Lab Report Essay

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Sheffield International College
AS1b SE FC
Lab report
Title:Lab report of the constant of elasticity Group:S204 Student ID:25436
Teacher:Simon Dale

Elasticity experiment report
Introduction:
In physics, elasticity is the tendency of solid elastomeric material, while after deformation recovering to the original shape. And spring is one of the elastic objects which can return to the original size when forces were removed. The constant of elasticity is the ratio of force and extension, which can describe how easily elastic materials can be stretched or compressed. Under the same forces, if the constant of elasticity larger, the extension is smaller,which means the spring is stronger,deduced from Hooke’s law : K=
To calculate F,this blow formula(Newton’s third law) was used, masses multiple the acceleration of gravity(used as 9.81m/s2) F=Mg
The constant of elasticity is a property of springs,every spring has its own the constant of elasticity,it cannot be influenced by masses of wights and extension,in fact,it determines extensions.The aim for this experiment is to find the constant of elasticity of 3 different springs.
Materials and method: The equipment was set up as Figure1 One end of the spring was hung with one side of the retort stand,the other end was hung with a mass hanger. The weight was added onto mass hanger,after one weight being added,the extension was measured by a ruler. The last procedure was repeated 10 times The whole procedures was repeated onto 3 different springs.

Figure1:Measuring extension of springs

Results:

Table1:The extension for springA
M/g
L1/mm
L2/mm
∆L=L1-L2/mm
10
14
15
1
20
14
16
2
30
14
18
4
40
14
20
6
50
14
22
8
60
14
23
9
70
14
25
11
80
14
26
12
90
14
27
13
100
14
29
15

Table 2:The extension for springB
M/g
L1/mm
L2/mm
∆L=L1-L2/mm
10
21
21
0
20
21
21
0
30
21
21
0
40
21
23
2
50
21
27
6
60
21
30
9
70
21
33
12
80
21
38
17
90
21
40
19
100
21
44
23

Table 3:The extension for springC
M/g
L1/mm
L2/mm
∆L=L1-L2/mm
10
20
20
0
20
20
21
1
30
20
22
2
40
20
25
5
50
20
29
9
60
20
32
12
70
20
35
15
80
20
39
19
90
20
42
22
100
20
45
25

Discussion: Calaulation To find out the gradient of the graph,which is the ratio of extension and masses in Figure2,two points were chosen to calculate it,the points are P1(60,9),P2(100,15). Gradient1 of Figure2= = =0.15 To find out the gradient of the Figure3,two points were chosen to calculate it,the points are P3(50,6),P4(70,12). Gradient of Figure3= = =0.30 To find out the gradient of the Figure4,two points were chosen to calculate it,the points are P5(40,5),P6(100,25). Gradient of Figure4= = =0.33
To calculate the constant of elasticity K,the below transformation was needed. Since F=K∆L So K= = Choose 0.07Kg as M,getting extension in this mass of each spring. So, K of springA= N/m =62.4N/m…