Elastic property of materials and Hooke’s law

Elasticity of Properties of Solids

A body is said to be elastic if it retains its original shape after undergoing stretch or compression. For wires or springs, it observed that if a gradual stretch force is applied, the extension e, produced at anytime is directly proportional to the force or load applied. But if you continue to apply load and it becomes too much for the string or wire, the extension will no longer be proportional to the load of force applied. Then the wire will lose its elasticity if more force is applied when it have reached is Elastic Limit.

Hooke’s Law

Hooke’s law states that provided the elastic limit is not exceeded, the extension, e, in an elastic material (e.g. wire) is proportional to the load or applied force, F.

F = ke

Where: F = tension acting on the spring.

e is extension = (l-lo); l is the stretched length and lo is original length.

k = the spring constant or stiffness

Elastic limit and yielding point

Elastic limit:  It is the maximum stretching force beyond which the stretched material would  not return to its original size or length when the force is removed. If a material returns to its original shape after forces are applied, it demonstrates elastic behavior

Yield point: it is the point at which elastic material lose its elasticity or elastic permanently and become plastic.If a material deforms from its original shape after forces are applied, it is a sign of plastic behavior.

Elastic
Elastic
Elastic
Elastic

Elastic Potential Energy

Elastic potential energy can be defined as the energy stored in strings or wire due to stretch or compression

“If the deformation caused is within the elastic limit, the work done in deforming the object is stored within it as potential energy. This is called (elastic) ‘strain energy’. When the applied force is removed the energy is released. The strain energy then performs work in changing the object and to its original state.”

It is given as

W=\frac { 1 }{ 2 } fe\quad 0r\frac { 1 }{ 2 } ke{ }^{ 2 }

where f = force applied, k = elastic constant, and e = extension

Stress, strain and Young’s Modulus

Stress:it is defined as the force per unit area of a material:

Stress = force/cross sectional area

where s = stress, F = force applied, A= cross sectional area

Units of s: Nm-2

Strain: it  is defined as extension per unit length:

Strain = extension/original length

where e = strain, lo = the original length, e = extension = (l-lo) and l = stretched length

Strain has no units because it is a ratio of lengths.

Young’s modulus, E : it be defined as

E = Stress/strain , Units of the Young modulus E: Nm-2

Questions

1.An elastic material extended by 1.0cm when a load of 20g hang from it , what additional load will be required  to cause further extension o 2.0cm

Solution, Using

F = ke where F=( 20/1000) * 10 =      0.2N,  K = ?, and e = 1.0cm = 0.01m

0.2 =0.01 k, then k =0.2/0.01 = 20Nm-1

Now let additional load = M, so the force exerted by new load is give as

(0.02 + M)*10 = k *0.03

0.2 + 10M = 20 * 0.03

10M =0.6 -0.2 = 0.4

10M = 0.4

M =0.4/10 =0.04Kg =40g

M =40g

  1. A spring of length 3m is extended by 0.01m, by force of 4N, what will be its length when the applied force is 12N.

Solution

From Hooke’s law, F= ke, where f = 4N, k =?, e = 0.01m

That is    4 = k *0.01

K = 4/0.01 =n 400Nm-1

When F = 12, k = 400Nm-1 and e =?  Then

12 = 400e

e = 12/400 = 0.03

e = 0.03m, therefore the length is

3m + 0.03m = 3.03m

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