Experiment 2 free fall physics
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Baruch College, CUNY *
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2001
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Mechanical Engineering
Date
Apr 3, 2024
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Mayra Veintimilla
PHY 2001- Professor Miraj Uddin
Lab Report #2
Free Fall
Objective: To measure the acceleration due to the force exerted by the earth. To find
the total force on a mass in mechanical equilibrium. Apparatus: Material Object
Drawing
Explanation Tape timer
A device that marks with graphite dots a paper strip in intervals of 40hz (40 dots a second) or 10hz (10 dots a second).
1 Meter Ruler
1 Meter Ruler: A measuring tool that counts to 0-100 cm and have divisions on millimeters of 10 mm for each
cm.
Tape
A strip of paper used to record the motion of an object over time
100g weight
A steel weight with a known weight normally used for generating a known force
200g weight
A steel weight with known weight normally used for generating a known force.
A force table This is a round, horizontal, circular table, with angles marked from 0 degrees to 360 degrees. It has a metal bolt in the center to keep a small ring (see below) from escaping. • A small circular plastic or metal ring.
Data: Weight/Attempt
Distance Between Dots
Amount of dots
Total Distance
cm/s^2
100g 1st
0-4 = 5cm
19
90 cm
797.784 cm/s^2
100g 2nd
1-5 = 5.42
19
97.5 cm
707.483 cm/s^2
100g 3rd
2-6 = 4.6
21
92 cm
908.642 cm/s^2
100g 4th
3-7 = 5.71
18
97 cm
859.834 cm/s^2
100g 5th
4-8 = 5.14 cm
19
92.5 cm
819.945 cm/s^2
100g 6th
5-9 = 5.38 cm
19
96.8 cm
774.4 cm/s^2
100g 7th
6-10 = 4.96 cm
20
94.2 cm
835.014 cm/s^2
100g 8th
7-11 = 5.28 cm
19
95 cm
760 cm/s^2
100g 9th
8-12 = 4.96 cm
20
94.2 cm
835.014 cm/s^2
100g 10th
9-13 = 4.3 cm
24
99 cm
550 cm/s^2
Weight/Attempt
Distance Between Dots
Amount of Dots
Total distance
cm/s^2
200g 1st
0-4 = 5.42 cm
19
97.5 cm
864.266 cm/s^2
200g 2nd
1-5 = 5.38 cm
18
91.5 cm
903.703 cm/s^2
200g 3rd
2-6 = 5.36 cm
19
96.5 cm
855.402 cm/s^2
200g 4th
3-7 = 5.79 cm
18
98.5 cm
972.84 cm/s^2
200g 5th
4-8 = 5.42 cm
19
97.5 cm
864.266 cm/s^2
200g 6th
5-9 = 5.36 cm
18
91.2 cm
900.741 cm/s^2
200g 7th
6-10 = 5.36 cm
21
95.5 cm
692.971 cm/s^2
200g 8th
7-11 = 4.78 cm
19
92 cm
817.285 cm/s^2
200g 9th
8-12 = 5.31 cm
19
95.5 cm
846.537 cm/s^2
200g 10th
9-13 = 4.62 cm
20
87.8 cm
702.4 cm/s^2
Journal: One pulley was initially positioned at 0°. The positions of the other two pulleys were 140° and 270°. These angles were referred to as. Every calculation was done in centimeters and seconds (s). The final findings must be expressed in cm/s^2. To generate a downward slope for the measurements, first assemble the tape timer and position it at a 90-degree vertical angle. After that, cut 20 1.2 m (or 120 cm) long pieces of tape. Next, attach the 100g weight with a strip and run it through the tape timer. After that, release the weight and configure the machine to run at 40 Hz.
Gather the strip and count the number of dots in 1 meter for every 100 centimeters.
Repeat 9 more times using the 100 g weight and 10 times using the 200 g weight. Once you're done, plot all the data on a graph and use the following formula to determine the average and cm/s^s2. Equation (2xD/(dots/total hz)^2). The following distances should also be measured: 0–4 dots, 1–5 dots, 2–6 dots, and so on, until you reach 8–12 dots.
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Calculations:
Conclusion: The purpose of this experiment was to familiarize us with acceleration and the relationship between mass and speed of an item. This is useful because it will tell us
how quickly things are moving and how far they can move in each length of time. To
do this, we learn how to adjust to various weights and time frames and how to use hz and cm to detect this in little items. Overall, this experiment taught us how to quantify acceleration and demonstrated how a seemingly insignificant detail can have an impact on the study's results.
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F(t)
Force_Crate
Known values:
Mass of Block
68 kg
TT
Hs μk
0
0.63 0.53 26°
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A crate is initially at rest on a sloped surface, when a force is applied: F(t)
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PHYS X PHYS X
印 PHYS X
PHYS X
POTPHYS X
PHYS X
E PHYS X E PHYS
top/semester2/physics%20for%20engineers/PHYS220_CH15_Lecture%20Notes_Problems%2015 19,15.29 S
(D Page view
A Read aloud
V Draw
Problem-15-19: page-475
A 0.500-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion
with an amplitude of 10.0 cm. Calculate the maximum value of its
(a) speed, and acceleration.
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(c) the time interval required for the object to move from.r50 to r5 8.O0 cm.
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