Hooke\'s Law PHY094 PDF

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PHY094 Lab Report Title:

HOOKE’S LAW Prepared for:

PN. NOR FARIDAH HANIM BINTI MAT JUNIT Prepared by: GROUP 3 NO.

NAME

STUDENT ID

1

ARINAH BATRISYIA BINTI AMAT SANO

2020842658

2

IFFAH BATRISYIA BINTI MOHD ADNAN

2020893054

3

ERYN ADRIANA BINTI MOHD JAFNI

2020463936

4

FARAH SU’AIDAH BINTI HAIRIN

2020620178

Date of Experiment: 18 August 2020 Date of Submission: 25 August 2020

TABLE OF CONTENT

PAGE

ABSTRACT

1

INTRODUCTION

1

METHODOLOGY

1

RESULTS AND ANALYSIS

2-3

SAMPLE CALCULATIONS

4

DISCUSSION

4-5

CONCLUSION

5

PRE-LAB QUESTIONS

6

POST-LAB QUESTIONS

7

ABSTRACT Two experiments were conducted to find the value of spring elasticity,k of two steel springs. The experiments were also performed to verify Hooke’s Law. The spring constant was determined by measuring the elongation of spring A and spring B when various masses are hung at the end of each spring. The resulting values of k are 22.731N/m and 4.9009N/m respectively. Our spring's behavior followed Hooke's law to within the limits of accuracy of the two experiments. INTRODUCTION An important property of solid is their ‘stretchiness’ which is called their elasticity. For example spring. Spring that does not have any exerted force acting on it or in equilibrium will have a natural length. When load is hanged at the end of the spring, it will stretch by an amount of x and pull downward. Fs, the restoring force will be in the opposite direction of x. As the amount of x increases, the force Fs increases in proportion. This relationship is known as ‘Hooke’s Law’. Hooke’s Law stated that the force F exerted by a spring is proportional and in the opposite direction to the displacement x of the end of the spring. This relationship can be expressed as:

F = -kx s

where k is spring constant. The value of k depends on how the spring was formed, the material it is made from and thickness of the wire. This experiment is done to see if the spring obeys Hooke’s Law and find the value of k of the spring. METHODOLOGY The apparatus needed to carry out this experiment are stand,ruler,scale pan,set of masses and two springs (spring A and spring B). The ruler was first attached to the stand securely by using the clamp and spring A was hooked to the arm of the stand beside the ruler. These steps were done to measure the spring’s length. Next,the initial length of spring A without the scale pan and masses is measured and the scale pan needs to be weight by using weighing scale.The scale pan then hooked onto spring A and the length of the spring is measured again to get the length of spring A with the scale pan.Then, 25g mass is added into the scale pan and the length of the spring with 25g mass is measured.This step is repeated with 50g mass,75g mass,100g mass,125g mass and 150g mass.Second reading of every length of the spring need to be measured to get average reading.Lastly,the elongation of the spring with various mass is measured by subtracting the average reading of the spring length with the initial length of the spring. These whole steps were then repeated using spring B.Two different springs which are spring A and spring B were used in this experiment to differentiate the elongation of two springs with different spring constants, k.

1

RESULTS AND ANALYSIS Spring A Length of spring, y (m) Mass, mmass (kg)

Total mass, (mpan+mmass) (kg)

Force, F=mg (N)

yn

0.000

0.000

0.000

0.000

0.005

0.025

Reading

Elongation of spring, ∆y (m)

Average reading, y n

 y  y n  yo

1

2

y0

0.069

0.067

0.0680

0.0000

0.049

y1

0.072

0.072

0.0720

0.0040

0.030

0.294

y2

0.082

0.085

0.0835

0.0155

0.050

0.055

0.539

y3

0.093

0.094

0.0935

0.0255

0.075

0.080

0.784

y4

0.104

0.102

0.1030

0.0350

0.100

0.105

1.029

y5

0.113

0.114

0.1135

0.0455

0.125

0.130

1.274

y6

0.123

0.125

0.1240

0.0560

0.150

0.155

1.519

y7

0.132

0.134

0.1330

0.0650

Spring B Length of spring, y (m) Mass, mmass (kg)

Total mass, (mpan+mmass) (kg)

Force, F=mg (N)

0.000

0.000

0.000

0.000

0.005

0.049

0.025

0.030

0.294

0.050

0.055

0.539

Reading 1

2

Average reading, y n

0.057

0.057

0.0570

0.000

0.067

0.069

0.0680

0.0110

0.117

0.118

0.1175

0.0605

0.167

0.168

0.1675

0.1105

yn y0 y1 y2 y3

Elongation of spring, ∆y (m)

2

 y  y n  yo

0.075

0.080

0.784

0.100

0.105

1.029

0.125

0.130

1.274

0.150

0.155

1.519

y4 y5 y6 y7

0.217

0.219

0.2180

0.1610

0.265

0.261

0.2630

0.2060

0.317

0.315

0.3160

0.2590

0.365

0.374

0.3695

0.3125

3

SAMPLE CALCULATIONS 1.

Gradient for graph of spring A

m  22.731 m k k1  22.731 2.

Gradient for graph of spring B

m  4.9009 m k k2  4.9009 DISCUSSION The experiment was started by hanging spring A on the stand and the initial length of the spring, yo is measured. and a scale pan is attached to it.Then,masses from 25g until 150g is added and the elongation of the spring is measured.The elongation of the spring which also known as the spring’s displacement is measured by subtracting the average length of spring with mass,yn, with the initial length of spring,yo.

y  y0  yn The mass of the scale pan is weighed to get the total mass ( M mass  M pan ). The elongation of the spring is measured once the spring is at rest to prevent errors.When the mass is attached to the spring there are 2 forces exerted on it which are gravity force and also the spring’s elastic restoring force.The gravity force acting on the mass is directed downward the mass while the spring’s elastic restoring force is directed upward the mass which is the opposite direction as the displacement.These two forces are equal. Therefore,the net force of mass when it’s on rest is equal to its weight (F=mg). The experiment was then repeated using spring B that has a different spring’s constant. The spring constant was calculated using the formula of Hooke’s Law which is (F = -kx). The spring constant for each value of displacement is the same within the experiment, which follows Hooke’s Law. The average spring constant is 13.816 N/m. A graph of force against elongation of spring resulted in the expected straight line showing that the spring obeys Hooke’s Law. The spring constant, k from the gradient which are 22.731 for spring A and 4.9009 for spring B. Based on the experiment, we observed that when force,F increases the elongation of spring will increase too. This statement is proved by plotting the graph. In the

4

graph, we knew that force,F is perpendicular to the elongation of spring because it forms a straight line graph. By referring to the graph, we knew that spring constant is equal to the gradient of the graph. The manipulated values in this experiment were the masses while the responding values were the elongation of the spring. CONCLUSION Based on the Hooke’s Law experiment that we have conducted in the lab session, the elongation of the spring is proportional to the applied force as in the formula (F=k∆y) which is F is the force, ∆y is the elongation of the spring and k as a constant of proportionality. If the applied force increases, the average elongation of the spring also increases. Thus, it presents a linear graph after we plotted the force,F against the elongation of the spring,∆y into the graph, and this lab proved our hypothesis certainly obeys Hooke’s Law.

5

PRE-LAB QUESTIONS 1.

When you apply a 52N force,a spring extends 13 cm.Assume the spring obeys Hooke’s Law. What is the spring constant (in N/m) for this spring ?

F = kx 52N = k (0.13m) 澠೎ k = th k = 400 N/m 2.

As above, when you apply a 52 N force,a spring extends by 13cm.How much energy was required to stretch the spring assuming you started from its unstreched length? How much energy (in Joules)must you use to stretch is another 13 cm ( from 13 cm to 26 cm)? h

PEs = 澠 kx2

x = 26 cm = 0.26 m

h

= 澠 (400N/m)(0.13m)2 =3.38 J 3.

PEs = = 13.52 J

h 澠

(400N/m)(0.26m)2

Suppose you have a mass m attached to a spring with constant k.The mass rests on a horizontal frictionless surface.Its equilibrium position is at x = 0.It is pulled aside a distance A and released.What is the speed of the mass as it passes the position x = A/2 (in terms of k,m, and A)? (KE + PEg + PEs)i = KE + PEg + PEs)f

h mv2i 澠 h 澠 h 澠

澠

h

h

h

h

mv2f = 澠kx2i - 澠kx2f h

h

h

mv2f = 澠k(A)2 - 澠k(澠A)2

h mv2f 澠 h

h

+ mghi + 澠kx2i = 澠mv2f + mghf + 澠kx2f

h

h

= 澠 2 - kA2 

mv2f =  2

Vf =

澠 

6

POST-LAB QUESTIONS 1.

Spring B used in this experiment obeys Hooke’s Law, determine the value of elongation, y of a spring if a total mass of 0.5 kg is attached to it? Show your calculation. F = mg F = (0.5)(9.8) F = 4.9 N F = kΔy 4.9 = (4.9009) Δy Δy = 1.0 m

2.

Spring C consists of spring A and spring B that connected in series. Calculate the elongation of spring C if a 30g mass was supported from the spring. F = mg  F=( 󰇜(9.8) h F = 0.294 N

k=

󰇛 香䁕ꀀ香 󰇜󰇛 香䁕ꀀ香 󰇜  香䁕ꀀ香  香䁕ꀀ香  󰇛澠澠th󰇜󰇛t󰇜

k= 澠澠tht k = 4.0849

F = kΔy 0.294 = (4.0849) Δy Δy = 0.0720 m

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