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AP PHYSICS B
Forces & Newton’s Laws
Teacher Packet
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Forces & Newton’s Laws
Objective
To review the student on the concepts, processes and problem solving strategies necessary to
successfully answer questions on forces and Newton’s laws.
Standards
Forces and Newton’s laws are addressed in the topic outline of the College Board AP* Physics
Course Description Guide as described below.
I. Newtonian Mechanics
B. Newton’s laws of motion (including friction and centripetal force)
1. Static equilibrium (first law)
2. Dynamics of a single particle (second law)
3. Systems of two or more bodies (third law)
F. Oscillations and Gravitation
4. Newton’s law of gravity
AP Physics Exam Connections
Topics relating to forces and Newton’s laws are tested every year on the multiple choice and in
most years on the free response portion of the exam. The list below identifies free response
questions that have been previously asked over forces and Newton’s laws. These questions are
available from the College Board and can be downloaded free of charge from AP Central.
http://apcentral.collegeboard.com.
2008
2007
2006
2005
2003
2000
2000
Free Response Questions
Question 2
2008 Form B Question 2
Question 1
2007 Form B Question 1
Question 2
2006 Form B Question 1
Question 1
2005 Form B Question 1
Question 1
2003 Form B Question 1
Question 1
Question 2
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Forces & Newton’s Laws
What I Absolutely Have to Know to Survive the AP* Exam
Force is any push or pull. It is a vector. Newton’s Second Law is the workhorse of the AP Physics B exam. It allows
you to write down mathematical relationships that are true. Thus, for a single body, if you pick any direction and sum
up all the positive and negative forces that act on the body along that line, the sum will equal the product of the
body’s mass and its acceleration along that line. A Free Body Diagram allows you to identify all of the forces acting
on a single body. Neglect one force or add a fictitious force on your FBD and you are in trouble.
•
•
•
Newton’s 1st Law: in an inertial frame of reference, an object in a state of constant velocity (including
zero velocity) will continue in that state unless impinged upon by a net external force. If ΣF=0, then a=0
and the object is at rest or moving at a constant velocity in a straight line. The converse is true also, if an
object is in a state of constant velocity (including zero velocity) then a=0 and ΣF=0.
Newton’s 2nd Law: A net force acting on a mass causes that mass to accelerate in the direction of the net
force. The acceleration (vector) is directly proportional to the net force (vector) acting on the mass and
ΣF
or ΣF = ma
inversely proportional to the mass of the object being accelerated. a=
m
Newton’s 3rd Law: For every action force, there exists an equal and opposite reaction force. Let’s say
you hit a table with your fist. Doing so, applies a force to the table which, if great enough will break the
table. Likewise, the table applies a force to your fist which, if great enough, will break your fist. The
size and direction of the force you apply must be equal and opposite the force the table applies to you.
Hence, the only way forces can be generated are in action/reaction pairs which occur on different
objects. If you try to apply 800 Newtons of force to a table that can only provide 600 Newtons of
reaction force back on you, you will never succeed. The table will break as soon as you exceed 600
Newtons, which is the maximum force it can apply to you.
Key Formulas and Relationships
ΣF = Fnet = ma
FW = mg
F f s max ≤ μs FN ( static )
F f k = μk FN ( kinetic )
FG =
Gm1m2
r2
ΣF = Sum of the forces is the Net Force
Newtons (N) =
kgm
s2
a = acceleration m = mass
FW = weight
g = acceleration due to gravity
Ff s max = maximum static frictional force
Ff k = kinetic frictional force
FN = normal force
FG = gravitational force
r = distance between the centers of two masses
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Forces & Newton’s Laws
Basic Kinds of Forces
Weight
Fg , W
Gravitational
FG
Fg = mg
Always directed toward the Earth’s center.
Force on a free falling body, if we neglect air friction.
mm
FG = G 1 2 2
r
A force of attraction between any two massive objects.
When the Earth is one of the two bodies involved, then the force felt by
the second body while positioned on the Earth’s surface will always be
directed toward the Earth’s center.
A force of support, provided to an object by a surface in which the object
is in contact.
Always directed perpendicular to and away from the surface providing
the support.
FN
Normal
FN , N
W
In the figure above, a box is supported by a table. The figure shows all
the forces acting on the box and is called a Free Body Diagram (FBD). If
a box, rests on a level table, then the FN = W = mg . Notice that the
normal force sometimes equals the weight but not always.
FN
y
x
mgcosθ
θ
W
If the box is placed on an inclined plane, then the FN = mg cos θ , the
component of the weight that is equal and opposite the normal force. For
the inclined plane above, the normal force and the weight are not equal
and not even in the same direction.
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Forces & Newton’s Laws
Friction is produced by the atomic interaction between two bodies as
they either slide over one another (kinetic friction) or sit motionless in
contact with one another (static friction).
F f s max ≤ μs FN (static)
Static Friction opposes the intended direction of relative sliding. The
static frictional force will only be as high as it needs to be to keep the
system in equilibrium. If successively greater and greater forces are
applied, the static frictional force will counter each push with a force of
equal and opposite magnitude until the applied force is great enough to
shear the bonding between the two surfaces. When you calculate f s or
F f s using the equation above, you are finding the maximum static
frictional force, one of an infinite number of possible frictional forces
that could be exerted between the two bodies. μs is a proportionality
constant called the coefficient of static friction. It is the ratio of the static
frictional force between the surfaces divided by the normal force acting
on the surface.
Friction
Ff , f
F f k = μk FN (kinetic)
Kinetic Friction or Dynamic Friction or Sliding Friction is always
opposite the direction of motion. The statement that kinetic friction is a
function of the normal force only (surface area is independent) is true
only when dealing with rigid bodies that are sliding relative to each
other. When you calculate f k or F f using the equation above, you are
finding the single, constant kinetic frictional force that exists between the
two bodies sliding relative to one another. No matter their velocity
(assuming heating does not alter the coefficient of kinetic friction) the
kinetic frictional force will always be the same. μk is a proportionality
constant called the coefficient of kinetic friction. It is the ratio of the
kinetic frictional force between the surfaces divided by the normal force
acting on the surface
WARNING! The two quantities f s and f k may look the same, but they
tell us different things. Kinetic friction is typically less than static friction
for the same two surfaces in contact. Note that the normal force
sometimes equals the weight but not always. When you draw a free body
diagram of forces acting on an object or system of objects, be sure to
include the frictional force as opposing the relative motion (or potential
for relative motion) of the two surfaces in contact.
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Forces & Newton’s Laws
FT is a force that is applied to a body by a rope, string, or cable.
FT is applied along the line of the string and away from the body in
question.
FT , T
Tension
FSubscript
Applied
Push me, pull you force that does not fall into one of the above
categories, for example, a friend shoves you. The magnitude of the force
is characterized by an F, with any subscript that makes sense to solve the
problem. Later in the year, you will encounter additional forces, like the
electric and magnetic forces.
Strategy on Force Problems
1. Take one body in the system and draw a Free Body Diagram (FBD) for it.
2. Choose x and y axes and place them beside your FBD. One axis must be in the direction of the
acceleration you are trying to find. If there is no acceleration, then ΣF = 0 .
3. If there are forces on the FBD that are not along the x and y directions, find their respective x
and y components.
4. Using Newton’s 2nd Law, sum the forces in the x direction and set them equal to ma x . If a
second equation is needed, sum the forces in the y direction and set them equal to ma y .
5. Repeat the above process for all the bodies in the system or until you have the same number of
equations as unknowns and solve the problem.
Effective Problem Solving Strategies
Free Body Diagram
(FBD)
A Free Body Diagram is normally depicted as a box showing all the forces
acting on the body. These forces are depicted as arrows. They don’t have to be
drawn to scale, but they should have a length that is appropriate for their
magnitude. Also, the force vectors do need to be directionally accurate and
labeled. Do not include components of the force vectors on your FBD.
When drawing Free Body Diagrams show only the force(s) that act on the body
in question and do not show forces that the body applies to other bodies. Also,
do not include velocity or acceleration vectors on your free body force
diagrams, since you will lose points for extraneous vectors on your FBD.
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Forces & Newton’s Laws
Example 1
A rope supports an empty bucket of mass 3.0 kg. Determine the tension in the rope when the bucket is (a)
at rest and (b) the bucket is accelerated upward at 2.0 m/s2.
Solution
Step 1 Draw a Free Body Diagram for each body in the system.
FT
W
Step 2 Choose x and y axes. In this case we could choose the y axes to be in the same dimension as the
tension and the weight.
Step 3 If there are off axes forces, then find the x and y components.
Step 4 Using Newton’s 2nd law, sum the forces in the y dimension and set equal to may.
(a) When the bucket is at rest, the net force is zero, so that the tension in the rope equals the weight of the
bucket.
ΣFy = ma y
FT − mg = 0
⎛ m⎞
FT = mg = ( 3.0kg ) ⎜ 10 2 ⎟ = 30 N
⎝ s ⎠
(b) When the bucket accelerates upward, the net force is may.
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Forces & Newton’s Laws
ΣFy = ma y
FT − mg = ma
FT = mg + ma = m ( g + a )
m⎞
⎛ m
FT = 3.0kg ⎜ 10 2 + 2.0 2 ⎟ = 36 N upward
s ⎠
⎝ s
Example 2
In the diagram below, two bodies of different masses (M1 and M2) are connected by a string which passes
over a pulley of negligible mass and friction. What is the magnitude of the acceleration of the system in
terms of the given quantities and fundamental constants?
M1
M2
Solution
Step 1: Draw a FBD for each body in the system.
There are two forces acting on each of the bodies: weight downward and the tension in the string upward.
The tension is distributed throughout the string. The pulley (negligible mass and friction) changes only the
direction of motion, not the tension, so the tension is the same on each side of the pulley. Our FBDs should
look like this:
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Forces & Newton’s Laws
T
T
W1
W2
Step 2 Choose x and y axes. In this case we could choose the y axes to be in the same dimension as the
tension and the weight for both masses.
Step 3 If there are off axes forces, then find the x and y components.
Step 4 Using Newton’s 2nd law, for each body sum the forces in the y dimension and set equal to may
and/or the x dimension. For m1 the tension is positive and the weight is negative since the acceleration is
upward.
ΣFy = m1a y
FT − W1 = m1a
For m2 the tension is negative and the weight is positive since the acceleration is downward.
ΣFy = m2a y
− FT + W2 = m2a
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Forces & Newton’s Laws
FT − W1 = m1a
− FT + W2 = m2a
W2 − W1 = ( m1 + m2 ) a
a=
W2 − W1
m g − m1 g ( m2 − m1 ) g
= 2
=
( m1 + m2 ) ( m1 + m2 ) ( m1 + m2 )
Example 3
A block of mass m rests on a horizontal table. A string is tied to the block, passed over a pulley, and another
block of mass M is hung on the other end of the string, as shown in the figure below. The coefficient of
kinetic friction between block m and the table is μk. Find the magnitude of the acceleration of the system in
terms of the given quantities and fundamental constants.
m
μk
M
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Forces & Newton’s Laws
Solution
Step 1: Draw a FBD for each body. Note that according to the FBD, the vertical acceleration of the block on
the table is zero, since the normal force is directed upward and the weight force is directed downward.
There is a horizontal acceleration for the block since the tension is greater than the frictional force.
FN
T
T
fk
W1
W2
Step 2 Choose x and y axes. In this case we could choose the y axes to be in the same dimension as the
tension and the weight for both masses. We could choose the x axis to be in the same dimension as the
frictional force and the tension pulling the mass to the right.
Step 3 If there are off axes forces, then find the x and y components.
Step 4 Using Newton’s 2nd law, for each body sum the forces in the y dimension and set equal to may and/or
the x dimension.
For block m the net force vertically is 0 since the block is accelerating to the right and not upward nor
downward. Summing the forces in the y dimension we find that the normal force equals the weight of the
block.
ΣFy = ma y
FN − W1 = 0
FN = W1 = mg
Summing the forces in the horizontal dimension, the direction of the acceleration for block m, we find that
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Forces & Newton’s Laws
ΣFx = ma x
T − f k = ma
For mass two, summing the forces in the vertical dimension, the vertical acceleration for block M, we find
that
ΣFy = M a y
−T + W2 = Ma
Adding the two equations together we determine the acceleration of the system
T − f k = ma
−T + W2 = Ma
W2 − f k = ( m + M ) a
a=
W2 − f k
Mg − μk mg
=
(m + M ) (m + M )
a=
( M − μk m ) g
(m + M )
Since f k = μk FN = μk mg
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Forces & Newton’s Laws
Example 4
Three blocks of mass m1, m2, and m3 are connected by a string passing over a pulley attached to a plane
inclined at an angle θ as shown below.
m2
m1
θ
m3`
θ
The coefficient of kinetic friction between blocks m2 , m1 and the table is μk. Assuming that m3 is large
enough to descend and cause the system to accelerate, determine the acceleration of the system in terms of
the given quantities and fundamental constants.
Solution
Step 1: Draw a FBD for each body. When drawing a FBD, be sure the orientation of the box representing
the body is the same as the actual orientation of the body in the problem. Note the weight force is always
directed straight downward and the normal force perpendicular to the surface.
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Forces & Newton’s Laws
FN2
T2
T1
T2
FN1
T1
f
f
W2
W1
W3
Step 2 Choose x and y axes. Place axes next to FBD. One axis must be in the direction of the acceleration
you are trying to determine.
Step 3 If there are off axes forces, then find the x and y components.
There is one force that is off axis, the weight force. The weight of each block must be broken into two
vectors parallel and perpendicular to the inclined plane.
m1 the parallel and perpendicular components of the weight force are
wx1 = F& = m1 g sin θ
w y1 = F⊥ = m1 g cos θ
m 2 the parallel and perpendicular components of the weight force are
wx 2 = F& = m2 g sin θ
w y2 = F⊥ = m2 g cos θ
Step 4 Using Newton’s 2nd law, for each body, sum the forces in the y dimension and set equal to may. Sum
the forces in the x dimension and set equal to max.
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Forces & Newton’s Laws
m1
vertical motion perpendicular to the incline plane
ΣFy = 0
FN − m1 g cos θ = 0
FN = m1 g cos θ
horizontal motion parallel to the incline plane
ΣFx = m1a
T1 − f k − F& = m1a
T1 − μk m1 g cos θ − m1 g sin θ = m1a
m2
vertical motion perpendicular to the incline plane
ΣFy = 0
FN − m2 g cos θ = 0
FN = m2 g cos θ
horizontal motion parallel to the incline plane
ΣFx = m2 a
T2 − T1 − f k − F& = m2a
T2 − T1 − μk m2 g cos θ − m2 g sin θ = m2 a
m3
vertical motion
ΣFy = m3a
−T2 + W3 = m3a
−T2 + m3 g = m3a
Looking at the motion along the axis of acceleration, up the incline plane
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Forces & Newton’s Laws
T1 − μk m1 g cos θ − m1 g sin θ = m1a
T2 − T1 − μk m2 g cos θ − m2 g cos θ = m2a
T2 − μk g cos θ ( m1 + m2 ) − g sin θ ( m1 + m2 ) = m1a + m2 a
T2 − ( μk g cos θ + g sin θ )( m1 + m2 ) = m1a + m2 a
−T2 + m3 g = m3a
T2 = m3 g − m3a
m3 g − m3a − ( μk g cos θ + g sin θ )( m1 + m2 ) = m1a + m2a
m3 g − ( μk g cos θ + g sin θ )( m1 + m2 ) = ( m1 + m2 + m3 ) a
a=
m3 g − ( μk g cos θ + g sin θ )( m1 + m2 )
( m1 + m2 + m3 )
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Forces & Newton’s Laws
Free Response
Question 1 (15 pts)
Two blocks, both of mass 0.5 kg, are connected to each other by a thin string which is
passed over a pulley as shown in the diagram. Block 1 sits on a rough horizontal part
where the coefficient of static friction is 0.6 and the coefficient of kinetic friction is 0.4.
Block 2 sits on a frictionless incline which forms an angle θ with the horizontal.
1
2
θ
A. On the diagram below, draw and label vectors to represent the forces acting on block
2.
1 point for each correctly drawn and
labeled vector (tension, normal, weight)
(3 points max)
T
N
1 point deducted for each extraneous
vector with a minimum score being 0
W
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Forces & Newton’s Laws
B. Determine the maximum angle θ the incline can make without the blocks sliding
down.
(6 points max)
When block 2 is not sliding, block 1 is not
sliding and the ΣF=0, so the maximum
angle means maximal frictional force.
1 point for a correct statement that
when the blocks are not sliding the
ΣF=0
Block 2
ΣF& = 0
1 point for a correct application of
NSL to block 2 such that the tension
equals the parallel component of the
weight
T − F& = 0
T = mg sin θ
1 point for a correct application of
NSL to block 1 such that the tension
equals the maximum frictional force
Block 1
ΣF& = 0
T − f s max = 0
1 point for a correct statement that the
maximum angle occurs when the
frictional force is maximum on block 1
using 0.6 for the coefficient of friction
T = f s max = μs FN
Thus
μs FN = mg sin θ
1 point for a correct statement that the
maximum frictional force on block 1
equals the parallel component of the
weight on block 2
μs mg = mg sin θ
sin θ = μs = 0.6
θ = sin −1 ( 0.6 )
θ = 37°
1 point for the correct answer
including correct units and reasonable
number of significant digits
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Forces & Newton’s Laws
C. Assume θ is 30º, find the magnitude of the acceleration of the system
1 point for any indication that there
are two forces acting on Block 1
parallel to the surface
(6 points max)
Block 1
ΣF& = ma
1 point for a correct application of
NSL to block 1 that includes the two
forces (tension and friction) and a non
zero acceleration
T − f k = ma
T − μk mg = ma
Block 2
1 point for a correct statement that the
frictional force is kinetic and using
0.4 for the coefficient of friction
ΣF& = ma
−T + F& = ma
−T + mg sin θ = ma
1 point for any indication that there
are two forces acting on Block 2
parallel to the surface
T − μk mg = ma
−T + mg sin θ = ma
mg sin θ − μk mg = 2ma
g sin θ − μk g
=a
2
⎛ m⎞
⎛ m⎞
°
⎜ 10 2 ⎟ ( sin 30 ) − ( 0.4 ) ⎜ 10 2 ⎟
s ⎠
⎝ s ⎠
a=⎝
2
m
a = 0.5 2
s
®
1 point for a correct application of
NSL to block 2 that includes the two
forces (tension and parallel
component of the weight) and a non
zero acceleration
1 point for the correct answer
including correct units and reasonable
number of significant digits
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Forces & Newton’s Laws
Question 2 (10 pts)
A block of mass m rests on an air table (no friction), and is pulled with a force probe,
producing the Force vs. Acceleration graph shown below.
A. Determine the mass of the block.
Newton’s 2nd law states that m =
Fnet
. This
a
ratio is the slope.
m = slope =
ΔF
9 N − 3N
=
= 1.5 kg
Δa 6 m/s2 − 2m/s2
1 point for using the slope of the F vs. a
graph to determine the mass or the
correct use of Newton’s second law
1 point for the correct answer including
correct units and reasonable number of
significant digits
The block is now placed on a rough horizontal board having a coefficient of static friction
μs = 0.2, and a coefficient of kinetic (sliding) friction μk = 0.1.
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Forces & Newton’s Laws
B. What is the minimum value of the force F which will cause the block to just begin to
move?
The block will just begin to move when the
force F overcomes the maximum static
frictional force:
Fmin = f s = μs FN = μs mg
Fmin = ( 0.2 )(1.5kg ) (10 m/s2 ) = 3 N
1 point for the correct equation for
the maximum static frictional force
1 point for the correct answer or an
answer consistent with part (A),
including correct units and reasonable
number of significant digits
The block rests on the rough horizontal board. One end of the board is slowly lifted until
the block just begins to slide down the board. At the instant the block begins to slide, the
angle of inclination for the board is θ.
C. Determine the relationship between the angle θ and the coefficient of static friction, μs.
(6 points max)
At the instant the block is just about to move,
the maximum frictional force directed up the
incline is equal and opposite to the parallel
component of the weight down the incline,
and the normal force is equal and opposite to
the perpendicular component of the weight.
ΣFx = 0
f s max − F& = 0
f s max = mg sin θ
ΣFy = 0
FN = mg cos θ
Thus
f s max
FN
=
1 point for a correct statement of net
force perpendicular to the plane is zero
or FBD indicating the same
1 point for a correct statement that the
maximum static frictional force equals
the parallel component of the weight
or FBD indicating the same
1 point for a correct statement that the
normal force equals the perpendicular
component of the weight or FBD
indicating the same
FN − F⊥ = 0
μs =
1 point for a correct statement of net
force parallel to the plane is zero or
FBD indicating the same
1 point for a correct statement that μs
is the ratio between the parallel and
perpendicular components of the
weight
mg sin θ
= tan θ
mg cos θ
1 point for the correct relationship
between θ and μs
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Forces & Newton’s Laws
Multiple Choice
Questions 1 – 2
m
2m
A block of mass 2m is suspended by a string, the other end of which is passed over a
pulley of negligible mass and friction and tied to a block of mass m.
1. The acceleration of the system is
1
a) g
3
1
b) g
2
2
c) g
3
d) g
4
e) g
3
ΣF = ma
Find net force on object
T
T
mg
2mg
The forces acting on each block are shown above. From
this, we write two net force equations, one for each block
(the smaller block is block 1) and apply ΣF = ma.
∑ F1 = T − mg = ma
∑F
2
A
= 2 mg − T = 2ma
These equations can be added together, eliminating T.
ma = T − mg
+2ma = 2 mg − T
ma + 2ma = 2 mg − mg
Simplifying and solving for a gives
g
a=
3
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Forces & Newton’s Laws
2. The tension in the string is
1
a) mg
3
1
b) mg
2
2
c) mg
3
d) mg
4
e) mg
3
ΣF = ma
Find net force on object
Use either net force equation from above and substitute for
a to solve for T
ma = T − mg
⎛ g ⎞ = T − mg
⎟
⎝3⎠
m⎜
T =
T =
mg
3
E
+ mg
4mg
3
F
θ
m
3. An object of mass m is pushed to the right by a force F that forms an angle θ with the
horizontal as shown. The object moves with constant velocity. The coefficient of kinetic
friction between the object and the surface is
a) F cos θ
b) F sin θ
cos θ
c)
sin θ
F cos θ
d)
mg
F cos θ
e)
F sin θ + mg
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Forces & Newton’s Laws
Resolve for vector into
perpendicular and
parallel components
ΣF = ma
Ff = μFN
Object moving with
constant velocity has
zero net force.
Resolve the force vector F into components and draw a
FBD. We know the net force in both directions is zero,
since the velocity is constant.
FN
Fx
Fy
mg
E
Analyzing the forces in the vertical direction, we can write
FN = mg + Fy = mg + F sin θ and Ff = Fx = F cos θ
We now apply the equation for kinetic friction:
Ff = μ FN
Substituting from our other two equations:
F cos θ = μ ( mg + F sin θ )
And solving for μ gives
F cos θ
μ=
( mg + F sin θ )
4. A 75 kg man and a 25 kg girl stand facing each other on a frictionless sheet of ice.
The girl exerts a force of 30 N on the man, causing him to accelerate to the north. The
magnitude and direction of the reaction force on the girl is
Magnitude
Direction
a)
10 N
north
b)
10 N
south
c)
30 N
north
d)
30 N
south
e)
90 N
south
Reaction forces are equal
and opposite
The force exerted by the girl on the man is 30 N to the
north, therefore the reaction force must be equal (30 N) and
in the opposite direction (south)
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D
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Forces & Newton’s Laws
Questions 5 – 6
T
1 kg
F
2 kg
A block of mass 1 kg is sitting on top of a block of mass 2 kg. The 1 kg block is tied to a
thin thread which extends horizontally to a wall where the other end is attached. The 2
kg block is pulled by an external horizontal force F with constant velocity to the right.
The coefficient of kinetic friction between the two blocks is .5 and the coefficient of
kinetic friction between the 2kg block and the surface is .8.
5. As the bottom block is pulled to the right, the tension in the thread T is most nearly
a) .2 N
b) 1 N
c) 5 N
d) 8 N
e) 15 N
Ff = μFN
Object moving with
constant velocity has
zero net force.
We know the net force on the top block is zero (it remains
at rest). Drawing the forces on the top block:
FN
Ff,k1
T
m1g
The force of kinetic friction opposes the relative motion of
the two blocks. Since the bottom block moves to the right,
friction is attempting to pull the top block to the right so the
two don’t move relative to each other. From the above
diagram, we can see that T = Ff,k1. Substituting and solving
for T gives (1 designates the top block, 2 the bottom)
T = Ff 1,k = μk ,1 FN ,1 = μk ,1m1 g
(
T = ( .5 )(1kg ) 10 m
s
2
C
)
T = 5N
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Forces & Newton’s Laws
6. The force F required to pull the bottom block with constant velocity is most nearly
a) 15 N
b) 21 N
c) 24 N
d) 29 N
e) 34 N
Ff = μFN
Object moving with
constant velocity has
zero net force.
The net force on the bottom block is zero as well (it moves
with constant speed). The forces on the bottom block are:
FN
Ff,k2
F
Ff,k1
m1g
m2g
D
In the y-direction: FN = m1 g + m2 g
In the x-direction: F = Ff ,k 1 + Ff ,k 2
Ff,k1 is already known to be 5 N, so we find Ff,k2 and
substitute to find F.
F = 5N + μ
(
F = 5N + μ
m g+m g
k, 2 N
k, 2 1
2
⎡
F = 5 N + ( .8 ) 10 m
⎢⎣
s
2
)
( 2kg + 1kg )⎤⎥
⎦
F = 5 N + 24 N
F = 29 N
7. An object moves with constant velocity while three forces act on it. Which of the
following must be true?
I. the three forces have equal magnitude
II. the vector sum of the three forces is zero
III. the forces must be perpendicular to the direction the object is traveling
a) I only
b) II only
c) I and II
d) I and III
e) II and III
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Forces & Newton’s Laws
Object moving with
constant velocity has
zero net force.
Since the object is moving with zero acceleration, it must
experience zero net force: the vector sum of the three forces
must equal zero. No other requirement is needed; the
forces can be of any magnitude and direction so long as
their vector sum is zero.
B
8. A 4 kg mass is hung vertically from an ideal spring, which causes it to stretch 8 cm. If
the 4 kg mass is replaced by a mass of 8 kg, how far will the spring stretch?
a) 2 cm
b) 4 cm
c) 8 cm
d) 16 cm
e) 64 cm
Hooke’s Law: Fs = -kx
Hooke’s Law shows that the stretch of a spring is directly
proportional to the force it exerts: hence double the force on
a spring and the distance it is stretched will double as well.
Thus, the 8 kg mass with stretch the spring 16 cm.
D
9. As a 40 N ball falls through Earth’s atmosphere, it experiences a drag force, which is
given by the expression FD = −bv , where b is known to be 5 kg/s. The terminal velocity
of the ball close to Earth’s surface is
a) 5 m/s
b) 8 m/s
c) 25 m/s
d) 40 m/s
e) 45 m/s
Object moving with
constant velocity has
zero net force.
Terminal velocity occurs
when FD = Fg
The force of drag increases with the velocity, reducing the
acceleration until the force of drag equals the weight. At
this point we can write
FD = Fg
bvT = 40 N
vT =
B
40 N
= 8m
s
kg
5
s
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Forces & Newton’s Laws
Questions 10 – 11
m
θ
An object of mass m is at rest on an inclined plane as shown above. The inclined plane
forms an angle θ with the horizontal. The coefficient of static friction between the object
and the plane is μs.
10. Which of the following diagrams correctly shows the forces acting on the object?
a)
b)
d)
Resolve forces on an
incline
Direction of forces
c)
e)
D correctly shows the forces. Weight always pulls an
object down, a normal force is always perpendicular to the
surface, and friction is always parallel to the surface. In
this case, friction points up the ramp because it opposes the
motion of the object sliding down the ramp.
D
11. The force of static friction acting on the object is
a)
b)
c)
d)
e)
μs mg
μs mg cos θ
μs mg sin θ
mg cos θ
mg sin θ
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Forces & Newton’s Laws
Resolve forces on an
incline
Static friction is variable
In this case, the relationship between static friction and the
coefficient of static friction cannot be used because it is an
inequality: 0 ≤ Ff , s ≤ μ s FN . We must determine the force
of friction by force analysis. We know that the net force is
zero (it is at rest), so parallel to the ramp we have:
Ff , s = Fg , x
Resolving the weight vector and substituting:
θ
E
Fg,y = Fgcosθ
Fg
Fg,x = Fgsinθ
Ff , s = mg sin θ
12. Three forces act on an object that is moving at constant velocity. Two of the forces
are shown on the diagram above. Which of the following correctly shows the third force
vector?
b)
a)
d)
Object moving with
constant velocity has
zero net force.
c)
e)
In order for the object to move with constant velocity, the
vector sum must be zero. Adding the vector shown in
answer choice A would accomplish this.
®
A
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Forces & Newton’s Laws
13. A 20 N bucket is at the bottom of a 9 m deep well and is initially at rest. The bucket
is then pulled to the top of the well by a string that exerts a constant tension force, such
that it reaches the top in 3.0 s. During this interval, the tension in the string is
a) 4 N
b) 6 N
c) 20 N
d) 24 N
e) 40 N
Newton’s Second Law &
Kinematics
Use kinematics to find the acceleration of the bucket:
1 2
x = x0 + v0 t + at
2
x=
1
2
at
2x
2
2 ( 9m )
= 2m 2
s
t
( 3s ) 2
The forces on the bucket are:
a=
2
=
D
And the net force equation is
∑ F = T − mg
Substituting Newton’s Second Law and solving for T:
ma = T − mg
(
T = ma + mg = 2kg 2 m
s
2
) + 20N
T = 24 N
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Forces & Newton’s Laws
14. Which of the following graphs shows an object whose net force is zero?
I.
x
x
II.
III.
v
t
t
t
a) I only
b) II only
c) I and II
d) I and III
e) II and III
Interpreting Motion
Graphs
Object moving with
constant velocity has
zero net force.
Zero net force means the velocity must be constant. Graphs
I and II show objects moving with constant velocity: graph
III shows an object with decreasing velocity. Thus, the
answer is C
C
Q
P
R
15. The diagram above shows the path of a projectile as it moves from left to right
without friction or drag. At which point along the projectile’s path is the acceleration of
the projectile the greatest?
a) P
b) Q
c) R
d) P and R
e) the acceleration is the same at all points
A projectile is free
falling due to gravity
The diagram shows a projectile: we know that the only
force acting on a projectile is gravity, which is constant.
Thus, the acceleration is the same at all points. The
projectile moves horizontally due to its initial velocity and
inertia.
®
E
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