Lesson 6: Elastic Properties of Solids (Hooke’s Law)

Unit 1: Elasticity
When a substance is deformed in any way, internal mutual forces (called stresses) are developed between adjacent parts of the substance. These forces tend to restore the substance to its equilibrum position.

Elasticity is the property by which an object changes its shape and size under the action of opposing forces and recovers its original configuration when the forces are removed. It is the ability of a body to regain its original shape when the deforming forces are removed. The opposite of elastic is inelastic or plastic.

Steel is a highly elastic even though it may be readily stretched. However, putty, lead dough, for which elastic limit is very small are plastic or inelastic materials.

Unit 2: Hooke’s Law
For a wire or spring, it is found that if a gradual stretching force, F, is applied, the extension, e, produced at any time is directly proportional to the force applied. If the applied force is removed, the wire should return to its original size and shape.

This situation however does not continue indefinitely. At a certain stage the wire becomes distorted. The load becomes too much for the wire to bear, and the extension will no longer be proportional to the applied force. The wire is said to have lost its elasticity, or to have reached its elastic limit.

When the stretching force is removed, it does not regain its original length. The law governing the elasticity of a wire or spring as stated in the Hooke’s law statement can therefore be derived as follows:

The Hooke’s law state that the extension, e, in an elastic material is proportional to the load or applied force, F provided the elastic limit is not exceeded.

F ∝ e

i.e. F = ke

Where k is the constant of the spring or wire, known as the stiffness or elastic constant.

If F is in newtons (N) and e is in metres (m), then the unit of k = \frac{F}{e} is in Nm-1

The elastic constant, k, is the force required to produce a unit extension. Hence it gives a measure of how strong a wire or spring is. Because the extension of a spring is proportional to force, the spring balance can be used to measure forces directly in newtons.

Unit 3: Experimental verification of Hooke’s Law and Determination of Elastic Constant of a spring

The apparatus is set up as shown in the figure below:

Without suspending any weight on the spring, the pointer reading Po is noted. With a mass, m, of say 0.10kg suspended (depending on how stiff the spring is), the reading P is noted and hence the extension (P – Po) = e is found. The experiment is repeated for another five values of m, increasing in steps of 100 grains. The results are tabulated.

A graph of m (in kilograms) against e (in metres) is plotted. Provided that the total 0.60kg mass does not permanently deform the spring, a straight line graph through the origin should be obtained. Hence Hooke’s law is verified.

If the slope of the graph is S, the force constant k is proportional to S. A mass of m (in kilograms) exerts a force of mg (in newtons).

k = \frac{F}{e} = \frac{mg}{e} = Sg

The graph of a strained elastic wire
The graph of load against extension for a wire which has been stretched gradually until it has exceeded its elastic limit is shown below:

Up to the elastic limit, Hooke’s law applies. For loads beyond this, the wire stretches permanently and if the load is removed the wire is found to have been distorted and does not return to its original length. Further increase in load result in a point being reached where a small increase in load produces a large extension. This is called the yield point. Finally, a point is reached where the wire cannot stand any further increase in load. At this point the wire breaks. This is known as the breaking point.

Unit 4: Energy Stored/Workdone on Elastic Spring
Work is done when an elastic material is stretched or compressed. If the force stretching the material is F newtons, then the work done is given by:
Work = average force x extension

i.e. Work = \left ( \frac{0+F}{2} \right ) x e = \frac{1}{2} Fe

but from hooke’s law, F = Ke; where e is the extension produced by force F.

i.e. Work = \frac{1}{2} Ke2

where work is in joules, e is in meters and k the elastic constant, is in Newton per meter

Unit 5: Elastic Potential Energy
The ability of the stretched or compressed elastic material to do work is called elastic potential energy. This is energy stored in the material as a result of its stretched or compressed state. Elastic potential energy is also given by:

W = \frac{1}{2} Fe = \frac{1}{2} ke2

Where F is the maximum stretching or compressing force, e is the extension or compression, and k is the force constant of the material.

Elastic potential energy is stored energy which can be transformed into other forms of energy. For example when you stretch the rubber of a catapult, and project a stone from it, the elastic potential energy stored in the rubber is transformed into the kinetic energy of the flying stone according to the law of conservation of energy.

Example:
A spring of natural length 3m is extended by 0.01m by a force of 5N, what will be its length when applied force is N10?

Solution:
From Hooke’s law; F = Ke
where F = 5N, e = 0.01
i.e. 5 = K x 0.01 = 0.01k
i.e. 5 = 0.01k
divide both sides by 0.01
\frac{5}{0.01} = \frac{0.01k}{0.01}

i.e. 500Nm-1 = k

i.e. k = 500Nm-1

Example:
A stone of mass 30g is released from a catapult whose rubber has been stretched through 6 cm. If the force constant of the rubber is 200Nm-1, calculate the velocity with which the stone leaves the catapult.

Solution:
Let the velocity of the stone be v ms-1
The elastic potential energy of the stretched rubber = kinetic energy of the stone
i.e. \frac{1}{2} ke2 = \frac{1}{2} mv2

30g = \frac{30}{1000} g = 0.03g; 6cm = \frac{6}{100} = 0.06m

i.e. \frac{1}{2} x \frac{200}{1} x (0.06)2 = \frac{1}{2} x 0.03 x v2

i.e. 0.36 = 0.015 v2
Divide both sides by 0.015

\frac{0.36}{0.015} = \frac{0.015v^{2}}{0.015}

i.e. 24 = v2
Take the square root of both sides
\sqrt{24} = \sqrt{v^{2}}
i.e. \sqrt{24} = v
i.e. v = \sqrt{24} = 4.89ms-1

Exercises:
A helical spring of unstretched length of 15cm hangs vertically on a rigid support. The length of the spring increases to 20cm when a load of 0.5kg is attached to its lower end.
(i) determine the new length of the spring if an additional load of 0.5kg is attached and the elastic limit has not been exceeded?
A.10cm B. 20cm C. 25cm D. 30cm

(ii) what load would be stretch spring to a length of 22cm?
A. 0.2kg B. 0.7kg C. 2.2kg D. 22.0kg

Unit 6: Young’s Modulus
Young’s modulus is the ratio of tensile stress to tensile strain provided the elastic limit is not exceeded.

i.e. E = \frac{tensile\: stress}{tensile\: strain}

the tensile stress is the ratio of force F (in newtons) applied to a wire to the cross-sectional area of the wire A, i.e. tensile stress = \frac{F}{A}

The tensile strain is the ratio the change in length of wire over original length of wire, i.e. tensile strain = \frac{e}{l_{o}}

We can now rewrite the equation of young modulus as E = \frac{F}{A} ÷ \frac{e}{l_{o}}
i.e E = \frac{F}{A} x \frac{l_{o}}{e} = \frac{Fl_{o}}{Ae}

i.e. Young’s Modulus = E = \frac{Fl_{o}}{Ae}

Example:
A copper rod of diameter 20mm and length 2.0m has a tensile force of 5kN applied to it. Determine
(a) the stress in the rod,
(b) by how much the rod extends when the load is applied.
Take the modulus of elasticity for copper as 96GPa.

Solution:

Exercise:
A bar of thickness 15mm and having a rectangular cross-section carries a load of 120kN. Determine the minimum width of the bar to limit the maximum stress to 200MPa. The bar, which is 1.0m long, extends by 2.5mm when carrying a load of 120kN. Determine the modulus of elasticity of the material of the bar.
ANS = 80 x 109

End of SS1 Physics 2nd Term Lesson Notes

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