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PHYSICS
The
Standard
Deviants®
Core
Curriculum
I n s t r u c t o r ’s Guide
Films for the
Humanities & Sciences
®
Introduction
The Core Curriculum Series is designed to be a complete treatment of major topics covered in
an introductory course on a given curricular subject. The series utilizes a unique educational
approach that combines the best of visual and tutorial elements. Through high-end graphics,
animation, and design techniques, academic concepts are broken down, thoroughly explained,
and demonstrated through the creative use of presentational devices and examples.
Each individual subject series consists of up to 10 programs of approximately 20 minutes each.
These programs can be integrated into the lesson plans for an entire course or as stand-alone
supplements for specific topics, as needed.
Each program is accompanied by an instructor’s guide that contains the following elements:
•
Program Outline
•
Key Terms and Formulas
•
Quiz
•
Solutions to Quiz
•
Suggestions for Instructors
The Program Outline lists the main topics covered in each program, and also relates each program to the overall subject series. For example, if you choose to view Program 3, you’ll also
see how Program 3 fits into the overall 10-part series.
Note: Occasionally during the show, you may notice a small heading in the corner of the screen
that makes reference to Part and Section divisions in the program. These headings refer to the
program in its original, extended format and do not correspond to the format of this series.
Copyright © 2001 Films for the Humanities & Sciences® A Films Media Group Company • P.O. Box 2053 • Princeton, NJ 08543-2053
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Program Outline
PROGRAM 1: NUMBERS, UNITS, SCALARS, AND VECTORS . . . . . . . . . . . . . . . . . . . page 6
Introduction to Physics
• Using Scientific Notation
Multiplying Exponents
Converting Numbers into Scientific Notation
• Using Numbers and Units in Physics
Basic Units
Conversion and Cancellation of Units
Scalars and Vectors
• What are Scalars and Vectors?
Magnitude
Direction
• Adding Vectors
Head-to-Tail Method
Component Method
• Multiplying Vectors
Dot or Scalar Product
Cross or Vector Product
PROGRAM 2: ONE-DIMENSIONAL KINEMATICS . . . . . . . . . . . . . . . . . . . . . . . . . . page 11
One-Dimensional Kinematics
• Basic One-Dimensional Quantities
Displacement
Velocity
Acceleration
• One-Dimensional Kinematic Equations
Instantaneous Velocity
Displacement
Final Velocity
PROGRAM 3: PROJECTILE MOTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 15
Two-Dimensional Kinematics
• Projectile Motion
First Projectile Motion Equation
Second Projectile Motion Equation
Third Projectile Motion Equation
Fourth Projectile Motion Equation
PROGRAM 4: CIRCULAR AND ROTATIONAL MOTION . . . . . . . . . . . . . . . . . . . . . . page 21
• Uniform Circular Motion
Centripetal Force
Centripetal Acceleration
Period
• Accelerated Circular Motion
Tangential Acceleration
Total Acceleration
3
• Rotational Motion
Rotating Through an Angle
Radians
Angular Velocity
Linear Speed
PROGRAM 5: LINEAR MOMENTUM AND NEWTON’S LAWS OF MOTION . . . . . . . page 26
Linear Momentum and Newton’s Laws of Motion
• Linear Momentum
The Principle of Conservation of Linear Momentum
Collisions
Final Momentum = Initial Momentum
• Newton’s First Law of Motion
Object in Motion Stays in Motion
Object at Rest Stays at Rest
Law of Inertia
• Newton’s Second Law of Motion
Force
F = ma
Weight
Equilibrium
Tension
• Newton’s Third Law of Motion
Equal But Opposite Force
Momentum and Newton’s Third Law
PROGRAM 6: FRICTION, WORK, AND ENERGY . . . . . . . . . . . . . . . . . . . . . . . . . . page 32
Friction
• What is Friction?
• Kinetic and Static Friction
Kinetic Friction Acts Opposite Motion
Normal Force
Static Friction Tries to Prevent Motion
Coefficient of Friction
Work and Energy
• Work
• Power
• Kinetic Energy
PROGRAM 7: GRAVITATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 38
Gravitation
• The Law of Universal Gravitation
• Kepler’s Laws of Planetary Motion
PROGRAM 8: HARMONIC MOTION AND WAVES . . . . . . . . . . . . . . . . . . . . . . . . . . page 43
Harmonic Motion
• Simple Harmonic Motion
• Behavior of Systems Undergoing Simple Harmonic Motion
• The Simple Pendulum
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Waves
• Wave Motion
• Wave Speed
• Types of Waves
PROGRAM 9: HEAT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 48
Heat
• Heat Transfer
• Changes of State
• Conduction, Convection, and Radiation
• Heat and Work
PROGRAM 10: THERMODYNAMICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . page 54
Thermodynamics
• The First Law of Thermodynamics
• The Second Law of Thermodynamics
• Heat Engines
• Entropy
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PROGRAM 1:
NUMBERS, UNITS, SCALARS, AND VECTORS
Key Terms and Formulas
Physics is the study of how things work.
Scientific notation is a way of writing really big or really small numbers so they’re easier to
work with by using exponents to express the number.
Kinematics is the study of motion.
Magnitude is the quantity of something, or how much of it there is.
Direction is the way something is going, like north, south, east, or west, and is often
described in degrees.
A scalar is a quantity that only has magnitude.
A vector is a quantity that has magnitude AND direction.
The sum of two vectors is called the resultant vector.
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Quiz
1. The density of water is 103 kg/m3 (or 1 gram per cubic centimeter). What is the mass of
1 liter of water? [A liter is equal to 10-3 m3.]
2. The mass of an atom is practically all in its nucleus. The radius of a uranium238 nucleus is
8.68 x 10-15 m. Since the atomic mass of 238U is 238 atomic mass units (amu) and 1 amu is
equal to a mass of 1.6605 x 10-27 kg, obtain the density of “nuclear matter.”
3. Can the difference of two vectors have a greater magnitude than the sum of the same two
vectors? Give an example that shows your answer.
4. Two vectors, 6 and 9 units long, form an angle (a) 0°‚ and (b) 60°. Find the magnitude and
direction of their vector sum.
5. Two vectors, 6 and 9 units long, form an angle (a) 0°, and (b) 60°. Find the magnitude of
their dot or scalar product.
6. Two vectors, 6 and 9 units long, form an angle (a) 0°, and (b) 60°. Find the magnitude of
their cross product. If the shorter vector lies along the +X-axis and the other vector is in the
XY-plane, determine the direction of the cross product.
7. Find the x and y components of a vector 15 units long when it forms an angle, with respect
to the +X-axis, of (a) 50°, (b) 130°, and (c) 230°.
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Solutions to Quiz
1. Since density is mass per unit volume, or ρ = m/V, then m = ρV. Substitution yields:
m = [103 kg/m3][1 liter][10-3 m3/liter] = 1kg.
2. Since density is mass per unit volume, or m/V, we need to find the mass and the volume of
the 238U nucleus. The mass is found by multiplying the number of nucleons by the atomic
mass unit, or:
m = 238 amu{1.6605 x 10-27 kg/amu] = 3.952 x 10-25 kg
Assuming we can think of a nucleus as a sphere, we find the volume from
V = (4/3)πr3 = (4/3) π[8.68 x 10-15m]3 = 2.739 x 10-42 m3
Dividing the mass by the volume, the result is
r = [3.952 x 10-25 kg] / [2.739 x 10-42 m3] = 1.443 x 1017 kg/m3.
3. Yes! The trick here is that we must remember that vectors have direction as well as magnitude.
For example, if two vectors have exactly opposite direction and have identical magnitude, their
sum is exactly zero, while their diff e rence is twice that of either one. More involved examples
can be found when the two vectors make an angle other than 180° with respect to each other
and/or do not have the same magnitude.
4. (a) This is easy. Think of the 6 unit vector lying across the +X-axis. Then the 9 unit vector
also is along the +X-axis and, using head to tail, we see the resultant vector lies along the
+X-axis and has a magnitude 15.
(b) Again think of the 6 unit vector lying across the +X-axis. Then draw the 9 unit vector
from the head of the 6 unit vector up an angle of 60°. The magnitude of their resultant is
less than 15 units and by measuring we could find that length. The angle the resultant
makes with the +X-axis could also be measured with a protractor. To do this problem mathematically, we can break the 9 unit vector into components as:
9x = 9cos60° = 9[0.500] = 4.500 units
9y = 9sin60° = 9[0.866] = 7.794 units
This then gives Rx = 6 + 4.5 = 10.5 units and Ry = 7.794 units. The square of the magnitude
of the resultant is then
R2 = (Rx)2 + R(y)2 = [10.5]2 + [7.794]2 = 170.996
Giving a magnitude of
R = 13.08 units
To get the angle, the tangent of the angle is given by tanθ= Ry/R = 7.794/10.5 = 0.7422
which gives us an angle of θ = 36.6°. We could also get the angle from cosθ = Rx/R =
10.5/13.08 = 0.8028, which also gives the angle as 36.6°.
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5. (a) First, we write the basic equation for the product: |A • B|= ABcosθ. For part (a) since
cos0° = 1.00, the dot product is simply 54.
(b) Since cos60° = 0.500, the scalar or dot product has a magnitude (6)(9)(0.500) = 27.
6. (a) First, we write the basic equation for the cross product: |A x B| = ABsinθ. For part (a)
since sin0° = 0, the magnitude in this case is zero and there is no vector.
(b) Since sin60° = 0.866, we have a magnitude of (6)(9)(0.866) = 46.76. To find the direction,
if we are "turning" the 6 unit vector "into" the 9 unit vector, the right hand rule has the
thumb pointing UP, which usually means in the +Z direction.
7. The X component of a vector is given by Ax = Acosθ and the Y component is given by Ay
= Asinθ. Knowing this makes this problem quite easy.
(a) Ax = 15cos50° = 15(0.6428) = 9.64 units; Ay = 15sin50° = 15(0.7660) = 11.49 units.
(b) Ax = 15cos130° = 15(-0.6428) = -9.64 units; Ay = 15sin130° = 15(0.7660) = 11.49 units.
(c) Ax = 15cos230° = 15(-0.6428) = -9.64 units; Ay = 15sin230° = 15(-0.7660) = -11.49 units.
These angles were chosen to show you that it is VERY important to consider the angle
of a vector.
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Suggestions for Instructors
1. There is considerably more to physics than collections of facts, although students sometimes
feel that is what gets them through a course. Indeed, there are many facts that are necessary
to be known in order to successfully negotiate physics, even at the most basic level. For
instance, a knowledge of the relative size of "things" is rather important. For this reason, the
first topic of consideration is units and numbers because, if you don’t measure objects, you
are not "doing" physics. The first two problems given are to look at the ordinary world and
then compare it with the subatomic world to show just how very different they are. Other
similar problems should be assigned: the size of the Sun [e.g., "How many Earths could fit
in the volume of the Sun?"], the distance between galaxies [e.g., "if it takes light eight minutes to travel from the Sun to the Earth, how large is our Milky Way galaxy across?"], or the
enormous number of atoms in even a small body [e.g., "Calculate the number of molecules
in one cubic centimeter of water."].
2. By the end of any reasonable course in physics, a student should be aware that there are
two levels of description of nature. One is macro and thus global and phenomenological,
corresponding to the world we perceive directly (sometimes with instruments). The other is
micro, and is structural; it is the domain of atoms and quantum theory. It is not necessary
to delve deeply into the micro level; however, the student should be aware that it does exist
and that "rules" that appear to be different from those followed in the macro world are
sometimes developed. Also it is important for the student to recognize that there are two
kinds of physical laws. One kind is fundamental, such as the conservation principles or the
laws of gravitation. The other kind is statistical, corresponding to the laws of friction, the gas
laws, Ohm’s law, etc.
3. The need to distinguish quantities that are purely numeric, i.e., scalars, from those physical
quantities that require direction as well, i.e., vectors, may be developed by some rather simplistic demonstrations using tools such as a pair of rulers, especially if they are of different
lengths. "Adding" two lengths can easily be shown to have numerous values as one ruler is
placed at the end of another and then rotated. Having a third ruler then extended from the
"tail" of the first to the "head" of the second, magnitude and angle change can be seen. It’s
even possible to "prove" the Pythagorean theorem this way (as Pythagorus did himself).
4. Introduction of the scalar and vector products is usually difficult for students with little threedimensional play in their previous experience. Encourage your students to look at some of
the geometrical relations that are reflected in these products. For instance, the vector product
has a numeric value equal to the area of the parallelogram created when the two vectors are
placed "tail to tail" to form two of the sides, while the remaining sides are a parallel pair. In
the case of the scalar product, the result is equal to the product of either one by the "vertical
shadow" of the other.
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PROGRAM 2:
ONE-DIMENSIONAL KINEMATICS
Key Terms and Formulas
One-dimensional kinematics focuses on motion along a straight line.
Displacement, or Δx, is the change in position of an object.
Average velocity, represented by vave, is the change in position, or displacement (Δx) of an
object divided by the change in time (Δt).
Average acceleration, represented by aave, is the change in velocity (Δv) over the change in
time (Δt).
Instantaneous velocity is simply the velocity of an object at a given instant in time.
Instantaneous acceleration is the acceleration of an object at a given instant in time.
The kinematic equations can only be used when we have an object traveling in a straight
line at a constant acceleration:
vf = vi + at
x = vit + 1/2 at2
vf2 = vi2 + 2ax
where vf is final velocity, vi is initial velocity, a is acceleration, x is position, and t is time.
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Quiz
1. Describe the procedure to measure the average velocity of a body in one-dimensional
motion.
2. What is the difference between average velocity and instantaneous velocity?
3. A body, at t = 0 seconds, has a velocity of 3 m/s and a constant acceleration of 4 m/s2 in the
same direction as the velocity. (a) What is the velocity of the body at t = 7 s? (b) What is the
distance covered after 7 seconds?
4. A ball is thrown straight up in the air with an initial velocity of 30 m/s. How high will it rise?
5. A car is moving along a street at a rate of 45 km/hr (12.5 m/s) when a red light flashes on
at an intersection 50 m ahead. If the driver’s reaction time is 0.7 s, what is the distance to
the intersection when the driver begins to apply the brakes?
6. If the driver in the car above decelerates the car uniformly at 2 m/s2 when the brakes are
applied, will the car stop safely?
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Solutions to Quiz
1. Find the position of the body at two times. Divide the change in position (i.e., final position
minus initial position) by the change in time (i.e., final time minus initial time). In symbols:
vave= [xf – xi]/[tf – ti] = Δx/Δt
2. Average velocity is defined as in answer #1 above; for instantaneous velocity we mean the
velocity at a specific time. Only in the case of uniform rectilinear (one-dimensional) motion
do the two have the same value at all times.
3. (a) a = 4 m/s2 = [vf – vi]/[tf – ti] = [vf – 3 m/s]/[7 s – 0 s]
vf – (3 m/s) = [4 m/s2][7 s] = 28 m/s
vf = 28 m/s + 3 m/s = 31 m/s.
(b) xf = xi + vit + 1/2at2 = 0 m + [3 m/s][7 s] + 1/2[4 m/s2][7 s]2
= 0 m + 21 m + 1/2[4 m/s2][49s2] = 21 m + 98 m = 119 m.
This could also be solved with the equation vf2 – vi2 = 2a(xf – xi) since we just found
the final velocity of the body. Using this, we have:
[31 m/s]2 – [3 m/s]2 = 2[4 m/s2][xf – 0 m] = 8 m/s2[xf]
961 m2/s2 – 9 m2/s2 = 952 m2/s2 = 8 m/s2[xf] or xf = [952 m2/s2]/[8 m/s2] = 119 m.
4. For this problem realize that if up is positive, then the acceleration (due to gravity) is down,
or negative. The previous equation, vf2 – vi2 = 2a(xf – xi), now becomes:
0 – vi2 = -2g(x f – 0)
And when we introduce the values of vi = 30 m/s and g = 9.8 m/s2, we have
-[30 m/s]2 = -900 m2/s2 = -2[9.8 m/s2]xf
xf = [-900 m2/s2]/[-19.6 m/s2] = 45.9 m.
5. First, distance equals velocity (uniform, because the driver hasn’t touched the brakes) times
time, or x = (12.5 m/s)(0.7 s) = 8.75 m. This is the distance covered before the driver starts
to apply the brakes. So, the distance remaining is 50 m – 8.75 m = 41.25 m.
6. Again, using vf2 – vi2 = 2a(xf – xi), we have:
0 – [12.5 m/s]2 = 2(-2 m/s2)(Δx)
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So, Δx = -[156.25 m2/s2]/[-4 m/s2] = 39.06 m. Whew! Barely 2 meters to spare!
Suggestions for Instructors
1. The most fundamental and obvious phenomenon we observe around us is motion. Blowing
air, waves in the ocean, flying birds, falling leaves—all of these are examples of motion.
Practically all imaginable processes can be traced back to the motion of certain particles or
objects. The Earth and other planets move around the Sun. The Sun, in turn, carries the
solar system around the center of our galaxy, the Milky Way. Electrons move within atoms,
giving rise to the absorption and emission of electromagnetic radiation. Electrons move within metal, producing an electric current. Gas molecules move in a random fashion, giving rise
to pressure and diffusion processes. Our everyday experience tells us that the motion of a
body is influenced by the bodies around it; that is, by its interactions with them. One role
of the physicist, and the engineer as well, is to discover the relation between motions and
the interactions responsible for them and then to arrange things so that useful motions are
produced. The rules (or principles) that apply to the analysis of all kinds of motion are
called mechanics and this is why most physics courses begin with its study. To carry out a
program in mechanics, however, we must begin by learning how to describe the motions we
observe; hence, one-dimensional kinematics (from the Greek, kinema) as our first step.
2. An object is in motion relative to another when its position, measured relative to the second
body, is changing with time [when the position doesn’t change, the object is at relative rest].
Both rest and motion are relative concepts, so it is rather necessary to define a frame of
re f e rence relative to which motion (or rest) is analyzed. Usually we choose the "fixed, flat"
Earth as that frame, and this works for many cases. However, this frame fails, miserably in
some outstanding instances, and such cases should be discussed with students. For instance,
using the Sun as the frame of reference, planetary motion becomes quite simple—which introduces such wonderful characters as Copernicus and Kepler. Also, Coriolis motion is "magic"
until students recognize that it’s simply a case of ignoring the rotation of the Earth on its axis.
3. Graphing position (displacement) versus time is a useful exercise for students. More of intere s t ,
however, is graphing velocity as a function of time for a number of reasons: A uniform velocity graphs as a straight horizontal line; a uniform acceleration results in a graph of velocity vs.
time as a straight line, whose slope is the value of the uniform acceleration; the area under the
curve between any two times is the displacement during that time interval. All the basic equations developed in the program can be "proven" from the geometry of these graphs—even the
displacement equation under uniform acceleration: xf = xi + vit + 1/2a(tf – ti)2.
4. Simple 1-D problems, where gravity is the uniform acceleration, help the student become
familiar with breaking down a problem into parts. This is a skill instructors should encourage students to develop because it can carry over to many other fields of endeavor.
5. To spark interest, get the latest data on the expanding universe from the Hubble telescope
information available at www.nasa.gov.
6. Even though the work here is in 1-D, continue to stress the vector nature of displacement,
velocity, and acceleration.
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PROGRAM 3:
PROJECTILE MOTION
Key Terms and Formulas
Two-dimensional kinematics is motion in a plane, as opposed to motion along a straight
line, like one-dimensional kinematics.
Projectile motion is when an object is tossed into the air. The object’s flight follows a
curved path.
Gravity is the force pulling down on the object (in the -y direction). The acceleration of
any object due to gravity, near the surface of the earth, is 9.8 m/s2.
The initial magnitude of the horizontal or x-component of a projectile is found using the equation
vix = vi cosθ
and the initial magnitude of the vertical or y-component of a projectile is found using the
equation viy = vi sinθ
To find the velocity in the y direction, use vy = viy – gt
and to find the horizontal velocity, use vx = vix
To find the distance that an object has traveled in the x direction, use xf = xi + vixt
To determine the displacement of an object in the y direction, use yf = yi + viyt – 1/2 gt2
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Quiz
1. In projectile motion the horizontal component of the projectile’s velocity remains constant.
Why?
2. For the ideal case of no air resistance, a projectile follows the path of a parabola. What
happens to the path when there is some air resistance?
3. Since a thrown baseball is a projectile, how is it possible to throw a curve?
4. A projectile is shot with a velocity of 70 m/s at an angle of 60° with the horizontal. Calculate
(a) the maximum height; (b) the time to reach the maximum height; (c) the total time to
again reach the ground; (d) the horizontal range.
5. A helicopter is flying at an altitude of 150 m with a horizontal velocity of 100 km/hr
(27.78 m/s). It drops a package of supplies to hit the blanket of a stranded hiker. Show
that the package should be released when the copter is a horizontal distance of 154 m
from the blanket.
6. The professor sits on a platform 15 m high. A student stands 30 m away from the base of
the platform with a ripe, red tomato. The student can toss the tomato with a speed of 50
miles per hour (22.3 m/s). Calculate the angle the tomato should be thrown if the student
REALLY wants to hit the professor.
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Solutions to Quiz
1. The "force of gravity" is straight down, so the only acceleration is straight down. If there’s
no acceleration in the horizontal direction, then there’s no acceleration [i.e., CHANGE in
velocity] in that direction. No acceleration, no change in velocity.
2. Lots of interesting things happen. Thinking of the components of the velocity, the vertical
component will slow down MORE than just due to gravity and so the projectile will not go
as high as in the ideal case; on the way down the vertical component will not speed up as
fast as it would without friction, so its value is considerably different than its original value
when it gets back to the ground; now the horizontal component slows down all during the
flight. So if you first draw the ideal parabola case, the real path will not go as high, will
peak at a shorter horizontal value and then curve down more sharply than the original, but
never vertical.
3. The only way possible is for a force to be applied to the ball in a direction perpendicular to
the plane that the ball would normally follow. A pitcher is able to have such a force be produced by spinning the ball as it is released. The aerodynamics of this is rather involved and
will take more analysis than we can spend time on here. Nevertheless, go throw curve balls!
4. Before we begin, break the velocity in the horizontal (x) and vertical (y) components:
vx = v0cos60° = (70 m/s)0.5 = 35 m/s
vy = v0sin60° = (70 m/s)0.866 = 60.62 m/s
(a) This is just like throwing a ball straight up with an initial velocity of 60.62 m/s. Gravity is
down and so we use vf2 = vi2 – 2gy or 0 = (60.62 m/s)2 – 2(9.8 m/s2)h where we have
used h instead of y because we are after the height when the projectile has no more
positive velocity in the vertical direction. Solving, we get:
h = (60.62 m/s)2/[2(9.8 m/s2)] = (3675 m2/s2)/(19.6 m/s2) = 187.5 m.
(b) To find the time, let’s use the equation vf = vi – gt or 0 = 60.62 m/s – (9.8 m/s2)t.
This gives t = [60.62 m/s]/(9.8 m/s2) = 6.18 s.
(c) If it takes 6.18 seconds to get up to the highest point, it will take the same amount
of time to go back down. The total time then is simply T = 2t = 2(6.18 s) = 12.36 s.
(d) This is really easy now because we know the horizontal velocity, 35 m/s, and the total
time the projectile is in the air, and so: Range = vxT = (35 m/s)(12.36 s) = 432.6 m.
17
5. If we knew how long it would take the package to fall, we could calculate the horizontal
distance it would travel in that time. That is, D = vxt. We know vx equals 27.78 m/s and
from the "answer" given, we see the time needed is D/vx = (154 m)/(27.78 m/s) = 5.54 s. So
we’ve turned the problem into "proving" that it takes 5.54 seconds to fall the 150 m and that
makes us realize the package WILL fall 150 m, starting from a zero vertical velocity. The
basic equation is yf = yi + vit – 1/2 gt2 where yf = 0 m (ground), yi = 150 m, vi = 0 m/s, and
g = 9.8 m/s2. This gives: 0 = 150 m + (0m/s)(t) – 1/2(9.8 m/s2)t2 or, rearranging:
t2 = [150 m]2/(9.8 m/s2) = 300 m/(9.8 m/s2) = 30.61 s2
The square root of that gives t = 5.53 s, and I guess we’ve proven it.
6. This could turn out to be a real messy problem (pun intended). We have to figure out the
angle the student should throw the tomato, so we’ll just call it θ for now. Then we have our
two equations: vx = v0cosθ and vy = v0sinθ where v0 = 22.3 m/s. Since we know the horizontal distance to the target (har har!) equals 30 m, using the horizontal velocity we can find
the time to target, namely: D = vxt, or
t = (30 m)/([22.3 m/s]cosθ) = (1.345/cosθ) s
Using the height equation, yf = yi + vit – 1/2 gt2 where yf = 15 m, yi = 0 m, vi is v0sinθ,
and t is given in the above equation, we can write:
15 m = 0 m + ([22.3 m/s]sinθ){(1.345/cosθ)s} – 1/2 (9.8 m/s2){(1.345/cosθ)s}2.
Now this IS a mess! Let’s see how to make it look better. Each term has the unit of meters,
so we can drop having to write units. Then, doing all multiplications in each term and multiplying by cos2θ, the above equation reads:
15cos2θ = 30sinθcosθ – 8.8680.
Such an equation is called transcendental, which means "we haven’t a clue how to solve it
in a straightforward manner." These equations appear in real life more often than the mathematicians and other scientists would lead you to believe, but having calculators and computers available allows us to get very close to a solution rather quickly. Here’s one way: First,
just for ease in calculation, rewrite the equation as
15 cos2θ – 30sinθcosθ – 8.868 = 0
Then we guess a value for θ [some people would say "approximate"] and see how close to
zero you get when you evaluate. For instance, guess θ = 30°. Substitution gives
15(cos30°)2 – 30sin30°cos30° + 8.868 = 15(0.866)2 – 30(0.5)(0.866) + 8.868
= 11.25 – 12.990 + 8.868 = 7.128
18
And that’s not too good a guess. Now try 40°:
15(cos40°)2 – 30sin40°cos40° + 8.868 = 15(0.766)2 – 30(0.643)(0.766) + 8.868
= 8.802 – 14.772 + 8.868 = 2.898
This is closer to zero, so let’s try 60°:
15(cos60°)2 – 30sin60°cos60°+ 8.868 = 15(0.5)2 – 30(0.866)(0.5) + 8.868
= 3.75 – 12.990 + 8.868 = -0.372
And now we’re pretty close. Notice also that the result has changed sign from a positive
result to a negative one; that means the solution is between 40° and 60° and a lot closer
to 60°. When we try 50°, we get 15(0.643)2 – 30(0.766)(0.643) + 8.868 = +0.294. So you
have the idea now. The correct angle is between 50° and 60°. By this iterative technique,
it is possible to get the angle down to within a tenth of a degree on only a few more tries.
19
Suggestions for Instructors
1. Projectile motion is a beginning step toward general curvilinear motion and starts the student
toward visualizing in a 3-D world. The reason projectile motion appears is that it allows
discussion of acceleration beyond the 1-D case; whenever the acceleration (recall it’s a vector) is constant, the resultant motion MUST be in a single plane. Not only that, the trajectory
of the motion is a parabola.
2. Many texts produce a series of rather involved equations for projectile motion, such as the
one for the range of the projectile. Unfortunately, this type of equation results in the student
considering physics as a collection of specialized "formulas" that must be memorized for success in the course. Try to emphasize that the equations developed are all based on the first
few they learned in the definitions for velocity and acceleration.
3. In today’s world of long-range missiles, space shuttles, and satellites, help the student realize
the basic projectile equations only "work" for a flat Earth where the acceleration of gravity is
constant, both in magnitude and direction. Ask your students what happens when you
throw a rock "really hard" horizontally from the top of a building—so hard the Earth begins
to "fall away" as the rock also falls. This was a concept Galileo had begun to work on and
Newton was very good at.
4. The last problem (#6) was chosen because it is not the standard kind you find in textbooks.
Because our students now have truly powerful tools at their disposal, they should be given
the opportunity to use them. After all, when a spaceship was sent to the Moon, the dynamics of the situation required a solution of the three-body problem, for which there is no satisfactory analytic solution. Let’s have students work in the real world.
20
PROGRAM 4:
CIRCULAR AND ROTATIONAL MOTION
Key Terms and Formulas
Uniform circular motion occurs when an object moves around a circle, with a radius, r,
at a constant velocity, v—like when you swing your key chain around your finger.
The force that allows an object to follow this curved path is called the centripetal force, Fc.
Centripetal force is directed towards the center of the circle.
The object’s acceleration is directed towards the center of the circle and is called the
centripetal acceleration, ac. It can be found using the equation
2
ac = v
r
Period, T, is the time it takes for the object to complete one orbit.
The velocity of an object traveling in uniform circular motion can be found using the equation
v = 2πr
T
Accelerated circular motion happens when an object moving around a circle changes the
magnitude of its velocity. This type of acceleration is called tangential acceleration, at.
Tangential acceleration is always in the direction that the object is moving and is always tangent to the circle. The total acceleration of the object is the vector sum of the tangential
acceleration and the centripetal acceleration. Its magnitude can be found using the equation
a2 = at2 + ac2
The term rotational motion describes objects that rotate about an axis, like a top.
The angular velocity of a rotating object, represented by the Greek letter little omega, ω,
is the angle, θ, through which it turns, over the time, t, it takes the object to turn. It can be
found using the equation
ω=θ
t
The axis of rotation is the axis about which all parts of the rotating body are moving.
21
Quiz
1. Why MUST the acceleration point toward the concave side of circular motion?
2. Why is the acceleration in uniform circular motion called "centripetal"?
3. Is it possible for a body to have centripetal acceleration but no tangential acceleration?
Is it possible to have tangential acceleration but no centripetal acceleration?
4. There is a relation in motion in a circle between angular velocity, ω, linear velocity, v, and
the radial distance, r, of the point with velocity v from the axis of rotation. But velocity and
displacement are vectors. Can you figure out a vector equation that could relate these quantities based on the equation ω = v/r ? [Hint: you didn’t learn any vector equation for dividing
but you did for multiplying.]
5. A large flywheel, whose diameter is 3 meters, is rotating at 120 rpm. Calculate: (a) the period
of rotation; (b) the angular velocity; (c) the linear velocity of a point on the rim;
(d) the centripetal acceleration of a point on the rim.
6. Find the magnitude of the velocity and the centripetal acceleration of the Earth around the
Sun. The radius of Earth’s orbit about the Sun is 1.49 x 1011 m and the period of its motion
is 3.16 x 107 s [i.e., 365.25 days].
7. The angular velocity of a flywheel increases uniformly from 20 rad/s to 30 rad/s in 5 s.
Calculate: (a) the tangential acceleration of a point 1.5 m from the axis of rotation; (b) the
centripetal acceleration at the start and at the end of the 5 seconds.
22
Solutions to Quiz
1. The acceleration must always point "in" because that is the DIRECTION in which the velocity
is changing. When the motion is uniform circular motion this is the easiest case to see; but
even when the motion is not, a few simple drawings of the vectors for the two velocities
pretty close together show you that final velocity vector MINUS first velocity vector always
gives a vector that points in.
2. The answer here is obvious for anyone who knows some Greek. For those of us who are
not up on our Greek, centri- clearly means center and the petal part is associated with the
concept of "seeking".
3. Yes, for the first part. Uniform circular motion is the case where this is possible. As long as
the speed of the body is unchanged, there is no tangential acceleration. For the second question, the answer is yes but only for the trivial case when the body moves in a straight line;
that is, one-dimensional motion.
4. This is kind of difficult, except that we can make up a direction for ω and so turn it into
a vector ω. Think of a circle or wheel rotating about an axis through its center. Now grab—
not really—the wheel with your right hand pointing the direction of rotation and point out
your thumb. That’s the direction we’ll give to vector ω. Now we have to take our relation
ω = v/r or rω = v and write it as a vector product equation. To get the magnitude correct,
r and ω must be perpendicular to each other. Great! The vector r is the one that goes from
the axis of rotation in the plane of the wheel out to the point on the rim, and the vector ω
points along the axis. All we need now is to make sure the direction of v comes out right.
And that can be done by making the equation v = ω x r. Not the other way around, because
then the velocity vector would point OPPOSITE to the direction of rotation [look at those
fingers on your right hand when they wrapped around the wheel].
5. Rather than talking through the solution of numerical problems, they will just be written out
as we would expect a student to do them. For instance:
120 rpm = [120rev/min]/[60sec/min] = 2 rev/sec = frequency
(a) But Period = 1/frequency, or Period = 1/[2 rev/sec] = 1/2 sec (per revolution)
(b) 120 rpm = [(120 rev/min)(2πrad/rev)]/[60 sec/min] = 4πrad/s = ω
(c) v = ωr = [4πrad/s][1.5m] = 6π m/s (= 18.85 m/s)
(d) ac = v2/r = [36π2 m2/s2]/[1.5 m] = 24π2 m/s2 (= 236.9 m/s2)
23
6. v = 2πR/T = 2π(1.49 x 1011 m)/(3.16 x 107 s) = 2.96 x 104 m/s
(= 29.6 km/s = 1.07 x 105 km/h = 66,300 mph)
ac = v2/R = [2.96 x 104 m/s]2/[1.49 x 1011 m] = 5.89 x 10-3 m/s2
7. ωf = 30 rad/s; ωi = 20 rad/s; r = 1.5 m; then:
vf = ωf r = [30 rad/s][1.5 m] = 45 m/s; vi = ωi r = [20 rad/s][1.5 m] = 30 m/s
(a) aT = Δv/Δt = [45 m/s – 30 m/s]/5 s = [15 m/s]/5 s = 3 m/s2
(b) ac(initial) = vi2/R = [30 m/s]2/1.5 m = 600 m/s2 (= 61g)
ac(final) = vf2/R = [45 m/s]2/1.5 m = 1350 m/s2 (= 140g)
24
Suggestions for Instructors
1. The program material may be extended to include angular acceleration, α, where α =
Δω /Δt just as linear acceleration, a, is defined as a = Δv/Δt. A comparable set of equations
may then be written with θ equivalent to x, ω equivalent to v, and α equivalent to a. For
instance, the equation ωf2 = ωi2 + 2αθ is correct, as is θf = θi + ωi t + 1/2 α t2.
2. So many "accelerations" may be confusing to the student. There is linear acceleration, a;
but in curvilinear motion we may have centripetal acceleration, ac, and tangential acceleration, aT, which are perpendicular to each other [therefore, ac is sometimes called "normal"
acceleration, aN—in circular motion it is easier to see this is true]. Most importantly, don’t
breathe "centrifugal acceleration"!
3. Rigid body rotation can be quite involved. Try to keep all discussion with rotation about
a principal axis. If students—or the text—extend you toward moment of inertia, be sure to
include a discussion of torque—τ = r x F and then τ = I α when you start dynamics (see
Program 5).
25
PROGRAM 5:
LINEAR MOMENTUM AND NEWTON’S LAWS OF MOTION
Key Terms and Formulas
According to the Principle of Conservation of Momentum, the final momentum of the system equals the initial momentum of the system. When two particles collide, the momentum of
the whole system, that is—of what happens before, during, and after the collision—is the same
before and after the collision. This is called the principle of the conservation of linear momentum, and gives us the relation p = mv.
Newton’s First Law of Motion states that an object at rest will remain in its state of rest
unless acted upon by an external force, and an object moving with uniform motion will remain
in uniform motion unless acted upon by an external force.
Newton’s Second Law of Motion states that the net force acting on an object equals the mass
of the object times its acceleration, or F = ma.
The unit of force is the Newton. One newton equals one kilogram meter per seconds squared.
Weight, w, is the force of gravity acting upon an object’s mass. So, w = mg.
Newton’s Third Law of Motion states that when an object exerts a force on another object,
the second object exerts an equal but opposite force on the first object.
When two particles collide,
m1v1 (after) + m2v2 (after) = m1v1 (before) + m2v2 (before)
26
Quiz
1. Is linear momentum a vector or scalar quantity? Why?
2. The principle of conservation of momentum says the momentum of a system remains
constant (or fixed). How is it then possible for the momentum of a body to change?
3. When the force on a body is zero, what is the change in its momentum?
4. What is the direction of the change in momentum of a body when there is a force on it?
5. If I apply a force F on a body, does the body exert a force on me? If so, how large is it
and in what direction? Explain then how a body can possibly move.
6. A log, with a mass of 45 kg, is floating downstream at 2.5 m/s. A 10 kg swan attempts to
land on the log, flying upstream at 2.5 m/s, but slides the length of the log and falls off the
end with a velocity of 0.5 m/s (upstream). Calculate the final velocity of the log downstream
after this weird interaction.
7. (a) Calculate the time needed for a constant force of 80 N to act on a body of 12.5 kg in
order to take it from rest to a speed of 72 km/h. (b) How long, and in what direction, must
the same constant force act on the body to return it to rest?
8. A cart having a mass of 1.5 kg moves along its track at 0.2 m/s until it runs into a fixed
bumper at the end of a track. What is its change in momentum and the average force
exerted on the cart if, in 0.1 s, it: (a) is brought to rest; (b) rebounds with a speed of 0.1 m/s?
(c) Discuss the conservation of momentum in these collisions.
27
Solutions to Quiz
1. Since linear momentum, p, is the product of a scalar, m, and a vector, v, it must be a vector
quantity. That means it has direction as well as magnitude.
2. The momentum any body gains (with respect to someone watching!!) must be at the
"expense" of some other body having its momentum changed by an equal AND OPPOSITE
amount (again with respect to the same someone!!). When a ball drops toward the Earth and
gains momentum, the Earth also gains momentum, equal and opposite to that gained by the
ball; but the mass of the Earth is so immense compared to that of the ball, its velocity is
hardly changed.
3. By the very definition of force, it is the change in momentum of a body per unit time. So, if
there is no force on a body, its momentum doesn’t change—in magnitude OR direction. To
go round in a circle, then, a force must be present.
4. Again, from the definition of force, F = Δp/Δt, the direction of the change in momentum
must be in exactly the same direction as the direction of the force. For instance, in circular
motion, the direction of the change in momentum is "centripetal"; so then is the direction
of the force.
5. The body must exert a force equal and opposite to the force I exert; otherwise it would be
impossible for us to remain at relative rest. If my hand is not in constant contact with the
body, I can’t apply the force. Besides, the force the body exerts on me is exerted ON ME,
not on the body, and so F = ma still works because F is all the forces on the mass m, not
the forces m exerts on other bodies.
6. Before the "collision":
plog = (45 kg)(2.5 m/s downstream) = 112.5 kg-m/s downstream;
pswan = (10 kg)(2.5 m/s upstream) = 25 kg-m/s upstream;
ptotal = plog + pswan = (112.5 – 25) kg-m/s downstream = 87.5 kg-m/s downstream
After the collision:
pswan = (10 kg)(0.5 m/s) upstream = 5 kg-m/s upstream;
28
Since ptotal = plog + pswan = 87.5 kg-m/s downstream, and this remains constant,
we can write that after the collision:
plog = ptotal – pswan = 87.5 kg-m/s downstream – 5 kg-m/s upstream
= (87.5 + 5) downstream = 92.5 kg-m/s downstream
With momentum equal to mass times velocity, we get the velocity of the log as:
vlog = plog /m = [92.5 kg-m/s downstream]/45 kg = 2.056 m/s downstream.
7. (a) 72 km/h = (72 x 103 m)/(3.6 x 103 s) = 20 m/s;
F = ma = 80 N = (12.5 kg)a
a = (80 N)/(12.5 kg) = 6.4 m/s2
vf = vi + at
t = (20 m/s – 0 m/s)/6.4 m/s2 = 3.125 s
(b) If the same constant force is applied in the opposite direction for 3.125 seconds, the
body will return to rest.
8. (a) Δp = mΔv = (1.5 kg)(0 m/s – 0.20 m/s) = -0.3 kg-m/s;
F = Δp/Δt = [-0.3 kg-m/s]/0.1 s = -3.0 N
[The minus sign means the force is opposite to the original momentum direction]
(b) Δp = mΔv = (1.5 kg)(-0.10 m/s – 0.20 m/s) = -0.45 kg-m/s;
F = Δp/Δt = [-0.45 kg-m/s]/0.1 s = -4.5 N
(c) The momentum of the cart is not conserved because the cart is not the entire system.
It interacted with the bumper in both cases and we would have to consider what happened to the whole rest of the system—bumper, Earth, etc.
29
Suggestions for Instructors
1. This program begins the investigation into why particles move the way they do. Why do
bodies near the surface of the Earth fall with constant acceleration? Why does the Earth orbit
the Sun along an elliptical path? Why do atoms bind together to form molecules? Why does a
spring oscillate when it is stretched? We want to understand these and many other motions
that we constantly observe around us. This understanding is not only important to our basic
knowledge of nature, but also for engineering and other practical applications. When we
understand how motions in general are produced, we are able to design machines and other
practical devices that move as we desire. The study of the relationship between the motion
of a body and the causes for this motion is called dynamics.
2. F rom daily experience, we know that the motion of a body is a direct result of its interactions
with other bodies around it. When a batter hits a ball, there is an interaction between bat and
ball. The path of a projectile is but a result of its interaction with the Earth. The motion of an
electron around a nucleus is the result of its interaction with the nucleus, and perhaps with
other electrons. Interactions are conveniently expressed quantitatively in terms of a concept
called f o rc e, where we have this intuitive notion of force, and of its strength, in terms of a
push or a pull such as the weight of a body or the pull of a magnet on iron fillings.
3. There are some easily perf o rmed experiments that support the law of inertia. A spherical ball
resting on a smooth horizontal surface will remain at rest unless acted upon. [If the surface is
not quite smooth, but has small variations in its surface, the ball will be found to roll about—
something that attracts the students’ attention when you assert initially that the tabletop is flat.]
We assume that the surface on which the ball is resting balances the interaction between ball
and Earth and hence the ball is essentially free of interactions. When the ball is struck, as in
billiards, it momentarily suffers an interaction and gains velocity (and therefore momentum),
but afterward is free again, moving in a straight line with the constant velocity it obtained
when it was struck. Since the ball does not go on indefinitely, because the surface is finite and
because of friction, we can begin to anticipate a discussion of this additional interaction, which
will be introduced in the next program.
4. The principle of the conservation of momentum holds for any number of particles that form
an isolated system, i.e., particles which are subject only to their own mutual interactions and
not to interactions with other parts of the world. For example, consider a hydrogen molecule
composed of two hydrogen atoms (there f o re, of two protons and two electrons). If the molecule is isolated so that only the interactions among these four particles have to be considere d ,
the sum of their momenta relative to an inertial frame of reference will be constant. Similarly,
consider the Solar System, composed of our Sun, the planets, and their satellites. If we could
neglect the interactions with all other heavenly bodies, the total momentum of the Solar
System, relative to an inertial frame of reference, would be constant.
No exceptions to this general principle (of conservation of momentum) are known. In fact,
whenever this principle seems to be violated in an experiment, the experimenter immediately
looks for some unknown or hidden particle, which remained unnoticed and which may be
responsible for the apparent lack of momentum conservation. It was this type of search that
led physicists to identify ("discover") the neutron, the neutrino, the photon, and many other
elementary particles.
30
5. Because of the existence of the principle of conservation of momentum, all three of Newton’s
Laws exist. It would be interesting to investigate with the student the rationale behind this
conservation principle—it’s not something we have worked out, like the gas laws or even
Newton’s laws. Rather, the conservation of linear momentum is due to the invariance of space
translation, which means that an isolated system, no matter where in the universe it may be
placed, will behave in the exact same manner. The other two dynamic conservation principles,
that of angular momentum conservation and energy conservation, are the result of similar
invariances.
6. When we look around, we see many diff e rent kinds of force. We apply a force on the
ground when we walk. We push and lift diff e rent objects. To stretch a string, we must apply
a force. The wind upsets a tree or pushes a sailboat by applying a force. The expansion of
gases in an internal combustion engine produces a force that causes an automobile, a boat,
or an airplane to move. Electric motors produce a force that moves objects. The above examples of physical forces occur in complicated systems, such as our bodies, a machine, the
Earth’s atmosphere, and on wires carrying an electrical current. We may say that these are
statistical f o rces because, when we analyze where these forces appear, we notice that they
involve bodies composed of large numbers of atoms. The question is then, if we break these
phenomena down into microscopic components, namely, molecules, atoms, electrons, and
their interactions, what do we discover as the interactions or forces? One of the very gre a t
achievements of the last few decades has been to reduce all forces observed in nature to a
very few basic or fundamental interactions. This means that statistical forces are simply manifestations of the fundamental forces when a very large number of particles are involved. The
four fundamental forces that we currently recognize are gravitational, electromagnetic, stro n g
(or nuclear), and weak.
31
PROGRAM 6:
FRICTION, WORK, AND ENERGY
Key Terms and Formulas
Friction is a phenomenon that impedes motion between two surfaces in contact.
There is kinetic friction when an object slides on a surface, and the interaction between the
object and the surface results in a force of friction that acts in a direction opposite to that of
the motion.
Static friction is the force between two stationary surfaces that inhibits motion.
The normal force is the force perpendicular to the touching surfaces, and is opposite in
direction to the object’s weight when the surface is horizontal.
The coefficient of friction, µ, is a measure of the amount of interaction between two surfaces
that results in a frictional force when an attempt is made to move an object.
Kinetic friction equals the coefficient of friction times the normal force, or, Ff = µkFN.
Static friction has a maximum value equal to the coefficient of friction times the normal force,
or Ff = µsFN.
Work is the dot scalar product of force and displacement, W = F • r. The unit of work is the
Joule, which is one Newton-meter.
Power is the rate at which work is done.
The Work-Kinetic Energy Principle encompasses the relationship between work and kinetic
energy, where net work on a body equals its final kinetic energy minus its initial kinetic energy.
32
Quiz
1. Is there a specific relation between the direction of the force of friction and the direction
of the velocity of a body? Is the direction of the force of friction on a body related to its
acceleration?
2. A body moves across a horizontal surface under an applied force. Describe the kind of
motion that results when the applied force is: (a) larger than; (b) equal to; (c) smaller than
the force of friction.
3. A body has a mass of 10 kg. The coefficient of friction between the body and a horizontal
surface is 0.35. Describe the motion when a horizontal force of 78 N is applied to the body.
4. Aristotle said different bodies fell at different rates. Galileo proved that all bodies fall with
the same acceleration. Under what conditions might both assertions be correct?
5. State the conditions under which the work done by a force is (a) zero; (b) positive;
(c) negative.
6. What happens to the kinetic energy of a particle when the work of the applied force is
(a) zero; (b) positive; (c) negative.
7. When I raise up a 10 kg (980 N) weight through a distance of 2 m, I perform (980 N)(2 m)
= 1960 N-m [1960 J] of work. But the kinetic energy of the body is still zero! Where did my
work go?
8. Calculate the work of a constant force of 12 N when the particle on which it acts moves
7 m if the angle between the directions of the force and the displacement is (a) 0°; (b) 60°;
(c) 90°; (d) 135°; (e) 180°.
9. In each of the above, calculate the power generated if the work is done in 5 s.
10. (a) Calculate the constant force required by the motor of an automobile, whose mass is
1500 kg, in order to increase its velocity from 4.0 km/hr (1.11 m/s) to 40 km/hr (11.11 m/s)
in 8 s. (b) Calculate the change in momentum and kinetic energy. (c) Calculate the work
done by the force. (d) Compute the average power of the motor.
33
Solutions to Quiz
1. To the best of our measuring ability, and in the basic physics class, the direction of the force
of friction is always opposed to the direction of motion—even opposed to the attempted
direction of motion for the case where the applied force isn’t large enough to overcome the
maximum force of static friction. Acceleration, as far as we can measure, has no effect on
the direction of the friction force. Incidentally, fluid or viscous friction, where the resisting
material more or less surrounds the object in relative motion, not only is always opposite to
the direction of motion but is usually some function of the velocity—and not just to the first
power (i.e., viscous friction is a nonlinear phenomenon).
2. (a) Since the applied force is greater than the force of friction, there will be a net force in
the direction of motion and so the body will accelerate in the direction of the net force.
(b) When the applied force is equal to the force of friction, the net force is zero and the
body will have constant motion [F = 0, a = 0, or v = a constant]. (c) If the applied force is
smaller than the force of friction, the net force will have a direction opposite to the direction
of motion and so the body will decelerate, finally stopping. Since the force of static friction
is always (well, almost always) larger than the force of kinetic friction, the body really stops!
3. The normal force pressing the body against the horizontal surface is its weight, mg, and
the force of friction—assuming the body will be moving—is
Ff = µmg = 0.35(10 kg)(9.8 m/s2) = 34.3 kg-m/s2 = 34.3 N
Since the horizontal force being applied is 78 N, we get a net positive force of
(78 – 34) N = 44 N and then using F = ma, we have a = (44 N)/(10 kg) = 4.4 m/s2.
4. Both assertions cannot be correct at the same time. However, the feather and penny in the
tube is a very good experiment to show how both Aristotle and Galileo can be right. When
we can ignore the frictional effects of surrounding materials, Galileo is shown to be correct
in that ALL objects will accelerate downward with the acceleration of gravity. But when the
air is let into the tube, we see that in the "real" world Aristotle was right. So it’s a case of
carefully stating the conditions of your experiment as well as the results. Too often what we
read in today’s newspapers or see and hear on television doesn’t completely give all the
facts, and so two sides have conflicting "results."
5. Work is a scalar product of force (a vector) and displacement (also a vector) and is therefore
written as W = F • r or W = Frc o sθ where θ is the angle between the direction of the force and
the direction of the displacement of the body on which the force acts [that’s a lot of words, but
let’s be very careful before we start using concepts!]. (a) The work done by a [nonzero] force
will be zero either when the body does not move, which means there must be some other
opposing force such as the force of static friction, or the value of cosθ is zero, which happens
only when the force and the displacement are perpendicular to each other; (b) The work is
positive when there is an actual displacement of the body and t h e value of cosθ is positive;
(c) The work is negative only when there is a displacement and c o sθ is negative—this usually
means that the force and the displacement are in somewhat opposite directions.
34
6. Since kinetic energy depends on the velocity [squared] of the body: (a) zero work results in
zero change in the kinetic energy; (b) positive work results in a positive change (increase)
in the body’s kinetic energy; (c) negative work produces a negative change (decrease) in the
kinetic energy of a body.
7. You performed 1960 J of positive work on the body because the direction of your force and
the direction of displacement were both up. Since the kinetic energy of the body did not
change, the net work done on the body must have been zero! What you forgot was that
there is a negative force due to "gravity" down, equal to mg. So, you did work on the body
and so did the Earth, or gravity, or whatever you want to call it. One way we get out of this
seeming paradox is to say your positive work increased the [gravitational] potential energy
of the body. This then lets us talk about potential energy and kinetic energy and the sum
of the two, which we call total energy.
8. W = F • r or W = Frc o sθ where θ is the angle between the direction of the force and
the direction of the displacement of the body on which the force acts. For all these cases,
F = 12 N and r = 7 m, so Fr = (12 N)(7 m) = 84 N-m = 84 J. Then:
(a) W = (84 J)cos0° = (84 J)(1.0) = 84 J
(b) W = (84 J)cos60° = (84 J)(0.5) = 42 J
(c) W = (84 J)cos90° = (84 J)(0.0) = 0 J
(d)W = (84 J)cos135° = (84 J)(-0.707) = -59.4 J
(e) W = (84 J)cos180° = (84 J)(-1.0) = -84 J
9. P = W/t, where P is in watts if work is in joules and time in seconds. So:
(a) P = (84 J)/(5 s) = 16.8 W
(b) P = (42 J)/(5 s) = 8.4 W
(c) P = (0 J)/(5 s) = 0 W
(d) P = (-59.4 J)/(5 s) = -11.9 W
(e) P = (-84 J)/(5 s) = -16.8 W
35
10. (a) F = ma = m(Δv/Δt) = (1500 kg)([11.11 m/s] – [1.11 m/s])/(8 s) = 1500 kg(10 m/s)/8s
= 1875 kg-m/s2 = 1875 N.
For later: a = F/m = 1875 N/1500 kg = 1.25 m/s2.
(b) Δp = m(Δv) = 1500 kg([11.11 m/s] – [1.11 m/s]) = 1500 kg(10 m/s) = 1.5 x 104 kg-m/s.
ΔEk = 1/2mΔv2 = 1/2(1500 kg)([11.11 m/s]2 – [1.11 m/s]2)
= 1/2(1500 kg)(123.456 m2/s2 – 1.234 m2/s2) = 750 kg(122.222 m2/s2) = 9.17 x 104 J.
(c) W = Fs, but we don’t have the distance covered. So, let’s calculate that using
vf2 = v02 + 2as, or s = [vf2 – v02]/2a
This gives
s = ([11.11 m/s]2 – [1.11 m/s]2)/2[1.25 m/s2] = (123.456 m2/s2 – 1.234 m2/s2)/2.5 m/s2
= (122.222 m2/s2)/2.5 m/s2 = 49.0 m.
So now, we use W = Fs = (1875 N)(49.0 m) = 9.17 x 104 N-m = 9.17 x 104 J. Please note
that this is exactly the change in the kinetic energy of the body—as we could have realized from the beginning and simply invoked the work–kinetic energy theorem!
(d) Power is work per unit time, P = W/t = [9.17 x 104 J]/8 s = 1.146 x 104 W (11.5 kW).
36
Suggestions for Instructors
1. When it comes to friction, it is important to stress the statistical nature of the friction force.
Simple experiments, such as placing a block on a relatively rough board and then raising the
board until the block slides, illustrate how the force of static friction increases until it reaches
its maximum value, which allows the coefficient of [static] friction to be experimentally
measured [= tanθ]. Also, by placing the block on the board in different manners, such as
sliding it back and forth initially or pressing down hard first, it becomes pretty clear that this
is a difficult property to determine with any accuracy. Kinetic friction can be studied—and
measured—by pulling a block along a horizontal surface using a spring calibrated to read
force (rather than grams, as so many are labeled!) and attempting to keep the velocity a
constant, which should keep the force constant.
2. The spring measuring device is also useful in illustrating work done on a body. You may
pull horizontally, at an angle, while the block slides horizontally, or pull up an inclined
board. The only real way to show work resulting in a change in kinetic energy is to drop a
body and try to measure the change in velocity [therefore, kinetic energy]. In the laboratory
a more sophisticated apparatus must be set up where first the kinetic coefficient of friction is
evaluated by having the block slide at a constant velocity [due to a set of "weights" that pull
horizontally on the block as they "fall"—the string over a pulley]; then a larger amount of
mass is added and the resulting acceleration measured [displacement from zero velocity in
a measured time interval], allowing final velocity to be calculated.
3. Have students perform a task that involves real physics-type work in differing time intervals
to have them understand the concept of power. Relate their actual values to sources of
power—particularly the "horsepower" of their automobiles and kilowatts of electric lights.
4. Potential energy only appears peripherally in this program. It would be useful to introduce
more information and some cases into class discussion.
37
PROGRAM 7:
GRAVITATION
Key Terms and Formulas
Gravitation is the attraction between every mass in the universe.
The Law of Universal Gravitation states that the attractive force between two masses is
proportional to the product of the masses of the two objects, over the square of the distance
between the objects. Or,
F=G
m1 m2
r2
The G in this equation is the gravitational constant and has a value of 6.67 x 10-11 N-m2/kg2.
Kepler’s First Law of Planetary Motion states that every planet has an elliptical orbit, with
the Sun at one focus.
Kepler’s Second Law of Planetary Motion states that an imaginary line that connects the
Sun to a planet sweeps out equal areas in equal times.
Kepler’s Third Law of Planetary Motion states that the square of a planet’s orbital period is
proportional to the cube of its semi-major axis, or
T2 =
38
1 GM 2 a
4 π2
3
Quiz
1. Why do all [massive] bodies attract each other the way described by Newton’s law of
universal gravitation?
2. If you were in a tightly enclosed, very well equipped laboratory, devise an experiment that
could prove you were on the Earth’s surface where the "acceleration of gravity" was 9.8
m/s2, rather than in a rocket ship in empty space accelerating at 9.8 m/s2.
3. Gravitation force falls off as 1/r2 and so is called an inverse square force. If the radius of the
Earth is 6.38 x 106 m, where the gravitational force on a 1 kg mass is 9.8 N, how far from
[the center of the] Earth must we be for the gravitational force on the 1 kg mass to equal
9.8 x 10-10 N?
4. How hard would you have to throw a stone straight up so that it would never come back?
(This speed is called the escape velocity from Earth.)
5. A satellite of mass m is going around the Earth in a circular orbit, with a radius r and a
speed v. [The centripetal force needed to keep a mass moving along a circular path is FC =
mv2/r.] If the entire centripetal force is supplied by the gravitational force, FG = GmM/r2,
where M is the mass of the Earth (i.e., FC = FG = mv2/r = GmM/r2), calculate the radius
["height"] such that the period of the satellite is exactly 24 hours.
39
Solutions to Quiz
1. There is no obvious reason why masses attract although many models, or ideas, have been
put forward. Kepler, with the help of Tycho Brahe, developed what we call Kepler’s Laws in
the late 1500’s and Newton collapsed them into a single idea in the late 1600’s. This concept
appears to be quite valid, with the possibility that it must be modified when Einstein’s ideas
concerning General Relativity are introduced, which says that a mass "warps" space, or vice
versa.
2. If you can come up with such an experiment, you are guaranteed a Nobel Prize because
you will have proven Einstein’s statement of the Principle of equivalence to be incorrect. The
proposed experiment is exactly what Einstein proposed to prove his point that gravitational
attraction caused by a mass is identical in all ways with an acceleration.
3. Using the law of gravitational force, we write:
F1 = G(ME[1 kg])/RE2 = 9.8 N, and
FX = G(ME[1 kg])/RX2 = 9.8 x 10-10 N
where the X stands for the new distance. There are a lot of common symbols in both
equations; so to solve with the least amount of math, let’s equate the same batch of symbols
from both equations:
G(ME[1 kg]) = RE2 [9.8 N] = RX2[9.8 x 10-10 N], or
RE2[9.8 N] = RX2[9.8 x 10-10 N]
Then, canceling the common 9.8 N on each side, we have RE2 = RX2[10-10] and taking square
roots gives RE = RX[10-5]. Therefore, RX = RE[105] = 6.38 x 1011 m, which is further than the
distance to the Sun.
4. To do this problem, we have to know what the gravitational potential energy of a mass is.
We can’t use PE = mgh because g is not a constant as we move away from Earth (or if we
go inside the Earth!). When two masses are infinitely far apart, the force of gravity is zero;
as they get closer to each other the force [of attraction] increases and so, to keep them apart
the direction of the force is opposite to their distance apart. This means that if we call gravitational potential energy zero when two masses are infinitely far apart, the value of their
potential energy is negative for any value less than infinity. What we discover is that the
relation is PE = -GM1M2/R; that is, it’s negative always and inversely proportional to the
distance—to the first power. So, with M1 = Earth’s mass, ME, and M2, the mass of what we
throw, we can write the total energy of a mass at the Earth’s surface:
1
40
/2M2v2 – GMEM2/RE = E
Then, when the mass reaches infinity, the potential energy will be zero. Let the kinetic energy
be zero also at that point. Therefore, if the total energy of the mass at the Earth’s surface is
z e ro and we throw the mass directly "up", it will reach infinity. So, let E = 0 and we can write:
1
/2M2v2 = GMEM2/RE
or, canceling out the common M2, multiplying by 2, and taking the square root, we have:
v = [2GME/RE]
1/2
= [2(6.67 x 10-11 Nm2kg-2)(5.97 x 1024 kg)/(6.38 x 106 m)]
1/2
v = [1.248 x 108 Nm/kg]
1/2
= 1.12 x 104 m/s [= about 40,000 km/hr].
5. Using the relation FC = FG = mv2/r = GmM/r2 and canceling the common m and one r,
we have v2 = GM/r. This is a relationship between any satellite orbiting a massive body in
a circular orbit. What we are looking for is a circular orbit with a period of 24 hours, which
equals (24 hours)(60 minutes/hour)(60 seconds/minute) = 8.64 x 104 seconds. Well, since
v = d/t, if we substitute 2πr for d and let t be the 24 hours, we have:
[2πr/t]2 = GM/r
or
[4π2r3]/GM = t2
Darn if that doesn’t look like one of Kepler’s Laws! So, we solve for r and get:
r = [GMt2/4π2]
1/3
1/3
r = [(6.67 x 10-11 Nm2kg-2)(5.97 x 1024 kg)(8.64 x 104 s)2/4π2]
= 4.22 x 107 m.
If you subtract the Earth’s radius, you then find how high this stationary satellite will sit
in the sky.
41
Suggestions for Instructors
1. When there is time, the introduction of angular momentum, L = r x p, when discussing planetary motion is of help in explaining why the motion is in a plane. Since the force of attraction between masses has a direction along the line joining them [a central force], it can be
shown that dL/dt must be zero, which results in a constant angular momentum, both in magnitude AND direction. If then the planets follow elliptical orbits, their distance to the Sun is
varying; there f o re, their speed must as well, such that the magnitude of L remains constant.
2. Students will become interested in this subject when artificial satellites, such as the orbiting
spacecraft of the U.S. and Russia, are used as examples. The Smithsonian Institution has a good
collection of such craft, and a Web site they may wish to visit http://www.nasm.si.edu/museum.
3. The first time the student becomes aware of the fact that the "interaction" between two
masses exists throughout all space, it may be possible to start a discussion about the gravitational field of a single mass. That is, we may say that a mass creates a gravitational field
about itself whose strength [a vector] is measured in terms of the force [a vector] per unit
mass at any point in space. That is, G = FG/m = -[GM/r2]ur, where ur is a unit vector in the
radial direction. When the electric interaction is studied, the analogy is usually not lost on
the student between electric field and gravitational field. Such a comparison allows for an
understanding of the relationship between electric and gravitational potential as well as electric and gravitational potential energy.
4. Here is it appropriate to mention that there are only four known fundamental forces: gravitational, electromagnetic, nuclear [also called "strong"], and weak. We know "everything" about
the gravitational case, a lot about the electromagnetic case, little about the strong interaction,
and hardly anything about the weak "force". There is a lot of room for new studies, ideas,
theories, and experimentation. That’s why physicists keep trying to build these very-highenergy machines.
42
PROGRAM 8:
HARMONIC MOTION AND WAVES
Key Terms and Formulas
Simple harmonic motion is the basic vibratory motion an object undergoes when it is
subjected to a restoring force that’s directly proportional to its displacement from equilibrium.
Hooke’s Law states that F = -kx, where k is the constant of proportionality, particular to every
object undergoing simple harmonic motion.
The period, T, of an object undergoing simple harmonic motion is the time it takes to complete one cycle.
Frequency, f, is the inverse of period, and is measured in Hertz.
Amplitude is the maximum displacement of an object undergoing simple harmonic motion
from its equilibrium position.
m
For an object oscillating on the end of a spring, T = 2π k
!
T = 2π!g
L
For a simple pendulum,
A wave is a periodic disturbance.
The bump which is sent in a wave is called the pulse.
The top of each wave is the peak; the bottom is called the trough.
Where the wave originally begins is called the equilibrium position, and the top of the peak
to the equilibrium position is the amplitude of the wave.
The distance between peaks is the wavelength, λ.
The frequency, f, is the number of waves that pass a certain point per second.
Periodic waves are waves that move ceaselessly and regularly.
The speed of a wave equals its frequency times its wavelength, or v = f λ.
Transverse waves occur when the particle motion is perpendicular to the wave motion.
Longitudinal waves occur when the particle motion is parallel to the wave motion.
Surface waves occur when the particles move in a circular motion, while the circle center
stays in one place.
For transverse waves on a stretched string, wave speed equals the square root of the tension
over the mass of the string divided by its length, or
T
v = m/L .
!
43
Quiz
1. How is the period of Simple Harmonic Motion [SHM] changed: (a) when the mass of the
particle is increased without changing the elastic constant; (b) when the elastic constant is
increased without changing the mass; (c) when the mass and the elastic constant are
changed by the same ratio?
2. Describe an experiment that will measure the spring [or elastic] constant of a spring.
3. The length of a pendulum clock is adjusted so that it gives the correct time at the Earth’s
surface. How would the length have to be adjusted if the clock were taken to the top of
a very high mountain?
4. When a man of mass 60 kg gets into a car, the center of gravity of the car lowers 0.3 cm
[3 x 10-3 m]. What is the net elastic constant of the springs of the car? Given that the mass
of the car is 500 kg, what is the period of vibration when it is empty and when the man is
inside?
5. By rocking a boat, surface waves are produced on a quiet lake. The boat performs 12
oscillations in 20 seconds, each oscillation producing a wave crest. It takes 6 seconds for
a given crest to reach the shore 12 m away. Calculate the wavelength of the surface waves.
6. A spring having a normal length of 1 m and a mass of 0.2 kg is stretched 0.04 m by a force
of 10 N. Calculate the velocity of propagation of longitudinal waves along the spring.
7. A steel wire having a diameter of 0.2 mm [2 x 10-4 m] is subject to a tension of 200 N.
Determine the velocity of propagation of transverse waves along the wire. Steel has a
mass density of 7.86 x 103 kg/m3.
44
Solutions to Quiz
1. The basic equation for the period of motion for a mass on a spring is T = 2π !m/k, where
m is the mass on the end of the spring and k is the elastic or spring constant. So, (a) if the
mass is increased, the period increases, but only by the square root, while (b) if the elastic
constant increases, the period decreases again by the square root. Then (c) if both change
by the same ratio, nothing happens to the period.
2. One experiment would be to hang the spring vertically with no mass attached. Then, after
noting (i.e., measuring) the position of the lower end, add a small mass and slowly allow
the spring to expand. Measure the new position of the lower end and then calculate the
extension of the spring, x, due to the addition of a force, F (= mg). The ratio of F/x will give
the spring constant. Of course, a good scientist would add further "weights" and create a
graph of total added force versus extension and get the value for k from the slope of the
curve. Alternatively, a computer program could be used to evaluate the least-squares fit of
the experimental data.
A second experiment could be done by using the SHM equation: add a mass and allow the
system to oscillate; measure the period of oscillation. Repeat for other masses and plot or
use your computer program.
3. The period of a pendulum clock is given by T = 2π!L/g, where L is the length of the pendulum and g is the acceleration of gravity. At the top of a mountain the value of g is somewhat
lower than at sea level. Therefore, recalling Problem 1 above, we should reduce the length of
the pendulum by the same ratio that g has been reduced.
4. F = -kx. Then k = [60 kg(9.8 m/s2)]/[3 x 10-3 m] = [588 N]/[3 x 10-3 m] = 1.96 x 105 N/m where
we have dropped the minus sign because we are not "doing" vectors. Using the period
equation:
T = 2π !m/k, for the two cases we get
T = 2π!500kg/1.96 x 105 N/m) = 2π!2.55 x 10-3 s2 = 0.317 s, and
T = 2π!(560kg)/(1.96 x 105 N/m) = 2π!2.86 x 10-3 s2 = 0.336 s.
5. If the boat makes 12 oscillations in 20 seconds, its frequency (of oscillation) is 12/[20 s]
= 0.6 Hz. Since the waves take 6 seconds to reach a shore 12 m away, their velocity is
[12 m/6 s] = 2 m/s. Then since v = f λ, the wavelength of these is
v /f = [2 m/s]/0.6 Hz = 3.33 m
45
6. This spring has a mass per unit length, m/L, of [0.2 kg]/1 m] = 0.2 kg/m. Its spring constant
is k = F/x = [10 N]/[4 x 10-2 m] = 2.5 x 102 N/m. Longitudinal waves on a spring must have a
value related to these properties. Using dimensional analysis, breaking newtons into its basic
units, we get N = kg[m/s2] and then k is 2.5 x 102 {kg[m/s2]}/m = 2.5 x 102 kg/s2. Just looking
at units, k divided by the mass per unit length gives units of {kg/s2}/{kg/m} = m/s2. If we
multiplied the length of the spring (one more time), we would get units of m2/s2 and we
could write an equation:
vlongitudinal = !kL/(m/L) = !kL2/m = !2.5(102)(1)2/0.2 = !1250 = 35.4 m/s.
7. The velocity of propagation of transverse waves along a stretched wire is given by
vtransverse = !T/(m/L)
Where T is the tension in the wire and m/L is the mass per unit length of the wire. We’re
given the tension but not the length density of the wire, so we have some work to do.
Suppose the wire is L long. It has a cross-sectional area of πr2 and so a total volume of
V = πr2L. Since we are given the mass per unit volume, ρ, we have the mass as m = ρV
= ρπr2L and so m/L = ρπr2. Remember, radius is half the diameter. So
vtransverse = !200N/(7.86 • 103kg/m3) π(1 • 10−4m)2 = !200N/π(7.86 • 10-5 kg/m)
=!8.099 • 105m2/s2 = 900 m/s.
46
Suggestions for Instructors
1. Some students find the use of a point in uniform circular motion and its projection [onto the
X-axis or, better, onto a screen] helpful in understanding SHM. Of course, if you are finding
that using trig functions slows things down it might be better not to introduce SHM as
x = Asiθt.
2. Adding acoustic SHMs to produce beats is a demonstration that students appreciate. Also,
adding SHMs on an oscilloscope screen so they are perpendicular gives patterns that begin
to uncover some of the mysteries of what a CRT can produce. Coupled oscillators are also
lots of fun to observe in demonstration.
3. Since real oscillators do not keep vibrating forever, some discussion of resistive terms is
useful. Most often the resistive term is chosen to be directly proportional to the velocity
[and opposed to the direction of motion] because the "second order differential equation"
is then linear (only composed of first-order terms, hence "linear") and is solvable by standard techniques taught in standard calculus courses. Of course, if students [and/or instructors] can program iterative equations, ANY type of resistive term can be added. For example,
with no resistance, SHM is:
F = ma = -kx, or m[dv/dt] + kx = 0, or m[d 2x/dt2] + kx = 0
which can be "programmed." Then a resistive force term is added so the equation is:
F = ma = -kx – Rv, or m[dv/dt] + Rv + kx = 0, or m[d2x/dt2] + R[dx/dt] + kx = 0
or any other such term to play with. Give the system some initial energy [by letting the
initial values of x and d2x/dt2 be zero and dx/dt some number]. Let the system "go"!
4. Experiments with wave interference, or superposition, intrigue students. In particular, setting
up standing waves in the laboratory gives students a good handle on the relationship
between frequency, wavelength, and velocity of propagation.
47
PROGRAM 9:
HEAT
Key Terms and Formulas
Heat is the energy that’s transferred from one object to another due to a difference in temperature; heat is measured in Joules.
The specific heat capacity of a substance is how much heat it takes to change one kilogram
of it by one degree Celsius.
Heat added or removed equals the mass of a substance times its specific heat capacity times
the change in temperature, or Q = mcΔT
The equilibrium temperature is the temperature at which, in an isolated system, no heat is
lost or gained by the system.
The latent heat, L, of a substance is the amount of heat needed to turn one kilogram of it
from one phase to another, or Q = mL.
The heat of fusion is the amount of heat required to change a substance from a solid into a
liquid.
The heat of vaporization is the amount of heat required to change a substance from a liquid
into a gas.
Conduction is heat transfer via the collision of molecules.
Convection is the transfer of heat due to the mass movement of a heated substance.
Radiation is the transfer of energy through space.
The change in work done by gases is equal to the product of the pressure and the change in
volume of the gas, or ΔW = pΔV.
48
Quiz
1. Why do diff e rent materials have diff e rent specific heat capacities? Why does the same material
have diff e rent specific heat capacities when it is in diff e rent states?
2. How do you measure temperature? Does the temperature of a molecule make any sense?
Is temperature a statistical concept?
3. How do you measure pressure? Does the pressure from a molecule make any sense?
Is pressure a statistical concept?
4. How does a microwave oven heat anything since there is no fire present? When you grill a
steak are you using conduction, convection, or radiation?
5. Assuming no heat is lost or gained to the outside world, will a 10 g [10-2 kg] ice cube at
-10°C melt completely in 100 cm3 [10-4 m3] of water originally at +10°C? If yes, what’s the final
temperature?
6. The rate at which a hot body cools depends directly on the temperature difference between
the hot body and the medium surrounding it (e.g., the room temperature). Assume an initial
temperature difference of 100°C and a rate constant such that in 1 minute the temperature
difference becomes 95°C. Estimate the temperature difference after 1 minute more; after 2
minutes more. How long would it take—make a rough guess—before the temperature difference is less than 1°C. [This phenomenon is called Newton’s law of cooling and, more or
less, answers the question about when you should add cream to your hot cup of coffee.]
49
Solutions to Quiz
1. Heat is a form of energy. There f o re, when a substance absorbs heat, it must increase its energy. Since the substance is "not moving" this energy must be within itself. This means that the
potential energy and the kinetic energy within the substance must increase. Since potential
energy is associated with change in position, as long as a substance doesn’t "grow" [ALL re a l
substances DO], then the internal kinetic energy must go up and that will be associated with
the molecular structure of the substance—and they’re all diff e rent. Ergo[??], the amount of
energy per unit mass and per temperature unit change is diff e rent for different substances.
For the same substance in different states the same kind of argument: diff e rent molecular
s t r u c t u re, diff e rent specific heat capacities.
2. Very interesting question. Simple answer: with a thermometer. But what’s a thermometer?
A thermometer is a device that uses some temperature-dependent property of some substance to allow for observation of changes in the internal kinetic energy of the substance.
For instance mercury [or red-tinted alcohol] expands when heated; so place some mercury in
a very thin tube and watch it expand. Stick the tube in a mixture of ice and water and when
the mercury level settles, mark the tube and call the temperature 0°C. Do the same thing as
the water boils and then not only do you have a thermometer, you also have a defined
temperature scale, arbitrary though it may be. No way are you going to be able to measure
the temperature of one molecule because it doesn’t have any internal energy—it’s all external kinetic energy [unless you look at it REAL close]. Since there is no way each and every
molecule of a substance will have identically the same amount of kinetic energy, the temperature of a sample of a material is an overall average of some kind. You are going to need
statistics of some kind to figure it out "exactly."
3. Pressure is a measure of force per unit area. So if you have a device with a known area on
which you can measure the force of the gas, you can divide the size of the force by that
area and measure pressure, by marking off the scale appropriately. No way will one molecule give you pressure; it’s just too small and will strike the device too infrequently. As a
matter of fact, if you think of a few million molecules [only] within the device, you can think
of them peppering the area like a bunch of "little balls" [bad analogy, but it helps] and you
can see the scale jiggle like crazy. More statistics are needed to make sense out of pressure.
Outer space doesn’t have a pressure—there’s only one molecule per cubic meter, on the
average!
4. A microwave oven sends out "microwaves," that’s how. Really, there’s an electronic tube
generating very short electromagnetic waves—wavelengths much longer than those of visible
light, however—of just the right size that they resonate the water molecules within the material to be heated. The resonance causes the molecules to increase their internal kinetic energy, which is then an increase in the temperature of the material. Voila, we’re cooked! When
you grill a steak, inside or out, you are using all three methods for heat transfer. The surface
is hot, the coals or fire radiate, and there is a bit—not much—of moving warm air to convect some heat.
50
5. To raise the temperature of the 10 g of ice to the melting point will require
Q[-10 to 0] = mciceΔT = [10-2 kg][1.946 x 103 J/kg°C][10°C] = 1.946 x 102 J
I looked up the heat capacity of ice so this part could be done. Next, to melt the ice we
need the heat of fusion for ice and that’s equal to 3.335 x 103 J/kg. For our ice cube:
Q[melt] = mL = [10-2 kg][3.335 x 105 J/kg] = 3.335 x 103 J
So just to get the ice cube melted requires 3335 + 194.6 = 3530 J. To reduce the temperature
of 100 g [100 cm3 has a mass of 100g] to 0° would require
Q[+10 to 0] = mcwaterΔT = [10-1 kg][4.186 x 103 J/kg°C][10°C] = 4.186 x 103 J
Therefore, since it would take only 3530 J to "heat up" and then completely melt the ice
cube, there was more than enough heat in the water to give up. So, we have to go back
and see what the temperature of the water was at the time the ice was all melted; that is,
when the water had given up 3530 J. We use the above equation but solve for ΔT instead:
ΔT = +10°C – T° = Q[+10 to T°]/[mcwater] = [3530 J]/[10-1 kg][4.186 x 103 J/kg°C] = 8.43°C
or
T° = 10 – 8.43 = 1.57°C. We are now at the point where 100 g of water is at 1.57°C and 10 g
of water is at 0°C. The 10 g will "warm up" a bit and the 100 g will cool down so that both
have the same final temperature. In other words, there will be a heat loss Q" by the 100 g
equal to [100 g]cwater[1.57°C – Tfinal] and an equal heat gain by the 10 g equal to
[10 g]cwater[Tfinal – 0°C]
or
[100 g]cwater[1.57°C – Tfinal] = [10 g]cwater[Tfinal – 0°C]
or
10[1.57°C – Tfinal] = [Tfinal – 0°C], which gives 11Tfinal = 15.7°C or Tfinal = 1.43°C.
6. Newton’s law of cooling has to be put into an equation form so we can use it. What it says
is that the heat loss per second is directly proportional to the temperature difference. So we
write:
ΔQ/Δt = K(T – Troom)
The information given was an initial temperature diff e rence of 100°C, so let’s say the ro o m
t e m p e r a t u re, which will remain constant, is 20°C. Then the initial temperature of the substance to cool is 120°C. This is so you can develop a plot of temperature versus time to get
some idea of what’s going on. Now with all these numbers, we can calculate the value of the
constant K f ro m
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K = (T – Troom)/[ΔQ/Δt] = (100°C)/[(5°C/1 min)] = 20 min,
where we are just using the temperature drop as the amount of heat loss per unit time,
ΔQ/Δt, because we aren’t given any particulars about what’s cooling. This is what we know:
Something is cooling and starts at a temperature of 120°C and after 1 minute its temperature
is 115°C. During that initial minute, the substance lost heat at the rate of [20 min][100°C] and,
all things being equal, will lose heat at the rate of [20 min][95°C] during the second minute,
or only 95% of what it lost the first minute. This means the temperature drop of the substance should be about 0.95[5°C] = 4.75°C and the new temperature will be 115°C – 4.75°C
= 110.25°C, giving a new temperature difference of 90.25°C. Using the same reasoning, during the next (third) minute the rate of heat loss will be [20 min][90.25°C], or 0.9025[4.75°C] =
4.287°C and temperature 110.25°C – 4.287°C = 105.96°C. This game [process] can be continued to find new lower temperatures. Plotting these values will prove very interesting. As you
can clearly see, it is going to take a lot longer than 20 minutes for the substance to cool to
room temperature because it loses less and less heat each unit of time.
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Suggestions for Instructors
1. Calorimetry, the calculation of heat transfer from one body to another, is usually a simple
algebraic process and as such is useful to help students improve mathematical skills.
However, there is not too much physics involved. More time can be spent on more difficult
ideas, such as what is a temperature scale and how temperature [absolute] is a measure of
the internal energy of a substance.
2. Once you have introduced the absolute temperature scale, your text [or curriculum] should
discuss internal energy of materials. Without necessarily going into the "kinetic theory of
matter" and a derivation, students should know that U = (3/2)kNT where k, the Boltzmann
constant, has been chosen to have its value so 3/2 remains as does N, the number of molecules of material. The other form, U = (3/2)nRT, where n is the number of moles of the
material and R is the gas constant, is sometimes useful. The critical point is that students
must understand that, under usual circumstances, the internal energy of a substance is directly proportional to its absolute temperature, whatever form the constant of proportionality
takes.
3. The reason we can write any "laws" in Heat and Thermodynamics is because of the statistical nature of these laws. When you get down to the level of being able to count individual
molecules, the best we can do is attempt to introduce some kind of statistical mechanics
and formulate concepts based on fundamental principles, such as conservation of energy,
and reasonable conditions, such as lowest energies get filled first. These are rather difficult
ideas, but sometimes they help students grasp the difference between the very few fundamental principles and the large body of laws that have been developed to explain various
phenomena.
4. Radiation, as one of the methods for heat transfer, allows you to get involved in a discussion
of the interaction of radiation with matter. This could be delayed for later in the course, but
a short introduction at this time will show the relevance of studies that lead to devices like
dental X-ray machines, microwave ovens, MRI scanners, etc.
5. The last problem for this program was chosen to allow for some possible use of computer
calculation and even a way to introduce the elements of calculus through a practical problem that needs solving in a more precise manner.
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PROGRAM 10:
THERMODYNAMICS
Key Terms and Formulas
The First Law of Thermodynamics states that the change in the internal energy of a system
equals the heat added minus the work done by the system, or ΔU = Q – W.
An isovolumetric process is when there is a gas and the volume is kept constant, but the
pressure changes.
An isobaric process is when the pressure is kept constant, but the volume changes.
In an adiabatic process, no heat is added or taken from the system.
The Second Law of Thermodynamics states that heat flows naturally from a hot object to
a colder one, and heat will not flow spontaneously from a cold object to a hot one. In other
words, the efficiency of a heat engine can never equal one hundred percent. A third way to
state the second law of thermodynamics is to say that the entropy of a closed system can never
decrease. It will always increase as heat is converted into work.
Heat engines turn heat energy into electrical or mechanical energy. The heat taken from the
hot reservoir equals the work done plus the heat rejected to the cold reservoir, and efficiency
is work done divided by heat in.
Entropy is the measure of disorder of a system, and the change in entropy equals heat change
of a system divided by the absolute temperature, or ΔS = Q/T.
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Quiz
1. What is the fundamental difference, in the thermodynamic sense, between work and heat?
2. Suggest how the temperature of a gas is kept constant during an isothermal expansion.
3. What is meant by a heat engine? What is the best way of increasing the efficiency of a heat
engine?
4. Why is it not possible to design an engine that has 100% heat efficiency?
5. When a system is taken from state A to state B along path ACB (see figure 1 below), the
system absorbs 80 J of heat and does 30 J of work.
(a) Given that the work done is 10 J, how much heat is absorbed by the system along path
ADB?
(b) The system is returned from state B to state A along the curved path. The work done
on the system is 20 J. Does the system absorb or liberate heat, and how much?
(c) Given that UA = 20 J and UD = 60 J, determine the heat absorbed in the processes AD
and AB.
6. A gas undergoes the cycle shown in Figure 2 below. The cycle is repeated 100 times per
minute. Determine the power generated by this "heat engine."
7. Explain why it is rather unlikely for the entropy of an isolated system to decrease.
Figure 1
p
C
Figure 2
p, kPa
B
A
30
20
A
0
55
10
D
V
B
0
0.2
C
0.4 0.6 0.8 1.0 V, m3
Solutions to Quiz
1. Work is a calculable quantity, by adding up all the pΔV terms; that is, the total amount of
work done can be measured off a pV diagram. Because the value of p may be changing as
you go from one V to another, we need a plan to do that sum. On the other hand, heat is
NOT calculable. In other words, both heat and work are forms of energy that can change
the energy of a system; work is "easy" to calculate while heat is not.
2. Isothermal means constant temperature. During an expansion, work is being done and, if
NO heat is added (Q = 0), then ΔU must decrease, which means the temperature must go
down since the absolute temperature is directly proportional to the internal energy. So, during an isothermal expansion an amount of heat must be added exactly equal to the amount
of work done. How you do this is your business—by conduction, convection, or radiation.
3. A heat engine is any device that converts heat [energy] into mechanical work [energy]. The
best way to increase the efficiency of a heat engine is to throw away as little heat as possible. This is done by making the ratio of the absolute temperature as great as possible.
4. Practically, trying to increase the temperature of the hot reservoir runs into containment
problems (materials begin to melt at high temperatures and they undergo fatigue faster).
Theoretically, it’s impossible to get to 0°K and it is very hard to maintain very cold temperatures at the cold reservoir for a long time—it’s expensive.
5. This problem is based on the first law of Thermodynamics: ΔU = Q – W, where ΔU is the
change in internal energy of, Q is the heat added to, and W is the work done by the system.
No matter how you go from one set of conditions (called the state of the system), to the
other—that is from A to B, the value of the change in internal energy is:
ΔU = 80 J – 30 J = 50 J.
(a) When we go from state A to state B by path ADB, ΔU = 50 J as we just found and
W = 10 J. Therefore, we have
ΔU = Q – W = 50 J = Q – 10 J, or Q = 50 J + 10 J = 60 J.
(b) Same thought process: ΔU = Q – W = -50 J = Q – (-20 J). Q = -50 J – 20 J = -70 J. This
means that the system liberated [gave off] 70 J of heat. We used -50 J for ΔU because the
path was from B to A.
(c) Given that UA = 20 J and UD = 60 J. Therefore, we have ΔUAD = UD – UA = 60 J – 20 J
= 40 J. When you go from state D to state B no work is done (because ΔV = 0 along
path DB). We were given that the total work on the path ADB is 10 J, and so all this
work is done along AD. So we write ΔUAD = QAD – WAD = 40 J = QAD – 10 J
Therefore QAD = 40 J + 10 J = 50 J.
56
6. The total work done by this heat engine each cycle is the area contained within the triangle.
Therefore:
Wcycle = 1/2 [0.8 m3 – 0.2 m3][3 x 104 Pa – 1 x 104 Pa] = 1/2 (0.6 m3)(2 x 104 Pa) = 6000 J
Since power is work per unit time, we have:
P = Wcycle[cycles/min][1 min/60 s] = [6 x 103 J][100 cycles/min][1 min/60 s] = 10 kW.
7. ΔS = Q/T, where ΔS is change in entropy, Q is the heat absorbed during this change, and T
is the (constant) absolute temperature during this change. Consider an isolated system composed of a totally sealed room containing "everything" to maintain the well-being of a few
humans, including trees, animals, an air-recirculating system, a waste recycling facility (all
kinds of waste), etc. Initially, this isolated system would be highly organized into a number
of very specific entities, such as plant life. But, to cook food it would be necessary to cut
down a tree and burn the wood. As time went on, the ultra-high organization of plants, minerals, and animals would become less organized—or more dispersed throughout the room.
Every time some action took place within the isolated system, there would be a ΔS that
would be positive, even in those cases where some parts would have a decrease in S. That
is, overall even though the energy within the system would remain conserved (unchanged),
the entropy would increase. As the system became less and less organized—more uniform in
nature (every fire reduces a highly organized group of molecules that made up the fiber of a
tree into a uniform pile of carbonized molecules that are all pretty much the same)—the
entropy increases. The end result is a system that can no longer increase its entropy and at
that point everything is pretty uniform—or in a state of chaos, if you prefer that term.
57
Suggestions for Instructors
1. The laws of thermodynamics are hardly understandable without the general equation of state
for gases, namely pV = kNT or pV = nRT, where k is the Boltzmann constant, N is the number of molecules of the material, n is the number of moles, R is the gas constant, and
T is the temperature of the material in kelvin. Recall that this equation of state produces
Boyle’s law when the absolute temperature is held constant (isothermal case); that is, pV =
a constant. When the pressure or the volume is held constant we get Charles’ law and GayLussac’s law, respectively. These empirical laws, along with the relation for the internal energy of a system, U = (3/2)kNT or U = (3/2)nRT, are sufficient to do most heat problems until
entropy must be defined.
2. Entropy is a rather difficult concept to explain. I suggest you use examples such as placing a
drop of ink into a glass of water. Initially the system is "structured" with all the ink in one
place. The entropy of the ink drop and water system is low. When the ink disperses
throughout the water and the ink-water system is uniform, the entropy is higher. Some
would say the second state is more chaotic than the first, while some insist the second state
is more uniform. So exactly what is meant by chaos must also be clearly defined. The natural tendency of an enclosed system is to go toward uniformity, or the highest state of
entropy (or the most probable state—that’s where statistics comes into play). If your students
would like to investigate an interesting experiment in an isolated system, have them look at
the Web site of the Biosphere II, located in Tucson, Arizona: www.bio2.com.
3. Drawing pV diagrams and placing on them various processes is an excellent exercise.
Isothermal pV lines are parabolas [because pV = constant], constant pressure lines are horizontal, and constant volume lines are vertical. Isentropic [i.e., Q = 0] lines are steeper than
isotherms and you’ll have to get into exponentials to get the right shape. Nevertheless, all
lines are interesting to draw. And don’t forget—the area enclosed within a closed path on a
pV diagram represents the total work done during the cycle.
58
Contributors
The Core Curriculum Series: Physics was designed by the Standard Deviants Academic Team,
including:
Edward Finn, Ph.D., Georgetown University
Barry Berman, Ph.D., George Washington University
Headed by:
David Sturdevant, Director of Writing, MFA, University of Louisville
with assistance from Margaret Packard, Films for the Humanities & Sciences.
59
Films for the Humanities & Sciences
A Wealth of Information. A World of Ideas.
PO Box 2053, Princeton, NJ 08543-2053
CALL 800-257-5126 • FAX 609-671-0266
10546
for use with
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