Download Chapter 7 Potential Energy and Energy Conservation

Chapter 7
Potential Energy and Energy
Conservation
7.1
Gravitational Potential Energy
So far the only form of energy that we defined was kinetic energy which is a
function of velocity squared (or speed squared)
1
K = mv 2 .
2
(7.1)
It turns out that it is also useful to define another form of energy which is a
function of position. This new physical quantity we will call potential energy.
The idea behind name is that it might be possible to store energy in the form
of potential energy by placing an object at certain position. The process of
storing potential energy would most certainly require work to be done on the
object, but later on the potential energy can be released.
The most familiar example of potential energy is the gravitational potential energy, when the work must be done on an object to lift it up. For
example, if on object has mass m and is lifted from height x1 to height x2 ,
then the work done by gravitational force is
!
"!
"
Wgrav = −mg î (x2 − x1 ) î = mgx1 − mgx2
(7.2)
Then it is convenient to define a gravitational potential energy
Ugrav ≡ mgx
(7.3)
and then the work done by gravitational force can be written as
Wgrav = −∆Ugrav = Ugrav,1 − Ugrav,2 .
78
(7.4)
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 79
Now if we assume that there are no other forces acting on an object, then
according to work-energy theorem
Wgrav = K2 − K1
(7.5)
and by combining (7.4) and (7.5) we get
Ugrav,1 − Ugrav,2 = K2 − K1
(7.6)
Ugrav,1 + K1 = Ugrav,2 + K2 .
(7.7)
or
The latter equation suggest that the total (mechanical) energy is unchanged
E ≡ K + Ugrav = constant.
(7.8)
More generally there might be other forces acting on a given system which
might do work to it and then the work-energy theorem would imply
Wother = Ef − Ei
(7.9)
Note that although the conservation of energy suggests that the total energy
of all systems should still be conserved, the mechanical energy of any one
system changes according to (7.9).
Conservation of energy is the first of many conservation laws. As well
as any other conservation law, the conservation of energy is due to certain
symmetries (in this case time shift symmetry) that play a central role in
modern physics.
Example 7.1 . You throw a 0.145 − kg baseball straight up, giving it an
initial velocity of magnitude 20.0 m/s. How high it goes?
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 80
Step 1: Coordinate system. It is sufficient to choose a 1D coordinate
system with x-axis pointing upwards and origin at the initial position of the
ball.
Step 2: Initial conditions. The initial velocity is
⃗vi = 20.0 m/s î,
(7.10)
1
Ki = mvi2
2
(7.11)
Ui = mgxi = 0 J.
(7.12)
initial kinetic energy is
and initial potential energy is
Step 3: Final conditions. Then from conservation of energy
Uf + Kf = Ui + Ki
1
mgxf + 0 = 0 + mvi2
2
and thus
xf =
(20.0 m/s)2
vi2
= #
$ = 20.4 m.
2g
2 9.8 m/s2
(7.13)
(7.14)
Note that we did not have to know the mass of the baseball.
Example 7.2. In Example 7.1 suppose your hand moves upward by
0.50 m while you are throwing the ball. The ball leaves your hand with an
upward velocity of 20 m/s. (a) Find the magnitude of the force (assumed
constant) that your hand exerts on the ball. (b) Find the speed of the ball at
a point 15.0 m above the point where it leaves your hand. Ignore air resistance.
Step 1: Coordinate system. We can choose the same coordinate system
as in Example 7.1.
Step 2: Free body diagram. There are two phases of motion, but we
only need the free-body diagram for the first phase with origin at the initial
position of the ball.
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 81
Initial condition for second phase / final condition for first phase: Velocity
is
⃗v2 = 20.0 m/s î,
(7.15)
1
1
K2 = mv22 = (0.145 kg) (20.0 m/s)2 = 29.0 J
2
2
(7.16)
kinetic energy is
potential energy is
#
$
U2 = mgx2 = (0.145 kg) 9.8 m/s2 (0.50 m) = 0.71 J.
(7.17)
total mechanical energy is
E2 = K2 + U2 = 29.7 J
(7.18)
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 82
Step 3: Apply Newton’s Laws / Work -energy theorem.
(a) For the first phase the (generalized) work-energy theorem implies
Wother = E2 − E1
F x2 = E2
E2
29.7 J
F =
=
= 59 N.
x2
0.50 m
(7.19)
(b) For the second phase the work-energy theorem implies
E3
K3 + U3
=
=
E2
E2
1 2
mv + mgx3
2 3
=
E2
%
v3 = ±
or
2 (E2 − mgx3 )
m
(7.20)
& #
#
$
$
2 59 N − (0.145 kg) 9.8 m/s2 (0.50 m + 15.0 m)
v3 =
= 10 m/s. (7.21)
(0.145 kg)
Note that speed must be a positive number and so we only picked a positive
solution from (7.20).
Example 7.3. A batter hits two identical baseballs with the same initial
speed and from the same initial height but at different initial angles. Prove
that both balls have the same speed at any height h is air resistance can be
neglected.
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 83
The total mechanical energy at the initial conditions is the same in both
cases, and from conservation of energy the mechanical energy at any moment
of time is also the same in both cases. Moreover when balls are at the same
hight the potential energies are the same and therefore the kinetic energies
and speeds are the same.
U1i + K1i
E1i
E1f
U1f + K1f
K1f
v1f
=
=
=
=
=
=
U2i + K2i from the same initial speed and height
E2i from definition of total energy
E2f from conservation of energy for each trajectory
U2f + K2f from definition of total energy
K2f from definition of potential energy
v2f from definition of kinetic energy
(7.22)
Example 7.6. We want to slide a 12 − kg crate up a 2.5 − m-long ramp
inclined at 30◦ angle. A worker, ignoring friction, calculates that he can do
this by giving it an initial speed of 5.0 m/s at the bottom of and letting it
go. But, friction is not negligible; the crate slides only 1.6 m up the ramp,
stops, and slides back down. (a) Find the magnitude of friction acting on
crate, assuming that it is constant. (b) How fast is the crate moving when it
reaches the bottom of the ramp?
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 84
Step 1: Coordinate system. Let’s choose x-axis to point horizontally and
y-axis to point vertically with origin at the initial state.
Step 2: Initial conditions. At the initial conditions velocity is
⃗v1 = 5.0 m/s î
(7.23)
1
1
K1 = mv12 = (12 kg) (5.0 m/s)2 = 150 J
2
2
(7.24)
U1 = mgy1 = 0
(7.25)
E1 = U1 + K1 = 150 J.
(7.26)
kinetic energy
potential energy
and total energy is
Step 3: Apply the work-energy theorem. There are two stages of motion. In
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 85
the first stage the work energy theorem implies
W1−2 = E2 − E1
−F s = U2 + K2 − E1
#
$
−F (1.6 m) = (12 kg) 9.8 m/s2 (1.6 m) sin (30◦ ) + 0 − 150 J
56 J
F =
= 35 N.
(7.27)
1.6 m
where F is the magnitude of the force of friction and −F s is the work done
by friction which is always a negative number.
In the second stage
W2−3 = E3 − E2
−F s = U3 + K3 − U2 − K2
#
$
1
− (35 N) (1.6 m) = 0 + (12 kg) v32 − (12 kg) 9.8 m/s2 (1.6 m) sin (30◦ ) − 0
& 2
v3 =
2 (94 J − 56 J)
= 2.5 m/s.
12 kg
(7.28)
where v3 is velocity of the crete when it reached the bottom of crate.
7.2
Elastic Potential Energy
Remember that the work done on a spring is given by
1
1
W = kx2f − kx2i
2
2
(7.29)
but so the work done by a spring is
1
1
Wel = kx2i − kx2f
2
2
(7.30)
Similarly to work done by gravity,
Wgrav = −∆Ugrav = mgyi − mgyf
(7.31)
the work done by a spring can be thought as a difference of the potential
energies
1
1
(7.32)
Wel = −∆Uel = kx2i − kx2f
2
2
with the difference that
Ugrav = mgy
(7.33)
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 86
but
1
(7.34)
Uel = kx2
2
where it is assumed that the origin is chosen at a position corresponding to
unstretched spring. It is sometime useful to combine both potential energies
together, to define a total potential energy
1
U = mgy + kx2
2
(7.35)
(where depending on the setup x and y may or may not be the same) and
also total mechanical energy as
E =U +K
(7.36)
We can now modify the work-energy theorem further by stating that if there
are other forces (not gravitational, nor elastic) doing work on the system
then the total mechanical energy of the system changes according to
Wother = Ef − Ei .
(7.37)
Example 7.7. A glider with mass m = 0.200 kg sits on a frictionless
horizontal air track, connected to a spring with constant k = 5.00 N/m. You
pull on the glider, stretching the spring 0.100 m, and releasing it from rest.
The glider moves back towards its equilibrium position (x = 0). What is its
x-velocity when x = 0.080 m?
Step 1: Coordinate system. Let’s choose x-axis to point to the right with
origin corresponding to equilibrium of spring.
Step 2: Initial conditions. At the initial conditions the velocity
⃗vi = 0
(7.38)
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 87
kinetic energy
(elastic) potential energy
1
Ki = mvi2 = 0
2
(7.39)
1
Ui = kx2i
2
(7.40)
Ei = Ui + Ki = Ui
(7.41)
and total energy
Step 3: Apply work-energy theorem. From the conservation of energy
Ef = Ei
Uf + Kf = Ui
1 2 1 2
1 2
kxf + mvf =
kx
2
2
2 i
(7.42)
and thus
&
&
#
$
k(x2i − x2f )
(5.00 N/m) (0.100 m)2 − (0.080 m)2
vf =
=
= ±0.30 m/s.
m
0.200 kg
(7.43)
where negative solution corresponds to the first time it reaches xf = 0.080 m.
Example 7.9. A 2000 − kg (or 19600 − N) elevator with broken cables
in a test rig is falling at 4.00 m/s when it contacts a cushioning spring at the
bottom of the shaft. The spring is intended to stop the elevator, compressing
2.00 m as it does so. During the motion a safety clamp applies a constant
17000−N frictional force to the elevator. What is the necessary force constant
k for the spring.
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 88
Step 1: Choose a 1D coordinate system with x-axis pointing upwards and
origin at the initial position (i.e. place of first contact of the elevator and the
spring).
Step 2: Initial conditions. At the initial state the kinetic energy is
Ki =
1
(2000 kg) (4.00 m/s)2 = 16.0 kJ
2
(7.44)
the potential energy is
1
Ui = mgxi + kx2i = 0
2
(7.45)
Ei = Ui + Ki = 16.0 kJ.
(7.46)
and the total energy is
Step 3: Final conditions are given by
Kf = 0
#
$
#
1
1
Uf = mgxf + kx2f = (2000 kg) 9.80 m/s2 (−2.00 m) + k (−2.00 m)2 = −39.2 kJ + 2.00 m2
2#
2
$
Ef = −39 kJ + 2.00 m2 k
(7.47)
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 89
From the work energy theorem
Wother = Ef − Ei
#
$
(17000 N) (−2.00 m) = −39.2 kJ + 2.00 m2 k − 16.0 kJ
−34.0 kJ + 39.2 kJ + 16.0 kJ
= 1.06 × 104 N/m.
(7.48)
k =
2.00 m2
7.3
Conservative and Nonconservative forces
We were able to define potential energy associated with work done by gravitational and elastic forces. All such forces are called conservative forces.
Work done by conservative forces:
• can always be expressed as difference between initial and final values
of a suitably defined potential energy
• it is reversible and is independent on the trajectory of the body, but
only on initial and final points
One might wonder if it is possible to do the same for all macroscopic forces
which would allow to rewrite the work-energy theorem as a simple law of conservation of energy. It turns out that it is not possible and there are other (or
non-conservative) forces for which it is not possible to define potential energy.
For example, frictional force or air resistance forces are nonconservative.
⃗ = C x ĵ,
Example 7.11. In a region of space the force of an electron is F
where C is a positive constant. The electron moves around a square loop in
⃗ during
the xy-plane. Calculate the work done on the electron by the force F
a counterclockwise trip around square.
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 90
There are four legs of straight displacements around a closed path and so
the total work is given by
'
( (L,0)
( (L,L)
( (0,L)
(
⃗
⃗
W = F (x, y)·dl =
Fx (x, 0)dx+
Fy (L, y)dy+
Fx (x, L)dx+
(0,0)
(L,0)
(L,L)
(7.49)
where
(
(L,0)
Fx (x, 0)dx = 0
(0,0)
(L,L)
(
Fy (L, y)dy = CL(L − 0) = CL2
(L,0)
(
(0,L)
Fx (x, L)dx = 0
(L,L)
(
(0,0)
Fy (0, y)dy = 0(0 − L) = 0
(7.50)
W = 0 + CL2 + 0 + 0 = CL2
(7.51)
(0,L)
and so
which is non-zero and thus the force is non-conservative.
The non-conservative forces cannot be described in terms of mechanical
potential energies, but one can still associate with them other energies such
as internal energy. For example the frictional force is non-conservative, but
when friction is applied to objects in contact the internal properties of objects
(0,0)
Fy (0, y)dy
(0,L)
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 91
change. In particular friction leads to increase in temperature (or internal
energy) which on the microscopic level corresponds to the change of kinetic
energy of individual molecules. But from a macroscopic point of view one
can think of work done by all non-conservative forces as a measure of change
of internal energies
Wother = −∆Uint
(7.52)
and the work energy theorem implies the conservation law of total energy
∆K + ∆U + ∆Uint = 0.
(7.53)
The big difference of the internal energy is that one cannot use it to do any
useful work. This follows from the so-called second law of thermodynamics
that we will see later in the course.
The conservation law of energy and the second law of thermodynamics
had passed a very large number of tests, but this does not stop people (usually
outside of academia or theoretical physicists) from trying to build perpetual
motion machines of the first kind (do work without input of energy) or of
the second kind (do work using internal energy).
7.4
Force and potential energy
Let’s look more closely at work done on a particle when the displacement is
small
W = Fx (x)∆x
(7.54)
which can also be expressed as a change in potential energy
W = −∆U.
(7.55)
By equating (7.54) and (7.55) we get
Fx (x)∆x = −∆U
(7.56)
or in the limit of small infinitesimal displacement
Fx (x) = − lim
∆x→0
For example
1
U = kx2
2
⇒
dU
∆U
=− .
∆x
dx
(7.57)
Fx (x) = −kx
(7.58)
Fx (x) = −mg.
(7.59)
or
U = mgx
⇒
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 92
This can be easily generalized to 3D with
dU
dx
dU
= −
dy
dU
= −
dz
= −
Fx = − lim∆x→0 ∆U
∆x
Fy = − lim∆y→0
∆U
∆y
Fz = − lim∆z→0 ∆U
∆z
or in terms of unit vectors
*
)
∂U(x, y, z)
∂U(x, y, z)
∂U(x, y, z)
⃗
î +
ĵ +
k̂ .
F=−
∂x
∂y
∂z
(7.60)
(7.61)
It is convenient to define a vector-like operator also known as gradient (often
pronounced as nabla)
)
*
⃗ ≡ ∂ , ∂ , ∂
∇
(7.62)
∂x ∂y ∂z
or
⃗ ≡ î ∂ + ĵ ∂ + k̂ ∂
(7.63)
∇
∂x
∂y
∂z
such that
⃗ = −∇U(x,
⃗
F
y, z).
(7.64)
For example the gravitational potential energy is
U = mgy
(7.65)
and so
⃗ = −∇(mgy)
⃗
F
*
)
∂
∂
∂
+ ĵ
+ k̂
(mgy)
= − î
∂x
∂y
∂z
∂ (mgy)
∂ (mgy)
∂ (mgy)
= −î
− ĵ
− k̂
∂x
∂y
∂z
= −mg ĵ.
(7.66)
Example 7.14.
A puck with coordinates x and y slides on a level,
frictionless air-hockey table. It is acted on by a conservative force described
by the potential-energy function
$
1 #
U(x, y) = k x2 + y 2 .
2
(7.67)
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 93
Find a vector expression for the force acting on the puck, and find an expression for the magnitude of the force.
From (7.64) we get
)
*)
*
$
1 # 2
⃗ = − î ∂ + ĵ ∂ + k̂ ∂
F
k x + y2
∂x
∂y
∂z
2
(7.68)
= −îkx − ĵky
and
F =
7.5
+
(kx2 ) + (ky)2 = k
,
x2 + y 2 .
(7.69)
Energy Diagram
A convenient way of describing physical systems is in terms of energy diagrams plotting the potential energy and total energy as a function of displacement. For example consider a harmonic oscillator whose potential energy is
1
U = kx2
2
(7.70)
E = K + U.
(7.71)
and total energy is
If there are no non-conservative forces, then the total energy can be represented by a constant line and potential energy by a parabola.
This is the so-called energy diagram whose main point is to study the possible
solutions qualitatively without having to solve equations of motion quantitatively. Moreover, the energy graph can be used to identify stable (unstable)
solutions corresponding to local minima (maxima) of the potential energy.
For the example of harmonic oscillator a solution with x = 0 and dx
=0
dt
corresponds to a stable equilibrium solution. Then one can study small perturbations around the equilibrium which is what is done in field theories
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 94
where k plays the role of mass-squared parameter. However, if we imagine that the spring constant is negative k < 0, then the solution x = 0 and
dx
=0 is unstable. This phenomena is known as tachyonic instability which in
dt
the case in field theories corresponds to negative mass-squares (or imaginary
masses in context of complex numbers).
For more general potentials you can have both stable and unstable equilibriums. For example:
Which statement correctly describes what happens to the particle when it is
at the maximum between x2 and x3 ?
1. The particle’s acceleration is zero.
2. The particle accelerates in the positive x-direction; the magnitude of
the acceleration is less than at any other point between x2 and x3 .
3. The particle accelerates in the positive x-direction; the magnitude of
the acceleration is greater than at any other point between x2 and x3 .
4. The particle accelerates in the negative x-direction; the magnitude of
the acceleration is less at any other point between x2 and x3 .
5. The particle accelerates in the negative x-direction; the magnitude of
the acceleration is greater than at any other point between x2 and x3 .
The total energy was defined as a sum of kinetic and potential energies which
CHAPTER 7. POTENTIAL ENERGY AND ENERGY CONSERVATION 95
are functions of position x and velocity
)
dx
E x,
dt
*
1
= m
2
dx
,
dt
)
dx
dt
e.g.
*2
1
+ kx2 .
2
(7.72)
but an even more fundamental quantity (called Lagrangian) is the difference
between kinetic an potential energies
)
*
) *2
dx
1
dx
1
L x,
= m
− kx.2
(7.73)
dt
2
dt
2
Note that for any given trajectory x(t) the Lagrangian L is nothing but a
function of time and thus can be integrated over some interval of time, i.e.
*
( tf )
dx
dt.
(7.74)
L x,
dt
ti
Clearly, for any trajectory x(t) the integral in 7.74(also known as action
functional) would be some real number and one might wonder what trajectory would produce the smallest (or largest) number? It turns out that such
trajectories corresponding to stable (or unstable) classical solutions which
is what you are most likely to observe in the lab. However, according to
quantum mechanics many other solutions are also possible although their
probabilities (or the so-called probability amplitudes) are suppressed exponentially,
! tf
dx
(7.75)
ei ti L(x, dt )dt
where the meaning of i will only be clear after you learn a (physicists version
of) quantum mechanics.