Title | Physics Lab #2 - Google Docs |
---|---|
Author | Steven Lim |
Course | General Physics I |
Institution | Rio Hondo College |
Pages | 5 |
File Size | 247.9 KB |
File Type | |
Total Downloads | 5 |
Total Views | 147 |
lecture notes...
Graph Stephanie Alaniz Steven Lim Shasha Tian Bosco
Lab 2: Free Fall Due: 2/28/2017
Graph Purpose of Lab The purpose of this lab is to show that an object under the influence of gravity alone accelerates at a constant rate. We do this by measuring the acceleration of the object as it falls and compare it to the actual value, which is given to us as 9.8 m/s 2
Equipment Demo Sparking Tape Apparatus Sparked Tapes, Masking tape Two-meter stick
Summary An object in free fall, or in other words, an object under the influence of gravity alone, accelerates at a rate of 9.8 m/s 2 , or “g”. By dropping an object using the sparking tape apparatus and recording the number of sparks the object creates as it falls, this experiment sought to prove the given rate of acceleration. As the object falls, it leaves behind dots on the sparking tape in intervals of 1/60 of a second. By measuring the distance between each dot as they grew increasingly further apart the closer the object came to the ground, we are able to roughly calculate the average velocity and the average rate of acceleration of the falling object.
Graph Data Sheet
Show work of dot 3: S=0.034m,
Graph Δs=S(dot3)-S(dot2)=0.034-0.022=0.012m, Vavg=60Δs=60x0.012=0.72 ΔV avg=V avg4-Vavg3=0.96-0.72=0.24 m/s Aavg=60xΔVavg =60x0.24=14.4 m/s2 Gexp= (A1+A2+……A27)/27=252/27= 9.33333 m/s 2 %error=100%x|9.33-9.8|/9.8=4.80%
Comments Based on the data we obtained, our calculations gave us a rate of acceleration of 9.33 m/s 2, giving us a 4.8% margin of error. Considering the fact that the results for any given experiment will always vary slightly with each repeated attempt, it is safe to assume that the theoretical expectations of the lab have been proven.
Answer of Question 1: d=1/2gt2 , 50=½ x 9.8 x t2 , so t = 3.1944 =3.19 second Final speed: v=gt =9.8x3.19=31.26 m/s
Answer of question 2: gmoon=1.63m/s 2 d=1/2gt2 , 50=½ x 1.63 x t2 , so t =7.8326=7.83second Final speed: v=gt =1.63x7.83=12.76m/s
Graph...