Lab 6 - Atwoods Machine and N2 Instructions
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Lab 6 – Atwood’s Machine & Newtons Second Law
PHY110 Lab – General Physics Lab I
Vernier/Video LAB
Objectives 1.
Measure the acceleration of the Atwood’s Pulley System using two different methods
2.
Determine the total mass of the system by plotting the Net force vs. Acceleration
3.
To verify Newtons Second Law stating that the Net force is proportional to the acceleration 4.
Determine the friction that acts on the system
Equipment List:
Vernier Software
Atwood’s Machine – Full Video
Vernier Data Files
o
Atwood’s Machine mass diff 4g
o
Atwood’s Machine mass diff 8g
o
Atwood’s Machine mass diff 12g
o
Atwood’s Machine mass diff 16g
o
Atwood’s Machine mass diff 20g
o
Atwood’s Machine mass diff 24g
Stopwatch (easy touch-based stopwatch)
Introduction Newtons Second Law of Motion can be modeled using a system called the Atwood’s Machine (as shown in Figure 1). It consists of two masses at the end of a thin string that passes over a pulley.
Figure 1 – Atwood’s Machine Apparatus
The difference in the two masses generates a net force on the system, causing the two masses to
accelerate. According to Newtons Second Law,
F
net = m a (1)
Where m is the mass of the system in kg, a is the acceleration in m/s
2
and F
net
is the net force in Newtons,
the acceleration of the system is directly proportional to the Net force of the system. The Net Force of the
system can be derived from the free-body diagram. Lets assume that m
2
is larger than m
1
. Therefore the
resulting free-body diagram is modeled in Figure 2 where m
2
is pulling the system downward.
Figure 2 – The free-body diagram of m
1
and m
2
, assuming m
2
is pulling the system downward. T represents the tension in the
string and the bottom force is the weight, or mg of each mass. The tension of the string is the same throughout, therefore the tension T
in the free-body diagram for m
1
is
the same as the tension in the free-body diagram for m
2
. The free body diagram gives the net force for
each mass
T
−
m
1
g
=
m
1
a
and m
2
g
−
T
=
m
2
a
(2)
One can add the two equations simultaneously to eliminate the Tension variable (also unknown in the
experiment) and to combine the equations into one
m
2
g
−
m
1
g
=
m
2
a
+
m
1
a
(3)
The expression can be reduced down to
(
m
2
−
m
1
)
g
=
(
m
2
+
m
1
)
a
(4)
Written this way, the expression models Newtons Second Law. The Net Force on the left is the
difference in weight and on the right, the m
is replaced by the combined masses (
m
2
+
m
1
)
and a
remains
the net acceleration of the system. In the lab setting the masses can be measured with a triple beam
balance or scale, and the acceleration can be measured either using kinematics or through Vernier Smart
Pulley photogate measurement. The pulley will impart some friction on the system. We can rewrite
equation 4 to incorporate the friction on the system with (
m
2
−
m
1
)
g
−
f
=
(
m
2
+
m
1
)
a
(5)
The friction can be added to the right side of the equation to model a linear relationship similar to y=mx +
b. In this case, the y-axis will be the Net force, (
m
2
−
m
1
)
g
, the x-axis will be a
. This leaves the slope as
the (
m
2
+
m
1
)
or the total mass of the system and the y-intercept will represent the frictional force implied
on the system.
(
m
2
−
m
1
)
g
=
(
m
2
+
m
1
)
a
+ f (6)
The acceleration in this lab will be measured two ways. The first way is through the kinematic equation
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Example
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900 mm
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The figure below shows a horizontal bar, of length 4.6 m, with forces acting on it. A 30 N force acts at its left end, point O, in a
direction down and to the left, 45° below horizontal. A 25 N force acts at its center, point C, in a direction up and to the right, 30° to
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25 N
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45°
C.
2.3 m
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4.6 m-
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magnitude
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dy
AddNo ding ong Pre lodurrial As,
Hangli Town Donggn city,China
732006020610…
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**I just need help with part G**
I have attached the problem below! Please view both attachments before answering. If you can please explain your answer so I can fully understand. Thank you so so much!
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PRE-LAB WORK: THE METHOD QUESTIONS
1. Create a force diagram of the car as it moves along the track.
2. Create a force diagram of the mass hanger as it moves towards the ground.
3. Create a force diagram of the system car and mass hanger as they move together.
4. How does the acceleration of the car compare to the acceleration of the mass hanger? Justify
your answer.
5. What causes the system to accelerate?
6. Do you expect the acceleration of your system to be smaller than, greater than or equal to the
acceleration due to gravity? Justify your answer.
7. Using Newton's Second Law and the diagram you created in question #3, give the acceleration of
the system as a function of the masses of the cart and the hanger.
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inspire Physics
PRACTICE Problems
ADDITIONAL PRACTICE
20. On Earth, a scale shows that you weigh 585 N.
a. What is your mass?
b. What would the scale read on the Moon (g = 1.60 N/kg)
21. CHALLENGE Use the results from Example Problem 3 to ar
would be exerted by the scale on a person in the following s
a. The elevator moves upward at constant speed.
b. It slows at 2.0 m/s² while moving downward.
c. It speeds up at 2.0 m/s² while moving downward.
d. It moves downward at constant speed.
In what direction is the net force as the elevator slows
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gpb.org/physics-motion Practice Problems
Date:
Work each of the following problems. SHOW ALL WORK.
1. The earth remains in orbit around the sun due to the force of gravity. How does the force of gravity exerted
by the sun on the earth compare to the force of gravity exerted by the earth on the sun?
2. Two objects exert a gravitational force of 4 N on each other.
a. If the mass of one object is doubled, what will be the new force of gravity between the two objects?
b. If both masses are doubled, what will be the new force of gravity between the objects?
c. If the masses do not change, but the distance between the objects is doubled, what will be the new force
of gravity between them?
d. If both the masses and the distance between the objects are doubled, what will be the new force of
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Block
Submit
Person
Ice
A person pushes a large block on a horizontal ice surface in a straight line to the right with constant speed, as shown above. The mass of the block is 10 kg and frictional forces between the block and the ice are negligible. However, the block has a wide cross-sectional area such that air resistance acting on the block cannot be neglected. The
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Force of person's push (N)
Constant speed of block (m/s) +0.05
20
1.25
40
2.51
60
3.73
80
5.00
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and Sally then take the doll to their friend Sue's house, which is 25 meters further east. The children take the doll back to Sally's house
and accidentally leave it there. What is the total displacement of the doll?
2
W
S
X
O 45 m, west
O 65 m, east and west
O 25 m, west
O 20 m, east
Save for Later
F2
#
3
E
D
80
C
F3
DOO
DOD
$
11
4
R
F
F4
V
%
5
T
G
F5
B
6
Y
H
MacBook Air
F6
N
&
7
U
J
8
M
F8
K
H
9
V
O
F9
L
O
F10
P
alt
Submit Answer
?
1
F11
{
[
+ 11
I
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c. Does tripling F triple Fappled?
d. Do Fpplied and Fy act in the same
direction? Explain why or why not.
2
4
8
FN (N)
6
2.
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D Present
rrange Tools Add-ons Help
Last edit was 41 minu.
Background
Layout-
Theme
Transition
A group of students calculated the acceleration of a
box moving across the table. They did this by
measuring the mass of the box and the force applied,
and used the formula F=mxa. Students discovered
that the actual acceleration was lower than what they
had calculated. What force might account for the
slower rate of acceleration compared to what their
calculations predicted? Explain your answer.
Write your response here:
eaker notes
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Exercise 1: Calculate me!
A 100-gram ball m1, and a 200-gram ball m2, connected by a rod with a length of 60 cm.
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Illustration:
A
Ace
m1
m2
B
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Last edit was 10 minutes agg
lormal text
Arial
11
BIUA
4.
C.slow down due to friction
5.
D.speed up due to momentum
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A. An outside force
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B. Momentum
C. Kinetic energy
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B.Force
C.Velocity
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A.10 Newtons
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m = mass of weight hanger
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Senior Physics Final
Your email will be recorded when you subrmit this form
Not 912615@oside.us? Switch account
* Required
Newton's Laws
An object with a mass of 40 kg is acted on by a net force of 680 N.
Determine the acceleration of the object. Enter your answer as a number
with no letters or units.
Your answer
This is a required question
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2 N+
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Solve the problem.
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and the bed of the truck is 0.6. You are not worried, because you are traveling on
road that
appears perfectly straight. Due to your confidence and inattention, your speed has crept
upward to 45 miles per hour. Suddenly you see a curve ahead with a warning sign saying,
"Danger: unbanked curve with radius of curvature 35 m". You are 15 m from the beginning of
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