Lab 2Lab 2- Kinematics.pdf142017 Lab 2 Kinematicsh.docx

Lab 2/Lab 2- Kinematics.pdf
1/4/2017 Lab 2: Kinematics
https://moodle.straighterline.com/pluginfile.php/72219/mod_res
ource/content/17/CourseRoot/html/lab004s001.html 1/20
Learning Objec뙕ves
Disᣊ�nguish between scalar and vector quanᣊ�ᣊ�es
Apply kinemaᣊ�c equaᣊ�ons to 1‐D and projecᣊ�le moᣊ�on
Predict posiᣊ�on, velocity, and acceleraᣊ�on vs. ᣊ�me graph
s
Calculate average and instantaneous velocity or acceleraᣊ�on
Determine that x and y components are independent of each oth
er
Relate velocity, radius, and ᣊ�me period to uniform circular m
oᣊ�on.
Explain the direcᣊ�on of acceleraᣊ�on during uniform circula
r moᣊ�on
1‐D Kinema뙕cs
1‐D kinemaᣊ�cs occurs when an object travels in
one dimension and can be described using words,
equaᣊ�ons and graphs. Linear mo뙕on describes
how an object will move horizontally or verᣊ�cally
with constant acceleraᣊ�on, how an object will

Figure 1: Pool balls in moᾷon demonstrate
1‐D kinemaᾷcs.
Figure 2: Line secant to the path of the
object.
travel if dropped from the side of a cliff, and the
path it will follow if thrown straight up into the air.
Keep in mind the moᣊ�on of an object is relaᣊ�ve to
the viewer. Even though you do not feel like you
are in moᣊ�on right now, you are on planet earth
that has rotaᣊ�onal moᣊ�on in addiᣊ�on to orbital
moᣊ�on around the sun. In almost all cases here
moᣊ�on will be relaᣊ�ve to the Earth.
Scalar and Vector Quan뙕뙕es
In physics, quanᣊ�ᣊ�es can be scalar or vector. The
difference between the two lies in direcᣊ�on.
Scalar quanᣊ�ᣊ�es include magnitudes, which are numerical
measurements. The distance an
object has traveled or the speed of an object is a scalar quanᣊ�
ty. Scalars do not take direcᣊ�on
into consideraᣊ�on and can be described with only a number a
nd a unit. For example,
somebody might say the temperature outside is 70°F. Seventy is
the magnitude, and
Fahrenheit is the unit; there is no direcᣊ�on associated with th
e quanᣊ�ty. Vector quanᣊ�ᣊ�es, on

the other hand, include magnitude and direcᣊ�on. The displace
ment from an object's iniᣊ�al
posiᣊ�on, velocity, and acceleraᣊ�on are vector quanᣊ�ᣊ�e
s. The direcᣊ�on of vectors can be
described as being in the posiᣊ�ve direcᣊ�on, in the negaᣊ�v
e direcᣊ�on, north, south, east, west,
leĀ, right, up, down, etc. One might describe an airplane's veloc
ity as 450 miles per hour due
west where both magnitude and direcᣊ�on are given. It is impo
rtant to disᣊ�nguish between
scalar and vector quanᣊ�ᣊ�es when trying to understand kine
maᣊ�cs.
Speed, Velocity, and Accelera뙕on
You may be familiar with speed outside of the physics classroo
m. When you drive in a car you
are traveling a distance over a certain amount of ᣊ�me: a speed
. How then is velocity different
from speed? Velocity (v) is a vector quanᣊ�ty described as the
rate at which an object's
posiᣊ�on changes divided by the ᣊ�me the object is in moᣊ�
on. Furthermore, the rate of change
in velocity plays an important role in physics. Accelera뙕on repr
esents the rate of change of an
object's velocity over ᣊ�me.
Speed, Average Velocity, and Average Accelera뙕on
Before we define speed, velocity, and acceleraᣊ�on, it is
important to disᣊ�nguish between distance and
displacement. Distance is how much “ground” an
object has covered; whereas, displacement is how far
an object has moved from its original posiᣊ�on. The
total change in an object's distance over ᣊ�me is
referred to as average speed. Distance and speed are

scalar quanᣊ�ᣊ�es. The average change in an object's
displacement over ᣊ�me is referred to as average
velocity, and the average change in an object's velocity
over ᣊ�me is referred to as average acceleraᣊ�on.
Displacement, velocity, and acceleraᣊ�on, as menᣊ�oned
before, are vector quanᣊ�ᣊ�es. When you are calculaᣊ�ng
the average velocity or acceleraᣊ�on of an object, you
Figure 3: Diagram showing the average
speed of a car. Speed is scalar.
Figure 4: Diagram showing the velocity of a
car.
are actually finding the slope of the line secant to the path betw
een the two points in ᣊ�me
(Figure 2).
Example 1: Speed
A car travels a distance of 800 meters over a ᣊ�me of
10 seconds. The car has an average speed of 80
meters per second (Figure 3).
Speed AVG =
distance
=
800 m

=
80 m
= 80 m/s
뙕me 10 s s
Example 2: Velocity
A car travels at 800 meters west. The car makes a U‐
turn and travels a distance of 800 meters east. It took
the car 20 seconds to complete its course. The
average velocity of the car is 0 meters per second
(Figure 4) because the car travels in opposite
direcᣊ�ons that cancel each other out.
VelocityAVG(vAVG) =
Δdisplacement
=
800 m + (‐800) m
=
0 m
= 0 m/s
뙕me 20 s 20 s
Even though the car experienced a lot of moᣊ�on in 20 seconds
, its posiᣊ�on did not change
over the course of its moᣊ�on.
Example 3: Accelera뙕on
A car travels from 0 m/s to 90 m/s in 15 seconds.
The car has an average acceleraᣊ�on of 6 m/s2.

Accelera뙕onAVG(aAVG) =
Δvelocity
=
90 m/s
=
6 m/s
= 6 m/s2뙕me 15 s s
Example 4: Instantaneous Velocity
A car is moving along a path whose displacement can be modele
d by funcᣊ�on x(t) = 2.0 m +
(4.0 m/s2)t2 (Figure 5‐leĀ). The car's velocity is the derivaᣊ�v
e of its displacement, v(t)= (8.0
m/s2)t (Figure 5‐right). The instantaneous velocity of the car at
3 seconds is 24 m/s.
Figure 5: The car's displacement over a period of ᾷme is graphe
d to the le뎌. The car's velocity over a period if
ᾷme is graphed to the right.
Displacement = x(t) = 2.0 m + (4.0 m/s2)t2
Velocity = (d/dt) x(t) = v(t) = (8.0 m/s2)t

v(3) = (8.0 m/s2) (3.0 s) = 240 m/s
Example 5: Instantaneous Accelera뙕on
A car is moving with a velocity that can be modeled by the func
ᣊ�on v(t)= (8.0 m/s2)t. The
car's acceleraᣊ�on is the derivaᣊ�ve of the velocity, a(t)= 8.0
m/s2 (Figure 6). The instantaneous
acceleraᣊ�on of the car at any point in ᣊ�me is 8.0 m/s2 beca
use in this case acceleraᣊ�on is not
dependent on ᣊ�me. In other words, the acceleraᣊ�on is const
ant. For example, the
acceleraᣊ�on at 5 seconds is 8.0 m/s2.
Figure 6: Acceleraᾷon over ᾷme of the car.
Velocity = v(t) = 8.0 m/s2)t
Accelera뙕on = (d/dt) v(t) = (8.0 m/s2)
α(5) = (8.0 m/s2)
Graphing 1‐D Mo뙕on
In science, we observe how one factor changes as a result of a c
hange in another. The effect
of one variable on another can be expressed as a funcᣊ�on. An
alternaᣊ�ve to using equaᣊ�ons
to describe moᣊ�on is to uᣊ�lize moᣊ�on graphs to visualize

the same relaᣊ�onships.
When we look at the speed vs. ᣊ�me graph of an object in free
fall (Figure 7), we observe a
linear relaᣊ�onship. In other words, for every increase in ᣊ�m
e, there is the same increase in
speed. Since the object is dropped from rest, the line starts at th
e origin (0,0). Since this
parᣊ�cular curve is a straight line, we know the slope (Δy/ Δx)
of the line is constant. On this
graph, the slope of the line represents acceleraᣊ�on. The steep
er the curve (the line) is, the
greater the acceleraᣊ�on. From the data presented here we kno
w that the acceleraᣊ�on
pictured in Figure 7 is constant.
Examining the curves on a velocity vs. ᣊ�me graph provides in
formaᣊ�on about the direcᣊ�on,
speed and acceleraᣊ�on of the moving object. Take a look at th
e series of linear moᣊ�on graphs
in Figure 7 and how they are interpreted.
Understanding the meaning of the shape and slope of a distance
vs. ᣊ�me graph is essenᣊ�al to
your understanding of linear moᣊ�on. Whether it's the speed or
direcᣊ�on that is changing, the
rate of acceleraᣊ�on will greatly influence the path of moᣊ�o
n. From driving your car to

shooᣊ�ng a gun, these concepts apply to many situaᣊ�ons enc
ountered in everyday life. Explore
the world around you and see how many applicaᣊ�ons of linear
moᣊ�on you can idenᣊ�fy!
Figure 7: Linear moᾷon graphs.
The Kinema뙕c Equa뙕ons
When you are at a red light and the light turns green, you hit the
gas acceleraᣊ�ng your car up
to the speed limit. During the period of ᣊ�me your foot is on th
e gas, your acceleraᣊ�on is close
to constant. This is the case for many different types of moᣊ�o
n. From interpreᣊ�ng the 1‐D
moᣊ�on graphs, acceleraᣊ�on is equal to the slope of velocity
versus ᣊ�me graphs: a = dv/dt.
Uᣊ�lizing calculus, the first kinemaᣊ�c equaᣊ�on can be der
ived using the following process:
(dv/dt) = a
(dv) = a ∙ dt
Integrate both sides of the equaᣊ�on.
∫ dv = ∫ a ∙ dt
∫ dv = a ∫ dt
If a is constant, it comes out of the integral.
v = at + C
The constant of integraᣊ�on is equal to the iniᣊ�al velocity at
ᣊ�me zero. The resulᣊ�ng equaᣊ�on

is the first kinemaᣊ�c equaᣊ�on.
vf = vo + at
Another integraᣊ�on results in the next kinemaᣊ�c equaᣊ�on
.
df = do + vot + ½at
2
These equaᣊ�ons are used to solve for the instantaneous posiᣊ
�on, velocity, or acceleraᣊ�on of
an object. They each relate two different variables and are sum
marized as the following:
df = d0 + v0t + ½ at
2 Relates distance and 뙕me
vf = v0 + at Relates velocity and 뙕me
vr
2= v0
2 + 2ad Relates velocity and displacement
where:
d is displacement which is equal to df ‐ d0
t is the amount of ᣊ�me the object is in moᣊ�on (s)

vf is the final velocity (m/s)
v0 is the iniᣊ�al velocity (m/s)
a is the acceleraᾷon (m/s2)
d0 is the iniᾷal posiᾷon (m)
df is the final posiᾷon (m))
Many equaᣊ�ons can be derived from the general equaᣊ�ons t
o find the above variables under
different condiᣊ�ons as long as all variables except the one bei
ng solved for are known. Let's
take a look at some examples. If the iniᣊ�al velocity of the obj
ect is equal to 0 m/s and iniᣊ�al
posiᣊ�on is equal to 0 m, the kinemaᣊ�c equaᣊ�ons simplify
to:
vf = at
df = ½ at
2
The following equaᣊ�on is useful to find the displacement an o
bject travels when gravity is the
only force acᣊ�ng on the object and the ᣊ�me is known. The a
cceleraᣊ�on of gravity, g, on earth
is about 9.8 m/s2:
d = ½ gt2
Click to Run
http://phet.colorado.edu/sims/moving-man/moving-man_en.jnlp

Figure 8: Skydivers experience free‐fall. To reduce the
effects of air resistance, skydivers orient their bodies
perpendicular to the ground. By doing this, they are
able to reach a terminal velocity of about 120 mph!
Accelera뙕on and Gravity
Gravity causes objects to accelerate
downward when falling. In reality, air
resistance decreases the effects of gravity on
the acceleraᣊ�ng objects unᣊ�l it balances the
force of gravity and there is no longer a
change in velocity — when the object reaches
its terminal velocity (Figure 8). However, when
we study linear moᣊ�on, it is convenient to
neglect air resistance and focus only on the
acceleraᣊ�on due to gravity (g), a state called
freefall. In all cases of free fall, g (9.8 m/s2) is
the acceleraᣊ�on due to the force of gravity
that can act in a posiᣊ�ve way by increasing the
speed of the object or negaᣊ�vely by
decreasing the speed. Thus, for every second
of free fall there is a change in velocity of approximately 10 m/
s. This leads us to the equaᣊ�on
for the acceleraᣊ�on of a falling object:
vf = gt
where v is the final velocity (m/s), g is the acceleraᣊ�on due to
gravity (9.8 m/s2), and t is the
elapsed ᣊ�me (s). Thus, velocity is a funcᣊ�on both of acceler
aᣊ�on and of how long the object is
acted on by that force.

Gravity is also responsible for slowing down an object thrown s
traight up into the air and
acceleraᣊ�ng it back to Earth. In fact, the object reaches a poin
t of zero velocity before it
changes direcᣊ�ons. This characterisᣊ�c behavior is typical o
f all mass in the presence of
gravity.
? Did You Know...
Physics is the study of how the world works. Currently, all
physics has been described by the interacᾷon of parᾷcles
in space at a certain ᾷme. We refer to this arena for
parᾷcles to interact and collide with each other "space‐
ᾷme". Recently, physicist a new kind of geometric shape
called the amplituhedron was discovered to simplify
calculaᾷons for colliding parᾷcles. Could this be the end
of "space‐ᾷme"?
Pictured to right: Amplituhedron
2‐D Kinema뙕cs

Projec뙕le mo뙕on takes into consideraᣊ�on objects that are mo
ving in two direcᣊ�ons at the
same ᣊ�me. Unlike linear moᣊ�on which only considers one d
irecᣊ�on, projecᣊ�le moᣊ�on
acknowledges both horizontal and verᣊ�cal moᣊ�on. Think ab
out two‐dimensional (2‐D)
moᣊ�on as two, independent, one‐dimensional (1‐D) moᣊ�ons
. The path of an object with
projecᣊ�le moᣊ�on can be described as curved. This is where
the concept of vectors comes into
play.
Vectors
As you learned previously, a quanᣊ�ty that
conveys informaᣊ�on about magnitude only is
called a scalar quanᣊ�ty. Vectors, like velocity,
describe magnitude and direcᣊ�on. Along with
detailing informaᣊ�on about the path of
moᣊ�on, vectors are also useful in physics
because they can be separated into
components.
An object with projecᣊ�le moᣊ�on has a velocity
that can be represented by a diagonal vector.
In Figure 9, the vector is labeled V. Vector V
can be resolved (broken down) into an
equivalent set of horizontal (x‐direcᣊ�on) and
verᣊ�cal (y‐direcᣊ�on) components, which are at
right angles to each other. The addiᣊ�on of the
components results in the magnitude of the

Figure 9: A vector can be broken down into
horizontal and verᾷcal components.
vector.
Calcula뙕ng the x‐component of a Vector
cosθ = Vx
V
Vx = V (cosθ)
Calcula뙕ng the y‐component of a
Vector
sinθ = Vy
V
Vy = V (sinθ)
Calcula뙕ng Magnitude and Direc뙕on
Unit Vectors
Unit vectors are helpful when describing direcᣊ�on for proble
ms that are in two or more
dimensions. They provide informaᣊ�on about the components
of a vector and have a
magnitude of 1. The unit vector î is directed in the posiᣊ�ve x‐
direcᣊ�on, and the unit vector ĵ is
directed in the posiᣊ�ve y‐direcᣊ�on. To describe vectors by
components, one can use unit
vectors, as shown below.

Rx = Rxî
Rx describes the enᣊ�re x‐component of the vector, and Rx des
cribed the magnitude of the x‐
component.
Ry = Ryĵ
Combining the x‐ and y‐ components we can find the vector R:
R = Rx + Ry
R = Rxî + Ryĵ
Figure 10: The path of a projecᾷle in the absence of
air resistance is a perfect parabola. The horizontal
component of velocity is 10 m/s throughout the
object's moᾷon.
Posi뙕on Vectors
To describe an object's locaᣊ�on in space, we use the posiᣊ�o
n vector. It is the vector that
points from the origin to the object's posiᣊ�on in space at one i
nstant. The posiᣊ�on vector can
be wriĀen using unit vectors.
r = xî + yĵ

Just as we learned in 1‐D Kinemaᣊ�cs, an object's posiᣊ�on, a
verage velocity, instantaneous
velocity, average acceleraᣊ�on, and instantaneous acceleraᣊ�
on are all important in describing
an object's moᣊ�on through space. In 1‐D Kinemaᣊ�cs averag
e velocity was found by taking the
change in the object's displacement over a period of ᣊ�me. Thi
s is very similar to average
velocity in 2‐D Kinemaᣊ�cs. Instead of using the change in the
object's displacement, we use
the change in the object's posiᣊ�on vector over a certain amou
nt of ᣊ�me.
Instantaneous velocity in 1‐D Kinemaᣊ�cs was found by taking
the derivaᣊ�ve of the object's
displacement. For 2‐D Kinemaᣊ�cs, instead of displacement, w
e take the derivaᣊ�ve of the
object's posiᣊ�on vector in 2 dimensions.
Average acceleraᣊ�on in 2‐D Kinemaᣊ�cs is virtually the sam
e as it is in 1‐D Kinemaᣊ�cs. We take
the change in the object's velocity vector over a certain amount
of ᣊ�me.
Projec뙕le Mo뙕on
Typically, a projec뙕le is any object which,
once projected, conᣊ�nues in moᣊ�on by its
own inerᣊ�a and is influenced only by the
downward force of gravity. This may seem
counter‐intuiᣊ�ve since the object is moving
both horizontally and verᣊ�cally, but gravity
(an applied force) acts only on the verᣊ�cal
moᣊ�on of the object. The term iner뙕a
describes an object's resistance to external

forces which could affect its moᣊ�on (both
velocity or direcᣊ�onality). When there are no
external forces acᣊ�ng upon an object, it will
conᣊ�nue to travel in a straight line at a
constant, linear velocity.
As shown in Figure 10, the projecᣊ�le with horizontal and ver
ᣊ�cal moᣊ�on assumes a
characterisᣊ�c parabolic trajectory due to the effects of gravity
on the verᣊ�cal component of
? Did You Know...
Cannons on Navy ba᧻le
ships can fire at objects
that are farther than
fi뎌een miles away. The
curvature of the Earth
makes it impossible to
see that far in the
distance. That means
that the cannons fire
projecᾷles at objects
they cannot physically
see!
moᣊ�on. If air resistance is neglected, there are no horizontal f
orces acᣊ�ng upon the projecᣊ�le,
and thus no horizontal acceleraᣊ�on. It might seem surprising,

but a projecᣊ�le moves at the
same horizontal speed no maĀer how long it falls!
Since projecᣊ�le moᣊ�on can be resolved into two independe
nt direcᣊ�ons, 1‐D kinemaᣊ�c
equaᣊ�ons can be applied to both components of the moᣊ�on
separately. The kinemaᣊ�c
equaᣊ�ons will allow you to solve for different aspects of a pr
ojecᣊ�le's flight, its height
(verᣊ�cal), range (horizontal), and ᣊ�me of flight. Applying 1
‐D kinemaᣊ�cs results in two sets of
equaᣊ�ons for 2‐D moᣊ�on:
Height Range
yf = v0,yt ‐ 1/2gt
2 xf = v0,xt
vf,y = v0,y ‐ gt
vf,y
2= v0,y
2 ‐ 2gy
The x and y subscripts for velocity
refer to the component of velocity
in the x and y direcᣊ�on. These two
sets of equaᣊ�ons (height and
range) also incorporate ᣊ�me
because the ᣊ�me of flight for the
projecᣊ�le moᣊ�on is the same for both the verᣊ�cal and hor
izontal moᣊ�ons. Noᣊ�ce that there
is only one equaᣊ�on for range, while there are three equaᣊ�o
ns for height! This is due to the
fact there is no acceleraᣊ�on in the horizontal direcᣊ�on. You

may oĀen be required to find the
ᣊ�me of flight using the height equaᣊ�ons in order to determi
ne the range of the projecᣊ�le.
Click to Run
http://phet.colorado.edu/sims/motion-2d/motion-2d_en.jnlp
Figure 11: With air resistance, the trajectory looks like
a "squashed" parabola, and the range of the object's
flight is noᾷceably affected.
Launch Angle
If the projecᣊ�le is fired at an angle, the range
is a funcᣊ�on of the iniᣊ�al launch angle, θ, the
launch velocity and the force of gravity. Using
algebra, you can derive the following
expression from the kinemaᣊ�cs equaᣊ�ons:
R = v2 sin(2θ)
g
It is important to remember that in many
cases, air resistance is not negligible (Figure
11) and affects both the horizontal and verᣊ�cal
components of velocity. When the effect of air
resistance is significant, the range of the projecᣊ�le is reduced
and the path the projecᣊ�le

follows is not a true parabola.
Circular Mo뙕on
Figure 12: Diagram of the relaᾷonship between ᾷme
period, velocity and radius during circular moᾷon.
Imagine you are driving in a car with the steering wheel turned
so that the car follows the
path of a perfect circle. If the speedometer read the same speed
the whole ᣊ�me, the moᣊ�on
of the car would be described as uniform circular mo뙕on.
Uniform Circular Mo뙕on
Uniform circular moᣊ�on is the moᣊ�on of an
object in a circle with constant speed, when
the object covers the same distance in each
instant of ᣊ�me.
Recall that the average speed is equal to the
distance traveled divided by the ᣊ�me. For one

revoluᣊ�on in a circle the distance traveled is
equal to the circumference of a circle. The
ᣊ�me to make one revoluᣊ�on is called a period.
The equaᣊ�on to find the average speed of an object traveling i
n uniform circular moᣊ�on is:
Vavg = 2πr
T
where T is the ᣊ�me period, and r is the radius of the circular p
ath the object travels. As the
radius increases, velocity increases. For example, imagine four
spherical objects moving in a
uniform circular path around a staᣊ�onary point (Figure 12). E
ach ball has the same ᣊ�me
period because they move around the circle in the same amount
of ᣊ�me, but the speed at
Figure 13: The change in velocity (Δv) is shown
in the small diagram to the right, and is the
sum of the two velocity vectors v1 and v2.
which each ball moves differs. As the radius gets larger, the spe
ed also increase in order to
cover greater distance in the same amount of ᣊ�me as the other
objects.
? Did You Know...

The word centripetal stems from the La뙕n meaning "toward cen
ter".
Centripetal Accelera뙕on
Newton's First Law states that an object in
moᣊ�on will stay in moᣊ�on unless acted on by
an external force. Figure 13 depicts an object
moving in a circular moᣊ�on at a constant
speed. We see that the direcᣊ�on of the
velocity vector is changing as the object
travels around the circular path. Therefore,
there must be an acceleraᣊ�on causing the
mass to change direcᣊ�on. What is the
direcᣊ�on of this acceleraᣊ�on? Average
acceleraᣊ�on is calculated by dividing the
change in velocity by ᣊ�me, so the acceleraᣊ�on is in the sam
e direcᣊ�on as the change in
velocity which is toward the center of the circle! This acceleraᣊ
�on is called centripetal
accelera뙕on. No maĀer what two velocity vectors you choose, t
he acceleraᣊ�on vector is
always perpendicular to the tangenᣊ�al velocity toward the cen
ter of rotaᣊ�on. Thus, an object
in circular moᣊ�on can always be thought of as acceleraᣊ�ng
toward the center of the circle,
even though the radius of rotaᣊ�on remains constant. The mag
nitude of centripetal
acceleraᣊ�on can be expressed in terms of the linear velocity a
nd the radius of rotaᣊ�on:
ac = v
2
R

The force causing centripetal acceleraᣊ�on is just large enough
to keep the object in its circular
path. This centripetal force for uniform circular moᣊ�on alters
the direcᣊ�on the object is
traveling in, but not the speed. Just like linear moᣊ�on, force i
s a measure of the mass in
moᣊ�on mulᣊ�plied by its acceleraᣊ�on. The magnitude of t
he force is related to the
acceleraᣊ�on of the object through the following relaᣊ�onshi
ps based on Newton's Second Law
of Moᣊ�on:
Click to Run
http://phet.colorado.edu/sims/rotation/rotation_en.jnlp
Figure 14: Swings at an amusement park
exhibit a circular path of moᾷon (white line).
Figure 15: Banked turns keep
cyclists from being pushed off a
track.
Let's analyze a real world example of centripetal
moᣊ�on. The seats on a swinging ride at an
amusement park twirl around the rotaᣊ�ng central
pole, and their moᣊ�on is described as a circle (Figure
14). The seats are connected to a chain, which

provides an (inward) centripetal force to keep the
swings from flying off in a straight line. Increasing the
velocity of rotaᣊ�on forces the chain out at a wider
angle to the point where the horizontal component of
its tension provides the necessary centripetal force
inward. Meanwhile, the verᣊ�cal component of the
tension must balance out the force of gravity
downward.
If you have ever been on a rotaᣊ�ng merry‐go‐round, you
have probably felt as if something is pulling toward the
outside edges, forcing you to hold onto the bar to keep you
from falling off. A common word that is oĀen used when
discussing circular moᣊ�on is centrifugal force (center fleeing)
,
which is the "ficᣊ�ᣊ�ous" force that seems to push a rotaᣊ�n
g
object away from the center of rotaᣊ�on. The aspects of this
"force" are really a misconcepᣊ�on: what you are feeling is the
centripetal force exerted by the bar to keep you from
traveling in a direcᣊ�on tangent to the rotaᣊ�on. The
centrifugal force is "felt" when the bar pulls you inward,
giving you the impression that a force is pushing you
outward. If you let go of the merry‐go‐round, your inerᣊ�a
(remember Newton's First Law) will keep you moving in a
direcᣊ�on tangent to the rotaᣊ�on (which is a straight line in
the direcᣊ�on of your velocity at that very moment), not
outward. To prevent this from occurring during sports
involving circular tracks and even on highways, the turns are ba
nked or inclined. This bank
provides a normal force on the object, which is perpendicular to
the track surface. The
horizontal component of the normal force plus the horizontal co
mponent of the force of
fricᣊ�on between the ᣊ�res and track are the source of centrip

etal force inward. If the angle of
the track and the speed of the vehicle are just right, no sideways
fricᣊ�on forces are required
to turn on the track (Figure 15).
Angular Mo뙕on
Consider a mass moving at a constant speed in a circular
path. Since the mass is not moving in a straight line, its
posiᣊ�on is described as the angle, θ, from the x‐axis or
angular posi뙕on (Figure 16). Angular posiᣊ�on is
Figure 16: A dark gray mass is moving in
a circular path at radius, R, and has
traveled an arc length of l.
expressed in radians (instead of degrees) and is related
to arc length, l, and the radius, R, of a circular path.
Rθ(rad) = l
Angular posiᣊ�on is posiᣊ�ve when measured
counterclockwise and negaᣊ�ve when measured
clockwise.
The arc length of a complete circle is equal to the
circumference (2Πr) and there are 360° in a complete
circle. Therefore, you can convert back and forth

between degrees and radians using the equaᣊ�on:
360° = 2πr rad
In linear moᣊ�on, velocity is an object's change in posiᣊ�on
divided by the ᣊ�me the object is in moᣊ�on:
△ d / △t
In circular moᣊ�on, angular velocity is the amount of rotaᣊ�o
n or revoluᣊ�on per unit of ᣊ�me.
Average angular velocity, ωavg, is the number of radians per se
cond (rad/s) over a period of
ᣊ�me and instantaneous angular velocity, ω, is the number or r
adians over an infinitesimal
period of ᣊ�me:
Average: ωavg = △ϑ / △t
Instantaneous: ω = dθ / dt
Average angular velocity can also be calculated by measuring th
e period of mo뙕on, P, which
is the amount of ᣊ�me it takes an object to make one full revol
uᣊ�on. Angular velocity is
mathemaᣊ�cally related to period as:
ωavg = 2πr / P
When an object rotates at a fixed radius, R, its linear velocity at
a given instant is always
tangent to its circle of rotaᣊ�on. For this reason, the term tang
enᣊ�al velocity refers to the
linear velocity of a rotaᣊ�ng object (as opposed to angular vel
ocity). A tangent line is one that
touches the circle at one point, but never intersects it. Therefore

, the direcᣊ�on of the
tangenᣊ�al velocity is always changing while the object makes
a revoluᣊ�on. Since the linear
distance traveled around one revoluᣊ�on is the circumference,
the tangenᣊ�al velocity of an
object in uniform circular moᣊ�on is the total distance, 2Πr, o
ver the ᣊ�me period, P:
v = 2πr / P = Rω
An object freed from its rotaᣊ�on will conᣊ�nue traveling in
a straight line tangent to the
previous circle of rotaᣊ�on because the linear velocity is alway
s tangent the circle.
In the case where the angular velocity is not constant, or non‐un
iform circular mo뙕on, the
angular acceleraᣊ�on, (rad/s2), is the rate of change of the ang
ular velocity, ω:
α = dω / dt
The linear kinemaᣊ�cs equaᣊ�ons easily translate into rotaᣊ
�onal kinemaᣊ�cs equaᣊ�on when
linear variables are replaced with rotaᣊ�onal ones:
© 2014 eScience Labs, LLC.All Rights Reserved
http://www.esciencelabs.com/

Lab 2/Lab2Kinematics.docx - Copy.docx
Lab 2 Kinematics PHY250L”
Student Name:
Kit Code (located on the lid of your lab kit): AC-65CS487
“Pre-Lab Questions”
1. “What is the acceleration of a ball that is vertically tossed up
when it reached its maximum height?”
0
“
2. “The displacement of a particle as it varies with time is given
by the equation
x(t) = (10.0 m/s) t + (2.50 m/s²) t². Find the particle’s
instantaneous velocity and instantaneous acceleration at t = 4.00
seconds.”
Click here to enter text.
3. “What does a positive and negative slope represent for a
velocity vs. time graph?”
This exemplifies an increase when positive and a decrease when
negative of an object's velocity in respect to time.
4. “You know that a car moves with a velocity that can be
modeled as v(t)= 4.0 m/s + (1.2 m/s²)t and that at t = 0 the car
has a displacement of 5.00 meters from the origin. What is the
position of the car at t = 4.0 seconds?”
5. “Derive the second kinematic equation by integration of the
first kinematic equation. Then derive the third kinematic
equation by using algebra to combine the first and second
kinematic equations.”
“Experiment 1: Distance of Free Fall”

“Table 1: Washer Free Fall Data”
“Trail”
“Drop Height (m)”
“Time (s)”
“1”
4.2 Meters
1.0
“2”
.88
“3”
.95
“Average”
.9433
“Description of Auditory Observations of Equally Spaced Hex
Nuts:” Consistent sound
“Description of Auditory Observations of Unequally Spaced
Hex Nuts:” Inconsistent sound
“Post-Lab Questions”
1. “Record your hypothesis from Step 1 here. Use evidence from
your results to explain if your hypothesis was supported or not.”
Varying distances will have a direct correlation with time. The
shorter the distance the smaller the value of time will be. Vise
versa also applies, the longer the distance, the larger the value
of time will be. I have observed this during the first exercise.
2. “What was the difference between the noise patterns for
equally spaced hex nuts compared to the unequally spaced hex
nuts?”
The equally spaced hex nuts gave off a consistent sound pattern,

while the inconsistently spaced hex nuts were the opposite, and
let off a sporadic sound pattern.
3. “If the noise patterns were different, explain why. If they
were similar, explain why.”
The sound was different on the unequally spaced nuts, due to
the less distance in between each nut. This cause a shorter time
period in which there wasn’t any sound produced, until the next
nut hit the floor.
4. “Using the time it took a single hex nut to reach the pan,
calculate the height from which it was dropped. Is this accurate
compared to your known height? Explain your conclusion.”
The time in which I had used the stopwatch to calculate the
drop distance was 1 second flat. This gave me a calculation of
4.9 Meters, or roughly 16 ft. In comparison to the actual
measurement of 13.9 ft or 4.2 meters, this is very close.
5. “A student ran this experiment, and instead of dropping the
hex nut, he threw it. This gave the nut a velocity of v(t) = (12
m/s²)t + 5 m/s. What is the hex nut’s displacement as a function
of time if its position at t = 0 seconds is 0 meters? If the hex nut
dropped 1.2 meters, how long did it take for the nut to reach the
ground?”
“Experiment 2: Distance Traveled by a Projectile”
1. “In one of your experiments, you will roll a marble down a
ramp to provide an initial horizontal velocity. Suppose you start
the marble at rest (v˳= 0 m/s) and it travels a distance of, d,
down the ramp. Use 1-D kinematics to predict the velocity of
the ball (vᶠ) at the bottom of the ramp. Hint: the acceleration of
the ball down the ramp is 9.81*sin(θ) m/s² where θ is the angle
of the ramp. Record your answer in variables (you will calculate
the velocity with magnitudes when you perform the

experiment).”
2. “Use the kinematic equations to derive a general equation for
the time it takes a ball dropped from rest at vertical height, h, to
reach the ground. Use this to write a general equation for the
distance travelled by a projectile that is rolling off a table of
height, h, with a horizontal speed of V˳ₓ.” Click here to enter
text.
3. “Prove that launching a projectile at 45° provides the largest
range. Write the range as a function of θ. Take the derivative of
the range with respect to θ and find the maximum angle.”
4. “A butterfly flies along with a velocity vector given by v =
(a-bt²) Î + (ct) ĵ where a=1.4 m/s, b=6.2 m/s³, and c=2.2 m/s².
When t= 0 seconds, the butterfly is located at the origin.
Calculate the butterfly’s position vector and acceleration vector
as functions of time. What is the y-coordinate as it flies over x
= 0 meters after t = 0 seconds?”
“Data:”
“Table 1: Range and Velocity of Projectile at Ramp Distance 1”
“Ramp Incline (degrees):” 25
“Ramp Distance (m):”33.5
“Trial”
“Measured Distance (m)”
“1”
42.5
“2”
41.5
“3”
41.
“4”
42.25

“Average”
41.8
“Ramp Distance (m):” 28.5
“Trial”
“1”
35.8
“2”
37
“3”
37.6
“4”
36.5
“Average”
36.7
“Ramp Distance (m): “23.5
“Trial”
“1”
32.5
“2”
32.5
“3”
33.3
“4”
33.5
“Average”
32.95
“Post-Lab Questions:”
5. “Use your predictions of velocity and range from the Pre-Lab
Questions and the data recorded from your experiment to

complete Table 4.”
“Table 4: Velocity and Range Data for all Ramp Distances”
Ramp Distance (m)
Calculated velocity (m/s)
Predicted Range (m)
Average Actual Range (m)
Percent Error
6. “How do your predictions compare to the observed data?
Explain at least two reasons for differences.”
7. “If you were to fire a paintball pellet horizontally and at the
same time drop the same type of paintball pellet you fired from
the paintball gun, which pellet would hit the ground first and
why is this so?”
They would both come into contact with the ground at the same
exact time. They both experience the same amount of downward
air resistance, therefore they both are approaching the ground at
the same exact rate.

8. “A marble slides down a wacky ramp with a velocity given
by v = (0.5t²-3.0t) Î + (0.33t³-0.6t) ĵ. At t = 3 seconds the
particle is shot off the ramp and behaves like a projectile. What
is the magnitude of the velocity of the marble when it leaves the
ramp? What is the marble’s acceleration vector while it is on
the ramp? If the marble falls for 6 seconds when it is shot off
the ramp at a 45° angle, what will be its displacement in the x-
direction?”
“Insert photo of your experimental setup with your name clearly
visible in the background:”
“Experiment 3: Squeeze Rocket™ Projectiles”
“Table 5: Projectile Data for Rockets and Different Launch
Angles”
Launch Velocity (m/s)
Initial Angle
Time (s)
Average Time (s)
Predicted Range (m)
Actual Range (m)
Average Range (m)
Range % Error
0.5
“90°”
1.34
1.42
1
.67
.905

17
“90°”
1.41
1
.83
33
“90°”
1.53
1
.99
1
“90°”
1.38
1
1.1
10

80
1.4
1.56
1.5
1.2
1.37
30
80
1.6
1.5
1.4
10
80
1.51
1.5
1.37
13

80
1.72
1.5
1.52
2
25
1.05
1.04
3
3.09
3.02
9
25
1.01
3
3.25
25
25
1.11

3
2.81
19
25
.98
3
2.92
8
1.28
1.05
1.25
2
1.98
2.2
2
60
1.15

2
2.48
48
60
1.22
2
2.25
25
60
1.36
2
2.1
10
1. “Which angle provides the greatest range? Which provides
the least? Based on your results, which angle should give you
the greatest ranged for projectile motion?”
The lower the angle the further the range as exemplified by the
data. The angle at which the projectile is launched has a direct

result on the distance at which it will travel.
2. “What role does air resistance play in affecting your data?”
Air resistance plays a huge role in the above data because of the
mass of the object itself. If the projectile had been more dense I
wouldn’t see the projectile sidewind as much as it did, however
it would take a bit more initial force to launch the projectile.
3. “Discuss any additional sources of error, and suggest how
these errors could be reduced if you were to redesign the
experiment.”
As stated above I would use an object more dense in structure,
due to the motion at which the projectile traveled during the
experiment this leaves room for some inconsistencies.
4. “How could kickers on a football team use their knowledge of
physics to better their game? List at least two other examples in
sports or other applications where this information would be
important or useful.”
A kicker on a football game could use knowledge of what angle
to kick the football at to make a precise kick in the field goal,
this way he could find exactly at what power and angle to kick
the ball at for the desired distance to the field goal. Another
practical use would be in golf. Having the ability to gauge
power, and angle for the desired distance in order make it onto
the putting green, or even a hole in one could be powerful.
5. “A student buys a high powered toy rocket gun and tries to
do this experiment. He decides to shoot the first rocket with an
initial angle of 0 degrees. The student knows that the rocket has
an initial velocity of 5 m/s when he shoots it off of the 3 meter
high table. The toy gun is able to give the rocket a horizontal
acceleration of (2.1 m/s³)t in the same direction as the initial
velocity. The vertical acceleration, directed downwards, is g.
Assume air resistance can be ignored. What is the horizontal

displacement of the student’s toy rocket?”
3 meters
“Experiment 4: Balancing Centripetal Force”
“Pre-Lab Questions:”
1. “In this lab, you will be rotating a mass on one side of a
string that is balanced by a second mass on the other end of the
string (Figure 5). Apply Newton’s Second Law of Motion to
mass 1, m₁, and mass 2, m₂, to solve for the period of mass 1.
Hint: assume m₁ = 4m₂. How is the centripetal force on m₁
related to the force of gravity on m₂.”
The centripetal force negates and counteracts the forces of
gravity on the end with more washers.
2. “Draw a free body diagram and solve for the centripetal
acceleration in terms of θ and g for one person riding on the
amusement ride in Figure 3.”
3. “The wheel of fortune is 2.6 meters in diameter. A contestant
gives the wheel an initial velocity of 2 m/s. After rotating 540
degrees, the wheel comes to a stop. What is the angular
acceleration of the wheel?”
4. “The angle that a spoke on a bicycle wheel has rotated
behaves according to function θ(t) = at² + bt where a = 0.6
rad/s² and b = 0.3 rad/s. Find the angular velocity of the spoke
as a function of time and the angular acceleration as a function
of time. Then find the instantaneous angular velocity and the
instantaneous angular acceleration at t = 3 seconds.”
“Data:”
“Table 1: Rotational Data”
Radius (m)
Time per 15 revolutions (s)

“Period (s)”
Expected Value
Percent Error (%)
“0.25”
5.96
.397
.5
10.3
“0.40”
7.6
.51
.75
24
“0.15”
4.94
.33
.4
7
5. “Compare your measured data to your predicted values with a
percent error calculation. Explain any difference with an error
analysis.”
6. “Draw a circle to represent the path taken by your rotating
mass. Place a dot on the circle to represent your rotating
washer. Add a straight line from the dot to the center of the
circle, representing the radius of rotation (the string). Now label
the direction of the tangential velocity and the centripetal
force.”
Insert photo of the circle with your name clearly visible in the
background:
7. “Use your data to calculate the average velocity, angular
velocity, and centripetal acceleration for the mass in each

radius.”
8. “Refer to the picture in Figure 3 again (picture in Pre-Lab
question #2). Before the apparatus begins to spin, the wires
connecting the swings to the top of the structure will be
completely vertical. Once the apparatus begins to spin the
swings move outward radially, but also upwards vertically.
From where does the force causing this vertical acceleration
come?”
9. “Refer to the picture in Figure 3. Imagine that the swings are
rotating around the center with a constant speed, and the wire
connecting the swings to the center pole is at a 45 degree angle.
The angular velocity of the center pole is then doubled. Does
this mean that the chairs’ velocities will increase by a factor of
two, less than two, or more than two? Explain your reasoning.”
10. “A uniform disc is rotating around a frictionless, vertical
axle that passes through its center has as radius of R=0.300m
and a mass of 25.0 kg. The disc rotates according to θ(t) = (2.20
rad/s²) t² + (5.63 rad/s) t. When the wheel has rotated 0.200 rev,
what is the resultant linear acceleration of any point on the
disc?”
Lab 3/Lab 3- Newton's Laws.pdf
1/4/2017 Lab 3: Newton's Laws

Learning Objec᧻ves
Formulate the law of iner᧻a
Relate force and accelera᧻on
Apply ac᧻on and reac᧻on pairs to forces
Draw and explain free body diagrams
Apply Newton's 2nd Law to the Atwood Machine
Introduc᧻on
The laws of physics that we know today were discovered and ha
ve been studied for centuries.
In the fourth century B.C. Aristotle proposed the general belief
that a force causes a constant
velocity. In addi᧻on to Aristotle, Isaac Newton, famous for his
Laws of Mo᧻on, based his work
on the discoveries and experiments of Galileo and Johannes Kep
ler. In order to reduce the
plague from spreading through the college where Newton was a
fellow, it was temporarily
closed. During this closure, Newton spent a few years in rela᧻v
e isola᧻on where he started to
Figure 1: The moĕon of moon as predicted
by Newton's First Law should follow the

doĥed path. Since the moon does not
follow the predicted path, there must be a
force acĕng on it.
formulate ideas on mathema᧻cs and physics. Newton was trying
to figure out what objects in
orbit and objects falling toward the Earth have in common. The
answer actually required a
new type of math: calculus! Newton formulated quan᧻ta᧻ve ex
plana᧻ons for the mo᧻on of
falling objects, orbi᧻ng objects, pulley systems, and much more
. These ideas were able to
explain all types of mo᧻on and can be broken down into three b
asic laws.
Force
While force can be described as a push or a pull, it is
more clearly defined as an ac᧻on that causes an object
to change its mo᧻on. Force is what causes the direc᧻on
of an object's velocity to change. Therefore, a net force
ac᧻ng on an object causes accelera᧻on.
Newton's First Law of Mo᧻on
Newton's First Law of Mo᧻on states that an object will
maintain its state of mo᧻on un᧻l acted upon by an
external force. In other words, an object will remain at
rest or move at a constant velocity un᧻l an outside
force acts upon it (Figure 1). Newton's First Law is also
called the Law of Iner᧻a. Iner᧻a is an object's tendency
to resist changes in mo᧻on (speed or direc᧻on). Ma台�er
has this property whether it is at rest or in mo᧻on.
When a net force on an object is applied, the object will acceler
ate in the direc᧻on of that

force. If the net force on the object equals zero then there will n
ot be a change in its
accelera᧻on. When discussing net force, the vector sum of the f
orces, the resultant, must be
examined.
The movement of planets around the sun is an example of iner᧻
a. Planets have a lot of mass,
and therefore a great amount of iner᧻a ‐ it takes a huge force to
accelerate a planet in a new
direc᧻on. The pull of gravity from the sun keeps the planets in
orbit; if the sun were to
suddenly disappear, the planets would con᧻nue at a constant sp
eed in a straight line,
shoo᧻ng off into space!
Vector Addi᧻on
Unlike addi᧻on of scalars, vector addi᧻on requires a geometric
al process where the tail of the
second vector is placed at the ᧻p of the first vector, �en refe
rred to as the tail to ᧻p method.
If an object is acted upon by force A and then by force B, it is s
ame as the object undergoing a
single force C (Figure 2). This value is known as the resultant o
r vector sum.
A + B = C

Figure 2: The addiĕon of vectors A and B to get
the resultant vector C.
Vectors can be broken up into components by using the proper᧻
es of right triangles
(discussed in Lab 5: 2‐D Kinema᧻cs and Projec᧻le Mo᧻on), an
d it offers a more general way
of adding vectors than the tail to ᧻p method. The x‐components
of each vector can be added
together, and the y‐components of each vector can be added tog
ether. Once we have both
the x‐ and y‐components, we can solve for the magnitude of the
vector by using the formula
|a| = √ax
2 + ay
2, for a general vector a. To get the direc᧻on of the vector you
can use Θ = tan‐
1 (ay/ax). This gives you the resultant of the forces ac᧻ng upon
an object. Finding the
resultant force, also known as the net force, is very important fo
r solving problems involving
Newton's Laws.
Mass vs. Weight
Newton observed a special rela᧻onship between mass and iner᧻
a. Mass is �en confused
with weight, but the difference is crucial in physics. While mass
is the measure of how much
ma台�er is in an object (how much stuff is there), weight is a m
easure of the force experienced
by an object due to gravity. Thus, weight is rela᧻ve to your loc
a᧻on. Your weight differs at the
Earth's core, the summit of Mount Everest, and especially in out

er space, when compared to
the Earth's surface. Conversely, mass remains constant in all the
se loca᧻ons. Mathema᧻cally,
weight, w, is the mass, m, of an object mul᧻plied by its acceler
a᧻on due to gravity, g:
w = mg
Newton's Second Law of Mo᧻on
Click to Run
http://phet.colorado.edu/sims/forces-and-motion-basics/forces-
and-motion-basics_en.jnlp
Newton also noted that the greater an object's mass, the more it
resisted changes in mo᧻on.
Therefore, he concluded that mass and iner᧻a are directly propo
r᧻onal: when mass
increases, iner᧻a increases. This predic᧻on produced Newton's
Second Law of Mo᧻on, an
expression for how an object will accelerate based on its mass a
nd the net force applied to
the object. This law can be summarized by the equa᧻on:
ΣF = ma
where ΣF is the sum of all forces ac᧻ng on the object, m is its
mass and a is its accelera᧻on.
The kilogram (kg) is the standard measurement for mass, and m

eter/second/second (m/s2) is
the standard measurement for accelera᧻on. The standard measur
ement for force is the
Newton, where 1 N = 1 kg∙m/s2. Comparing this equa᧻on to the
first one (w = mg) helps
reinforce the difference between mass and force (such as weight
).
Free body diagrams (as seen in Figure 3) are a useful tool for so
lving problems related to
Newton's Second Law of Mo᧻on. They allow you to iden᧻fy an
d draw all of the forces ac᧻ng
on an object. Objects can be represented by simple shapes like c
ircles and squares. Forces are
represented by solid arrows. It is helpful to visualize the net for
ce ac᧻ng on an object by
labeling all objects and forces included in the free body diagram
.
Since force and accelera᧻on are both vector quan᧻᧻es, Newton
's Second Law can be wri台�en
as a vector equa᧻on.
ΣF = ma
It can also be broken up into x‐ and y‐ components (for three di
mensional problems there is
also a z‐component).
ΣFx = max
ΣFy = may
Figure 3: Free body diagram of a block hanging form a beam by
a string.

Newton's Third Law of Mo᧻on
Newton's Third Law of Mo᧻on states that for
every ac᧻on (force) there is an equal and
Figure 3: Newton's Third Law of Moĕon explains why
you move backwards when you throw a ball on ice.
opposed reac᧻on. Imagine standing on an ice
ska᧻ng rink and holding a ball (Figure 3).
Ini᧻ally you are not moving. You throw a ball
and as you do, you start to move backwards
on the ice. When you exerted a force on the
ball it exerted a force on you equal in
magnitude and opposite in direc᧻on. Even
when you walk, you push against the ground,
and it pushes right back!
Newton's three laws of mo᧻on govern the
rela᧻onship of forces and accelera᧻on. There are many applica
᧻ons of Newton's Laws in your
everyday life. To get that last bit of ketchup from the bo台�le, y
ou shake the bo台�le upside‐
down, and quickly stop it (with the lid). Consider riding in a car
. Have you ever experienced
iner᧻a while rapidly accelera᧻ng? Thousands of lives are saved
every year by seatbelts, which
are safety restraints that protect against the iner᧻a that propels
a person forward when a car

comes to a quick stop.
Lab 3/Lab3NewtonsLaws.docx - Copy.docx
Lab 3 Newton’s Laws PHY250L
Student Name: Click here to enter text.
Kit Code (located on the lid of your lab kit): Click here to enter
text.
“Use the free body diagram of the pulley (Figure 5) to answer
the Pre-Lab Questions.”
1. “Draw free body diagrams for M₁ and M₂.”
“Insert photo of diagram with your name clearly visible in the
background:”
2. “Apply Newton’s Second Law to write the equations for M₁
and M₂. You should get two equations with tension in the string,
weight for each mass and accelerations for each mass (a₁ and
a₂).”
F1=M1*A1 / F2=M2*A2
3. “Post-Lab Question 2 results in two equations with three
unknowns! A third equation is required to solve the system.
What is the third equation?”
4. “In the diagram to the right, there is a 5 kg block of ice on a
slippery slope positioned 25° from horizontal. What is the
acceleration of the block as it slides down the ramp?”

5. “An object is acted upon by a force that can be modeled by
F(t) = 5.6 N î + 2.4 N ĵ. The object has a mass of 3.0 kg and
starts at rest. Calculate the velocity v(t) as a function of time.”
“Experiment 1: Graphing Linear Motion”
“Table 1: Motion of Water Observations”
“Motion”
“Observations”
“a”
water sloshed to one side
“b”
Water eventually remained at rest
“c”
Water sloshed to the left and then to the right
“d”
water sloshed forwards
“Table 2: Observations After Flicking Notecard Off of Cup”
“Trial”
“Observations”
“1”
Washer dropped to the bottom of the cup
“2”
“3”
“4”
“5”
1. “Explain how your observations of the water and washer
demonstrate Newton’s law of inertia.”
For the washer, the only force acting upon it is gravity, and

friction. The friction moved the washer slightly in the direction
the index card flew, but gravity overtook and brought the
washer to the bottom of the cup. For the water, the only forces
acting on the body of water was gravity and the edges of the
container.
2. “Draw a free body diagram of your containers of water from
the situation in Part 1 Step 4d. Draw arrows for the force of
gravity, the normal force (your hand pushing up on the
container), and the stopping force (your hand accelerating the
container as you stop). What is the direction of the water’s
acceleration?” Forward
background:”
3. “Can you think of any instances when you are driving or
riding in a car that is similar to this experiment? Describe two
instances where you feel forces in a car in terms of inertia.”
When the brakes are applied abruptly. Your body remains in
motion as the vehicle slows down causing you to feel
constrained when the seatbelt slows your body down.
4. “You and your friend both preform 3b of Part 1 of this
experiment. You walk at 0.5 m/s, but your friend walks at 1.5
m/s. Which container of water will experience a greater net
force?”
My friends container will experience greater net force due to
the higher amount of intertia
“Experiment 2: Newton’s Third Law and Force Pairs”
“Table 3: Force on Stationary Springs”
“Force on Stationary 10 N Spring Scale (N)”
5
“Force on Stationary 5N Spring Scale (N)”
5

“Table 4: Spring Scale Force Data”
“Suspension Set Up”
“Force (N) on 10 N Spring Scale”
“Force (N) on 5 N Spring Scale”
“0.5 kg Mass on 10 N Spring Scale”
.5
“0.5 kg Mass with String on 10 N Spring Scale”
.5
“0.5 kg mass, string and 5 N Spring Scale on 10 N spring scale”
.8
.7
“0.5 kg mass, string and 5 N Spring Scale on 10 N spring scale
on Pulley”
.6
.7
1. “How did the magnitude of the forces on both spring scales
compare after you moved the 10 N spring scale?”
The forces acted upon each scale were the same.
2. “How did the magnitude of the forces on both spring scales
compare after you move the 5 N spring scale?”
The forces acted upon each scale were the same.
3. “Use Newton’s Third Law to explain your observations in
Questions 1 and 2.”
For each action there is an equal and opposite reaction. So each
force applied reacted to each scale equally.
4. “Compare the force on the 10 N spring scale when it was
directly attached to the 0.5 kg mass and when there was a string
between them.”

there was no difference in the amount of force, however there
was a slight increase due to the mass of the string itself.
5. “Compare the force on two spring scales in Steps 5 and 6.
What can you conclude about the tension in a strong?”
Tension on a string will eventually reach the point where it’s
taught, and hold the amount of weight that is applied to it.
6. “Olympic spring Usain Bolt set the world record in the 100
meter dash with a time of 9.58 seconds. The physics behind his
performance are impressive. Physicists have modeled his
position as a function of time to be as follows:”
“where A=110 m/s, B=12.2 m/s, k=0.9 1/s. Using this
information, find his acceleration when t = 0 seconds. If his
mass was 86 kg at the time of the race, what was his force at t =
0 seconds. What was the force from the ground on Bolt at t = 0
seconds?”
“Experiment 3: Newton’s Third Law and Force Pairs”
“Table 5: Motion Data”
“Mass of 15 Washers (kg)”
.040
“Average of Mass of Washer (kg)”
.00266
“Procedure 1”
“Height (m):” .365
“Trial”
“Time(s)”
“1”
1.25
“2”

1.13
“3”
1.09
“4”
1.06
“5”
1.31
“Average”
1.17
“Average Acceleration (m/s²)”
.266
“Procedure 2”
“Height (m):” .365
“Trial”
“Time(s)”
“1”
.78
“2”
.69
“3”
.66
“4”
.72
“5”
.65
“Average”
.7
“Average Acceleration (m/s²)”
.744
1. “Draw a free body diagram for M₁ and M₂ in Procedure 2.
Draw force arrows for the force due to gravity acting on both
masses (Fᵍ₁ and Fᵍ₂) and the force of tension (Fᵀ). Also draw
arrows indicating the direction of acceleration, a.”

background:”
2. “Use Newton’s Second Law to write an equation for each of
the free body diagrams you drew in Question 1. Be sure to use
the correct signs to agree with your drawings. Solve these four
equations for the force of tension (Fᵀ). Your answer should be
written in variable form.”
3. “Set the two resulting expressions for the force of tension
equal to one another (as long as the string does not stretch, the
magnitude of the acceleration in each equation is the same).
Replace Fᵍ₁ and Fᵍ₂ with M₁ and M₂, respectively. Solve the
resulting equation for a. Then, go back to Questions 2 and solve
for the Fᵀ.”
4. “Calculate the acceleration for the two sets of data you
recorded and compare these values to those obtained by
measuring distance and time using percent error. What factors
may cause discrepancies between the two values?”
5. “Calculate the tension in the string for the falling washers.
From these two values, and the one where the masses were
equal, what trend do you observe in the tension in the string as
the acceleration increases? Show all calculations.”
6. “Using the theoretical acceleration found in Question 4 for
Procedure 1, find the velocity of the block as a function of time
by integration.”

Lab 4/Lab 4- Friction.pdf
1/4/2017 Lab 4: Friction
Learning Objec᧻ves
Explore and explain the difference between sta᧻c and kine᧻c fr
ic᧻on
Determine the dependence of the force of fric᧻on on the normal
force
Apply the force of fric᧻on to objects on an incline
Introduc᧻on
Newton's Laws of Mo᧻on can help iden᧻fy forces ac᧻ng on ob
jects that can not be seen with
the unaided eye and would go unno᧻ced without them. Fric᧻on
al force and normal force are
two forces that can be inferred from an object's mo᧻on or lack t
hereof.
Normal Force

Figure 1: Free body diagram
of an object at rest.
Figure 2: Surfaces have microscopic imperfec笕�ons
that contribute to fric笕�on.
When you place a book on a table, why doesn't it fall through
the table? The force due to gravity (Fg = mg) never stops
pulling down on the object, yet it remains sta᧻onary on the
table. Newton's Second Law states the net force on an object
must be zero if the object is at rest. Therefore, there must be
another force ac᧻ng on the object that is opposing the force
of gravity. We call this force the normal force (Figure 1).
Similar to tension, normal force is also passive and will only
push back as hard as the applied force. Another important
characteris᧻c is that the normal force only pushes (applies a
force) perpendicular to its surface.
Fric᧻on
Newton's First Law states that an object in
mo᧻on will remain in that state of mo᧻on
un᧻l an outside force acts on it. If you roll a
ball across the floor or slide a book across a
table, you will see them eventually come to
rest. Similar to normal force and gravity, there
seems to be another force at play that stops
the rolling ball. The force that opposes the
direc᧻on of mo᧻on and causes objects to come to rest is called
fric᧻onal force. Fric᧻on exits
between two solid surfaces no maᘀ er how smooth they may see
m to the naked eye. If you
zoom in on the surfaces, you will observe microscopic bumps th

at impede the mo᧻on of an
object (Figure 2). Unlike the normal force, the fric᧻onal force a
lways acts parallel to the
surface.
Sta᧻c vs. Kine᧻c Fric᧻on
An object does not always move when you exert a horizontal for
ce. For example, have you
ever tried pushing a car that has broken down and not been able
to get it in mo᧻on? Sta᧻c
fric᧻on keeps an object at rest. If you push with a greater force
and the mass s᧻ll has not
changed posi᧻on, sta᧻c fric᧻on increases. Once you push hard
enough to put the car in
mo᧻on, you have overcome the maximum amount that sta᧻c fri
c᧻on can push back. Sta᧻c
fric᧻on can be modeled by the inequality Ff ≤ μsFN, where Ff i
s the force of fric᧻on, μs is the
coefficient of sta᧻c fric᧻on, and FN is the normal force. The le
ss than or equal to sign
indicates that the force of sta᧻c fric᧻on can range from zero to
a maximum value. The mass
will begin to move on the surface once the applied force is great
er than μsFN,.
Once the object starts to move, an external force must be applie
d in order to keep the object
moving to overcome fric᧻on due to an object sliding over a surf

ace. This type of fric᧻on is
called kine᧻c fric᧻on. The equa᧻on for kine᧻c fric᧻on is Ff =
μkFN where μk is the coefficient
of kine᧻c fric᧻on. No᧻ce that this is an equa᧻on and not an in
equality. As long as a mass is
sliding across a surface, the force due to kine᧻c fric᧻on will st
ay the same. In order for a mass
to speed up, the applied force must be greater than the fric᧻ona
l force, otherwise it will slow
down. In order to maintain a constant velocity, the applied force
must be equal to the force of
kine᧻c fric᧻on.
Figure 4: The coordinate axis is 笕�lted so that the x‐axis is
parallel to the surface upon which the object is in mo笕�on.
Figure 3: With the help of gravity,
skiers can overcome the force of
sta笕�c fric笕�on.
Fric᧻on on Inclines
The accelera᧻on of objects on inclines is due to gravity, yet
the accelera᧻on points along the surface of the incline. When
you ski, the gravita᧻onal force on you is able to overcome the
force of sta᧻c fric᧻on and allow you to move along the incline
of the mountain (Figure 3). When solving physics problems
with objects on an incline, it is o᧻�en beneficial to ᧻lt the

coordinate axis such that the x‐axis points along the incline
and the y‐axis points perpendicular to the surface (Figure 4).
This means that the normal force points along the y‐axis and
the force of gravity is now in both the x and y direc᧻on.
Click to Run
http://phet.colorado.edu/sims/motion-series/forces-and-
motion_en.jnlp
Click to Run
http://phet.colorado.edu/sims/the-ramp/the-ramp_en.jnlp
Lab 4/Lab4Friction - Copy.docx
Lab 4 Friction PHY250L
text.
1. “If the normal force perpendicular to a surface, what happens
to the magnitude of the normal force on an object as the angle
of the incline is increased?”
The magnitude of the normal force when theta is increased
results will show a decrease in normal force.

2. “Applying Newton’s Second Law and the equation for static
friction (F = μsN)₁ prove that the coefficient of static friction
(μs) is related to the minimum angle, θ, that causes the block to
slip (Figure 5) by the equation μs = tan(θ).”
When the angle (theta) is increased, friction is lessened due to
lesser normal force, causing the block to slide faster.
3. “A person applied a horizontal force to a crate of mass, m,
that caused the crate to move at a constant velocity (Figure 6).
Show that the relationship between the applied force and the
normal force is FA = μFN.”
Applied force and frictional force are equal. The relationship is
true because the scenario states that there is a constant velocity.
“Experiment 1: Static Friction and Mass on an Inclined Plane”
“Table 1: Wooden Block Incline Data” “Table 2:
Metal Washer Incline Data”
“Trial”
“Angle”
“1”
20
“2”
23
“3”
21
“4”
22
“5”
20
“Trial”
“Angle”
“1”
30
“2”
33

“3”
32
“4”
30
“5”
31
1. “Using the result from Pre-Lab Question 2, calculate the
coefficient of friction for each angle in Table 1 and Table 2.
Find the average value for the coefficient of friction for wood
and metal on cardboard.”
2. “Comment on your coefficients of static friction for wood
and metal. If they are different, why do you think they are
different? If they are the same, why do you think they are the
same?”
“Experiment 2: Static Friction vs. Kinetic Friction”
“Table 3: Peak Static Frictional Force”
Total Mass (kg)
Trial 1 (N)
Trial 2 (N)
Trial 3 (N)
Average (N)
.5
2.6
2.4
2.2

2.4
.75
3.0
2.8
2.8
2.86
1
3.2
3.2
3.4
3.26
“Table 4: Kinetic Frictional Force”
Total Mass (kg)
“Trial 1 (N)
“Trial 2 (N)
Trial 3 (N)
Average (N)
.5
1.6
1.6
1.6
1.6
.75
2.2
2.2
2.2
2.2
1
2.8
2.8
2.8
2.8
Calculate the average force from Trials 1, 2, and 3 for static and
kinetic friction.

Record the averages in Tables 5 and Table 6.
Average Applied Force (N)
Normal Force (N)
Average Applied Force (N)
Normal Force (N)
“Table 5: Static Frictional Force”
Table 6: Kinetic Frictional Force
1. “From the result from Pre-Lab Question 3, the relationship
between the applied force and the normal force is FA = μFN.
When the data for applied force vs normal force is plotted, the
slope of the graph is equal to the coefficient of friction. Plot

graphs of average force vs normal force from the data in Table
5 and Table 6.”
Please draw a diagram to include on this document.
background:””
“Insert photo of experimental setup with your name clearly
visible in the background:””
Lab 5/Lab 5- Conservation of Energy Material.pdf
1/4/2017 Lab 5: Conservation of Energy
Learning Objec�ves
Relate energy to work
Calculate the amount work done by a force
Compare and contrast types of energy
Apply The Law of Conserva�on of Energy to poten�al and kine
�c energy
Introduc�on
You may be familiar with the common usage of the word energy

. For example, you may grab
an energy drink to get through the night when cramming late at
night for a test the next
morning, If you're an athlete, you may reach for an energy bar t
o help you finish a game.
Energy is one of the central concepts in science; it has been use
d to explain many natural
phenomena. But what is energy, exactly? Energy is defined as th
e ability to do work. The
amount of energy an object has equals the amount of work it has
the ability to do. Think of
Figure 1: Work done on a tray.
energy as the currency for performing work. For example, 100 J
oules of energy is required to
do 100 Joules of work.
Work Done by a Constant Force
Work is done when a force causes a mass to move a distance. Th
e unit for work is called the
Joule (J), which is defined as a force of one Newton ac�ng over
one meter. Quan�ta�vely, the
defini�on of work is:
Work = F∙d
where F is the force applied, and d is the displacement. Both for

ce and displacement are
vector quan��es, and the opera�on in the equa�on above is kn
own as the dot product. It is
one of the forms of vector mul�plica�on, which yields a scalar
quan�ty. This equa�on for
work is used for a constant force along straight line displaceme
nt. We can simplify the
equa�on above to be:
Work = F||∙d
In this equa�on, F|| is the amount of force applied along or agai
nst the displacement, d. We
can also write this equa�on as:
Work = Fd cosθ
where F is the magnitude of the force, d is the displacement,
and θ is the angle between the direc�on of the force and
displacement. The direc�on factor is a cri�cal aspect of work
to understand. For example, a waiter exerts a force to hold a
serving tray at a steady height (Figure 1). Is work done on the
tray? It may be surprising that the answer is no. Although it
takes energy to keep the tray raised, and the waiter is moving
across the room, the direc�on of the applied force is not the
same as the direc�on of mo�on. Therefore, there is no work
done on the tray by the force applied to it.
Work Done by a Variable Force
Forces are not always constant and may change with an object’s
loca�on. Springs are a good
example: as a spring is stretched, it must be pulled harder to kee
p the spring stretched out.
When the force is only varying in the x‐component and has a str
aight line displacement,

mathema�cally, this can be solved using integra�on:
where F(x) is the variable force integrated with respect to posi�
on from its original posi�on,
x0, to its final posi�on, xf. Graphically, this is equivalent to cal
cula�ng the area under the
curve of the force versus posi�on graph (Figure 2).
Figure 3: Wind turbines.
Figure 2: The area under the plot equals the amount of work don
e.
If the force is variable and the displacement is along a curved p
ath, work can be found by
using the following equa�on:
where F is a variable force integrated from one point to another
with respect to an
infinitesimally small displacement vector (dl) that lies tangent t
o the path of its posi�on. Once
again the dot product is present but can be simplified in certain
cases where Φ is the angle
between F and dl.
Types of Energy
Energy is one of the most important concepts in
physics, and comes in a variety of forms ‐

chemical, gravita�onal, elas�c, electric, and
rota�onal energy to name a few. Of these many
forms, energy can generally be categorized as
kine�c or poten�al energy. Kine�c energy (KE) is
the energy associated with the state of mo�on of
an object. Wind energy is an example of kine�c
energy (Figure 3). The molecules of gas in the air
are in constant mo�on, providing energy that can
be harnessed by wind turbines like the ones above
that convert it into electric energy. The faster the
object is moving, the greater its KE. The rela�onship between k
ine�c energy (KE), mass (m)
and velocity (v) is:
KE = ½ mv2
In contrast, poten�al energy (PE) is the stored
energy associated with the posi�on of an
object. This type of energy exists when there is
some kind of force that returns an object to its
Figure 4: Water stored in the Shasta dam in
California for hydroelectricity.
original posi�on a�er being displaced. For
instance, an object li�ed above the ground has
gravita�onal poten�al energy because it will
accelerate to the ground when released. Water
stored in a dam for hydroelectricity genera�on

is also a form of poten�al energy. When the
valves open to allow the water to flow, the
gravita�onal poten�al energy of the water is
converted to kine�c energy (Figure 4). The more
massive the object and the greater the height,
the more energy the object has when it falls to
the ground. The equa�on to determine
gravita�onal poten�al energy (PEgravity) is:
PEgravity = mgh
where m is the mass of the object, g is the gravita�onal force (9
.8m/s2), and h is the height
above the Earth's surface.
There are �mes when poten�al energy can be found, and the fo
rce on the object is needed.
By taking the deriva�ve of poten�al energy with respect to pos
i�on, this gives us the force on
the object. In one dimension the equa�on is:
for this equa�on U is poten�al energy measured in Joules. Fro
m this equa�on, we can also
relate work to poten�al energy:
W = ‐△U
Another example of poten�al energy is a rubber band. A stretch
ed rubber band has the
poten�al to return to its original length, doing work in the proc
ess (force applied over a
distance). In this sense, poten�al energy is o�en referred to as t
he energy of posi�on, as it is
energy that depends on how far an object is removed from a pos
i�on of equilibrium.
Similarly, a spring has elas�c poten�al energy that increases as

it is stretched, and also
depends on the material it is made from. With increased s�ffnes
s in springs comes the ability
to store more energy and do more work.
? Did You Know...
As the cyclist rides through the mountain trail, chemical energy
is converted into ki �c
energy and thermal energy (you get hot riding a bike). As the cy
clist climbs the mountain,
the �c energy in the bicycle and the cyclist will be conver
ted to other forms of energy.
�on between the �res and ground causes �c ener
gy to transform to heat energy.
If a cyclist approaches a hill with enough speed, no more peddli
ng will be necessary to
reach the top because the �c energy will be converted to
�al energy. If the
cyclist needs to stop and applies the brakes, �c energy is
dissipated through �on
as heat energy.
Work Energy Theorum
Work is related to the object's displacement as can be seen by th
e equa�ons discussed
earlier. However, work is also dependent upon the change in an
object's speed. If an object is

speeding up then work is posi�ve, if the object is slowing down
then work is nega�ve, and if
there is no change in the object's speed then there is no work do
ne. The work‐energy
theorem says that:
Work = K2 ‐ K1 = △U
This means that the change in a object's kine�c energy equals t
he work done by the force on
the object. Remember that kine�c energy equals 1/2mv2.
Power
Power is the rate at which work is done, so power is analyzing t
he work being done over �me.
Just as there is average velocity and instantaneous velocity, ther
e is average power and
instantaneous power. To find average power, you look at the cha
nge in work over a �me
period:
For instantaneous power, you take the deriva�ve of work with r
espect to �me:
The standard unit of power is a wa� (W) where 1 W is equal to
1 J/s. Power can also be found
by looking at the force ac�ng upon an object and the object's ve
locity. In this case, average
power can be found by the given equa�on:

Where F|| is the component of the force that acts parallel to the
displacement vector and
vaverage is the average velocity. To find the instantaneous pow
er by looking at force and
velocity, the dot product can again be used:
Conserva�on of Energy
Energy can change forms, such as poten�al energy transforming
into kine�c energy. The law
of conserva�on of energy states that the energy of an isolated s
ystem cannot be created or
destroyed; it can only change forms or be transferred from one o
bject to another. Consider a
system composed of a ball and the Earth. If a ball is sta�onary t
hree meters above the
ground, it has gravita�on poten�al energy. When the ball falls
due to the force of gravity
(remember Earth is part of our system), the kine�c energy of th
e ball right before it hits the
ground is equal to the stored gravita�onal poten�al energy of t
he ball at three meters above
the ground. When the ball bounces back up, the kine�c energy
will be converted back into
poten�al energy. Since the total energy of the system is equal t
o the original poten�al energy
at three meters, the maximum height the ball can ever bounce to
is three meters. A ball
rarely bounces back to the original height because some energy
goes into hea�ng the air and
ground through air resistance and contact with the ground.
As long as all energies are accounted for, the sum of the energie
s is the same at any moment.

This is a very useful principle because now the energy of an obj
ect at any two moments in
�me can be compared! Let's examine this quan�ta�vely using t
he ball and Earth system.
Problem: A ball of mass, m, is dropped from rest from a height,
h, of 3 meters. What is the
ball's speed a�er it has fallen 2 meters?
Solu�on: If the ground is chosen as a reference height (where t
he height equals zero meters),
the two states to consider are when the ball is at rest at 3 meters
above the ground, and a�er
is has fallen 2 meters, or it is located 1 meter above the ground.
The total energy of the first
state, E1, is equal to the total energy of the second state, E2:
E1 = E2
Energy converted to heat by air fric�on will be ignored. This m
eans that the only energies the
ball can have during the drop is kine�c and poten�al energy:
KE1 + PE1 = KE2 + PE2
When the kine�c energy states and poten�al energy states are r
eplaced with their
equivalents the following equa�on can be derived:
½ mball v1
2 + mball gh1` = ½ mball v2
2 + mball gh2
No�ce the mass of the ball is in every term and can be divided
out of the equa�on. In E1, the

ball is dropped from rest which means v1 = 0 m/s.
gh1` = ½ v2
2 + gh2
When the speed of the ball 1 meter above the ground (v2) is sol
ved for, the following
equa�on can be derived and solved:
Lab 5/Lab 5- Conservation of Energy.pdf
Experiment 2: Conservaꭍ�on of Energy‐ Data Analysis
In this experiment you will apply your knowledge of Conservaꭍ
�on of Energy to a given data
set. Table 2 contains data for the posiꭍ�on of a ball with mass 0
.5 kg that has been dropped
from 5 meters from rest at given ꭍ�me points. In order to test th
e conservaꭍ�on of energy,

the potenꭍ�al energy and kineꭍ�c energy of the ball need to be
calculated.
Materials
Table 2
Procedure
1.
To calculate the kineꭍ�c energy, the velocity of the ball needs t
o be known. From
the given table, only posiꭍ�on is provided. To find velocity, a
method called the
"leap‐frog” method can be used to approximate the velocity usin
g the posiꭍ�on and
ꭍ�me points just before and a⚽er:
v= Δx
Δt
The first velocity point has been done for you as an example:
v =
Δx
=
x3 ‐ x1 =
0.10 m‐ 0.00 m
= ‐ 0.40 m/s
Δt t3 ‐ t1 0.496 s ‐ 5.00 s

2.
Use this method to calculate the other velociꭍ�es. Record the ve
lociꭍ�es in Table 5.
Table 2: Dropped Ball Data
Time (s)
Ball Posiꭍ�on
(m)
Ball Velocity
(m/s)
Potenꭍ�al Energy
(J)
Kineꭍ�c Energy
(J)
Total Energy
(J)
0.00 5.00
0.05 4.99
0.10 4.96
0.15 4.89
0.20 4.78

0.25 4.69
0.30 4.54
0.35 4.40
0.40 4.22
0.45 4.00
0.50 3.80
0.55 3.50
0.60 3.26
0.65 2.93
0.70 2.60
0.75 2.23
0.80 1.88
0.85 1.46
0.90 1.05
0.95 0.58
1.00 0.11
Lab 5/Lab5ConservationofEnergy.docx - Copy.docx
Lab 5 Conservation of Energy PHY250L
text.
1. “The force due to a spring is variable and is quantified by F =
kx, where k is the spring constant and x is the displacement.
Given the graph of the force versus displacement graph for a
spring in Figure 5, write an equation for the amount of work
done by the spring:”
a. “The work done by the spring integration”

b. “The work done by the spring by calculating the area under
the curve.”
2. “Consider the ball example in the introduction when a ball is
dropped from 3 meters. After the ball bounces, it raises to a
height of 2 meters. The mass of the ball is 0.5 kg.”
a. “What is the speed of the ball right before the bounce?”
b. “How much energy was converted into heat after the ball
bounced off the ground? (Hint: Thermal Energy (TE) will now
need to be included in your conservation of energy equation and
you will now need to know the mass of the ball).”
c. “What is the speed of the ball immediately after the ball
bounces off the ground?”
3. “A 2 kg book is sitting on a horizontal glass table top. A
woman pushes the book across the table with a horizontal force
F. The book’s position changes as a function of time given by
x(t) = (2.9 m/s³) t³- (3.6 m/s²) t². What is the book’s velocity at
t=1.2 seconds? Calculate the magnitude of the force at t=1.2
seconds. Find the work done by the force during the first 1.2
seconds of its motion.”
4. “Propeller-driven airplanes have engines that develop a thrust
(a forward force on the airplane) of 15,000 N. What is the
instantaneous power produced by the aircraft when it is flying
at 166.67 m/s?”

“Experiment 1: Work Done by a Spring”
“Table 1: Spring Scale Force Data”
“Force (N)”
Distance, x (m)
ForceAverage (N)
Δ Distance, Δx (m)
Work (J)
“0”0
“0”
“0.01”
1
“0.01”
“0.01”
2
“0.02”
“0.01”
3
“0.03”

“0.01”
4
“0.04”
“0.01”
5
“0.05”
“*Note, you will finish completing Table 1 in the Post-Lab
Questions section.”
1. “Create a Force vs. Displacement (stretch) graph.”
“Insert photo of graph with your name clearly visible in the
background:”
2. “Using the result of Pre-Lab Question 1, calculate the work
done by the spring.”
3. “The work done by the spring can be broken down by the
work done by each 1 cm stretch. Fill in the rest of Table 1 to
calculate the average force applied by the spring over each 1 cm
stretch.”
4. “Calculate the work done in each segment and determine the
total work done by adding all of the segments together. How
does this compare to the work done by the spring calculated in
Post-Lab Question 2?”

5. “A student finds a spring that does not obey Hooke’s Law
and stretches it 5.0 cm. The force that he had to use in order to
stretch the spring is modeled by Fx= 5.4 N + (-2.3 N/m²) x² +
(4.3 N/m³) x³. How much work was required for this task?”
“Experiment 2: Conservation of Energy – Data Analysis”
“Table 2: Dropped Ball Data”
“Time (s)”
Ball Position (m)
Ball Velocity (m/s)
Potential Energy (J)
Kinetic Energy (J)
Total Energy (J)
“0.00”
“5.00”
“0.05”
“4.99”
“0.10”
“4.96”
“0.15”
“4.89”

“0.20”
“4.78”
“0.25”
“4.69”
“0.30”
“4.54”
“0.35”
“4.40”
“0.40”
“4.22”
“0.45”
“4.00”

“0.50”
“3.80”
“0.55”
“3.50”
“0.60”
“3.26”
“0.65”
“2.93”
“0.70”
“2.60”
“0.75”
“2.23”

“0.80”
“1.88”
“0.85”
“1.46”
“0.90”
“1.05”
“0.95”
“0.58”
“1.00”
“0.11”
1. “Graph the potential energy, kinetic energy, and total energy
of the ball.”
“Insert photo of graph with your name clearly visible in the

background:”
2. “Describe the shape of each graph.”
3. “What are the limitations of using the leap-frog method?”
4. “An object is moving horizontally with no vertical
movement, and it has a mass of 0.20 kg. Its potential energy
function can be described as U(x) = (3.8 J/m²) x² — (2.1 J/m) x.
What is the force on the object? What is the magnitude of the
acceleration at x = 0.44 meters?”

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