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Unformatted text preview: CHAPTER THE LAW OF
CONSERVATION OF
ENERGY You see, therefore, that living force [kinetic energy] may be converted into
heat, and that heat may be converted into living force, or its equivalent
attraction through space. All three, therefore — namely, heat, living force, and
attraction through space (to which I might also add light, were it consistent
with the scope of the present lecture) — are mutually convertible into one
another. in these conversions nothing is ever lost. The same quantity of heat
will always be converted into the same quantity of living force. We can
therefore express the equivalency in definite language applicable at all times
and under all circumstances. James Prescott Joule, ”On Matter, Living Force, and Heat” (1847) And in each of these decades [the 19503 and 19603] more oil was
consumed than in all of man's previous history combined. < President Jimmy Carter (18 April 1977) 13.1 TOWARD AN IDEA OF ENERGY The law of conservation of energy is a fundamental law of physics. No matter what you
do, energy is always conserved. The total amount of energy in the universe is, has been,
and always will be the same as it is right now. So why do people tell us to conserve
energy? Evidently the phrase conserve energy has one meaning to a scientist and quite
a different meaning to other people, for example, to the president of a utility company
or to a politician. What then, exactly, is energy? 245 246 THE LAW OF CONSERVATION OF ENERGY The notion of energy is one of the few elements of mechanics not handed down to
us from Isaac Newton. The idea was not clearly grasped until the middle of the nineteenth
century. Nevertheless, we can find its germ even earlier than Newton. The essence of
the idea of the conservation of energy can be seen in the incredibly fertile experiments
that Galileo performed with balls rolling down inclined planes. It is astonishing how many results Galileo squeezed out of his simple experiments.
Bodies fall much too fast to be timed by the crude water clocks of the seventeenth century,
but by slowing down the falling motion with his inclined planes, Galileo showed that
uniformly accelerated motion was a part of nature. That alone was an achievement to
crown him as a genius. ' Galileo, of course, did much more besides his inclined—plane experiments, but he
also did much more with his experiments. He arranged that once a ball had finished
rolling down one inclined plane it would proceed to roll back up another, which could
be more or less inclined than the first. Here he discovered a suggestive fact: no matter
what the incline of the first plane and no matter what the incline of the second, the ball
would finally come to rest on the second plane at the same vertical height above the table
as that at which it had started on the first. He concluded that if the second plane were horizontal the ball would continue rolling
with the same speed forever. In other words, he discovered the law of inertia, which we
discussed in Chapter 4. Once we have grasped the idea of inertia, we can easily see why the ball starts up
the second inclined plane after rolling down the first. But that does not tell us why it
always reaches the same vertical height at which it started. The ball almost seems to
remember its origin. We prefer to say that something is conserved, rather than remem
bered. The name we give to the conserved quantity is energy. When Galileo lifted a ball from the table to the height of its starting point, as illustrated
in Fig. 13.1, he endowed it with a form of energy called potential energy. Energy that
a body has by virtue of its location is called potential energy. Then Galileo released the
ball and it started rolling, picking up speed. By the time it reached the bottom of the
incline, it was at the level of the table. If it previously had potential energy because of
its height above the table, that energy was now gone. In place of the potential energy,
the ball was in motion. The potential energy had not been lost, but rather had been
transformed into another form. Energy of motion is called kinetic energy. As the ball continued, ascending the second inclined plane, it slowed down. It was
losing kinetic energy, but in return it was regaining energy of position, potential energy,
as its height increased. When the ball finally came to rest, all its kinetic energy had been Figure 13.1 Galileo’s experiment with inclined planes and rolling
balls. 13.2 WORK AND POTENTIAL ENERGY 247 transformed back into potential energy. That happened when the ball’s height above the
table was precisely what it was at the beginning of the experiment. Earlier we would have described the same experiment in different terms. We would
have said: To lift the ball, Galileo applied a force opposing gravity. When he released
the ball, the force of gravity made it roll down the plane. At the bottom of the plane, its
inertia made the ball roll back up the second plane. We have a valid description without
the idea of energy, so why talk about energy? We need the idea of energy because it expresses one fact our old description didn’t
prepare us for: the ball ends up at the same height it started at, never any higher, and,
ignoring air resistance and friction, never any lower. Something is the same at the end
of the ball’s motion as it was in the beginning. That something is energy. The idea of energy was invented precisely because it is conserved. Then why do
politicians and gas company executives tell us to conserve energy? Before answering this
burning question, we need precisely and quantitatively to define energy. 13.2 WORK AND POTENTIAL ENERGY The word work is used to describe the transfer of energy from one thing to another. We
say that when Galileo lifted a ball onto the inclined plane, he was doing work on it. The
clue to the definition of work is that we have two descriptions of Galileo’s inclined—plane
experiment, both of which must be correct: Galileo applied a force opposing gravity,
lifting a ball to a certain height; in terms of energy, he did work on the ball, thereby
giving it a certain potential energy equal to the work done. By comparing these two
views, we see that work is related to both force and distance. Our intuition suggests that it takes twice as much work to lift a ball twice the weight
to the same height, and that it takes three times as much work to lift a given ball three
times as high. That indicates how work is related to both force and distance. The simplest
relation is to define the work W done by a constant force F to be the product of the force
times the distance h through which it acts. Thus, for a constant force, W = Fh. (13.1) For example, if you lift an object of mass m in such a way that it doesn’t accelerate,
then the force is equal to the weight mg of the object. Since mg is constant, the work
done to raise the object up to height h is W = mgh. What if the force is not constant? Suppose a force that changes with position acts
on a particle moving in a straight line. Say the force is F (2) (in the direction of the line)
when the particle is at 2 (as shown in Fig. 13.2). How much work is done by this variable
force in moving the particle from 2 = a to any other position 2 = x? The work will depend on x and we denote it by W(x). We haven’t defined W(x) yet
but we can give an argument which suggests that a reasonable definition is (13.2) This is a deﬁnition, so it cannot be proved, but it can be motivated as follows: 248 THE LAW OF CONSERVATION OF ENERGY F01) F (Z) F(X)
a z x Figure 13.2 Work done by force F(z) moving an object from 2 = a
to z = x is W(x) = I: F(z) dz. First, we want W(a) = 0 since no work is done if we don’t move the particle at all.
Now suppose we move the particle from the position 2 = x to a nearby position 2 =
x + h, where h is a small positive number. The work done by the force from x to x + h is the difference
W(x + h) — W(x). We can express this same work in another way. Over the interval from x to x + h, the
force F (2) will change, but it has some average value, which we can call F(E). If we
treat the force as though it had the constant value F (2) over the interval from x to x + h,
then the work it does is F (E)h, the product of force and distance. Therefore W(x + h) — W(x) = F(7)h
or, dividing by h, W(x + h) — W(x) h = F(Z). Now we let h shrink to zero on both sides of the equation. The left—hand side becomes
a'W/dx and the righthand side becomes F (x), so in the limit we find dW — = F . dx (x)
In other words, the derivative of the work is F (x). Therefore, by the second fundamental
theorem of calculus, we have W(x) — W(a) = J: F(z) dz. Since W(a) = 0, this gives us Eq. (13.2). Note that if the force is constant, the integral is the constant force times the length
of the interval, which agrees with (13.1). The dimensions of work are those of force times distance. The SI unit of work is
one newtonmeter, called one joule (after a British physicist whom we shall encounter
later) and abbreviated l J. In the cgs system the unit of work is one dynecentimeter, or
1 erg. In the British system, one footpound is the basic unit of work and has no other
name. Using conversions between fundamental units, you can verify that 1 J = 107 erg = 0.738 ft lb. An important remark is appropriate at this point: The integral 13.2 WORK AND POTENTIAL ENERGY 249 W(x) = [X F(z) dz is the work done by the force F (2) acting on a particle in moving it from 2 = a to z = x.
The negative of this quantity, —W(x) = —J’x F(z) dz, is called the work done against the force F (2). For example, suppose a car is rolling
slowly along a road and a man pushes backward on it with a constant force of 50 1b,
bringing it to rest within a distance of 10 ft. The work done by the car on the man is
500 ft lb, a positive quantity. But the work done by the man on the car is — 500 ft lb,
a negative quantity because the force and the displacement are in the opposite directions. Now that we have a definition of work, we need to understand its connection with
energy and why energy is conserved. To do so, let’s think about Galileo’s experiment.
To get the ball rolling, Galileo first lifted it up to a certain height and in doing so
performed work on it, endowing it with a certain amount of potential energy. The work
he did and the potential energy of the ball are each equal to mgh. Once he released it,
the ball rolled down the plane. Its potential energy, equal to mg times its height above
the table, was converted into kinetic energy. As it rolled up the second incline, the kinetic
energy was converted back into potential energy. This continued until its original potential
energy mgh was restored. Thus it rolled to a height equal to that at which it started, and
where it again had the same potential energy as it did initially. The ball would continue to roll up and down the inclined planes without Galileo’s
attention. So once Galileo lifted the ball, once he did work on it, he was out of the
picture. We no longer need to consider Galileo, but rather we can focus entirely on the
energy and motion of the ball. Let us reiterate: work is the name we use to describe a process in which energy is
transferred from one part of the universe to another. Here the first part of the universe
is Galileo, and the second system is the ball. Energy entered the system in the form of
potential energy from the universe outside of it by means of work done on it. Work is
merely a bookkeeping device to keep track of transfer of energy from one thing to another. We identify the work done to lift an object without accelerating it as the transfer of
potential energy U to the object: U = W. (13.3)
It follows that in Galileo’s experiment, the ball receives an amount of potential energy
U = mgh. Aside from depending on the mass of the ball, the potential energy depends only on
where it is, which here is its height h above the table. This connection is plausible because
when the ball rolls up the second plane it reaches the same height as it started from, h,
where it has exactly the same potential energy. Thus energy defined in this way is indeed a conserved quantity.
So far our definition of work as a force times displacement in the same direction appears feasible. But what happens if the force and resulting displacement are not in the 250 THE LAW OF CONSERVATION OF ENERGY same direction? For example, when gravity works on a ball on Galileo’s inclined plane,
the force of gravity is always vertically downward, but the ball is displaced not straight
downward, but along the inclined plane. How do we handle this? Since gravity should do exactly as much work on the ball moving it down the plane
as Galileo did to put it up there, we can find the answer. As shown in Fig. 13.3, mg is
the force and r is the displacement along the inclined plane. The work done by gravity
to bring the ball to the bottom of the inclined plane must be equal to mgh. From the
figure we see that h = r cos 6, so we can write the work done as W = mgh = mgr cos 6. Figure 13.3 Work done by gravity is mgr = mgh. Taking another look at Fig. 13.3, we realize that mg cos 6 is the component of the
weight in the direction of the displacement r. We already have a notation prepared exactly
for this purpose, the dot product. As you recall from Chapter 5, the dot product is equal
to the product of the component of one vector along a second vector times the magnitude
of the second vector. In summary, we describe the work done here as W = Fr = mgr = mgrcos 6.
In general, the work done by a constant force F acting through a displacement r is W = Fr. (13.4) Example 1
Show that the work done against gravity depends only on the vertical distance an object
is moved. Suppose we wish to move an object from point A to point B shown in the figure. U3 3‘
Oo—v—p—>—>—>o ._>—.—>—p—p 13.2 WORK AND POTENTIAL ENERGY 251 We can think of moving the object from A to B by first moving it horizontally, a distance
x to point C, then vertically a distance h to B. The work done from A to B is simply the
sum of the work from A to C and from C to B: WAB = WAC + was Because the force we apply to the object is vertically upward and equal to mg, Eq. (13.1)
tells us that WAC = mgx cos 90° = 0, WCB = mgh cos 0° = mgh.
Therefore the total work done is simply WAB = mgh. In fact, the work done in moving from A to B will always be mgh, no matter what path
is taken. If the path is replaced by vertical and horizontal steps, the work done along
any horizontal path is zero. The work done against gravity depends only on the difference
in vertical height between the initial and final points. Example 2
What amount of potential energy is stored in a spring that is stretched from equilibrium to a distance d? We’ll use Eq. (13.2) to calculate the work, which in turn is equal to the potential
energy stored in the spring. When we stretch a spring a distance x, the spring exerts a
force F = —k.x. (10.2) Therefore, to stretch it, we need to apply a force F = + kx, which is, of course, in the
same direction as the displacement of the spring. Thus the work we do is d
W=I kxdx=§kd€
0 where we know the value of the integral from Chapter 7. Since the potential energy stored
in the spring is equal to the work done on it, we have = £de The potential energy is the same whether the spring is stretched or compressed. Questions 1. If you were to hold a book with an outstretched arm, you would soon tire and
claim that it is hard work to keep the book there. According to Eq. (13.1), are you
doing work on the book? 252 THE LAW OF CONSERVATION OF ENERGY 2. Suppose that two forces, F1 and F2, act on an object, with F1 = 10 N, F 2 = 16
N, such that the two forces are acting in opposite directions. What is the total work
done on the object when it moves a distance of 2 m under the action of these
forces? 3. Two springs A and B are identical except that A is stiffer than B (larger spring
constant). On which spring is more work done if (a) they are stretched by the same
force, (b) they are stretched by the same amount? 4. If work represents the energy transferred to a system from the outside, where does
the energy come from that enables you, for example, to lift a ball onto a table? 5. A gardener pushes a lawn mower with a force of 20.0 N at an angle of 370 down
from the horizontal for a distance of 15.0 m. Calculate the amount of work done by
the gardener. 6. A force F = (3.0 N)i — (7.0 N)j moves an object through a displacement r =
(4.0 m)i + (3.0 m)j + (2.0 m)k. Find the amount of work done on the object by
this force. 7. The force exerted on an object is described by F = F 0(x/b — 1), where F 0 and b
are constants. Find the work done as the object moves from x = 0 to x = 3b by (a) plotting F (x) and graphically determining the area under the curve,
(b) evaluating the integral analytically. 8. The work done to stretch a certain spring a distance of 1.0 cm from equilibrium is
0.2 J. If 0.2 J additional work were done on the stretched spring, how far from
equilibrium would the spring be stretched? 13.3 THE LAW OF CONSERVATION OF ENERGY When the ball reaches the bottom of Galileo’s incline, where h = 0, it will have lost all
of its potential energy U, converting it to kinetic energy K. We know exactly how large
K will be: it will be equal to U. But K is the energy of motion; it should depend not on
where the ball was, as U does, but on how fast it is going. What exactly is the connection
between kinetic energy and speed? Let’s imagine an experiment slightly different from Galileo’s. Suppose we don’t start
with an inclined plane. Instead we have a smooth block on a smooth table, at height
h = 0. If we push the block horizontally, we set it in motion; the force merely overcomes
the block’s inertia. While we are pushing on the block, we keep it on the table, so h and
therefore the potential energy remain equal to zero. However, suppose we cleverly arrange
that by the time the block reaches the second incline, as shown in Fig. 13.4, it has the
same speed v0 it would have acquired in sliding down an incline starting at height h. F Figure 13.4 Sliding block on inclined planes. 13.3 THE LAW OF CONSERVATION OF ENERGY 253 Since the kinetic energy should depend on the block’s speed, and we are arranging
things so that the speed is the same as it would have been had it slid down the plane, K
must be equal to U, which is equal to the work W. In other words, the work we have
done pushing the block horizontally must be the same as W. But how can we express
that in terms of speed? We don’t even know how much force in the horizontal direction
we need to get the block up to the right speed. We do know, however, that F = —. (135) As we push the block, we do work over some period of time, so it is reasonable to
think of the work done per unit time (known as power) rather than how much work we
do at a given place. Thinking of the work W as a function of time t we use the chain
rule to get a relation between F and dW/dt: dW_ dex dx =F — .
dt dx dz dt
But dx/dt is the block’s speed 12. Therefore we have
F
_ = 1).
dt Using Newton’s second law, F = m dv/dt to substitute for F, we obtain dW dv
__ = my—
dt dz
Because d(v2)/dt = 2v dv/dt we can write this last result as
dW d
__ = __(% m‘UZ)
dt dt This means that the time rate of change of work is the same as the rate of change
of the quantity— —mv2. In other words, the total work we do by the time we get the block
to the base of the second incline is— 2.mvz Since the block started at rest, 1/ = 0, the
change is slimpzlyé —mv0, where v0 is the speed it finally acquires; hence We want this to be equal to the work it takes to lift the block,
W = U = mgh, which implies that _ 1 2
mgh— E m'Uo. We’ve succeeded in finding a quantitative expression for the conservation of energy
applied to our own Galileolike experiment. We have used a sliding block instead of a
rolling ball so that we could ignore the fact that some energy is necessary just to get a
ball rotating. When we originally lifted the block a distance h above the table, our muscles
did work on it and gave the block a potential energy equal to mgh. That initial amount
of energy remains in the system even though it can change into a mixture of both potential 254 THE LAW OF CONSERVATION OF ENERGY and kinetic energy. When the block has reached the bottom of the incline, its potential
energy has been completely transformed into kinetic energy, which we can now quan
tiatively express as émvzz (13.7) This formula for kinetic energy was derived for a sliding block having the speed v it
would have acquired by sliding down an incline under the constant force of gravity.
Using this example as motivation, we now deﬁne the kinetic energy K of any body of
mass m moving with speed 1) by the equation 2 K = %mv .
The foregoing analysis which showed that
dt 2 _ d; ' is valid not only for a sliding block but for the work done by any variable force F acting
in a fixed direction. The work done against this force is —W, the potential energy U imparted to the body. Since K = 5sz2 and W = — U, Eq. (13.8) states that £__ﬂ
dt dt ’
or
d
d7 (U + K) = 0. (13.9) Therefore the quantity U + K remains constant in time. This quantity, is called the total energy of the particle. Both the kinetic energy K and the potential
energy U can change as the particle moves, but their sum — the total energy — remains
constant. This is the law of conservation of energy. It is described mathematically by
Eq. (13.9). Let’s see what the law tells us about the sliding block. Let h denote the maximum
or initial height above the table, and let 2 be an arbitrary height, as indicated in Fig.
13.5. Then mgz is the potential energy when the block is at height 2, and émv2 is the
kinetic energy. The conservation law says that mgz + émv2 = const. By saying that the lefthand side is constant, we mean that its value does not change in
time. Since this quantity is the same anywhere along the block’s motion, we can determine 133 THE LAW OF CONSERVATION OF ENERGY 255 Figure 13.5 Conservation of energy applied to Galileo's experiment. the constant by evaluating the lefthand side at any point where we know all the quantities.
For example, if we take the initial point where z = h, we know that v = 0, so the value
of the lefthand side is simply mgh, which is the constant. Thus we can write mgz + émv2 = mgh. (13.10) From this statement of conservation of energy, we can ﬁnd the speed of the block at any height 2. None of this depends in any way on how steeply the planes are inclined. In fact, the
planes could be vertical; then we would have a freely falling body. Regardless of whether
a body is slipping down an inclined plane or freely falling, if it starts at height h, its
speed 1) can be found from (13.10), which when solved for I) gives 1) = \/2g(h — z). The speed depends only on the vertical distance fallen. Does it agree with the law of
falling bodies?
Recall that for a body in free fall, _ l 2
S — 5g!
and
v = gt. If we eliminate t, we get v = \/2gs, where s is the distance fallen and is just h — z. The law of conservation of energy predicts
exactly the same speed as the law of falling bodies. This comparison helps to confirm
that the way we defined work by Eq. (13.1) is a sensible choice; it leads to a quantitative
expression of total energy, E = U + K, which is conserved. The law of conservation of energy is useful for analyzing motion not only on inclined
planes and for falling bodies but also for many other phenomena such as pendulums and
springs. The following is a generalization of the procedure we used to apply the law of
conservation of energy: (a) Define the system.
(b) Write down the total energy of the system at the point, say A, where you want to determine some unknown quantity (like speed or height): E A = U A + K A. 256 THE LAW OF CONSERVATION OF ENERGY (c) Find another point, say point B, where you know everything about the object’s
motion and write down the total energy at that point: EB = UB + KB. ((1) Conservation of energy implies that E A = EB; equate the two energies and
solve for the unknown quantity. So far, we know two types of potential energy which U can represent. One is the
potential energy a mass m has due to its height 2 above the earth’s surface: Ugravity = ng. The other is the potential energy stored in a spring either stretched or compressed by an
amount x: kxz. 1
Uspring E The following examples apply the law of conservation of energy to various problems. Example 3
A 2.0kg block on a horizontal, frictionless surface is pressed against a spring of spring constant 1.5 X 103 N/m. The spring is compressed a distance of 8.0 cm. When released,
what speed will the block acquire? From Example 2, we know that the potential energy stored in a spring compressed
by a distance d is = lkd2
2
where k is the spring constant. The law of conservation of energy tells us that the total potential and kinetic energy of the system, which here consists of the block and spring,
remains constant: E=K+U=§kx2+émv2. Let’s apply this to the following two points: point A, where the spring is compressed an
amount (1 and the block has 1) = O; and point B, where the spring has no potential energy
and the block has speed 2/; E A = EB implies that ékd2 + O = O + émvz,
from which we can solve for the speed 1): v = (k/m)“2d = [(1.5 X 103 N/m)/(2.0 kg)]“2(0.08 m) = 2.2 m/s. 13.3 THE LAW OF CONSERVATION OF ENERGY 257 Example 4
A pendulum bob is pulled aside from the vertical through an angle 9 and released. Find
the speed of the bob and the tension in the string at the lowest point of the swing, assuming L = 0.3 m, 6 = 30°, andm = 0.5 kg. Since we anticipate using the law of conservation of energy, let’s measure all vertical
heights (and therefore potential energy) from the lowest point of the swing, point B. Now
the total energy of the bob at point A, just before being released, is EA = mgz, where z is the vertical height above B. Using geometry, we see that z = L — L cos 0
= L(l — cos 6), and so the initial energy is EA 2 mgL(l — cos 9).
The total energy at B is purely kinetic energy because we have chosen 2 = 0 there, so EB = lmvz. Therefore the law of conservation of ener , E A = EB, im lies
2 B gy p mgL(1 — cos 0) = émvﬁ. Solving for the speed v3 we find
vB = \/2gL(1 — cos 6). Substituting the numbers, this turns out to be 0.9 m/s.
To find the tension in the string — a force — we need to use Newton’s second law. Applying the second law with the freebody diagram for the bob at its lowest point, we
see that a combination of weight and tension causes the bob to move in a circle of radius L.
T _ 2
T a  uB/L
mg Therefore, we have T — mg = mvﬁ/L, 258 THE LAW OF CONSERVATION OF ENERGY which implies
T = mg + mvﬁ/L. Substituting for v3 from our law of conservation of energy, we get
T = mg + 2mg(l — cos 0) = mg(3 — 2cos 6). Numerically the tension turns out to be 6.2 N when 0 = 0°.
A combination of the law of conservation of energy and Newton’s laws provides a powerful method for attacking a variety of problems in classical physics. Example 5
A toy dart gun consists of a spring that when compressed 0.05 m can project a 20g rubber dart vertically upward to a height of 3.0 In. What is the spring constant? Here we have a dart and a spring, each of which can have energy. Initially the system
(dart and spring) has potential energy stored in the spring. At its highest point, the dart
possesses only gravitational potential energy and the spring, no longer compressed, has
no potential energy stored in it. The law of conservation of energy implies that at these
two points the energy must be the same. Let’s formulate mathematically what we’ve just
stated verbally. Initially the energy of the system is EA = Us + Ud + Kd, being the
sum of the potential of the spring (Us 2 ékxz), the gravitational potential energy of the
dart (in general U = mgz, but initially z = 0), and the kinetic energy of the dart (initially
Kd = 0). At the highest point, we have EB = US + Ud + Kd = 0 + mgh + 0. EA
= EB implies lat2 = mgh. %
Solving for k, we find
k = 2mgh/x2 = 2(0.02 kg)(9.8 rn/sz)(3.0 m)/(0.05 m)2
= 4.7 X 102 N/m. Questions 9. Suppose you were redesigning a roller coaster ride so that for added thrill, the
roller coaster will be moving twice as fast at the bottom of the hill. How many
times higher would you need to make that hill to achieve this? 10. A truck at rest on the top of a hill is allowed to roll down, and at the bottom it
has a speed of 5 km/h. If it is allowed to roll down the hill again, but this time
starting with an initial speed of 3 km/h what will be its speed at the bottom of the
hill? 13.3 THE LAW OF CONSERVATION OF ENERGY 259 11. Three identical blocks slide down the frictionless surfaces shown below after being
released from the same height A: 12. 13. (a) How do the speeds of the blocks at point B compare?
(b) Which block do you think reaches point B first? Why? A mass attached to a vertical spring is gently lowered to its equilibrium position, at which the spring is stretched a distance d. If the same object is attached to the same vertical spring but allowed to fall instead, what maximum distance does this
stretch the spring? A block of ice slides down the frictionless incline shown below and compresses
the spring. Taking the mass of the block to be 1.5 kg, the spring constant to be
3.0 X 102 N/m, and the distance the block slides down the incline before striking
the spring as 1.2 In, find the maximum distance the spring is compressed. 14. A spring with spring constant 3.0 X 103 N/m and length 0.10 cm propels a small
(0.5kg) block up a frictionless plane inclined at 37°. If the spring were initially
compressed 0.6 cm, how far up the plane from the point of release would the
block travel before coming momentarily to rest? When the block returns and hits
the spring, how much will it compress the spring? 15. A pendulum bob of mass m hanging at the end of a string of length L is struck so that it has speed v. What must v be in order for the string to go slack (tension
becomes zero) at the top of the swing (180° from where the bob began)? 260 THE LAW OF CONSERVATION OF ENERGY 13.4 HEAT AND ENERGY Early attempts to formulate the idea of energy conservation ran into a rigid prejudice:
Newton’s laws embodied all truth about nature; if a law could not be based on Newton’s
mechanics, it was not physics but mysticism. Nevertheless, a dozen scientists in the first
half of the nineteenth century proposed in some form that energy is conserved. Two of
those were James Prescott Joule, a practicalminded brewery owner working in England,
and Herman von Helmholtz, a German romantic. Through their work, the law of con
servation of energy became scientifically respectable. But to achieve this, they had to
confront the apparent nonconservation of energy in the world. There are numerous phenomena for which energy — at least in the form of kinetic
and potential energy — appears not to be conserved. A box sliding across the ﬂoor
eventually comes to rest; a rubber ball dropped from some height bounces less and less
until it too eventually comes to rest. However, if we could look at a sliding box or
bouncing ball closely enough to see atoms and molecules, we would better understand
what is happening. For example, as a box slides along the ﬂoor, atoms and molecules
of the box and floor interact and are pulled from their equilibrium positions. However,
the atoms aren’t free to move very far; the electric forces pull them back to equilibrium
position. When they spring back, they hit the adjacent atoms and set them moving, and
so on. This motion goes on inside the box and inside the table. Energy is not really lost;
it’s converted into kinetic and potential energy of atoms and molecules. Thermal energy is the name we give to energy in the form of hidden motion of atofns
and molecules. Technically, heat is transferred between a system and its environment as
a result of temperature differences. But the term heat is often used generically to encom
pass thermal energy as well. There are many processes in nature that turn kinetic energy,
the organized energy of the motion of an entire large body, into thermal energy, the
unorganized motion of atoms. Friction is one example; viscosity is another. In an au
tomobile engine or a steam engine, the opposite process occurs: heat is turned into work. If it weren’t for the fact that energy can turn into heat, it would have been easier to
discover the law of conservation of energy. But in the nineteenth century, our idea of
atoms and molecules in constant motion had not yet developed. Instead, before the
discovery of the law of conservation of energy, there was a different theory. Heat was
thought to be a kind of ﬂuid, called caloric. In this theory, heat itself was a conserved
quantity. The caloric theory was not idle speculation but a detailed mathematical theory. A
caloric theorist would describe how an iron rod in a fire becomes hot by saying that
caloric is ﬂowing from the fire into the iron rod; he would know exactly how much heat,
or caloric, was required to bring the rod to a given temperature. If the hot rod were
immersed in water, a caloric theorist would say that some of the caloric leaks out of the
metal into the water, thereby warming up the water; by applying the law of conservation
of caloric, the precise rise in temperature of the water could be predicted. The missing
element, the barrier to discovering the law of conservation of energy, was the fact that
it is possible to change work or kinetic energy into heat and vice versa. The credit for the law of conservation of energy goes not to the first of the many
people who discovered it, but to the last because that person pinned it down so well that
it didn’t need to be rediscovered. James Prescott Joule, the son of a wealthy English
brewer, is said to have become interested in heat by a desire to develop more efficient 13.4 HEAT AND ENERGY 261 engines for the family brewery. In experiments conducted between 1837 and 1847 in the
brewery and at his own expense, Joule established the law well enough that it never had
to be discovered again. Joule made careful measurements of exactly how much work turned into heat. The
invention of the steam engine had led the way to measuring energy changes. Almost from
their beginning, steam engines were rated according to their duty, which referred to how
heavy a load an engine could lift by using a given supply of fuel. Joule used such a
practical engineering approach to determine first if electric motors could be made eco
nomically competitive with steam engines (and improve the brewery) and later to quantify
the relation between work and heat. Joule’s famous experiments involved an apparatus, as shown in Fig. 13.6, in which
slowly descending weights turned paddle wheels in a container of water. As the weights
fall and paddles turn, the water temperature rises. Through precise comparisons of the
work done by the weights and the rise in temperature of the water, Joule uncovered the
relationship between heat and work. The unit of heat is the calorie, which is the amount
of heat required to raise the temperature of 1 g of water 1°C. Joule found that this unit
of heat is related to units of energy according to 1 calorie = 4.18 J. II. M )WERY L i" 1‘ ._ / ’14 1.3 JVJLEJ .. (may! 14/491 KIT”.
m :1 .zc c 11 azazazryfgﬁm,/ 16%,1 g4
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wW/rv 4.: 11.1; I ’fv‘vj‘zy”, «~44 2 Figure 13.6 Joule’s apparatus for measuring the conversion of work
into heat. 262 THE LAW OF CONSERVATION OF ENERGY (Of course, he didn’t call them joules — that came later. A food calorie is equal to 1000
calories.) Begin a practical man and possessing a limited mathematical background, Joule was
content with experimenting in his lab. Despite his irrefutable experiments, many physicists
of the time wanted a rigorous, mathematical theory of the law of conservation of energy.
Such respectability was bestowed by the German physiologist and physicist, Herman von
Helmholtz. In a paper published in 1847, Helmholtz showed, much as we did in Section
13.3, that the law of conservation of energy follows from Newton’s laws. With his proof
in hand, Helmholtz went on to apply the law to various cases, such as gravity, electricity,
and friction. We began our discussion with the observation that in many cases it appears that
energy is not conserved; such cases involve friction, viscous force, or other means of
dissipating energy into heat. Joule and Helmholtz showed us that energy is always
conserved and that it can be transformed from one form into another. M Example 6
A ball of mass 0.5 kg is thrown vertically upward with an initial velocity of 20 m/s and reaches a height of 15 In. How much energy is dissipated by air resistance?
The ball begins with kinetic energy émv2 and winds up with potential energy mgh.
The difference between these is the energy dissipated, émv2 — mgh = %(0.5 kg)(20 m/s)2 —— (0.5 kg)(9.8 m/s2)(15 m) = 26 J. Example 7 A spring with spring constant 5.0 X 102 N/m, initially compressed 0.07 m, fires a 0.4
kg block across a rough horizontal surface. If the coefficient of friction between the block
and surface is 0.5, how far from its starting point will the block travel before coming to
rest? M1“ We can think of the block as doing work against the force of friction. This work is
then dissipated into heat. The work done against friction as the block moves through a
displacement x is, by definition [Eq. (13.2)], WAB = f~x = pkmgx. Initially the block has only potential energy ékd2 from the spring. Substituting into our
energy equation, we have %ka’2 = ukmgx, 13.5 A FINAL WORD 263 so we can solve for x and obtain
)6 = kd2/(2Mkmg), which, when values are inserted, gives 0.6 m. Questions 16. How much work does a 60kg athlete do in one pushup, which raises then lowers
the body by 0.4 m? How many calories does this represent? 17. Suppose that in Joule’s experiment a lkg mass falls through a distance of 0.5 m.
By how many degrees would this work raise the temperature of l g of water? 18. Starting at a height of 25 m, a sled of mass 20 kg slides down a hill. If the sled
starts from rest and has a speed of 15 m/s at the bottom of the hill, calculate the energy dissipated by friction along its path. 19. A halfton pickup truck moving at 24 m/s runs into a giant haystack and pene—
trates 3 m into it. A similar halfton truck with a halfton load of grain runs into the same haystack. How far will this truck travel into the haystack? 20. A 25g bullet fired at 400 m/s penetrates 10 cm into a block of wood, which re
mains stationary. (a) What is the energy dissipated by friction between the block and bullet?
(b) What is the average force exerted on the block by the bullet? 21. A driver traveling at 30 km/h slams on the brakes and skids 10 m before stopping.
If the car were moving at 90 km/h, how far would it skid? 22. A 0.4—kg block is given a speed of 1.5 m/s as it is projected up a plane inclined at
30°. If the coefficient of friction between the block and plane is 0.6, how far up
the plane will the block slide? 13.5 A FINAL WORD Once you understand something, it’s almost impossible to put yourself in the position of
not understanding it and trying to figure out how people thought about the problem before.
Up to the time when Joule did his experiments, presumably heat itself was the conserved
quantity and couldn’t be created out of work. But how could people for thousands of
years before Joule not realize that by rubbing their hands together they could warm them?
Even more to the point, by the 18305, long before Joule’s experiments, railroads were
strung across Europe. The burning of coal in locomotives eventually set the train in
motion — converting heat into work. How could people living at this time, riding on
railroads, not have believed that it was possible to turn heat into work? It was not through lack of trying to understand, as we can see from the thoughts of
a young French military engineer, Nicolas Leonard Sadi Carnot. Although Carnot died
at the age of 32 from scarlet fever, he is one of the important figures of nineteenth 264 THE LAW OF CONSERVATION OF ENERGY century science. Carnot discovered what we now call the second law of thermodynamics,
which is perhaps the most profound law in physics. He succeeded in doing this without
knowing the law of conservation of energy, which is the first law of thermodynamics.
Yet it may have been easier to discover the second law of thermodynamics without
knowing the first law. Carnot reasoned by analogy. His idea of how heat worked was by analogy to a water
wheel. As water runs down and over a water wheel, it makes the wheel turn, yet the
water is conserved. That is, to get work out of the water doesn’t require using up water.
Instead water falling from a large height to a lower height turns the water wheel. Carnot
thought that heat worked in exactly the same way. He thought that caloric, starting out
at high temperature, was capable of doing work on the way down to low temperature,
analogously to water ﬂowing over the water wheel. His analogy also suggested the idea
that came to be the second law of thermodynamics: once heat runs downhill from high
to low temperature, it will not run uphill again, back up to the high temperature. That
seemed obvious from the water wheel analogy. A piece of coal has potential energy stored in it in the form of chemical bonds. The
process of combustion, say, in a steam locomotive, turns that energy directly into heat
at high temperature. Some of that heat is able to drive a piston, producing work and
setting the locomotive in motion. Eventually all that heat, including the part turned into
work, winds up as heat once again, after the locomotive goes through its frictional
processes and so on. But the heat is no longer at the high temperature of the burning
coal, but rather at the low temperature of the air outside. The net resut of all this, aside
from getting you from one place to another, has been to turn potential energy into high
temperature heat and finally into lowtemperature heat. And once that’s done, it is no
longer possible to get that heat back into the form of the original potential energy so it
can be used again. That’s what the second law of thermodynamics says: sometimes things
happen that can never be undone, no matter what. Although Carnot could not know that, unlike water, some of the heat actually is
transformed into work before changing back into heat again, that did not prevent him
from discovering the second law of thermodynamics. Energy from the sun, at very high temperature, is stored temporarily in coal, oil,
and other fossil fuels. We can release it as heat at high temperature (but lower than the
temperature of the sun) and, whether we use it for useful work or not, it always winds
up as low temperature heat (heat at ambient temperature). All of that energy is perfectly
conserved, not a bit is ever lost. But the value of the energy has been lost — it has become
useless. The world has run down a little bit, and it never will be wound up again. This is the real nature of our energy crisis. We are not using up energy, we are just
transforming it into useless forms. 13.5 A FINAL WORD 265 1
i Figure 13.7 James Prescott Joule. (Courtesy of the Manchester
Literary and Philosophical Society.) ...
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 Fall '07
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