LAB Report Hooke\'S LAW-pages-deleted PDF

Preview

Summary

Download LAB Report Hooke'S LAW-pages-deleted PDF

Description

Abstract This experiment was performed to determine the value of spring elasticity,k of two different springs which are spring A and spring B.Spring elasticity was measured through its elongation of the spring which influenced by the total mass of scale pan and slotted weight.First we investigated the relationship between the force applied to a spring and the displacement of the spring from its rest length (yo).we hung various masses from the spring A and spring B and measured the vertical displacement.we compared the elongation from the two spring which have different stiffness.From that, we found the mass was increased in propotionally with the elongation of the spring.Our spring behaviour confirm follow the Hooke’s Law Fs=-kx.

INTRODUCTION The main objective of this experiment is to show Hooke’s Law of spring, calculate the total energy absorbing in the spring. As an equation, it can be written as: F = -kx where F is the force applied to the spring (N). where x is the displacement of the spring (m). where k is the spring constant, the rate at which the spring is displaced (N/m).Hooke’s law applies, as long as the material is within it’s elastic limit.Once a sufficient amount of force has been applied, so as to extend the material beyond it’s elastic limit, the material enters it’s plastic region. With the material in it’s plastic region, the force applied causes permanent displacement of the material. An experiment is to be undertaken to determine the behaviour of three materials, in relation to Hooke’s law. Two of the materials within their elastic regions and one in it’s plastic region. The results will then be analyzed, interpretting them to determine what is happening physically and the differences between the materials, and presented with a conclusion.

METHODOLOGY

Figure 1

Firstly, set up all the apparatus needed (stand with graduated scale,scale pan, spring and set of masses) as shown in the Figure 1. Then, the length , y₀ of Spring A without the scale pan and masses is measured. Secondly, the mass of scale pan, mpan, is weighted and hooked into the spring. Then, the length, y1 of the spring is measured.Thirdly, 25g mass,mmass was added into the scale pan and the length, y2 of the spring is measured and recorded. The process is repeated using different mass = 50g,75g, 100g, 125g and 150g.The whole procedures is repeated to get the second reading then the average reading is calculated. Lastly ,using the same set up, repeat all the above steps using Spring B.

RESULT AND ANALYSIS Table 1.1

Mass

Total mass

Force,

mass

(mpan+mmass)

F=mg

(kg)

(kg)

Length of spring, y(m)

Elongation of spring, ∆y (m)

Reading

(N)

Yn

1

2

Average reading ,Ῡn

∆Ῡ= Ῡn- Ῡo

0.000

0.0000

0.0000

Y0

0.080

0.080

0.0800

0.0000

0.000

0.0049

0.0480

Y1

0.082

0.082

0.0820

0.0020

0.025

0.0299

0.2930

Y2

0.090

0.093

0.0915

0.0115

0.050

0.0549

0.5380

Y3

0.100

0.102

0.1010

0.0210

0.075

0.0799

0.7830

Y4

0.110

0.112

0.1110

0.0310

0.100

0.1049

1.0280

Y5

0.121

0.122

0.1215

0.0415

0.125

0.1299

1.2730

Y6

0.131

0.131

0.1310

0.0510

0.150

0.1549

1.5180

Y7

0.141

0.142

0.1415

0.0615

Table 1.2

Mass

Total mass

Force,

mass

(mpan+mmass)

F=mg

(kg)

(kg)

Length of spring, y(m)

Elongation of spring, ∆y (m)

Reading

(N)

Yn

1

2

Average reading ,Ῡn

∆Ῡ= Ῡn- Ῡo

0.000

0.000

0.000

Y0

0.080

0.082

0.0810

0.0000

0.000

0.0049

0.0049

Y1

0.084

0.085

0.0845

0.0035

0.025

0.0299

0.7350

Y2

0.101

0.101

0.1010

0.0200

0.050

0.0549

0.9800

Y3

0.119

0.119

0.1190

0.0380

0.075

0.0799

1.2250

Y4

0.136

0.135

0.1355

0.0545

0.100

0.1049

1.4700

Y5

0.154

0.154

0.1540

0.0730

0.125

0.1299

1.7150

Y6

0.172

0.173

0.1725

0.0915

0.150

0.1549

1.9600

Y7

0.190

0.189

0.1895

0.1085

ANALYSIS

GRAPH FORCE APPLIED VS. ELONGATION OF SPRING 1.6 0.0615, 1.51802 1.4 0.051, 1.27302

Force Applied, (N)

1.2

0.0415, 1.02802

1 0.8

0.031, 0.78302

0.6

0.021, 0.53802

0.4 0.0115, 0.29302 0.2 0

0.002, 0.04802 0, 0 0 0.01

0.02

0.03

0.04

0.05

Elongation of spring, y(m)

GRAPH 1 = SPRING A

0.06

0.07

GRAPH FORCE APPLIED VS. ELONGATION OF SPRING 1.6 0.1085, 1.51802

1.4

0.0915, 1.27302

Force Applied,(N)

1.2 0.073, 1.02802

1 0.8

0.054, 0.78302

0.6 0.038, 0.53802 0.4 0.02, 0.29302 0.2 0

0.0035, 0.04802 0, 0 0 0.02

0.04

0.06

Elongation of spring, y(m)

GRAPH 2 = SPRING B

0.08

0.1

0.12

CALCULATE THE GRADIENT OF THE GRAPH

SPRING A (GRAPH 1) y2-y1/x2-x1 =1.27302-0.53802/0.051-0.021 =0.735/0.03 =24.5 N/m

SPRING B (GRAPH 2) y2-y1/x2-x1 =1.27302-0.53802/0.0915-0.0380 =0.735/0.0535 =13.7 N/m

SAMPLE CALCULATION

1. Total mass hung on spring = mpan+mmass = 0.0049kg + 0.025 kg = 0.0299 kg # The mass of scale pan is added to the mass of load as it will also affect the extension of spring as it also has weight. 2. Force applied on the spring, F= W W= mg = (0.0299 kg)(9.8 m/s) = 0.29302 N # The mass is converted into weight as weight is also a form of force. 3. Average reading = (reading 1 + reading 2)/2 =(0.08m+0.08m)/2 = 0.08 m # The reading of length is taken twice and calculated as average to reduce human error. 4. Elongation of spring,Δy = Δy= y n- y 0 = 0.082 m – 0.080 m = 0.002 m # The length of spring aer the force is supplied is subtracted with the length of spring in its equilibrium state to obtain the net extension of spring produced

DISCUSSION This experiment was conducted to investigate the applicability of Hooke’s Law by measuring the spring constant using two different spring. In part 1, the first spring was hung vertically with a mass pan attached to the end of the spring then masses from 25g to 150g were added. The final length of the spring was measured by a ruler once it came to rest. In part 2, the second spring was hung vertically like the first spring. A mass pan attached at the end of the spring and different masses were added. The final length was measured. In this state, two equal and opposite forces acted on the hanging mass which is gravity directed downward and the spring’s elastic restoring force directed upward, in the opposite direction of displacement. Using Hooke’s Law (F=-kx), a spring constant was calculated for each spring. The spring constant for each value of displacement are the same, within experimental uncertainty which verifies Hooke’s Law. The average spring constant of spring A is (24.5) N/m while spring B is ( 13.7 ) N/m A graph of forces, F versus the elongation of the spring,Δy was plotted and resulted in the expected straight line in the range of different forces and is consistent with Hooke’s Law. The slope of this line represent the spring constant,k which obeys to the average value of the spring constant.The greater the value of k , the steeper the slope was. The intercept for the best fit straight line intersects is close to origin which consistent with Hooke's law. The sources of error in this experiment are due to the precision of meter ruler’s location for measurement and the accuracy of the slotted masses and the mass of mass pan. The meter ruler was set up verticallly behind the spring. The elongation of the spring was measured relative to the base of the mass pan/hanger. Some effort was made up to keep the spring remain at rest to sight the measurement correctly. Thus, it was necessary to view the meter ruler at a different /slight angle.This sighting was required for each measurement when different masses were added, and the displacement was the difference between the location and the reference. Hence, this systematic error due to parallax error should be minimal. Both experimental spring constants were larger than the theoretical spring constant. This elastic capacity of the spring might have increased due to the new condition of the springs. Both springs are still new and unused which means they were still at their best performance capability to stretch and have greater elastic force compared to the old spring which have been repeatedly used in the past. As a result, the experimental spring constants are higher than the theoretical spring constants since elastic force correspons to a larger spring constant since displacement and restroring force are proportional so an increase in one componet lead to an increase to the other component which as result give a larger k value. This systematic error affected the average result. Another minor cause of error could have come from frictional forces between the spring and air result to an increase in resorative force of the spring and leads to higher k values because they are directly proportional.

CONCLUSION In conclusion, the results obtained from the experiment confirm that Hooke’s Law is true. It is shown by the Δy from the spring A and spring B. From graph 1 and graph 2 we can see that if the total mass was increased the elongation was also increased. The elasticity of the spring could affect the outcome of the result where spring B which is more elastic than spring A distinguish the higher elongation of the spring.Every spring has different spring constant, k which results in different elongation of spring. The greater the spring constant, the smaller the elongation of the spring. The Hooke’s Law statement that the force needed to extend or compress a spring by some distance is proportional to that distance is justified in this experiment.

s

REFERENCES https://www.baylor.edu/physics/doc.php/110769.pdf https://en.wikipedia.org/wiki/Hooke%27s_law https://www.youtube.com/watch?v=VBV-fSAcoEs

POST LAB QUESTION

1. F= K Δy

(0.5)(9.80) = 13.7 Δy Δy = 4.9÷13.7 Δy = 0.3577 m

2. Spring A F= kx (30÷ 1000)(9.8)= 24.5 x x= 0.012 m

Spring B F=kx (30÷1000)(9.8)= 13.7x x=0.0215m Total elongation = 0.012+0.0215 = 0.0335m...