Now let's disconnect the spring from the wall and attach the spring to the block so that the spring
sticks out horizontally from the block and again push the block up against the wall, compressing
the spring, and release the block from rest. Let our system be the block plus spring. The block
goes sliding off as before, this time with the spring attached. The wall does no energy-transfer
work on the system because the part of the wall that is exerting the force on the system is not
moving—there is no displacement. However, we find something useful if we calculate the
integral of the vector force (exerted by the wall) dot vector infinitesimal displacement of the
center of mass of the system—loosely stated, force times distance moved by center of mass. I'm
calling that "something useful" the center of mass pseudo-work experienced by the system. It's
useful because our Newton’s Law derivation shows that quantity to be equal to the change in the
center of mass kinetic energy of the system. In this case the system gained some center of mass
kinetic energy even though no energy was transferred to it. How did that happen? Energy that
was already part of the system, energy stored in the spring, was converted to center of mass
kinetic energy.
So what is the subtle difference? In both cases we are, loosely speaking, calculating force times
distance. But in the case of energy transfer work, the distance is the distance moved by that
element of the agent of the force that is in contact with the victim at exactly that point where the
force is being applied, whereas, in the case of center of mass work, the distance in “force times
distance” is the distance moved by the center of mass. For a particle, there is no difference. For
a truly rigid body undergoing purely translational motion (no rotation) there is no difference.
But beware, a truly rigid body is an idealized object in which no bit of the body can move
relative to any other bit of the body. Even for such a body, if there is rotation, there will be a
difference between the energy transfer work and the center of mass pseudo-work done on the
object. Consider for instance a block at rest on a horizontal frictionless surface. You apply an
off-center horizontal force to the block for a short distance by pressing on the block with your
finger. The work you do is the integrated force times distance over which you move the tip of
your finger. It will be greater than the integrated force times the distance over which the center
of mass moves. Some of the work you do goes into increasing the center of mass kinetic energy
of the rigid body and some of it goes into increasing the rotational kinetic energy of the rigid
body. In this case the energy transfer work is greater than the center of mass pseudo-work.
Concluding Remarks
At this point you have two ways of calculating the work done on an object. If you are given
information about the force and the path you will use the “force times distance” definition of
work. But if you are given information on the effect of the work (the change in kinetic energy)
then you will determine the value of the change in kinetic energy and substitute that into the
work energy relation, equation 24-2:
W = K
∆
to determine the work (or center of mass pseudo-work as applicable). There is yet another
method for calculating the work. Like the first method, it is good for cases in which you have
information on the force and the path. It only works for certain kinds of forces, but when it does
work, to use it, the only thing you need to know about the path is the positions of the endpoints.
This third method for calculating the work involves the potential energy, the main topic of our
next chapter.
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Chapter 25 Potential Energy, Conservation of Energy, Power
25 Potential Energy, Conservation of Energy, Power
The work done on a particle by a force acting on it as that particle moves from point A to point B
under the influence of that force, for some forces, does not depend on the path followed by the
particle. For such a force there is an easy way to calculate the work done on the particle as it
moves from point A to point B. One simply has to assign a value of potential energy (of the
particle1) to point A (call that value U ) and a value of potential energy to point B (call that value
A
U ). One chooses the values such that the work done by the force in question is just the negative
B
of the difference between the two values.
W = − (U − U )
B
A
W = − U
∆ (25-1)
U
∆ =U −U is the change in the potential energy experienced by the particle as it moves from
B
A
point A to point B. The minus sign in equation 25-1 ensures that an increase in potential energy
corresponds to negative work done by the corresponding force. For instance for the case of near-
earth’s-surface gravitational potential energy, the associated force is the gravitational force, a.k.a.
the gravitational force. If we lift an object upward near the surface of the earth, the gravitational
force does negative work on the object since the (downward) force is in the opposite direction to
the (upward) displacement. At the same, time, we are increasing the capacity of the particle to
do work so we are increasing the potential energy. Thus, we need the “−“ sign in W = − U
∆ to
ensure that the change in potential energy method of calculating the work gives the same
algebraic sign for the value of the work that the force-along-the path times the length of the path
gives.
Note that in order for this method of calculating the work to be useful in any case that might
arise, one must assign a value of potential energy to every point in space where the force can act
on a particle so that the method can be used to calculate the work done on a particle as the
particle moves from any point A to any point B. In general, this means we need a value for each
of an infinite set of points in space.
This assignment of a value of potential energy to each of an infinite set of points in space might
seem daunting until you realize that it can be done by means of a simple algebraic expression.
For instance, we have already written the assignment for a particle of mass m for the case of the
2
universal gravitational force due to a particle of mass m . It was equation 17-5:
1
G m m
U = −
r1 2
1 The potential energy is actually the potential energy of the system consisting of the particle, whatever the particle
is interacting with, and the relevant field. For instance, if we are talking about a particle in the gravitational field of the earth, the potential energy under discussion is the potential energy of the earth plus particle and gravitational
field of the earth plus particle. For accounting purposes, it is convenient to ascribe the potential energy to the
particle and that is what I do in this book.
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Chapter 25 Potential Energy, Conservation of Energy, Power
2
N ⋅ m
in which
11
G is the universal gravitational constant G
6 67 ×
−
= .
10
and r
2
is the distance that
kg
particle 2 is from particle 1. Note that considering particle 1 to be at the origin of a coordinate
system, this equation assigns a value of potential energy to every point in the universe!
The value, for any point, simply depends on the distance that the point is from the origin.
Suppose we want to find the work done by the gravitational force due to particle 1, on particle 2
as particle 2 moves from point A, a distance r from particle 1 to point B, a distance r from
A
B
particle 1. The gravitational force exerted on it (particle 2) by the gravitational field of particle 1
does an amount of work, on particle 2, given by (starting with equation 25-1):
W = − U
∆
W = − (U − U )
B
A
G m m G m m
W = − −
1
2
−
1
2
r
−
r
B
A
1
1
W = G m m
1
2
−
r
r
B
A
The Relation Between a Conservative Force and the Corresponding
Potential
While this business of calculating the work done on a particle as the negative of the change in its
potential energy does make it a lot easier to calculate the work, we do have to be careful to
define the potential such that this method is equivalent to calculating the work as the
force-along-the-path times the length of the path.
Rather than jump into the problem of finding the potential energy at all points in a three-
dimensional region of space for a kind of force known to exist at all points in that three-
dimensional region of space, let’s look into the simpler problem of finding the potential along a
line. We define a coordinate system consisting of a single axis, let’s call it the x-axis, with an
origin and a positive direction. We put a particle on the line, a particle that can move along the
line. We assume that we have a force that acts on the particle wherever the particle is on the line
and that the force is directed along the line. While we will also address the case of a force which
has the same value at different points along the line, we assume that, in general, the force varies
with position.
Remember this fact so that you can find the flaw discussed below. Because we
want to define a potential for it, it is important that the work done on the particle by the force
being exerted on the particle, as the particle moves from point A to point B does not depend on
how the particle gets from point A to point B. Our goal is to define a potential energy function
for the force such that we get the same value for the work done on the particle by the force
whether we use the force-along-the-path method to calculate it or the negative of the change of
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Chapter 25 Potential Energy, Conservation of Energy, Power
potential energy method. Suppose the particle undergoes a displacement ∆x along the line under
the influence of the force. See if you can see the flaw in the following, before I point it out: We
write W = F ∆x for the work done by the force, calculated using the force-along-the-path times
the length of the path idea, and then W = − U
∆ for the work done by the force calculated using
the negative of the change in potential energy concept. Setting the two expressions equal to each
U
∆
other, we have, F ∆x = − ∆U which we can write as F = −
for the relation between the
∆x
potential energy and the x-component of the force.
Do you see where we went wrong? While the method will work for the special case in which the
force is a constant, we were supposed to come up with a relation that was good for the general
case in which the force varies with position. That means that for each value of x in the range of
values extending from the initial value, let’s call it x , to the value at the end of the displacement
A
x + ∆x, there is a different value of force. So the expression W = F ∆x is inappropriate. Given
A
a numerical problem, there is no one value to plug in for F, because F varies along the ∆x.
To fix things, we can shrink ∆x to infinitesimal size, so small that, x and x + ∆x are, for all
A
A
practical purposes, one and the same point. That is to say, we take the limit as ∆x → 0. Then
our relation becomes
∆U
F = lim
x
∆ →
−
x
0
∆x
which is the same thing as
∆U
F = − lim
x
∆x→
0 ∆ x
U
∆
dU
The limit of
that appears on the right is none other than the derivative
, so:
∆x
dx
dU
F = −
( (25-2)
x
dx
To emphasize the fact that force is a vector, we write it in unit vector notation as:
dU
F = −
i ( (25-3)
dx
Let’s make this more concrete by using it to determine the potential energy due to a force with
which you are familiar—the force due to a spring.
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Chapter 25 Potential Energy, Conservation of Energy, Power
Consider a block on frictionless horizontal surface. The block is attached to one end of a spring.
The other end of the spring is attached to a wall. The spring extends horizontally away from the
wall, at right angles to the wall. Define an x-axis with the origin at the equilibrium position of
that end of the spring which is attached to the block. Consider the away-from-the-wall direction
to be the positive x direction. Experimentally, we find that the force exerted by the spring on the
block is given by:
F = − k xi ( (25-4)
where k is the force constant of the spring. (Note: A positive x, corresponding to the block having
been pulled away from the wall, thus stretching the spring, results in a force in the negative
x direction. A negative x, compressed spring, results in a force in the +x direction, consistent with
dU
common sense.) By comparison with equation 25-3 (the one that reads F = −
i ) we note that
dx
the potential energy function has to be defined so that
dU = k x
dx
This is such a simple case that we can pretty much guess what U has to be. U has to be defined
such that when we take the derivative of it we get a constant (the k) times x to the power of 1.
Now when you take the derivative of x to a power, you reduce the power by one. For that to
result in a power of 1, the original power must be 2. Also, the derivative of a constant times
something yields that same constant times the derivative, so, there must be a factor of k in the
potential energy function. Let’s try U = k x2 and see where that gets us. The derivative of kx2 is
2kx. Except for that factor of 2 out front, that is exactly what we want. Let’s amend our guess
1
by multiplying it by a factor of
, to eventually cancel out the 2 that comes down when we take
2
1
dU
the derivative. With
2
U =
k x we get
= k x which is exactly what we needed. Thus
2
dx
1
2
U =
k x (25-5)
2
is indeed the potential energy for the force due to a spring. You used this expression back in
chapter 2. Now you know where it comes from.
We have considered two other conservative forces. For each, let’s find the potential energy
dU
function U
F = −
that meets the criterion that we have written as,
i .
dx
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Chapter 25 Potential Energy, Conservation of Energy, Power
First, let’s consider the near-earth’s-surface gravitational force exerted on an object of mass m,
by the earth. We choose our single axis to be directed vertically upward with the origin at an
arbitrary but clearly specified and fixed elevation for the entire problem that one might solve
using the concepts under consideration here. By convention, we call such an axis the y axis
rather than the x axis. Now we know that the gravitational force is given simply (again, this is an
experimental result) by
F = − g
m j
where the mg is the known magnitude of the gravitational force and the −j is the downward
direction.
Equation 25-3, written for the case at hand is:
dU
F = −
j
dy
For the last two equations to be consistent with each other, we need U to be defined such that
dU = g
m
dy
For the derivative of U with respect to y to be the constant “mg ”, U must be given by
U = mgy (25-6)
and indeed this is the equation for the earth’s near-surface gravitational potential energy. Please
verify that when you take the derivative of it with respect to y, you do indeed get the magnitude
of the gravitational force, mg.
Now let’s turn our attention to the Universal Law of Gravitation. Particle number 1 of mass m
1
creates a gravitational field in the region of space around it. Let’s define the position of particle
number 1 to be the origin of a three-dimensional Cartesian coordinate system. Now let’s assume
that particle number 2 is at some position in space, a distance r away from particle 1. Let’s define
the direction that particle 2 is in, relative to particle 1, as the +x direction. Then, the coordinates of
particle 2 are (r,0,0). r
is then the x component of the position vector for particle 2, a quantity that
we shall now call x. That is, x is defined such that x r
= . In terms of the coordinate system thus
defined, the force exerted by the gravitational field of particle 1, on particle 2, is given by:
G m m
1
2
F = −
i
2
x
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75
Chapter 25 Potential Energy, Conservation of Energy, Power
which I rewrite here:
G m m
1
2
F = −
i
2
x
Compare this with equation 25-3:
dU
F = −
i
dx
Combining the two equations, we note that our expression for the potential energy U in terms of
x must satisfy the equation
dU
G m m
1
2
=
2
dx
x
It’s easier to deduce what U must be if we write this as
dU
−2
= G m m x
1
2
dx
For the derivative of U with respect to x to be a constant (Gm m
) times a power (−2) of x, U
1
2
itself must be that same constant (Gm m
) times x to the next higher power (−1), divided by the
1
2
value of the latter power.
−1
G m m x
1
2
U =
−1
which can be written
G m m
U
1
2
= −
x
Recalling that the x in the denominator is simply the distance from particle 1 to particle 2 which
we have also defined to be r, we can write this in the form in which it is more commonly written:
G m m
U
1
2
= −
(25-7)
r
This is indeed the expression for the gravitational potential that we gave you (without any
justification for it) back in Chapter 17, the chapter on the Universal Law of Gravitation.
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Chapter 25 Potential Energy, Conservation of Energy, Power
Conservation of Energy Revisited
Recall the work-energy relation, equation 24-2 from last chapter,
W = ∆K ,
the statement that work causes a change in kinetic energy. Now consider a case in which all the
work is done by conservative forces, so, the work can be expressed as the negative of the change
in potential energy.
− U
∆ = ∆K
Further suppose that we are dealing with a situation in which a particle moves from point A to
point B under the influence of the force or forces corresponding to the potential energy U.
Then, the preceding expression can be written as:
− (U − U ) = K − K
B
A
B
A
−U + U = K − K
B
A
B
A
K + U = K + U
A
A
B
B
Switching over to notation in which we use primed variables to characterize the particle when it
is at point B and unprimed variables at A we have:
K + U = K ′+ U ′
Interpreting E = K + U as the energy of the system at the “before” instant, and E′ = K ′ + U ′ as
the energy of the system at the “after” instant, we see that we have derived the conservation of
mechanical energy statement for the special case of no net energy transfer to or from the
surroundings and no conversion of energy within the system from mechanical energy to other
forms or vice versa. In equation form, the statement is
E = E′ (25-8)
an equation to which you were introduced in chapter 2. Note that you would be well advised to
review chapter 2 now, because for the current chapter, you are again responsible for solving any
of the “chapter-2-type” prob