physics lab report experiment 2b

Upload: rich-garr

Post on 03-Apr-2018

220 views

Category:

Documents


0 download

TRANSCRIPT

  • 7/29/2019 Physics Lab Report Experiment 2b

    1/7

    Georgia Southern University

    Laboratory Report in Physics

    Course/Section: PHYS 2211A/ 1113A

    Experiment 2 Date: 24 January 2013

    Experiment Title: Acceleration of a Freely Falling Body

    Name: Richard Garrison

    Partners Names: John Demere and Robert Branch

    Abstract:

    Instructors Comments:___________________________________________________________

    _____________________________________________________________________

    _____________________________________________________________________

    _____________________________________________________________________

    _____________________________________________________________________

    _____________________________________________________________________

    Instructors Name:_______________________ Grade:________________________

  • 7/29/2019 Physics Lab Report Experiment 2b

    2/7

    Abstract

    Acceleration under the influence of gravity plays a crucial role in physics as it is important to be

    able to predict and manipulate objects in free fall. In order to measure the acceleration of a

    freely falling body under the influence of gravity an experiment was set up to calculate it in two

    different ways. The first being done by dropping a projectile and calculating its acceleration and

    the second being done by calculating the projectiles velocity, graphing the results, and finding

    its slope (the slope of the best fit line is acceleration). A timing mechanism was set up to

    measure the time it took for a steel ball to strike a pressure sensitive timer from varying

    heights. Acceleration was then calculated by manipulating the following equation. The

    experiment returned the following results. The experiment would produce an average

    acceleration rate of 9.78 m/s^2. Although this is not the accepted value of acceleration the

    slope of the graph which displays the results confirm that 9.8 m/s^2 is in fact an accurate value.

    The standard deviation for the experimental acceleration values was 0.12, while the percent

    error was 0.20%.

    Data

    Table 1. Determining Acceleration

    Trails Time

    (seconds)

    Acceleration

    (m/s^2)

    1 0.38 s 13.85

    2 0.44 s 10.33

    3 0.32 s 19.534 0.32 s 19.53

    5 0.41 s 11.89

    Distance the Ball Fell = _____________1.00____________m

    Average Acceleration = ______________15.03______________m/s^2

    Table 2. Velocity/Time Relations

    Trials Time(seconds)

    Distance(m)

    Vfinal(m/s)

    1 0.38 1.00 5.26

    2 0.28 0.85 6.07

    3 0.25 0.70 5.60

    4 0.25 0.55 4.40

    5 0.18 0.40 4.44

  • 7/29/2019 Physics Lab Report Experiment 2b

    3/7

    6 0.15 0.25 3.34

    7 0.12 0.10 1.67

    Calculations

    Acceleration can be calculated by rearranging the following equation.

    Since yo and vo are zero according to the experiment the following calculation can be used to

    determine acceleration from the data previously gathered.

    , and therefore .Computing values to be entered in Table 1 gives us,

    Trial 1

    Trial 2

    Trial 3

    Trial 4

    Trial 5

    Solving for the average acceleration gives us,

  • 7/29/2019 Physics Lab Report Experiment 2b

    4/7

    Calculating for standard deviation gives us,

    Computing for percent error gives us,

    Computing values to be entered in Table 2 gives us,

    Trial 1

    Trial 2

    Trial 3

    Trial 4

    Trial 5

    Trial 6

  • 7/29/2019 Physics Lab Report Experiment 2b

    5/7

    Trial 7

    Based on the slope of the Vfinal vs Time Graph, the percent error is as follows, Questions

    1. What is the significance of the slope and y intercept of Vfinal vs Time graph? Give valuesof both.

    After making a graph of my results a plotting a best fit trendline I was able to determine the

    slope to be 9.8011. This is significant because the accepted acceleration rate of any object nearEarth under the influence of gravity is 9.8 m/s^2. It is also significant because the equation for

    acceleration is

    The slope is acceleration because a slope is

    . The y intercept of the graph is -0.0056. I suspectthat this is actually the initial velocity, but it is so negligible that it can be assumed that the

    initial velocity is 0.

    2. Are the graphs linear, parabolic, or hyperbolic? (a) Vfinal vs Time (b) Distance vs TimeBoth graphs appear to be of linear behavior as the graphs represent equations for a constant

    acceleration body.

    Exercise

    On the surface of the moon the acceleration due to gravity is about 1/6 the value on the Earth.

    An object weighing 12 N weighs how much on the moon?

    The commonly accepted rate of acceleration on Earth is 9.8 m/s^2 so converting a 12N object

    to the moons possible rate would be given in the following calculations

  • 7/29/2019 Physics Lab Report Experiment 2b

    6/7

    is the acceleration due to gravity on the moon. Keep in mind that theobject has a mass of 1.22kg found by manipulating the equation F=ma (

    ). Therefore onthe moon the object weighs .Conclusion

    By conducting the experiment I calculated that the average acceleration rate due to

    gravity was actually 9.78 m/s^2 as compared to the accepted value of 9.8 m/s^2. Although

    possible error in the experimental value is attributed to faults within the timing mechanism, it

    should be noted that the slope of the Vfinal vs Time graph is actually 9.8011 m/s^2. This

    concludes that although there is slight error in the results, they are still precise enough to prove

    that acceleration due to gravity is closer to the accepted value. The standard deviation for the

    experimental acceleration values was 0.12, while the percent error was 0.20%. The percent

    error for the slope of the Vfinal vs Time Graph was only 0.01%

  • 7/29/2019 Physics Lab Report Experiment 2b

    7/7

    Graphs

    y = 9.8011x - 0.0566

    R = 0.9979

    0

    0.51

    1.5

    2

    2.5

    3

    3.5

    4

    4.5

    5

    0 0.1 0.2 0.3 0.4 0.5

    Vfinal(m/s)

    Time (seconds)

    Vfinal vs Time

    Trials

    Linear (Trials)

    y = 2.9253x - 0.3927R = 0.9728

    0

    0.2

    0.4

    0.6

    0.8

    1

    1.2

    0 0.1 0.2 0.3 0.4 0.5

    Distance(meters)

    Time (seconds)

    Distance vs Time

    Trials

    Linear (Trials)