Summary

Hooke’s Law states that force is directly proportional to their elongation,

F = kx

The value of k is known as the spring constant and sometimes can be stated as –k rather than +k. This value (negative or positive) is depends on the displacement. If the spring is compress the value of k will be in negative value and if the spring is stretch the value of k will be in positive value. The force is known as restoring force as this force show you this reading is depends on the value k. The greater the value of k, the greater the restoring force. While for the x, is also known as the displacement (meter). To find either the value of force (F), spring constant (k) or the displacement (m) it is important to know which formula that you should use. If the spring is in vertical condition (up side down) the formula that we should use is,

mg = kx (k can be negative or positive)

And so if the spring is set in horizontal position (left to right) the formula that we should use is,

ma = kx (once again the k value can be positive or negative)

Summary From Internet

A string lengthens slightly when you stretch it. Similarly, we have already discussed how an apparently rigid object such as a wall is actually flexing when it participates in a normal force. In other cases, the effect is more obvious. A spring or a rubber band visibly elongates when stretched. Common to all these examples is a change in shape of some kind: lengthening, bending, compressing, etc. The change in shape can be measured by picking some part of the object and measuring its position, x. For concreteness, let's imagine a spring with one end attached to a wall. When no force is exerted, the unfixed end of the spring is at some position xo. If a force acts at the unfixed end, its position will change to some new value of x. The more force, the greater the departure of x from xo.

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m / Defining the quantities F, x, and xo in Hooke's law.

Back in Newton's time, experiments like this were considered cutting-edge research, and his contemporary Hooke is remembered today for doing them and for coming up with a simple mathematical generalization called Hooke's law:

F ≈ pk(x - xo). [Force required to stretch a spring; valid for small forces only]

Here k is a constant, called the spring constant that depends on how stiff the object is. If too much force is applied, the spring exhibits more complicated behavior, so the equation is only a good approximation if the force is sufficiently small. Usually when the force is so large that Hooke's law is a bad approximation, the force ends up permanently bending or breaking the spring. Although Hooke's law may seem like a piece of trivia about springs, it is actually far more important than that, because all solid objects exert Hooke's-law behavior over some range of sufficiently small forces. For example, if you push down on the hood of a car, it dips by an amount that is directly proportional to the force.

Introduction

In mechanics, and physics, Hooke's law of elasticity is an approximation that states that the extension of a spring is in direct proportion with the load added to it as long as this load does not exceed the elastic limit. Materials for which Hooke's law is a useful approximation are known as linear-elastic or "Hookean" materials.

Hooke's law is named after the 17th century British physicist Robert Hooke. He first stated this law in 1676 as a Latin anagram, whose solution he published in 1678 as Ut tensio, sic vis, meaning:

‘As the extension, so the force’

For systems that obey Hooke's law, the extension produced is directly proportional to the load:

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where:

x = the distance that the spring has been stretched or compressed away from the equilibrium position, which is the position where the spring would naturally come to rest (meters),

F = the restoring force exerted by the material (Newtons), and

k = the force constant (or spring constant). The constant has units of force per unit length (newtons per meter).

When this holds, we say that the behavior is linear. If shown on a graph, the line should show a direct variation. There is a negative sign on the right hand side of the equation because the restoring force always acts in the opposite direction of the x displacement (when a spring is stretched to the left, it pulls back to the right).

The most commonly encountered form of Hooke's law is probably the spring equation, which relates the force exerted by a spring to the distance it is stretched by a spring constant, k, measured in force per length.

The negative sign indicates that the force exerted by the spring is in direct opposition to the direction of displacement. It is called a "restoring force", as it tends to restore the system to equilibrium. The potential energy stored in a spring is given by

Which comes from adding up the energy it takes to incrementally compress the spring. That is, the integral of force over distance.

(Note that potential energy of a spring is always non-negative)

Objective

To investigate and verify Hooke’s Law

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Theory

In this experiment you will examine the motion of an object moving under the

influence of a force that dependent on position. Since the force changes as the particle moves

the kinematic equations are not appropriate to describe the motion. Many different position

dependent forces exist, and motion under the influence of these forces can result in motion

that is far from simple. However, this experiment will be restricted to the study of motion

under the influence of the force which depends linearly on position: the force from a spring.

Springs generates a force which is directly proportional to their elongation and are described

by Hooke’s Law,

F= kx ……………………………… [5.1]

Where,

F= the force

x= the elongation

k= spring constant (proportionally constant)

Where F is the force, x is the elongation and k is proportionality constant, called the

spring constant. The minus sign in the Hooke’s Law expresses the fact that the force is

always directed opposite its displacement. A force which has a direction opposite to the

displacement is called restoring force, since it lends to restore the objects to its original

resting place. The spring constant, k, depends on the material and dimensions of the actual

spring.

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Procedures

1) A light spring is attached to the clamp of a retort stand as shown in figure 5.1. A mass

hanger is suspended from the spring.

2) The distance, d is measured between the clamp and bottom of the mass hanger.

3) A 50 g mass is added to the hanger. The distance, d is measured from the clamp to the

bottom of the hanger. The extention of the spring is given by D – d. The result

recorded in a table.

4) Another 50 g mass is added to the hanger and the extention of the spring is measured.

This step is repeated until about six 50 g masses on the hanger.

Apparatus

Retort-stand

Mass Hanger

Pin

Spring

Ruler

SET UP

d D

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Results

Length, d = 0.255 m

Mass, m (kg)

Force, F (N)

Distance from the clamp to the hanger Extension, x (m)

(x = D – d)Initial length, d (m) Final length, D (m)

0.02 0.1962 0.255 0.320 0.065

0.04 0.3924 0.255 0.384 0.129

0.06 0.5886 0.255 0.446 0.191

0.08 0.7848 0.255 0.510 0.255

0.10 0.9810 0.255 0.575 0.320

0.12 1.1772 0.255 0.640 0.385

F = mg

F = Force (N)

m = Mass (Kg)

g = Gravity (9.81 ms-2)

x = D – d

x = Extension

D = Final Length

d = Initial Length

F = kx

F = Force (N)

x = the elongation

k = spring constant (Nm-1)

1m = 100 cm

Calculations

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Reading 1 (0.02 Kg)

a) Force, F ;

F = mg

= (0.02 Kg) (9.81 ms-2)

= 0.1962 N

b) Extension, x ;

x = D – d

= [0.320 – 0.255] m

= 0.065 m

c) Spring Constant, k ;

F = kx

k = F

x

= 0.1962 N

0.065 m

= 3.018 Nm-1

Reading 2 (0.04 Kg)

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a) Force, F ;

F = mg

= (0.04 Kg) (9.81 ms-2)

= 0.3924 N

b) Extension, x ;

x = D – d

= [0.384 – 0.255] m

= 0.129 m

c) Spring Constant, k ;

F = kx

k = F

x

= 0.3924 N

0.129 m

= 3.042 Nm-1

Reading 3 (0.06 Kg)

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a) Force, F ;

F = mg

= (0.06 Kg) (9.81 ms-2)

= 0.5886 N

b) Extension, x ;

x = D – d

= [0.446 – 0.255] m

= 0.191 m

c) Spring Constant, k ;

F = kx

k = F

x

= 0.5886 N

0.191 m

= 3.082 Nm-1

Reading 4 (0.08 Kg)

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a) Force, F ;

F = mg

= (0.08 Kg) (9.81 ms-2)

= 0.7848 N

b) Extension, x ;

x = D – d

= [0.510 – 0.255] m

= 0.255 m

c) Spring Constant, k ;

F = kx

k = F

x

= 0.7848 N

0.255 m

= 3.078 Nm-1

Reading 5 (0.10 Kg)

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a) Force, F ;

F = mg

= (0.10 Kg) (9.81 ms-2)

= 0.9810 N

b) Extension, x ;

x = D – d

= [0.575 – 0.255] m

= 0.320 m

c) Spring Constant, k ;

F = kx

k = F

x

= 0.9810 N

0.320 m

= 3.066 Nm-1

Reading 6 (0.12 Kg)

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a) Force, F ;

F = mg

= (0.12 Kg) (9.81 ms-2)

= 1.1772 N

b) Extension, x ;

x = D – d

= [0.640 – 0.255] m

= 0.385 m

c) Spring Constant, k ;

F = kx

k = F

x

= 1.1772 N

0.385 m

= 3.058 Nm-1

Discussions & Question

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1) Plot a graph of a force causing the extension against the extension, x. what is the

gradient of this graph?

Points is ( 0.130 , 0.40) and ( 0.255 , 0.78 )

Gradient = y2 – y1

x2 – x1

= 0.78 – 0.40

0.255 – 0.130

= 0.38

0.125

= 3.04

2) How much work is done in stretching the spring through 0.1 m? find this from the

graph.

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Work is done in stretching the spring through 0.1 m is 0.23 N

W = Fs

= (0.32 N) (0.1m)

= 0.032 J

3) How much energy is stored in the spring when it is extended 0.1m?

From 2 when x = 0.1 m, F = 0.32 N

Substitute in to equation,

F = kx

k = F

x

= 0.32 N

0.1 m

=3.2 Nm-1

PEelastic = ½ kx2

= ½ (3.2) (0.1)2

= 0.0115 J

= 1.6 × 10 -2 J

4) Describe any errors that may occurred during this experiment

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Throughout this experiment, the results and calculations seem to turn out well as expected but not very well enough. Some of the results are approximately to be correct. It is because one major factor, which cannot be avoided, is the elasticity of the spring. This normally occurred in this kind of experiment probably because the spring has not been changed from time to time as either it should be or it has already worn out. That is why my results are affected.

Besides that, the other factor is because of the edge of the ruler was worn out. This too had affected the results that have been obtained. Other than that, when we were carrying out this experiment, we did not use the exact mass, 50g. We had used smaller mass than that because if we did not, the spring will not have the strength to hold on. It may ruin the elasticity of the spring, which it is already in the first place.

Finally yet importantly, the last factor is a human factor. For example, when

conducting this experiment, one should not stand near the apparatus that has already

been set up. What had happened was one of my course mates accidentally knocking

the hanger that contained a mass and the mass fell off. We had to start all over again

from square one.

5) What can you conclude from your results about the relationship between the extension

of a spring and the force causing the extension.

Based on my results and from the graph, we can conclude something that commonsensical and judgment. When more mass added, the force value also increase. Hence, when the force is add up, the extention too will increased. In other words, for systems that Hooke's law, the extension produced is directly proportional to the load :

F = kx

or

F = - kx

Conclusions

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For the conclusion, our experiment is successful agree to the objective of this

experiment to investigate and verify Hooke’s Law. To be conclude, spring stretched when

force is being applied on positive direction, F = kx. The spring retain its original position by

restoring force, F = -kx. F is directly proportional to x if the x is not too great until the spring

reached its limit of elasticity. We can conclude also the force, N acting on the spring is

proportional to the extension, x from the graph we construct. The longer the extension makes

it to yield high value of force, N. Therefore, the obeys hooke’s Law from this experiment.

Recommendations

There are several important factors that require to be taken accounted from this experiment.

Because the experiment is emphasizing on the Hooke’s law would be one of the main

contributors to the failure of the experiment. Some of the proper actions that should be taken

to allow the experiment to succeed are such as:

Check the apparatus before use for any faultiness so that the measurements taken are

accurate.

Take at least three readings in order to get accurate readings.

Try our best to avoid parallax error.

Make sure the apparatus used such as retort stand is stable and not faulty.

Make sure that the initial reading same as the beginning. In the other words, the

spring can restore its original position after certain amount of force applied.

Placed carefully the mass on the spring so that the reading can be taken more

accurately and easily.

Make sure that the spring is in good condition during the experiment.

References

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Mr. Rudiyanto lecturer for CHE 175 ( note )

Mrs. Nur Asma lecturer for laboratory CHE 175

6th Edition, Physics Giancoli, Douglas C. Giancoli

http://wikieducator.net/hookeslaw/introduction.html

http://en.wikipedia.org/wiki/Hooke%27s_law

http://schools.matter.org.uk/Content/HookesLaw/introduction.html

Serway, Raymond A.; Jewett, John W. (2003). Physics for Scientists and Engineers.

Brooks/Cole. ISBN 0-534-40842-7.

Tipler, Paul (1998). Physics for Scientists and Engineers: Vol. 1 (4 th ed.). W. H.

Freeman. ISBN 1-57259-492-6.

Wylie, C. R. (1975). Advanced Engineering Mathematics (4th ed.). McGraw-Hill.

ISBN 0-07-072180-7.

Appendices

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