Design Lab

[Awi]

Hi! I am doing my final design lab in physics but I have no idea what to do it on? After three design labs my creativity is all gone. I would appreciate some ideas!

[cielo]

Purpose: The purpose of this lab is to calculate the acceleration due to gravity from launching an object in the horizontal direction. Gravity is the vertical force that acts downwards on an object in motion. We will be able to figure out the acceleration due to gravity using the data collected for the horizontal direction. The equation that will be used to calculate this is d=v_it+(1/2)a(t²).

Hypothesis: The acceleration due to gravity value which I will calculate will be close to 9.8 m/s². The universally accepted value for the acceleration due to gravity is considered 9.8 m/s². The force of gravity will have an effect on the object since all objects in motion on earth are affected by gravity. My calculated value will not be exactly -9.8 m/s² because there might be a few errors in our process and calculations. The independent variable in this lab is the initial velocity of the object. The dependent variable is the horizontal displacement, which depends on the initial velocity. The horizontal displacement is how far the object travels in the x component. The control variable is the vertical displacement, which happens to be the height of the table in this case.

Materials:

·         Box (used as the object)

·         String

·         Table

·         2 meter sticks

·         Lab Probe device

·         Motion sensor

·         Netbook

·         Tape

 

 

 

 

 

 

 

 

 

 

 

Procedure:

1.      The lab probe was connected to the motion sensor and the netbook

2.      Logger Lite was opened on the netbook to ensure the motion sensor was working

3.      The motion sensor was place on the table, behind the box

4.      A string was attached to the inside of the box using tape

5.      Two meter sticks were placed on the floor in front of the table to measure the horizontal displacement

6.      The puller jerked  on the object in a horizontal direction, parallel to the meter sticks

7.      The spotter spotted where the box first landed and took the measurement on the meter stick

8.      The displacement and initial velocity were copied on to Excel

9.      Steps 6-8 were repeated five more times

Results:

Vertical displacement (height of the table): 75 ± 0.5 cm

Horizontal displacements:

·         41 ± 0.5 cm

·         34 ± 0.5 cm

·         38 ± 0.5 cm

·         33 ± 0.5 cm

·         36 ± 0.5 cm

·         46 ± 0.5 cm

Velocity:

(Trial 1) Position

Velocity

0.2563925

5.74983542

0.449673

5.56572592

0.5091835

6.034

0.558404

6.67498264

(Trial 3) Position

Velocity

0.304584

5.502053472

0.523418

5.593758333

0.577269

6.3412125

(Trial 5) Position

Velocity

0.350546

3.22134167

0.4066265

5.39343681

0.6161995

8.15958889

 

 

(Trial 2) Position

Velocity

0.308014

5.17143958

0.5102125

3.08767

0.542969

3.61055139

 

(Trial 6) Position

Velocity

0.256564

4.956

0.308357

8.14109931

 

(Trial 4) Position

Column2

0.3064705

2.01679236

0.348831

5.66736042

 

                                                        

 

 

 

Calculations/Analysis:

Uncertainty of each velocity:

·         (V_max-V_min)/2

·         (5.98765-5.17144)/2 = ± 0.408105 cm

·         Same method for all trials

Average uncertainty of velocities:

·         Total uncertainties/no. of uncertainties

·         (0.408105 + 0.419605 + 0.4957 + 0.762 + 0.699 + 0.3136)/6 = ± 0.4626cm

Slope of uncertainty lines to find the uncertainty for the slope:

·         1^st line end points: lowest velocity point – 0.5, highest velocity point + 0.5

·         2^nd line end points: highest velocity point +0.5, lowest velocity – 0.5

·         m= (y₂-y₁)/(x₂-x₁)

§  1^st line: (45.5 – 33.5)/(6.543 – 3.84) = 4.439

Uncertainty of the general slope:

·         (1^st m – 2^nd m)/2

·         (5.17943 – 4.439512)/2 = ± 0.369965 s

Slope of a general line:

·         Best fit line: 2 points (3.84, 32) and (6.543, 43)

·         m = (y₂-y₁)/(x₂-x₁)

·         (43 – 32)/(6.543 – 3.84) = 4.34827 s

Finding g                

·         d = v_it+(1/2)a(t²)

o   d= 75 ± 0.5 cm

o   V_i = 0

o   A=g

o   T= d/v (m)

o   h = 0 + (1/2)(g)(m²)

o   h = (1/2)(g)(m²)

o   g = (2h/g)^1/2

o   (2(75 cm))/(4.24827 sec)² = 8.3116 m/s²

·         Uncertainty of g:

o   U. of g: [(2(u. of g)/g) + (u. of h/h)]g

o   [(2(0.36997)/4.29827) + (0.5/75)]x8.3116

o   U. of g = + or – 1.50305 m/s²

Acceleration due to gravity: 8.3 ± 1.5 m/s²