Download Assessing to Learn in the Classroom Sample Questions

Assessing to Learn in the Classroom
Sample Questions
Ian Beatty
William Gerace
William Leonard
University of Massachusetts-Amherst
Question Topic
Page
Electromagnetism: Electric Forces and Fields
2
Electromagnetism: Charge Distribution and Electric Fields
5
Forces; Constrain Forces; Static Friction
8
Forces Acting on an Object
10
Kinematics Acceleration
12
Work and Force
15
Work Dependence on Force and Distance
17
Work Dependence of Component Force
18
Forces Relating to Acceleration
21
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Assessing to Learn in the Classroom Sample Questions
Question 1a
Description: Electromagnetism: electric forces and fields
The diagrams below show two uniformly charged spheres. The charge on the right sphere
is 3 times as large as the charge on the left sphere. Which force diagram best represents
the magnitudes and directions of the electric forces on the two spheres?
1.
2.
3.
4.
5.
Commentary
This question should be paired with Question 1b, presented one after the other and then
discussed jointly. The commentary on 1b addresses this question as well.
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Question 1b
Description: Electromagnetism: electric forces and fields
The diagrams below show two uniformly charged spheres. The charge on the right sphere
is 3 times as large as the charge on the left sphere. Each arrow represents the electric field
at the center of one sphere created by the other. Which choice best represents the
magnitudes and directions of the electric field vectors created by one sphere at the
location of the other sphere?
1.
2.
3.
4.
5.
Commentary
This question should be paired with Question 1a; this discussion addresses both.
Purpose: The purpose of these questions is to sensitize you to the distinction between
electric fields and electric forces, and to how they are related.
Discussion: According to Newton's Third Law, the force exerted on the left sphere by the
right one must be equal in magnitude and opposite in direction to the force exerted on the
right sphere by the left. (4) is therefore the only possible answer to 1a.
According to Coulomb's law, the electric field created by a sphere is proportional to the
amount of charge it has. Note that in S1b, the field at the center of each sphere is created
by the other sphere. The sphere with the smaller charge will therefore have a larger
electric field acting on it, so (5) is the only possible answer to 1b.
How can the electric fields have different magnitudes when the forces have the same
magnitudes? Because the force caused by an electric field is proportional to the charge of
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Assessing to Learn in the Classroom Sample Questions
the object being acted upon as well as to the magnitude of the electric field. Each force
therefore depends on the product of both charges; the fields, however, each depend on
only one charge.
Key Points:
•
Be careful not to confuse electric fields with the electric forces they cause.
•
Electric fields depend only on the charge creating them, whereas electric forces
depend on the charge being acted upon as well.
•
Mind your qs and Qs: when an equation depends upon a charge, make sure you
understand which charge is meant.
For Instructors Only
We recommend presenting 1a and 1b back-to back, with no discussion or revelation of the
"right" answer between the two.
The question pair is an example of the "compare and contrast" tactic: by juxtaposing two
very similar questions, we sensitize students to the one feature that differs, helping them
to distinguish and hone related concepts; simultaneously, we help them develop their
understanding of how the concepts are related, clustering the knowledge.
1b also reveals students' tendency to throw the variable "Q" around without being careful
about whether it represents the acting or acted-upon charge. By letting students make an
error because of this and then explicitly bringing the tendency and its harmful
consequences to students' conscious awareness, we help them learn to overcome it.
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Question 2
Description: Electromagnetism: charge distributions and electric fields: qualitative
reasoning.
All charged rods have the same length and the same linear charge density (+ or –). Light
rods are positively charged, and dark rods are negatively charged. For which arrangement
below would the magnitude of the electric field at the origin be largest?
1.
2.
3.
4.
5.
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6.
7.
8. Impossible to determine.
Commentary
Purpose: To develop your ability to reason qualitatively about electric fields. No
equations or numbers are necessary, only a basic understanding of vector addition,
symmetry, superposition, and the direction of electric fields caused by positive or negative
charges. The ability to reason qualitatively about electric fields, and in fact about anything
in physics, is crucial to "understanding" and to solving problems.
Discussion: Each rod will cause an electric field at the origin that points either directly
away from (for positive rods) or towards (for negative rods) the center of the rod. (Any
other orientation would violate symmetry.) Since each rod has the same magnitude of
charge, all such fields will have the same magnitude. Now, all you need to do is draw
electric field vectors at the origin due to each rod, and then add up those vectors to get the
total. Note that components of vectors will frequently cancel.
For example, in (1), the two rods cause electric fields that are down and to the right, and
up and to the right; adding these together results in a net field that is directly to the right,
as the horizontal components add and the vertical components cancel. (2) is identical to
(1), because the negative rod on the top right in (2) causes a field towards itself that is
exactly the same as the field away from the bottom left rod in (1).
In (3), the fields due to the two rods cancel exactly: same magnitudes, opposite directions.
The same argument applies to (5) and (7).
(6) can be thought of as a copy of (2) rotated by 90° so that its net field points downward,
superposed with (added to) a mirror image of the same thing (the left half of the diagram)
with a net field that also points downward. Thus, the total field for (6) must have a
magnitude twice as large as that of (2). Similar arguments can be used to compare (6) to
(1) and to (4). By this kind of logic, you can deduce that arrangement (6) must produce
the greatest net field magnitude without ever using an equation.
Key Points:
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•
Graphical representations (such as vector diagrams) and qualitative reasoning can
be powerful tools for solving problems without using equations.
•
The concepts of symmetry and superposition can help you deduce much about a
physical situation.
•
To find the electric field of a complex charge arrangement, you can break it into
simpler pieces, find the electric field of each, and then add up the fields to find the
total field.
For Instructors Only
Students will only pay attention to qualitative reasoning and to graphical representations
as thinking tools if they see problems where these approaches are clearly superior. This is
such a problem.
This question also reveals the degree of student's comfort and facility with vector
addition, superposition, and symmetry. The instructor should devote significant time to
this problem if students reveal weakness here; asking students to draw or describe the
actual vector arrows for each of the rods in each situation, and visually combining them
into a resultant, can be helpful.
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Question 3
Description: Forces: constraint forces: static friction.
A mass of 5 kg sits at rest on an incline making an angle of 30° to the horizontal. If μs =
0.7, what is the friction force on the block?
1. 43.3 N, down the incline
2. 25 N, up the incline
3. 10 N, down the incline
4. 30.3 N, up the incline
5. None of the above
Commentary
Purpose: This problem will help you to understand how to determine the value of the
static friction force. Static friction is a "constraint" force — a force that isn't defined by a
formula, but rather takes on whatever value is necessary (within limits) to satisfy
Newton's Second Law. The normal force and tension force are other constraint forces. We
sometimes call them "procedural" forces because one must resort to a procedure, rather
than a specific equation, to determine their value.
Discussion: Many students will say that the magnitude of the friction force is the
coefficient of friction μ times the normal force N (which must itself be determined
procedurally, though that is straightforward here). That would be correct for kinetic
friction, where fk = μkN; however, here we are dealing with static friction, where fs ≤ μsN.
Because of the inequality, it only gives us an upper limit on the magnitude; the actual
value will be whatever is required to keep the block stationary on the plane, i.e. to satisfy
Newton's Second Law with zero acceleration.
If you work out the numbers, (2) turns out to be the best answer: the component of the
block's weight parallel to the plane is 25N (downhill), and no other forces have a parallel
component, so the static friction force must be 25N up the plane. You don't even need to
determine the normal force, unless you want to check whether friction is strong enough
here to hold the block in place. (If the angle were larger than about 35°, it would not be.)
Key Points:
•
Constraint forces like normal, tension, and static friction forces cannot generally be
calculated from a formula, but will take on whatever value is required to satisfy a
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physical constraint on the object (such as "the block doesn't slide" or "the block
doesn't fall through the plane").
•
The "equation" for static friction, fs ≤ μsN, only gives the maximum possible value
of the force, not the actual value.
For Instructors Only
Choice (1) is the magnitude of the normal force, but pointing down the incline. (2) is
correct. (3) is a random distracter. (4) will be popular: it is the maximum static friction
force (μsN), which students will choose if they look for a formula to blindly plug into.
This problem can further the misconception that N = mg cos(θ) is a general formula for
determining the normal force. In order to combat that and enhance students' appreciation
of constraint forces, we suggest following this question with one in which the normal force
must be determined and is not equal to mg cos(θ). During discussion, you might ask
students what the normal force on he block in this problem is, and then ask them what it
would be if a second, smaller block were perched on top of it, or if a string were pulling
vertically upward on it.
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Question 4
Description: Forces: identifying forces acting on an object.
A block of mass m is on a rough surface, with a spring attached and extended. As the block
moves up the incline a small distance, how many forces are exerted on the mass?
1. 1 force
2. 2 forces
3. 3 forces
4. 4 forces
5. 5 forces
6. 6 forces
7. 7 forces
8. More than 7 forces
9. None of the above
10. Impossible to determine
Commentary
Purpose: An important step in many physics problems is identifying all the (relevant)
forces acting on an object. This problem provides an opportunity to practice that skill and
to consider some of the ambiguities and points of confusion that can arise.
Discussion: The question has many defensible answers. What matters is not which
answer you pick, but the validity of your reasoning. The most extreme answer is "an
uncountable number of forces, one for each atom outside the mass interacting with each
atom inside the mass." So (8) is not wrong, but this way of thinking is not useful: it does
not help us to analyze or understand the situation.
To identify the forces on an object, we identify all the objects interacting with it. In this
case, the object is touched by the plane, string, spring, and air, and the Earth pulls on it
gravitationally. The plane, string, and spring each exert a contact force on the object, and
the air exerts both buoyancy and drag (air resistance), so we can say there are six forces
acting: answer (6). (Warning: when you study electricity and magnetism, you will
encounter other forces that act "at a distance" the way that gravity does.)
However, we often separate the contact force of a surface into its two components and
treat them as distinct forces (normal and friction), so we can identify seven forces on the
object: normal (due to the plane), kinetic friction (also due to the plane), tension (due to
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the string), elastic (due to the spring), buoyancy (due to the air), air drag (also due to the
air), and gravitation (due to the Earth): answer (7).
But, buoyancy and air drag are generally negligible, so to solve a physics problem we
would usually ignore them an identify the other five as the ones we must consider. That
means answer (5) is defensible, and this would probably be the "expected" answer on a
physics exam, since these are the five we need to use to solve a typical problem. (If we
treated the force of the plane as one contact force rather than as distinct friction and
normal forces, we would have an argument for answer (4).)
So, one can make a plausible argument for answers (4), (5), (6), (7) or (8). What matters
is not which answer you choose, but whether your reasoning is cogent and defensible.
You should not count the "net force" along with the others: the net force is the vector sum
of all of the individual forces, and is not a separate force that needs to be considered or
accounted for.
Key Points:
•
To identify the forces acting on an object, identify all the objects that contact it;
each one will exert a force. If the object has mass and the problem is not in "outer
space", the Earth will also exert a gravitational force.
•
Some contact forces are split into "pushing" and "sliding" components and treated
as two separate forces, a normal one and a frictional one.
For Instructors Only
Part of learning to do physics is learning to make the "right" assumptions — the
assumptions that experts do. This is a process of acculturation as well as of understanding
physical laws, and we can help students along by explicitly discussing the assumptions we
make and the ways we interpret questions. This question demonstrates a powerful tactic
that involves presenting students with a question with multiple defensible answers, and
then discussing the various answers and demonstrating the reasoning an expert might use
and the standard assumptions she might make. We strongly recommend downplaying the
notion of "right" and "wrong" here in favor of cogency.
When introducing forces to your students, we suggest identifying all valid forces, even if
you will routinely neglect some of them in the future.
Most experienced teachers and students consider the normal and friction forces separate
and independent, even though they are simply components of one contact force exerted
by the incline. Students need to understand this if the advice to "count forces by
identifying all objects interacting with the body" is to be useful.
Some students will include other, "ridiculous" forces such as the gravitational attraction of
the Moon, the Sun, the planets, etc. This is a good opportunity to make a distinction
between "wrong", "right", and "right but not useful".
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Question 7
Description: Kinematics: acceleration: everyday usage vs. physicists' definition.
Source: Leonard P151
Consider the following situations:
•
a car slowing down at a stop sign
•
a ball being swung in a circle at constant speed
•
a vibrating string
•
the Moon orbiting the Earth
•
a skydiver falling at terminal speed
•
an astronaut in an orbiting space station
•
a ball rolling down a hill
•
a person driving down a straight section of highway at constant speed with her foot
on the accelerator
•
a molecule in the floor of this room
In how many of the situations is the object accelerating?
1. One of the situations.
2. Two of the situations.
3. Three of the situations.
4. Four of the situations.
5. Five of the situations.
6. Six of the situations.
7. Seven of the situations.
8. Eight of the situations.
9. Nine of the situations.
10. None of the situations.
Commentary
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Purpose: This question is used to introduce the concept of acceleration, explore
students' pre-existing definitions of the term, and contrast "everyday" usage of the word
with the way physicists use it.
Discussion: In physics, "acceleration" means the rate of change of velocity with respect
to time. The more the velocity changes during a certain interval (say, 1 second), the larger
the acceleration is.
There are three ways in which the velocity can change: (1) the speed increases; (2) the
speed decreases; or (3) the direction of motion changes. If any one of these occurs, then
the object is said to be "accelerating". In other words, unlike everyday usage,
"accelerating" does not necessarily mean "speeding up".
When a car is slowing down, we say it is accelerating. We generally avoid the phrases
"deceleration" and "negative acceleration" as unnecessary and unhelpful. We avoid
"negative acceleration" because an object can have a negative value for acceleration and
still be speeding up — for example, if it is moving in the negative direction.
When a ball is moving in a circle, its direction of motion is constantly changing, so it is
accelerating even if its speed is constant.
A vibrating string (except for its endpoints) is accelerating at most instants in time,
though it happens so quickly that we cannot perceive the changes in velocity. The only
time it is not accelerating is when its speed is maximum. When the string stops and
reverses direction at the extremes of its motion, its acceleration is largest.
The Moon orbiting the Earth is moving in a circle, so it is accelerating.
A skydiver moving at terminal speed has a constant velocity and thus zero acceleration.
An astronaut in an orbiting space station may not be moving relative to the space station,
but the station is not a proper frame of reference since it is accelerating as it orbits the
Earth. (In this course, a proper frame of rereference has no acceleration.) The Earth is a
better frame of reference, and the astronaut is moving in a circle around it, so she is
accelerating.
A ball rolling down a hill generally speeds up.
A person (or car, or any object) moving at constant speed along a straight line has
constant velocity, and thus zero acceleration. This is true even though we must press the
"accelerator" pedal to maintain constant velocity.
A molecule in a solid that is not at absolute zero temperature is vibrating, so like the
vibrating string it is accelerating most of the time. The solid as a whole may still be at rest.
Key Points:
•
An object's acceleration is the rate at which its velocity is changing.
•
If an object's speed or direction of motion changes, we say it is "accelerating".
•
We avoid using the phrases "deceleration" and "negative acceleration" to mean
"slowing down".
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For Instructors Only
A nice feature of this style of question ("how many of the following...") is that even if a
student answers the "correct" number of objects (7), there are 36 different combinations
to arrive at this answer, so there is really no need to focus any attention at all on the
correctness of students' multiple-choice answers. Rather, class discussion should proceed
immediately to considering each situation independently and applying the definition of
acceleration.
We encourage students to wrestle with their own definitions at this stage. This question
may be used prior to any formal presentation of acceleration, since everyone already
"knows" the term; the disagreements that typically arise prepare students to appreciate
and better understand the formal definition.
It is not too early to mention forces, at least informally. Many students already have some
ideas about force that can help them sort out what you are trying to explain regarding
acceleration. For example, in the cases that have nonzero acceleration, identify the
force(s) causing it; in cases with zero acceleration, discuss how any forces present balance
out.
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Question 8a
Description: Work: integrating with force ideas.
Source: Leonard P151
Two blocks with mass m2 > m1 are pulled as shown along horizontal surfaces, each at a
constant speed v. Which tension force does more work moving the block a distance D
along the surface?
1. T1
2. T2
3. Neither; both do zero work
4. Neither; both do the same amount of negative work
5. Neither; both do the same amount of positive work
6. Impossible to determine without knowing the masses
7. Impossible to determine without knowing the coefficients of friction
8. Impossible to determine without knowing the speed v
9. Impossible to determine for some other reason
Commentary
Purpose: To integrate force ideas and work ideas.
Discussion: In each case, the block is not accelerating, so the net force is zero. Thus, the
tension force must be equal in magnitude to the kinetic friction force, and the normal
force must be equal in magnitude to the weight of the block. The kinetic friction force is
proportional to the normal force (fk = μkN). The normal force is larger in case 2, since the
mass (and hence the weight) is larger.
The work done is the tension multiplied by the distance D. The distances are the same, so
the question becomes, "Which tension force is larger?" Since the accelerations are zero,
this is equivalent to asking "Which friction force is larger?" If we assume that the surfaces
are the same and the blocks are made of similar material, then the coefficients of friction
will be the same. With this assumption, case 2 will have a larger friction force, which
means T2 is larger and therefore does more work. So, answer (2) is defensible.
On the other hand, since we are not told anything about the coefficients of friction, answer
(7) is defensible.
Students might think that no work is being done on the blocks, since the kinetic energy is
not changing. However, the question asks about the work done by a particular force, not
the net work.
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Students might ignore friction. Since the blocks are moving at constant velocity, friction
must be present to oppose the tension force.
(Note that if the tension were being applied at an upward angle, the problem would
become considerably more complicated.)
Key Points:
•
The work done by a particular force is different from the total work done on an
object.
•
The work done by a particular force depends on the displacement of the object and
the component of the force along the direction of motion.
•
Even though we are not given numerical values for the masses, speed, etc., we can
still reason through to an answer under certain assumptions.
•
There is more than one defensible answer to this question. Make sure your answer
is consistent with your chosen assumptions, and make sure you are aware of the
assumptions you are making.
For Instructors Only
This is the first of three related questions designed to explore student reasoning about
work, the normal force, and the kinetic friction force. Here we highlight the reasoning
needed to understand work. Even though we are talking about work and energy ideas, we
still need to be able to analyze situations using forces; we cannot forget all that useful
stuff!
Note that it is common to say that an object moving at constant speed has "no
acceleration". It is more proper to say "zero acceleration", since the object still has an
acceleration; its value is zero.
Yes, this question is ambiguous: should students assume that the coefficients of friction
are the same? Rather than a liability, this ambiguity is a strength of the question: it
focuses students' attention on the significance of that quantity and the implications of
assuming one way or another. We suggest not mentioning anything about the coefficients
of friction before or while students answer the question; instead, let disagreement among
students raise the topic naturally. Then, stress that neither assumption (and
corresponding answer) is "right"; rather, the point is the implications of the choice and
the cogency of subsequent reasoning.
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Question 8b
Description: Work: definition of work: dependence on force and distance, not velocity
or time.
Two blocks with mass m2 > m1 are pulled as shown along rough horizontal surfaces with
the same coefficient of friction. Both are pulled at constant speed, but v2 < v1. Which
tension force does more work moving the block a distance D along the surface?
1. T1
2. T2
3. Neither; both do zero work
4. Neither; both do the same amount of negative work
5. Neither; both do the same amount of positive work
6. Impossible to determine without knowing the masses
7. Impossible to determine without knowing the speeds
8. Impossible to determine for some other reason
Commentary
Purpose: To integrate force ideas with work ideas, exploring the definition of work and
the quantities it does and does not depend upon.
Discussion: There are four forces on each block: (1) gravitation, down; (2) normal, up;
(3) tension, left; and (4) kinetic friction, right. Since the vertical component of
acceleration is zero, the normal force balances gravitation and is therefore larger for the
heavier block (case 2). Since the coefficients of friction are the same, and fk = μkN, the
friction force is larger in case 2 as well. Since neither block is accelerating, the tension
force must balance the kinetic friction force, so the tension force is larger in case 2. Both
blocks move through the same distance, so according to the definition of work, T2 does
more work.
How about the sign? Both forces do positive work, even though they are pointed to the
left. The displacement is also to the left, so the component of tension along the direction
of motion is positive. Also, think about what would happen without any friction: Both
blocks would speed up, which means tension (the only force doing work) is doing positive
work.
The result does not depend on the speed of the block, since work depends on force and
displacement (not velocity or time). Block 2 takes more time to move a distance D, but
that does not affect the work.
Key Points:
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•
Work is positive when the component of force along the direction of motion is
positive, and negative when the component is negative. Negative work does not
mean work in the negative (leftward) direction; it means work due to a force that
opposes the object's motion.
•
Work depends on force and displacement, not velocity or time. Because forces are
related to acceleration, work is also related to acceleration.
•
Two objects moving at constant velocity have the same acceleration (zero), even
when they have different speeds.
For Instructors Only
This is the second of three related questions using similar situations. In the first, we
considered the impact of different masses (and left the coefficients of friction ambiguous).
In this, we have equal masses but vary the block speeds.
Students commonly think work is a vector, whose sign is used to indicate direction; such a
misconception should be elicited, made explicit, and dispelled.
A forward reference to the concept of power may be of benefit here.
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Question 8c
Description: Work: applying the definition: dependence on component of force.
Two identical blocks are pulled as shown along rough horizontal surfaces with the same
coefficient of friction. Both are pulled at constant speed v. Which tension force does more
work moving the block a distance D along the surface?
1. T1
2. T2
3. Neither; both do zero work
4. Neither; both do the same amount of negative work
5. Neither; both do the same amount of positive work
6. Impossible to determine without knowing the mass m
7. Impossible to determine without knowing the speed v
8. Impossible to determine without knowing the angle of T2
9. Impossible to determine for 2 of the reasons above
10. Impossible to determine for some other reason
Commentary
Purpose: To integrate force ideas with work ideas, highlighting the role of a force's
components in determining the work it does.
Discussion: There are four forces on each block: (1) gravitation, down; (2) normal, up;
(3) tension, right/up; and (4) kinetic friction, left. The normal force and the vertical
component of tension balance gravitation, so the normal force is larger in case 1. Since the
coefficients of friction are the same, and fk = μkN, the friction force is larger in case 1 as
well. Since neither block is accelerating, the horizontal component of the tension force
must balance the kinetic friction force, so this component is larger in case 1. The
displacement is horizontal, so the work done depends only on the horizontal component
of tension. Therefore, T1 does more work.
T2 might be larger than T1, depending on the angle of the rope, but this is not relevant to
the question asked. It is not the larger force that does the larger amount of work, but the
force with the larger component along the direction of motion.
Key Points:
•
The work done by a force depends on the component of that force along the
direction of motion. A large force does a small amount of work when the
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Assessing to Learn in the Classroom Sample Questions
component along the direction of motion is small. And when the component is
zero, the work is zero.
•
The normal force is not always equal to the weight of the object. The normal force
is a "constraint" force, which means it will balance all forces with components
perpendicular to the surface.
For Instructors Only
This is the last of three related questions using similar situations. In the first, we
considered the impact of different masses (and left the coefficients of friction ambiguous);
in the second we had equal masses but varied the block speeds. In this, masses and speeds
are equal, and we instead vary the angle and perhaps magnitude of the tension force.
This question highlighs the fact that the normal force is a constraint force (rather than as
a force with an empirical law). It also highlights the fact that only the component of a
force along the direction of motion contributes to the work done.
Some students may not notice that the masses of the blocks are the same. (In the two
previous cases, the lower block was more massive than the upper one.) Rather than
heading this off, let it happen and then come to light during the discussion. This is of
greater long-term benfit, helping students learn to be more careful in the future.
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Ian Beatty, William Gerace and William Leonard, University of Massachusetts-Amherst
Question 9
Description: Forces: relating to accelerations: tension as a constraint force, qualitative
reasoning.
Source: Leonard P151
Consider the three situations below, labeled A, B, and C. Ignore friction.
After each system is released from rest, how do the tensions in the strings compare?
1. A = B = C
2. B = C < A
3. A = C < B
4. A < B < C
5. A < C < B
6. B < A < C
7. B < C < A
8. C < A < B
9. C < B < A
10. Impossible to determine
Commentary
Purpose: To reason qualitatively about forces and accelerations, and to recognize that
tension is a "constraint force" with a situation-dependent magnitude.
Discussion: It is common for students to think that the tension is equal to 20N in all
three cases. However, this is impossible, otherwise the net force on each hanging mass
would be zero, and it would not accelerate.
In A, the 2kg block does not move or accelerate, so the tension in the string balances
gravitation, which is 20N.
In B, both blocks speed up. This means that the tension is the string must be smaller than
the weight of the hanging block, so that there is a net force acting downward on the 2kg
(hanging) block.
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Assessing to Learn in the Classroom Sample Questions
Likewise, in C, both blocks speed up, so the tension in the string must be smaller than the
weight of the hanging block.
So we need to ask "In which case, B or C, is the acceleration larger?" because the case with
the larger acceleration will have the larger net force and thus the smaller tension.
Intuitively, the acceleration of the blocks in B must be largest, B would have the smallest
tension. Thus, B < C < A = 20N.
We can also use extreme case reasoning to distinguish B and C without solving any
equations. Imagine the 1kg block in B is only 1 gram (smaller than a dime), and imagine
that the upper 2kg block in C is 2000kg (as big as a car). In this case, the hanging block in
B would fall at nearly 10m/s/s and the tension would be almost zero. And the hanging
block in C would barely accelerate at all and the tension would be just slightly less than
20N.
Key Points:
•
Tension is a "constraint" force, which means there is no formula or empirical law to
determine its value. Instead, the tension adjusts to the given situation, taking on
whatever (positive) value is required to keep the string a fixed length.
•
You do not have to solve these situations in order to compare the tensions.
Qualitative reasoning is all you need to do the comparison. Actually solving for the
tensions is much more work, and you might not even believe the result when you
get it. Reasoning through to an answer can increase your understanding.
•
"Extreme case reasoning" — imagining the situation with exaggerated quantities —
is a useful technique for making qualitative comparisons and developing intuition
for a situation.
For Instructors Only
This style of problem — comparing the value of a quantity across similar situations — is
powerful for drawing attention to particular features and for demonstrating the utility of
qualitative reasoning. The instructor should strongly resist students' desire for
quantitative calculations! Extreme case reasoning may be new and uncomfortable to
students, but this is a good situation to engage them in it.
Making connections to other constraint forces, such as the normal and static friction
forces, can be helpful.
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