  m g hA  m g hC
2
vA 
2 g  hC  hA 
potential
barrier
potential
well
Example 7.7. H2
Near the bottom of the potential well of H2, U = U0 + a ( x  x0 )2 ,
where U0 = 0.760 aJ, a = 286 aJ / nm2 , x0 = 0.0741 nm.
( 1 aJ = 1018 J )
What range of atomic separation is allowed if the total energy is 0.717 aJ?
Turning points:
E  U  U 0  a  x  x0 
x  x0  
E  U0

a
x0  0.074 nm

2
0.717 aJ   0.760 aJ 
  0.0123 nm
2
286 aJ / nm
x  0.0864 nm to 0.0618 nm
Force & Potential Energy
Force ~ slope of potential curve
U  F  r   F x
( x along direction of F )
F 
U
x
 U    dU
F  lim  

x  0
dx
 x 
U    F  dr
GOT IT?. 7.4.
Below is the potential energy curve for an electron in a microelectronic device.
(a)Find the point where the force on the electron is greatest.
(b) Find the rightmost position possible if the electron has total energy E1 .
(c) Find the leftmost position possible if the electron has total energy E2 & starts out to the right of D.
(d) Find a point where the force on the electron is zero.
(e) Find a point where the force on the electron points to the left.
In some cases, there may be multiple answers.
(B)
(E)
(C)
(A,D)
(B,E)
```
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