LAB REPORT 2
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Dean Taylor 6/5/23 PHY 133 L69 TA: Chamathka Thotamuna Wijewardhana Force and Acceleration
Introduction: The goal of this lab is to use the sensors of the IO Device and an understanding of Newton’s Second Law in order to determine the relationship of force and acceleration. By doing this, the mass of the IO Device can be determined. The following equations will be used: ࠵?
!"#
= ࠵?࠵?
࠵?
$
= ࠵?࠵?
࠵? =
࠵? ࠵? ࠵?
࠵?
%
Through several trials, an understanding of the relationship between force and acceleration can be developed and data gathered can be organized such that meaningful conclusions can be drawn. It is expected that that as force increases, acceleration does as well. Procedure: “Time and Magnitude Section” (Part 1) 1.
Ensure the device is paired and on 2.
Attach the hook screw onto the IO Device and place the device face up 3.
Begin recording data and hit the device by the screw (aim to only push the device in the “y direction”) 4.
Press stop once ample data has been collected Part 2
1.
Attach the plate screw to the IO Device 2.
Attach the long spring to the screw 3.
Use a heavy object to hang the other end of the spring off of a table or other inclined height 4.
Begin recording and pull the device down (in the Y direction) such that it being oscillating 5.
Press stop after a reasonable amount of time passes 6.
Use gathered data to create a parametric plot Results:
Figure 1: “Time and Magnitude” Section Data (without zoom) Figure 2: “Time and Magnitude” Section Data (with zoom applied)
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Figure 3: “Finding the Known Value of Mass” Section Average Force Data Figure 4: “Finding the Known Value of Mass” Section Average Acceleration Data Average σ Average Force (Figure 3 Data) -1.830 N 0.053 N Average Acceleration (Figure 4 Data) -9.820 m/s^2 0.010 m/s^2 Figure 5: Valuable data extracted from Figure 3 and 4
Acceleration (m/s^2) Force (N) 2.078 1.425 8.772 2.827 12.206 3.015 15.520 3.881 22.185 5.067 Figure 6: Peak Values for Force and Acceleration Figure 7: Increasing Force and Acceleration Peaks from “Quantitative Measurement” Section Figure 7: One Peak from “Quantitative Measurement” Section (With Zoom)
Figure 8: Data from Figure 6 in Excel Graph (Trendline Set to Run Through Origin) Slope of Graph = 0.2462 Figure 9: Parametric Plot of Data from Oscillating IO Device
X Y Point 1 -12.6402 m/s^2 -2.3368 N Point 2 -1.2581 m/s^2 -6.6806 N Figure 10: Two Data Coordinate Points from Parametric Plot
Calculations: ࠵? = ࠵?࠵?
-1.830 N = m (-9.820 m/s^2) ࠵? =
−1.830 N
−9.820 m/s^2
= 0.186 ࠵?
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࠵? =
࠵?
%
− ࠵?
&
࠵?
%
− ࠵?
&
=
−1.2581
࠵?
࠵?
%
− (−2.3368
࠵?
࠵?
%
)
−6.6806 ࠵? − (−12.6402 ࠵?)
= 0.181002 = 0.181
F= (0.246g)(12.206 m/s^2)=3.00 N Measured Value: 3.015 N Error Analysis: ࠵? A
࠵?
࠵?
D = E
F
࠵?
’
࠵?
G
%
+ F
࠵?
(
࠵?
G
%
=
I
A
0.053 N
−1.830 N
D
%
+ J
0.010
m
s
%
−9.820
m
s
%
K
%
= 0.02898
0.02899/0.2462= 0.117749797 x100=11.77% Certainty for Part 1 0.02899/0.181=0.1601657459 x100=16.02% Certainty for Part 2 % Difference (Mass) Part 1: (accepted value – experimental value)/ ((accepted + experimental)/2) x 100%=( 0.2462-
0.186)
/(( 0.2462+
0.186
)/2)= 0.2785747…x100= 27.86% difference Part 2: (accepted value – experimental value)/ ((accepted + experimental)/2) x 100%=( |0.181 −
0.186|)
/(( 0.181+
0.186
)/2)=0.0272479… x100= 2.7% difference Discussion: As evidenced by Figure 1
and Figure 2
, it is apparent that force applied and acceleration do occur at the same time. It is worth mentioning, however, that following acceleration in the positive direction, there is a negative acceleration seen (likely due friction, or others causes of deceleration). Furthermore, it can best be evidenced by Figure 1
that as force increases acceleration does as well. In assessing Figure 1, it is also made clear that as force increases alongside acceleration, the duration of the negative acceleration following the positive spike also increases. As evidenced by much of the data, the calculated slope of the “quantitative measurement” section and the slope from the parametric plot are quite close with the values being 0.186g and 0.181g respectively. These values exhibit a percent difference of only 2.7% and have a higher, albeit quite low percent certainty, than that of part 1 with a value of 16.02%. Part 1 displays much less promising results with a percent certainty of 11.77% and a percent difference of 27.86%. It appears this low percent certainty is likely a result of error such as poor calibration of sensors, misreads in data, or random error. Ways to combat this would be to ensure the device is properly calibrated prior to each experiment and repeated trials prior to using any data. Throughout the experiment, it is clear the original hypothesis was largely supported, though some outliers in the data complicate a clear quantitive assessment of mass. Link to Video of Proof: https://drive.google.com/file/d/1323bmX6GE2UmA5oF67A6xNHd29YohG9z/view?usp=
drivesdk
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圖|& 品凸?
Time
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dx
Center of Mass of Rectangle 2
5.000
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Jain@
IFI
... N
ot
rot
***lad
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4.000
Velocity of Center of Mass of Rectangle 2
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Ve
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VVy
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w 30
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w..
MI
Ve
母100
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Vo
... rad/s
+
EAcceleration of Center of Mass of Rectangle 1
Ax Ax
A Ay
AUJAI
Ae
--- m/s^2
... m/s^2
-- m/s^2
.-- rad/s^2
3.00
Aø
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PLEASE ANSWER NUMBER 14.MECH 221-KINEMATICS: PLEASE GIVE DETAILED ANSWER AND CORRECT ANSWERS. I WILL REPORT TO BARTLEBY THOSE TUTORS WHO WILL GIVE INCORRECT ANSWERS.
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5. Two cables are tied to the 2.0 kg ball shown below. The ball revolves in a horizontal
circle at constant speed. (Hint: You will need to use some geometry and properties of
triangles and their angles!)
60°
1.0 m
60°
© 2013 Pearson Education, Inc.
(a) For what speed is the tension the same in both cables?
(b) What is the tension?
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Wind Farms
Consider the arrangement of three wind turbines in the following schematic in which wind
turbine C is in the wakes of turbines A and B.
Given the following:
- Uo = 12 m/s
A
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-XBC = 200 m
- z = 60 m
- Zo = 0.3 m
U.
-r, = 20 m
B
- CT = 0.88
Compute the total velocity deficit, udef(C) and the velocity at wind turbine C, namely Vc.
Activate Windows
Go to Settings to activate Windows.
Wind Farms (Example Answer)
5:43 PM
A 4)) ENG
5/3/2022
I!
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yourself. This question should only be answered if you are between 61 and 90 in the student list
(#61-90).
F
Bearing
☑ ☑
☑ ☑
Shaft
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Required information
The cutaway of the gun barrel shows a projectile moving through the barrel. The projectile's exit speed is vs = 1675 m/s
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NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
L
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Honor code
16 nours agO• Chemistry Hign Scnoo1
You are currently in a stable orbit 9000 km
(1km=1000m) from the center of planet Proxima
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According to the calculations done by NASA, Planet Proxima Centauri B has
a mass of approximately 7.6x10^24kg.
Based on fuel consumption and expected material aboard the ship, your
colony ship lander has a mass of approximately 1.1x10^5 kg.
1. Using Newton's Law of Universal Gravitation, determine what the force of
gravity on your ship at that orbit be using the early measurements
2.Using Newton's second law determine the acceleration of your lander due
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3.Upon arrival, you measure the actual force of gravity to be 6x10^5 N,which
ic difforent from vourcalclated vauein cuection 1 ldentify 2cnocific
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Coefficient of friction; 0.4
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Subject: Mechanical Measurements
Do not copy online solutions. It's different value
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Using MATLAB
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For the Following question Graph all 4 : [I just need all 4 graphs and please explain and make clean solution]
Position vs time
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Question : A 550 kilogram mass initially at rest acted upon by a force of F(t) = 50et Newtons. What are the acceleration, speed, and displacement of the mass at t = 4 second ?
a =(50 e^t)/(550 ) [N/kg]
v = ∫_0^t▒(50 e^t )dt/(550 )= v_0 +(50 e^t-50)/550=((e^t- 1))/11
x = ∫_0^t▒(e^t- 1)dt/(11 )= x_0 +(e^t- t - 1)/(11 )
a(4s)=(50*54.6)/550= 4.96[m/s^2 ]
v(4s)=((e^4-1))/11= 4.87[m/s]
x(4s)=((e^4- 4 - 1))/11= 4.51 [m]
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- Part III Capstone Problem Interactive Physics - [Lab7Part3.IP] Eile Edit World View Object Define Measure Script Window Help Run StoplI Reset 圖|& 品凸? Time Length of Spring 22 6.00 dx Center of Mass of Rectangle 2 5.000 Tension of Rope 3 Jain@ IFI ... N ot rot ***lad Split 4.000 Velocity of Center of Mass of Rectangle 2 Vx Vx V Vy MM Ve - m/s m/s 3.00 *** m/s Vo ..* lad/s 2 00 Center of Mass of Rectangle 1 1.000 tol rot *.* rad EVelocity of Center of Mass of Rectangle 1 Vx Vx VVy M 0.000 -m/s w 30 m/s w.. MI Ve 母100 *** m/s Vo ... rad/s + EAcceleration of Center of Mass of Rectangle 1 Ax Ax A Ay AUJAI Ae --- m/s^2 ... m/s^2 -- m/s^2 .-- rad/s^2 3.00 Aø Mass M1 = 2.25 kg is at the end of a rope that is 2.00 m in length. The initial angle with respect to the vertical is 60.0° and M1 is initially at rest. Mass M1 is released and strikes M2 = 4.50 kg exactly horizontally. The collision is elastic. After collision, mass M2 is moving on a frictionless surface, but runs into a rough patch 2.00…arrow_forwardVibrationsarrow_forwardPlease answer by own do not copy paste from other answer.arrow_forward
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