Elastic properties of solids

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 k is the constant of the spring or wire, known as the stiffness or elastic constant. F is represented in Newtons (N) and e in metres (M). Therefore the SI unit is Nm^-1.

Young’s modulus

    F/A = e/l₀ (provided elastic limit is not exceeded).

The expression F/A is known as the tensile stress (Nm^-2) and e/l is known as tensile strain (no units). 

Young modulus is therefore defined as:

E= tensile stress/ tensile strain

Source: www.golabz.eu 

Source: www.britannica.com 

Verification of Hooke’s law

Graph of load against extension

Source: www.freexamacademy.com 

Energy stored or work done on springs and elastic strings

          Work= average force × extension

      =( O+F )/2 × e

      = 1/2Fe

But from Hooke’s law, F=Ke

Where e is the extension produced by force F. Work= ½ ke²

Where work is in joules, e is in metres and the elastic constant is in Newton per metre. 

Elastic potential energy

Potential energy is defined as the energy possessed by a body by virtue of its state or position. This potential to do work arises from the fact that work is done in stretching or compressing the elastic material. Thus, the ability of the stretched or compressed elastic material to do work is called elastic potential energy. Elastic potential energy is expressed as:

     W= ½ Fe

        = ½ ke²

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

Past questions

1.The tendon in a man’s leg is 0.01m long. If a force of 5N stretches the tendon by 2.0×10^-5m, calculate the strain on the muscle. (Jamb 1998)

A. 5×10⁶

B. 5×10²

C. 2×10^-3

D. 2×10^-7

Answer: C

Solution: strain = extension/ original length

 2×10^-3/0.01 = 2×10^-3

2. The diagram above shows the force- extension curve of a piece of wire. The energy stored when the wire is stretched from E to F is ? (Jamb 2002)

A. 1.5×10^-2J

B. 7.5×10^-1J

C. 7.5×10^-3J

D. 2.5×10^-3J

Answer: D

Solution: Energy= ½ Fe=½ ×(0.2-0.1)×(0.1-0.05) = 2.5×10^-3J

3. An elastic material has a length of 36cm when a load of 40N is hung on it and a length of 45cm when a load of 60N is hung on it. (Jamb 2005)

A. 18cm

B. 12cm

C. 20cm

D. 15cm

Answer: A

Solution: f=Ke; k=F/e = 200/9

But e = F/K = 400/200/9 = 400/1×9/200 =18cm

4. If a load of mass 10N stretches a cord by 1.2cm, what is the total work done? (Jamb 2009)

A. 6.0×10^-2J

B. 7.6×10^-2J

C. 1.8×10^-2J

D. 6.6×10^-2J

Answer: A

Solution: W= ½ Fe = ½ × 10×1.2/100

=0.06 ~ 6×10^-2J

5. Define (I) Elasticity (ii) Young’s modulus (III) Force constant. (Wassce 2000)

Answer: (I) Elasticity is the ability of a body to regain its normal shape and size after undergoing a stretch or compression.

(II) Young’s modulus can be defined as the ratio between stress and strain = stress/ strain

(III) Force constant is the ratio of force to that of the extension produced when load is exerted on elastic material. F= e (f=force, e= extension); F=Ke; k=f/e; k= force constant.

6. State Hooke’s law of elasticity. (Wassce 2001)

Answer: Hooke’s law of elasticity states that provided the elastic limit is not exceeded, the extension, e, is directly proportional to the force or load applied on to it. F=e; F=Ke

7. The diagram above represents the graph of stress against strain for an elastic wire. The point Q on the graph is the? (Wassce 2016)

A. Elastic limit

B. Breaking point

C. Yield point

D. Proportional limit

Answer: B

8. A wire is gradually stretched until it snaps. Sketch a load-extension graph for the wire and on the graph indicate the 

A. Elastic limit

B. Yield point

C. Maximum load

D. Breaking point

Answer: