Part b1)Mass (kg)Velocity (m/s)Force (N)Acceleration
(m/s2)Time to come to rest (s)stopping distance (m)
Lab 4--Part b1)
An object with a mass 'm' is moving with an initial speed 'v'and
is acted on by a single force ‘F’ in the opposite direction of its
motion. Use Excel to determine how long it will take the object
to come to rest and how far the object travels until it stops..
i) If the mass is doubled, what is the effect on the time?, on the
stopping distance?
ii) If the initial velocity is doubled, what is the effect on the
time?, on the stopping distance
input: mass, initial velocity, force
output: acceleration, time to come to rest, stopping distance
Part b2)Mass (kg)Fx (N)Fy (N)ax (m/s2)ay
(m/s2)2000050000100000time
(s)vx(m/s)vy(m/s)v(m/s)x(m)y(m)d
(m)00.511.522.533.544.555.566.577.588.599.510
Lab 4--Part b2)
A rocket ship, with mass m=40,000kg, and engines mounted
perpendicularly in the x and y directions, fires both rockets
simultaneously. The engine oriented in the x-direction fires for
3s and shuts off. The engine oriented in the y-direction fires
for 7s and shuts off. The force from the engine in the x-
direction is 50,000N and the force from the engine in the y-

direction is 100,000N. Make a scatter plot of the y-position of
each particle as a function of the x-position, showing the
trajectory of the rocket.
Use Excel to determine the following:
i) While the engines are firing, what is the acceleration of the
rocket in the x and y directions?
ii) After 7s, what is the velocity of the rocket in the x and y
directions?
iii) After 7s, what is the speed of the rocket?
iv) After 7s, how far has the rocket travelled in the x-direction?
How far has it travelled in the y-direction?, After 10 s?
v) After 7s, what is the displacement of the rocket? After 10 s?
Is the displacement of the rocket the same as the distance
travelled? Explain.
vi) If the mass of the rocket is doubled, what happens to the
displacement?
Output: ax, ay, vx, vy, x, y, d
Rocket Trajectory
x
y
Part a1)Mass (kg)Force (N)Acceleration
(m/s2)105010100205020100

Lab 4--Part a1)
Use Excel to determine the acceleration for an object with mass
'm' being pulled by a constant,
horizontal force (F) on a flat, frictionless surface.
i) What happens to the acceleration if the magnitude of the
force doubles?
ii) What happens to the acceleration if the mass of the object
doubles?
iii) What happens to the acceleration if both the mass and the
force are doubled?
Input: mass and force
Output: acceleration
Part a2)Mass (kg)Angle
(degrees)μkμsf_s(max)f_kF_Wsin(q)Acceleration
(m/s2)Accelerating or
Stationary?400.20.5450.20.54100.20.54150.20.54200.20.54250.
20.54260.20.54270.20.54280.20.54290.20.54300.20.54350.20.5
4400.20.54450.20.54500.20.510500.20.54900.20.5
Lab 4--Part a2)
Use Excel to determine the acceleration for an object with mass
'm' sliding down a surface inclined at an angle θ (between 0 and
90 degrees) above the horizontal. The surface has a coefficient
of kinetic friction μk and a coefficient of static friction μs.

Note 1: Within the calculation, before taking sine or cosine, the
angle must be converted to radians.
Use the built in function 'RADIANS' for this purpose
Note 2: the coefficient of static friction must be overcome in
order for the object to start sliding. Therefore, you must
determine if the force pulling the object down the incline is
greater than the force of static friction. If it is, the acceleration
can be determined. Otherwise, the object will remain stationary
and the acceleration reading will be erroenous.
Use the built in function 'IF' for this purpose.
If the object is accelerating, have the IF command output
"Accelerating".
If the object is not accelearting, have the IF command output
"Stationary".
Once it has been determined that the object is accelerating,
determine the magnitude of the acceleration.
Create a scatter chart of acceleration versus angle of incline for
angles from 0 to 50 degrees.
i) Does the mass of the object affect the acceleration? Explain.
ii) What happens when θ=90 degrees? Is the acceleration
affected by the coefficients of friction?
Input: m, θ, μk, μs
Output: Accelerating or Stationary?, magnitude of the

acceleration
acceleration versus angle of incline
0 5 10 15 20 25 26 27 28 29 30 35 40
45 50
angle (degrees)
acceleration (m/s^2)
Part a3)Mass (kg)Weight (N)Acceleration (m/s2)Scale Reading
(N)
Lab 4--Part a3)
An object, with mass m=100kg, is sitting on a scale in an
elevator. Use Excel to determine the reading on the scale (in
Newtons) for the following 4 situations:
i) elevator accelerating upward at +3 m/s2
ii) elevator moving at constant speed
iii) elevator accelerating downward at -3 m/s2
iv) elevator accelerating downward at -9.8 m/s2
Input: acceleration of the elevator
Output: Reading on the scale

Lab 4—Newton’s Second Law of Motion
Background
Newton’s Second Law of Motion
Newton‟s Second Law of Motion relates the total force on an
object to the resultant acceleration.
Mathematically, it can be expressed as
where m is the „mass‟ or „inertial mass‟ of the object and a is
the acceleration.
On the left hand side of this equation, the forces are being
added together, using Σ. Thus, if there
is more than one force acting on an object, it is the resulting
sum that determines the
acceleration. On the right hand side of the equation, is the
effect of the force: a net force results
in an acceleration that is proportional to an objects mass.
For two dimensional problems, the vector equation above can be
rewritten as two scalar
equations

where now the forces are accelerations are scalar quantities.
In the SI system, we use the following units for mass and
acceleration: [m] = kg, [a] = m/s
2
.
From Newton‟s Second Law, therefore, the units for a force are
[F] = (kg•m/s
2
). In honor of
Isaac Newton, we call a (kg•m/s
2
) a newton (N) and express forces in newtons.
Gravitation
With gravity, every particle in the universe with mass attracts
every other particle with mass.
This was first postulated by Isaac Newton, and it later became
known as Newton‟s Law of
Universal Gravitation:
Every particle of matter in the universe attracts every other
particle with a force that is directly
proportional to the product of the masses of the particles and
inversely proportional to the square
of the distance between them

Mathematically, this can be expressed as
2
21
r
mGm
where m1 and m2 are the objects gravitational masses
r is the distance between the center of masses
G is Newton‟s gravitational constant, 2211 /1067.6 kgmNG
For any object with mass „m‟ near the surface of the Earth, we
can write
gmm
r
Gm
r
mGm
F
E
E

21
where
rE is the radius of the Earth
mE is the mass of the Earth
Since mE and rE are known quantities, they can be combined
with „G‟, leading to
Since we call the local force of gravity on an object due to the
earth the object‟s weight, we can
write
Normal Force
The normal force arises from two solid objects in direct contact
and it is always a push. An
object resting on a surface is supported by a normal force and a
surface has a normal force
exerted on it by an object it supports. The normal force is
called „normal‟ because it‟s always
perpendicular to or normal to a surface.

For an object resting on a scale for example, the reading on the
scale is the normal force exerted
by the object, on the scale. In other words, the scale reads how
hard the object is pushing on it
(the normal force).
Kinetic and Static Friction
Frictional forces act to resist the motion of objects. The contact
force between two bodies can
often be represented by the normal force + frictional force. The
normal force acts perpendicular
to surface and the frictional force acts parallel. The normal and
frictional forces are thus always
perpendicular
Frictional force can be categorized as either „kinetic‟ or
„static‟. The kinetic friction force (fk )
acts when a body slides over a surface. Its magnitude increases
as the normal force increases. In
many cases the magnitude of the kinetic friction force can be
expressed (approximately) as

where µk is the coefficient of kinetic friction and „n‟ is the
magnitude of the normal force
The coefficient of kinetic friction is a unit-less quantity
between 0 and 1 and characterizes how
„slippery‟ the interface between two materials is. µk is smaller
(closer to 0) for more „slippery‟
surfaces and larger (closer to 1) for less „slippery‟ surfaces.
The static friction force( fs ) acts when there is no relative
motion between two surfaces. And
like the force of kinetic friction, its magnitude also increases as
the normal force increases. In
order to move a body, the static friction force must be overcome
first, and then the force of
kinetic friction acts.
fs is a variable force, meaning it can take on a range of values,
and can be expressed
(approximately) as
where µs is the coefficient of static friction and „n‟ is the
magnitude of the normal force.

As with the coefficient of kinetic friction, the coefficient of
static friction is a unit-less quantity
between 0 and 1 and characterizes how „slippery‟ the interface
between two materials is. The
Note that after fs is overcome, fk is less (since μk<μs). So,
once an object breaks loose
(overcomes static friction), it slides more easily.
Inclined plane with friction:
Newton‟s Second Law is a useful tool to analyze an object
sliding down an incline. For an
object with a mass „m‟ sliding down a surface inclined at an
angle θ above the horizontal,
we can draw a free body diagram and coordinate system, like
the one below
θ
m
with the x-axis oriented parallel to the surface and the y-axis
oriented perpendicular.
Assuming the object is already sliding, we can apply Newton‟s

Second Law to the „x‟ and „y‟
directions separately to find the acceleration is the x-direction
(down the incline).
y-direction
x-direction
cossincossin
cossinsin
sinsin
kk
kkW
x
kWfWxx
ggg
m
mgmg
m

nF
a
nFFFmaF
As mentioned above, the object will not slide until the force of
static friction is first overcome.
Applying the same analysis, this requires that
Thus, when this condition is met, the force of gravity acting
down the incline is greater than the
force of static friction, and the object begins to slide.
θ
Ff
Fgx

Fgy
Fg
x
y
n
This laboratory exercise is divided into a part (a) and a part(b).
At your instructor’s
discretion, you may be required to work each part together or
separately.
Part a1)
Apply Newton‟s Second Law directly to determine the
acceleration of an object, given a mass
and a net force. Use Excel to calculate the acceleration for any
force and mass entered by the
user. Since there is only one force acting, Newton‟s Second
Law reduces to .
Part a2)
Using Excel and the equations above, analyze an object on an
inclined plane. First, determine if
the component of the gravitational force acting down the plane
is enough to overcome the static

frictional force. To do so, use the IF command
IF(logical_test, value_if_true, value_if_false)
For the logical test, use
For the value if true, type “Stationary”
For the value if false, type “Accelerating”
Allow the user to input the mass, angle, and coefficients of
friction and have Excel determine if
the object is accelerating, and if so, what the magnitude of the
acceleration is.
To do so, you will need to have Excel convert from degrees to
radians using the function
RADIANS. Note that this expression can be inserted directly
into a mathematical statement. So,
for example, to calculate the cosine of an angle initially given
in degrees, we can convert it to
radians and calculate the cosine in a single step:
=COS(RADIANS(angle))
Part a3)
Using Excel and the equations above, determine the apparent
weight of an object under
acceleration. In this case, we are considering an object resting

on a scale in an elevator. If we
consider the object alone, there are two forces acting on it: the
force of gravity and the normal
force from the scale.
Object
n
From Newton‟s third law of motion, the normal force from the
scale on the object is equal and
opposite to the normal force from the object on the scale. Since
the scale reads the normal force
from the object, we can solve for the normal force exerted on
the object and know the reading on
the scale.
Applying Newton‟s second law to the object, we have
∑
Note: when applying this equation, g is +9.8 m/s
2

(not -9.8m/s
2
) and „a‟ is negative if the
elevator is accelerating downward and positive if it‟s
accelerating upwards.
Allow the user to input the acceleration of the elevator and have
Excel determine the reading on
the scale.
Part b1)
Using Excel and the equations above, determine how long it
takes an object to come to rest if it‟s
acted on by a force in the opposite direction of its velocity.
From Newton‟s Second Law, we can solve for the acceleration
„a‟ of an object with mass „m‟,
affected by a single force „F‟
Recall that for an object under constant acceleration, its
velocity as a function of time is
where vi is the initial velocity. Thus, using the two equations
above, we can solve for the
stopping time „ts‟ where v = 0.

Note: when using this equation, vi and F must be in opposite
directions (opposite signs) or the
object will never come to rest. From the stopping time, the
stopping distance can be calculated
from the formula:
Fg
Part b2)
Using Excel and the equations above, determine the
displacement of an object acted on by forces
in both the „x‟ and „y‟ directions. As mentioned above, we can
apply Newton‟s second law to
determine the accelerations in the „x‟ and „y‟ directions
separately
Once we know the accelerations in the „x‟ and „y‟ directions,
we can apply our kinematic
equations to solve for the velocity and displacement as a

function of time.
and
with similar equations for „y‟.
Recall though that the kinematic equations are only valid when
the acceleration is constantThe
acceleration is constant for „x‟, from 0 to 3 secondsbut for „x‟,
at t=3s, the acceleration changes
(to zero). Thus from 3s to 10s, the displacement in the x-
direction is
where vi is the speed in the x-direction after 3 seconds. The
total displacement in the x-direction
is the displacement from 0s to 3s plus the displacement from 3s
to 10s. The acceleration in the y-
direction is constant from 0s to 7s and then is zero from 7s to
10s.
Deliverables
See the syllabus for due date information. Place all your work
inside the Week 4 Lab
Template. Be sure to follow all instructions carefully.
Save your ONE Excel file using the filename format
"lastname_firstinitial_week4lab".

For example, if you are Albert Einstein and you are submitting
your Week 4 lab, the
filename should be "einstein_a_week4lab". Submit your
assignment to the Lab4
dropbox located on the silver tab at the top of this page.
Tutorials: Entering formulas in Excel and creating graphs
For more information on using Excel, please watch the
following tutorials from the Microsoft
Excel 2007 website:
Excel 2007 support/training: http://office.microsoft.com/en-
us/training/CR010047968.aspx
1) “Get to know Excel 2007: Enter formulas”
2) “Charts 1: How to create a chart in Excel 2007”
”
http://office.microsoft.com/en-us/training/CR010047968.aspx
TemplateTime35m/s @ 55°35m/s @ 25°t (s)x (m)y (m)vx
(m/s)vy (m/s)speed (m/s)x (m)y (m)vx (m/s)vy (m/s)speed
(m/s)00.00.00.00.00.250.50.7511.251.51.7522.252.52.7533.253.
53.7544.254.54.7555.255.55.756
x-position vs time
55 degrees 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2

2.25 2.5 2.75 3 3.25 3.5 3.75 4 4.25 4.5 4.75 5
5.25 5.5 5.75 6 0 25 Degrees 0 0.25 0.5
0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25
3.5 3.75 4 4.25 4.5 4.75 5 5.25 5.5 5.75 6 0
Time (s)
Horizontal Position (m)
y-position vs time
55 Degrees 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2
2.25 2.5 2.75 3 3.25 3.5 3.75 4 4.25 4.5 4.75 5
5.25 5.5 5.75 6 0 25 Degrees 0 0.25 0.5
0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3 3.25 0
Time (s)
Height (m)
Trajectory
55 Degrees 0 0 25 Degrees 0 0
x-position
y-position
Lab 3
Two identical projectiles are launched from the Earth's surface
Projectile #1 has an intial velocity of 35 m/s and is launched at
an angle of 55 degrees above horizontal.
Projectile #2 also has an initial velocity of 35 m/s and is
launched at an angle of 25 degrees above horizontal.
Assume the ground is perfectly flat and both projectiles are
launched from ground level (y=0m), where x=0m.
Neglecting air resistance, do the following:
i) Determine the x-position and y-position of both projectiles as
a function of time, from t=0s

until they hit the ground (y=0m). In each case, include only one
value where y<0.
Note that one projectile will hit the ground before the other, so
one set of values needs to be calculated beyond the
other.
ii) Make a scatter plot of the x-position versus time for each
projectile, on the same graph (started to the right).
Approximately, what is the range of each projectile?
Compare this value with the range equation, R=(vi2•sin2θ)/g
where vi is the inital speed of the projectile,
θ is the launch angle and 'g' is the local gravitational constant
(=9.8m/s2 near Earth's surface).
iii) Make a scatter plot of the y-position versus time for each
projectile on the same graph (started to the right).
Approximately, what is the maximum height of each projectile?
Compare this value with the height equation, h=(vi2•sin2θ)/2g
where vi is the inital speed of the projectile,
θ is the launch angle and 'g' is the local gravitational constant
(=9.8m/s2 near Earth's surface).
How long does it take each projectile to reach it's maximum
height?

iv) Make a scatter plot of the y-position of each particle as a
function of the x-position, showing the trajectory of
the projectiles.
Mathematically, how do we describe the shape of this graph?
v) Determine the velocity in the x and y directions, along with
the speed of both projectiles as a function of time,
from t=0s until they hit the ground (y=0m). As with step i), in
each case, include only one value where y<0.
What happens to the velocity in the y-direction when the
projectiles reach their maximum height?
What is true about the velocity in the x-direction for both
projectiles?
What is true about the speed of both projectiles at the instances
when they are launched and when they hit the ground?
Does the launch angle affect the speed of the projectiles when
they hit the ground? Explain.

Lab 3—Projectile Motion
Background
For an object undergoing constant acceleration in one
dimension, we developed equations that
describe the position and velocity of the object as a function of
time.
∆ ∆ and ∆
If we assume the object starts at position xi=0 at time ti=0, the
equations can be simplified as
and
where ‘x’ is the final position of the object and ‘v’ is the final
velocity.
Since these equations are true for any direction, they can be
applied to the y-direction in a two-
dimensional motion problem, assuming the acceleration in the
y-direction is also constant. Thus
we have,
and
For an object with an initial speed vi and initial velocity
directed at an angle θ above the +x axis,
the x and y components of the velocity vector are

∙ and ∙
Since the velocity has components in both the x direction and
the y direction, the speed of the
object at any time can be found from the Pythagorean Theorem:
| |
Projectile motion problems are a specific case of two-
dimensional motion problems. Most
projectile problems take place on the Earth’s surface. To solve
the problems in the simplest
terms, we ignore air resistance. Furthermore, we assume the
projectile does not travel far from
the surface. Otherwise, the acceleration in the y-direction
would not be constant. Under these
conditions, the components of the acceleration are constant, and
shown below
and
where ‘g’ is the local gravitational acceleration near the surface
of the Earth.
Inserting these values into our general equations, we have
and

Inserting our expressions for the initial velocities, we now have
∙
and
∙
At this point, we will assume our projectile is travelling over a
flat surface. Without this
assumption, we would need to consider whether our projectile
might hit a hill or even a mountain
or land at a higher or lower point. From our equation for y, we
can set y = 0, solve for t. This
will be the time it takes the projectile to hit the ground. If we
plug this expression into the
equation for x, we will have an expression for the horizontal
distance the object travels before it
strikes the ground as a function of the initial speed, launch
angle and local gravitational
acceleration. This is the so called range equation.
0
1
2
→
1 2⁄
g
v
R i

Realizing that the expression for t in the equation above is the
time it takes for the projectile to
hit the ground, ½ of this value will be the time it takes to reach
its maximum height. We can
then derive an expression for the maximum height by plugging
the time to reach the maximum
height into our expression for y:
1
2
∙
1 2⁄
→
∙
1
2
→
∙
2
Modeling in Excel
Using the equations above, develop an Excel spreadsheet to
determine the velocity and position
of projectiles as a function of time. To do so, you will need to

have Excel convert from degrees
to radians using the function RADIANS. Note that this
expression can be inserted directly into a
mathematical statement. So, for example, to calculate the
cosine of an angle initially given in
degrees, we can convert it to radians and calculate the cosine in
a single step:
=COS(RADIANS(angle))
For all of the steps below, consider two projectiles, both
launched with the same initial speed,
but at different angles.
Projectile #1 has an initial velocity of 35 m/s and is launched at
an angle of 55 degrees above
horizontal.
Projectile #2 also has an initial velocity of 35 m/s and is
launched at an angle of 25 degrees
above horizontal.
Answer the questions in the text box in Excel by typing
complete sentences directly under the
question. Change the color of your text to dark blue to highlight
your answers.
i. Determine the x-position and y-position of both projectiles as
a function of time, from t =
0 s until they hit the ground (y = 0 m). In each case, include
only one value where y < 0.
Note that one projectile will hit the ground before the other, so
one set of values needs to
be calculated beyond the other.

ii. Make a scatter plot of x versus t, estimate the range of the
projectile and compare this
value with the range equation shown above.
iii. Make a scatter plot of y versus t, estimate the maximum
height of the projectile and
compare this value with the height equation shown above.
iv. Make a scatter plot of y versus x. This shows the trajectory
of the projectile, or the path
traced out by the projectile, through space.
v. Determine the components of the projectile’s velocity and
speed as a function of time
Again, note that one projectile will hit the ground before the
other, so one set of values
needs to be calculated beyond the other.
Deliverables
See the syllabus for due date information. Place all your work
inside the Week 3 Lab
Template. Be sure to follow all instructions carefully.
Save your ONE Excel file using the filename format
"lastname_firstinitial_week3lab".
For example, if you are Albert Einstein and you are submitting
your Week 3 lab, the

filename should be "einstein_a_week3lab". Submit your
assignment to the Lab3
dropbox located on the silver tab at the top of this page.
Tutorials: Entering formulas in Excel and creating graphs
For more information on using Excel, please watch the
following tutorials from the Microsoft
Excel 2007 website:
Excel 2007 support/training: http://office.microsoft.com/en-
us/training/CR010047968.aspx
1) “Get to know Excel 2007: Enter formulas”
2) “Charts 1: How to create a chart in Excel 2007”