AP Physics II Summer Assignment
IMPORTANT NOTE: This assignment is not for a grade. You don’t have to turn it in however the first class
that you have AP Physics II you are going to take a test from this summer assignment. So, information and
AP Physics II, Summer Assignment
1
the questions in this packet are extremely important for the test. If your score is below 80 then you are going
to get a “0” from a major grade assignment. Good luck!
1. SCIENTIFIC NOTATION:
The following are ordinary physics problems. Write the answer in scientific notation and simplify the units (π=3).
AP Physics II, Summer Assignment
2
4.5  102 kg

2.0  103 kg s2
a.
Ts  2
b.
2
3.2 109 C 9.6 109 C

9 N m 
F   9.0 10


2
C2 
 0.32m 

𝐹=_______________
c.
1
1
1


2
Rp 4.5  10  9.4  102 
𝑅𝑝 =_______________
d.
K max  6.63 1034 J  s 7.09 1014 s  2.17 10 19 J 
𝐾𝑚𝑎𝑥 =_______________
e.
 
𝛾=_______________
f.
K
g.
1.33 sin 25.0  1.50 sin 
𝑇𝑠 =_______________


2.25  108 m s
1
3.00  108 m s


RP 

1




1
6.6×102 kg 2.11×104 m s
2

2

𝐾=_______________
𝜃=_______________
2. SOLVING EQUATIONS:
Often problems on the AP exam are done with variables only. Solve for the variable indicated. Don’t let the
different letters confuse you. Manipulate them algebraically as though they were numbers.
AP Physics II, Summer Assignment
3
1 2
kx
2
a.
K
, x  _______________
b.
Tp  2

g
, g  ______________
c.
Fg  G
m1m2
r2
, r  _______________
d.
mgh 
1 2
mv
2
, v  _______________
e.
x  xo  vot 
1 2
at
2
o I
2 r
,r 
_______________
,d 
_______________
, t  _______________
f.
B
g.
xm 
h.
pV  nRT
, T  _______________
i.
sin  c 
n1
n2
, c  _______________
j.
qV 
m L
d
1 2
mv
2
,v 
_______________
3. TRIGONOMETRY
AP Physics II, Summer Assignment
4
Using the generic triangle to the right, Right Triangle Trigonometry and
Pythagorean Theorem solve the following. Your calculator must be in degree
mode.
a.
 = 55o and c = 32 m, solve for a and b.
_______________
b.
 = 45o and a = 15 m/s, solve for b and c.
_______________
c.
b = 17.8 m and  = 65o, solve for a and c.
_______________
d.
a = 250 m and b = 180 m, solve for  and c.
_______________
e.
a =25 cm and c = 32 cm, solve for b and .
_______________
f.
b =104 cm and c = 65 cm, solve for a and .
_______________
4. DRAWING RESULTANT VECTORS
Draw the resultant vector using the parallelogram method of vector addition.
AP Physics II, Summer Assignment
5
Example
b.
d.
a.
c.
e.
Draw the resultant vector using the tip to tail method of vector addition. Label the resultant as vector R
Example 1: A + B
B
B
A
A
R
-B
Example 2: A – B
A
B
A
f.
R
X+Y
g.
T–S
h.
P+V
X
Y
T
S
P
V
C–D
i.
C
D
AP Physics II, Summer Assignment
6
A vector in two dimensions may be resolved into two component vectors acting along any two mutually
perpendicular directions. The figure shows the vector F and its x and y vector components:
Fx = F cos 
Fy = F sin 
a. Draw and calculate the components of the vector F (250 N, 235o)
b. Three ropes are tied to a stake and the following forces are exerted. Find the resultant force.
AP Physics II, Summer Assignment
7
5. GRAPHS
A. Reading: Motion Maps
A motion map represents the position, velocity, and acceleration of an object at various clock readings.
Suppose that you took a stroboscopic picture of a car moving to the right at constant velocity where each image revealed the
position of the car at one-second intervals.
This is the motion map that represents the car. We model the position of the object with a small point. At each position, the
object's velocity is represented by a vector.
If the car were traveling at greater velocity, the strobe photo might look like this:
The corresponding motion map has the points spaced farther apart, and the velocity vectors are longer, implying that the car is
moving faster.
If the car were moving to the left at constant velocity, the photo and motion map might look like this:
More complicated motion can be represented as well.
Here, an object moves to the right at constant velocity, stops and remains in place for two seconds, then moves to the left at a
slower constant velocity.
Consider the interpretation of the motion map below. At time t = 0, cyclist A starts moving to the right at constant velocity, at
some position to the right of the origin.
Cyclist B starts at the origin and travels to the right at a constant, though greater velocity.
At t = 3 s, B overtakes A (i.e., both have the same position, but B is moving faster).
AP Physics II, Summer Assignment
8
B. Graph Analysis
A graph is one of the most effective representations of the relationship between two variables. The independent
variable is usually placed on the x-axis. The dependent variable is usually placed on the y-axis. It is important for
you to be able interpret a graphical relationship and express it in a written statement and by means of an algebraic
expression.
Graph Shape
Written Relationship
Modification
Required to liberalize
graph
Algebraic
representation
When you state the relationships, tell how y depends on x (e.g., as x increases, y …).
AP Physics II, Summer Assignment
9
Draw qualitative graphs of x-versus-t, v-versus-t, and a-versus-t.
vo = 0
x=0
x
v
a
t
t
t
vo = 0
x=0
x
v
a
t
t
t
AP Physics II, Summer Assignment
10
vo ≠ 0
x=0
x
v
a
t
t
t
vo = 0
x=0
x
v
a
t
t
t
AP Physics II, Summer Assignment
11
6. FREE-BODY-DIAGRAMS
A free-body-diagram (FBD) is a vector diagram that shows all the forces that act on an object whose motion is
being studied.
Draw a FBD for each situation below following these directions:
- Choose a coordinate system defining the positive direction of motion.
- Draw arrows to represent the forces acting on the system.
1. Object lies motionless.
7. The object is motionless.
2. Object slides at constant speed without friction
8. The object is motionless.
3. Object slows due to kinetic friction.
4. Object slides without friction.
9. The object is motionless.
5. Static friction prevents sliding.
10. The object is falling (no air resistance).
6. An object is suspended from the ceiling.
AP Physics II, Summer Assignment
12
7. UNIFORM CIRCULAR MOTION
Uniform circular motion is motion in which there is no change in speed, only a change in direction.
Centripetal Acceleration
An object experiencing uniform circular motion is continually accelerating. The position and velocity of a particle
moving in a circular path of radius r are shown at two instants in the figure.
When the particle is at point A, its velocity is represented by vector v1. After a time interval t, its velocity is
represented by the vector v2. The acceleration is given by:
a
v v2  v1

t
t
The change in velocity v is represented graphically in the figure below:
The centripetal acceleration is given by: ac 
v2
r
Units: m/s2
v is the linear speed of a particle moving in a circular path of radius r.
The term centripetal means that the acceleration is always directed toward the center. The velocity and the
acceleration are not necessarily in the same direction; v points in the direction of motion which tangential to the
circle. v and a are perpendicular at every point.
The period T is the time for one complete revolution. So the linear speed can be found by dividing the period into
the circumference: v 
2r
T
Units: m/s
Another useful parameter in engineering problems is the rotational speed, expressed in revolutions per minute
(rpm), revolutions per second (rev/s) or Hz (s-1). This quantity is called the frequency f of rotation and is given by
the reciprocal of the period.
f 
1
T
Centripetal Force
The inward force necessary to maintain uniform circular motion is defined as centripetal force. From Newton's
Second Law, the centripetal force is given by:
mv 2
Fc 
r
Units: Newtons (N)
The centripetal force is not a 'special' kind of force. The centripetal force is provided by the force that keeps the
object in a circle, this is called the centripetal force requirement.
AP Physics II, Summer Assignment
13
8. WAVES AND SOUND
A wave is, in general, a disturbance that moves through a medium. (An exception is an electromagnetic wave,
which can travel through a vacuum. Examples are light and radio waves.) A wave carries energy, but there is no
transport of matter. Examples of mechanical waves include water waves, waves on a string, and sound waves.
There are two types of mechanical waves: transverse and longitudinal.
Types Of Waves
In a transverse wave, the particles of the medium move at
right angles to the direction of motion of the wave. The top
part of the wave is called the crest while the portion of the
wave below the equilibrium position is called the trough.
In a longitudinal wave the particles of the medium move back and forth, parallel to the direction of motion of the
wave.
The region where the particles of the medium are close
together is called a compression (or condensation) while
the region where the particles are farther apart is called
an expansion (or rarefaction). In analyzing longitudinal
waves, it is convenient to compare a region of
compression with the crest of a transverse wave and an
expansion with the trough of a transverse wave.
Water waves are a combination of longitudinal and transverse waves.
Parts Of A Wave
The wavelength λ (lambda) of a periodic wave is the
distance between adjacent wave crests.
We can find the velocity of a wave by relating it to the
velocity equation:
v
x 
 f
t T
Units: m/s
The velocity of a wave depends on the properties of the medium through which it travels. For a transverse wave on
a string, the velocity is given by the following equation:
v
FT
m/ L
Units: m/s
FT is tension in the string and m/L is the mass per unit length of the string (linear density)
AP Physics II, Summer Assignment
14
Measurements show that the wavelength of a sound wave in a certain material is
18.0 cm. The frequency of the wave is 1900 Hz. What is the speed of the sound wave?
A horizontal cord 5.00 m long has a mass of 1.45 g.
a. What must be the tension in the cord if the wavelength of a 120 Hz wave is 60 cm?
b. How large a mass must be hung from its end to give it this tension?
Standing Waves In Strings
At certain frequencies standing waves can be produced in which the waves seem to be standing still rather than
traveling. This means that the string is vibrating as a whole. This is called resonance and the frequencies at which
standing waves occur are called resonant frequencies or harmonics.
As shown in the diagram below, a standing wave is produced by the superposition of two periodic waves having
identical frequencies and amplitudes which are traveling in opposite directions. In stringed musical instruments, the
standing wave is produced by waves reflecting off a fixed end and interfering with oncoming waves as they travel
back through the medium.
The points of destructive interference (no vibration) are called nodes, and points of constructive interference
(maximum amplitude of vibration) are called antinodes.
First Harmonic or Fundamental
1
L  1
2
Second Harmonic or First Overtone
L  2
Third Harmonic or Second Overtone
3
L
AP Physics II, Summer Assignment
2
3
15
The lowest possible frequency that can produce a standing wave in a string fixed at both ends is called the first
harmonic. This frequency is also referred to as the fundamental frequency or first mode of vibration. The next
possible frequency is the first overtone. This frequency is also referred to as the second harmonic or second mode
of vibration. The frequencies of the harmonics are consecutive whole number multiples of the first harmonic.
The positions of the nodal points are designated by the letter N and the antinodal points by the letter A. The length
of the string is represented by L. The frequency of the particular harmonic can be determined as follows:
fn = n f'
Where n is the harmonic and f' is the fundamental frequency.
A metal string is under a tension of 88.2 N. Its length is 50 cm and its mass is 0.500 g.
a. Find the velocity of the waves on the string.
b. Determine the frequencies of its fundamental, first overtone and second overtone.
A string 2.0 m long is driven by a 240 Hz vibrator at its end. The string resonates in four segments. What is the
speed of the waves on the string?
9. ELECTROSTATICS
Electric Charge
There are two types of electric charge, arbitrarily called positive and negative. Rubbing certain electrically neutral
objects together (e.g., a glass rod and a silk cloth) tends to cause the electric charges to separate. In the case of the
glass and silk, the glass rod loses negative charge and becomes positively charged while the silk cloth gains
negative charge and therefore becomes negatively charged. After separation, the negative charges and positive
charges are found to attract one another.
If the glass rod is suspended from a string and a second positively charged glass rod is brought near, a force of
electrical repulsion results. Negatively charged objects also exert a repulsive force on one another. These results
can be summarized as follows: unlike charges attract and like charges repel.
Conservation Of Electric Charge
In the process of rubbing two solid objects together, electrical charges are not created. Instead, both objects contain
both positive and negative charges. During the rubbing process, the negative charge is transferred from one object
to the other and this leaves one object with an excess of positive charge and the other with an excess of negative
charge. The quantity of excess charge on each object is exactly the same.
The law of conservation of electric charge: "The net amount of electric charge produced in any process is zero."
Another way of saying this is that in any process electric charge cannot be created or destroyed, however, it can be
transferred from one object to another.
The SI unit of charge is the coulomb (C).
1 C = 6.25 x 1018 electrons or protons
The charge carried by the electron is represented by the symbol -e, and the charge carried by the proton is +e.
AP Physics II, Summer Assignment
16
e = 1.6 x 10-19 C and melectron = 9.11 x 10-31 kg while mproton = 1.672 x 10-27 kg
A third particle, which carries no electrical charge, is the neutron. mneutron = 1.675 x 10-27 kg.
Experiments performed early in this century have led to the conclusion that protons and neutrons are confined to
the nucleus of the atom while the electrons exist outside of the nucleus. When solids are rubbed together, it is the
electrons that are transferred from one object to the other. The positive charges, which are located in the nucleus, do
not move.
Insulators, Semiconductors And Conductors
An insulator is a material in which the electrons are tightly held by the nucleus and are not
free to move through the material. There is no such thing as a perfect insulator, however examples of good
insulators are: glass, rubber, plastic and dry wood.
A conductor is a material through which electrons are free to move through the material. Just as in the case of the
insulators, there is no perfect conductor. Examples of good conductors include metals, such as silver, copper, gold
and mercury.
A few materials, such as silicon, germanium and carbon, are called semiconductors. At ordinary temperature, there
are a few free electrons and the material is a poor conductor of electricity. As the temperature rises, electrons break
free and move through the material. As a result, the ability of a semiconductor to conduct improves with
temperature.
Induced Electric Charge
If a negatively charged rod is brought near an uncharged electrical conductor, the negative charges in the conductor
travel to the far end of the conductor. The positive charges are not free to move and a charge is temporarily induced
at the two ends of the conductor. Overall, the conductor is still electrically neutral and if the rod is removed a
redistribution of the negative charge will occur.
If the metal conductor is touched by a person’s finger or a wire connected to ground, it is said to be grounded. The
negative charges would flow from the conductor to ground. If the ground is removed and then the rod is removed, a
permanent positive charge would be left on the conductor. The electrons would move until the excess positive
charge was uniformly distributed over the conductor.
10. COULOMB’S LAW
Coulomb’s Law states that two point charges exert a force (F) on one another that is directly proportional to the
product of the magnitudes of the charges (q) and inversely proportional to the square of the distance (r) between
their centers. The equation is:
F k
q1q2
r2
k = 9x109 N. m2/C2
The value of k can also be expressed in terms of the permittivity of free space (εo):
k
1
4 o
 9x109 N. m2/C2
The proportionality constant (k) can only be used if the medium that separates the charges is a vacuum. If the region
between the point charges is not a vacuum then the value of the proportionality constant to be used is determined by
dividing k by the dielectric constant (K). For a vacuum K = 1, for distilled water K = 80, and for wax paper K =
2.25
AP Physics II, Summer Assignment
17
Problem-Solving Strategy
1. Draw and label a figure indicating positive and negative charges along with the given distances.
2. Draw the force of attraction or repulsion on the given charge on a neat, labeled FBD.
3. Find the resultant force. Important: Do not use the signs of the charges when applying Coulomb's law!
Two charges q1 = -8 μC and q2= +12 μC are placed 120mm apart in the air. What is the resultant force on a third
charge q3 = -4 μC placed midway between the other charges?
Three charges q1 = +4 nC, q2 = -6 nC and q3 = -8 nC are arranged as shown in the figure. Find the resultant force on
q3 due to the other two charges.
AP Physics II, Summer Assignment
18
11. ELECTRIC FIELD
An electric field is said to exit in a region of space in which an electric charge will experience an electric force. The
magnitude of the electric field intensity is given by:
E
F
q
Units: N/C
The direction of the electric field intensity at a point in space is the same as the direction in which a positive charge
would move if it were placed at that point. The electric field lines or lines of force indicate the direction. The
electric field is strongest in regions where the lines are close together and weak when the lines are further apart.
The electric field intensity E at a distance r from a single charge q can be found as follows:
E
kq
r2
Units: N/C
When more than one charge contributes to the field, the resultant field is the vector sum of the contributions from
each charge.
E
kq
r2
Units: N/C
The electric field intensity between two plates is constant and directed downward. The magnitude of the electric
field intensity is 6x104 N/C. What are the magnitude and direction of the electric force exerted on an electron
projected horizontally between the two plates?
AP Physics II, Summer Assignment
19
What is the electric field intensity at a distance of 2 m from a charge of -12 μC?
Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart, as shown in the figure. Determine the electric field
a. At point A and b. At point B.
AP Physics II, Summer Assignment
20
12. ELECTRICITY
Electric Current
The electric current I is the rate of flow of charge Q past a given point on an electric conductor.
I
Q
t
Units: C/s = Ampere (A)
The direction of conventional current is always the same as the direction in which positive charges would move,
even if the actual current consists of a flow of electrons.
Electromotive Force
A source of electromotive force (emf) is a device that converts, chemical, mechanical, or other forms of energy into
the electric energy necessary to maintain a continuous flow of electric charge.
Ohm’s Law
"For a given resistor at a particular temperature, the current is directly proportional to the applied voltage."
R
V
I
Units: V/A = ohm (Ω)
Four devices are commonly used in the laboratory to study Ohm’s law: the battery, the voltmeter, the ammeter and
a resistance. The ammeter and voltmeter measure current and voltage respectively. The following symbols are used
in electric circuits:
AP Physics II, Summer Assignment
21
Electric Power And Heat Loss
The rate at which heat is dissipated in an electric circuit is referred to as the power loss.
P= V I =12 R =
V2
R
Units: watts (W)
A current of 6A flows through a resistance of 300 Ω for 1 hour.
a. What is the power loss?
b. How much heat is generated in loss?
Resistivity
The resistance of a wire of uniform cross-sectional area is determined by:
o The kind of material
l
o The length
R
Units: ohms (Ω)
o The cross-sectional area
A
o The temperature
Where ρ is the resistivity of the material in Ω.m, l is the length in m, and A is the cross-sectional area in m2.
What is the resistance of a 20 m length of copper wire with a diameter of 0.8 mm?
(ρ= 1.72x10-8 Ω.m)
Resistors In Series
- In a series circuit, the current is the same
at all points along the wire.
IT = I1 = I2 = I3
- An equivalent resistance is the resistance of
a single resistor that could replace all the resistors
in a circuit. The single resistor would have the same
current through it as the resistors it replaced.
RE= R1 + R2 + R3
- In a series circuit, the sum of the voltage drops equal the voltage drop across the entire circuit.
VT = V1 + V2 + V3
AP Physics II, Summer Assignment
22
Two resistances of 2 Ω and 4 Ω respectively are connected in series. The source of emf maintains a constant
potential difference of 12 V.
a. Draw a schematic diagram with an ammeter and a voltmeter.
b. What is the current delivered to the external circuit?
c. What is the potential drop across each resistor?
Parallel Circuits
- In a parallel circuit, each resistor provides a new path for electrons to flow. The total current
is the sum of the currents through each resistor.
IT = I1 + I2 + I3
- The equivalent resistance of a parallel circuit decreases as each new resistor is added.
1
1
1
1



RE R1 R2 R3
- The voltage drop across each branch is equal to the voltage of the source.
VT = V1 = V2 = V3
AP Physics II, Summer Assignment
23
Kirchhoff’s Laws
Kirchhoff developed two laws related to circuits.
Law 1. Conservation of Charge:
The sum of the currents entering a junction is equal to the sum of the currents leaving that junction.
I in  I out
Law 2. Conservation of Energy:
The sum of the emfs around any closed current loop is equal to the sum of all the IR drops around that loop.
  IR
A junction refers to any point in the circuit where two or three wires come together.
Emf And Terminal Potential Difference
Every source of emf (ε) has an inherent resistance called internal resistance represented by the symbol r. This
resistance is a small resistance in series with the source of emf. The actual terminal voltage VT across a source of
emf with an internal resistance is given by:
VT = ε - I r
Units: Volts (V)
The total applied voltage to the circuit in the figure is 12 V and the resistances R1, R2 and R3 are 4, 3 and 6 Ω
respectively.
a. Determine the equivalent resistance of the circuit.
b. What is the current through each resistor?
AP Physics II, Summer Assignment
24
MORE PROBLEMS
1. A 0.5-kilogram object rotates freely in a vertical circle at the end of a string of length 2
meters as shown above. As the object passes through point P at the top of the circular path,
the tension in the string is 20 newtons. Assume g = 10 meters per second squared. a)On the
following diagram of the object, draw and clearly label all significant forces on the object
when it is at point P. b)Calculate the speed of the object at point P.
2. A student pulls a box of books on a smooth horizontal floor with a force of 100 N in a direction of angle 37
above the horizontal. If the mass of the box and the books is 40.0 kg, what is the acceleration of the box?
9. A ball of mass 0.5 kilogram, initially at rest, is kicked directly toward a fence from a point 32 meters away, as
shown above. The velocity of the ball as it leaves the kicker's foot is 20 meters per second at an angle of 37° above
the horizontal. The top of the fence is 2.5 meters high. The ball hits nothing while in flight and air resistance is
negligible. a) Determine the time it takes for the ball to reach the plane of the fence. b) Will the ball hit the fence? If
so, how far below the top of the fence will it hit? If not, how far above the top of the fence will it pass?
AP Physics II, Summer Assignment
25
3. A 200-g block connected to a light spring for which the force constant is
5.00 N/m is free to oscillate on a horizontal, frictionless surface. The block is
displaced 5.00 cm from equilibrium and released from rest, as in Figure (A)
Find the period of its motion. (B) Determine the maximum speed of the block.
4. Given this lovely circuit: (a) Find the equivalent resistance for
this circuit. (b) Find the current supplied by the battery. (c) Find
the current through the 65.0  resistor.
15.0 V
58.0 
75.0 
45.0 
65.0 
35.0 
5. Three resistors are arranged in a circuit as shown above. The battery has an
unknown but constant emf E and a negligible internal resistance. a) Determine the
equivalent resistance of the three resistors. The current I in resistor R3 is 0.40
ampere. b)
Determine the emf E (Voltage) of the battery. c) Determine the
potential difference across resistor R1 d) Determine the power dissipated in resistor
R1. e) Determine the amount of charge that passes through resistor R3 in one
minute.
AP Physics II, Summer Assignment
26
6. An electric field of 900 N/C is produced by a charge of 4  1011 C. For this field strength, what is the distance to
the charge? (k = 9  109 Nm2/C2)
7. Two small objects, each with a charge of -4.0 nC, are held together by a 0.020 m length of insulating string as
shown in the diagram below. The objects are initially at rest on a horizontal, nonconducting frictionless surface.
The effect of gravity on each object due to the other is negligible.
(a) Calculate the tension in the string.
(b) Illustrate the electric field by drawing electric field lines for the two objects on the following diagram.
The masses of the objects are m1 = 0.030 kg and m2 = 0.060 kg. The string is now cut.
(c) Calculate the magnitude of the initial acceleration of each object.
(d) On the axes below, qualitatively sketch a graph of the acceleration a of the object of mass m2 versus the
distance d between the objects after the string has been cut.
(e) Describe qualitatively what happens to the speeds of the objects as time increases, assuming that the objects
remain on the horizontal, nonconducting frictionless surface.
AP Physics II, Summer Assignment
27
8. A hollow tube of length 10m open at both ends as shown, is held in midair. A tuning fork with a frequency 100
Hz vibrates at one end of the tube and causes the air in the tube to vibrate at its fundamental frequency.
a. Determine the wavelength of the sound. b. Determine the speed of sound in the air inside the tube. c. Determine
the next higher frequency at which this air column would resonate.
The same tube is submerged in a large, graduated cylinder filled with water. The tube is
slowly raised out of the water and the same tuning fork, vibrating with frequency 100 Hz,
is held a fixed distance from the top of the tube.
d. Determine the height h of the tube above the water when the air column resonates for
the first time.
9. Two metal spheres that are initially uncharged are mounted on insulating
stands, as shown above. A negatively charged rubber rod is brought close to,
but does not make contact with, sphere X. Sphere Y is then brought close to
X on the side opposite to the rubber rod. Y is allowed to touch X and then is
removed some distance away. The rubber rod is then moved far away from
X and Y. What are the final charges on the spheres?
Sphere X
Sphere Y
A) Zero
Zero
B) Negative
Negative
C) Negative
Positive
D) Positive
Negative
E) Positive
Positive
10. A block of mass m is accelerated across a rough surface by a force of magnitude
F that is exerted at an angle  with the horizontal, as shown. The frictional force on
the block exerted by the surface has magnitude f. What is the acceleration of the
block?
(A) F/m
(B) (Fcos)/m
(C) (F–f)/m
(D) (Fcos–f)/m
(E) (Fsin–mg)/m
AP Physics II, Summer Assignment
28