When a force, F, is applied to a string of wire, an extension, e, is produced and the more force applied, the more extension produced in the wire, which shows that the extension produced in the wire is directly proportional to the force applied.

If the force is removed, the wire or string returns to its original size or shape.

The ability of a wire/spring to return to its original size or length or shape after an applied force is removed is called elasticity.

An elastic material is one that regains its original size after the distorting force has been removed.

The law governing elasticity was investigated by Robert Hooke and this is known as Hookeâ€™s law which states that:

The extension, *e*, produced in an elastic material is directly proportional to the applied force, *F*, provided the elastic limit is not exceeded.

\( \scriptsize F \propto e \)

F = Ke

K is a constant known as **elastic constant**.

â‡’ \(\scriptsize k = \normalsize \frac{F}{e} \\ = \normalsize \frac{Newton}{metre}\)

The unit of elastic constant is Newton per meter or Nm^{-1}.

An elastic constant is defined as the force required to produce a unit extension.

An elastic constant is the maximum stress which a material can attain before permanent deformation. When an elastic limit of a solid is exceeded, the material can no longer retain its original size or shape when the applied force is removed. A small force will produce a large extension.

### Example 1:

If a force of 0.9 N stretches an elastic spring by 15 cm, find the elastic constant of the spring.

**Solution:**

Force = 0.9N

e = 15cm

convert e to m

â‡’ \( \frac{15}{100} \\ = \scriptsize 0.15m\)

From Hooke’s law

*F* = *ke*

make k the subject of the formula:

â‡’ \(\scriptsize k = \normalsize \frac{F}{e} \)

â‡’ \(\scriptsize k = \normalsize \frac{0.9}{0.15} \)

â‡’ \(\scriptsize k = 6 \: Nm^{-1} \)

### Example 2:

A material of length 150 *mm* is stretched by a force of 7.4 *N* to 175 *mm*. Calculate the elastic constant of the material.

**Solution:**

F = 7.4 N

extension, e = 175 mm – 150 mm = 25mm

convert to m

â‡’ \( \frac{25}{1000} \\ = \scriptsize 0.025 m\)

*F* = *ke*

â‡’ \(\scriptsize k = \normalsize \frac{F}{e} \)

â‡’ \(\scriptsize k = \normalsize \frac{7.4}{0.025} \)

â‡’ \(\scriptsize k = 296\: Nm^{-1} \)

### Example 3:

A force of 3 N stretches an elastic material by 40mm. What additional force will stretch the material 45mm? Assuming the elastic limit is not exceeded.

**Solution:**

If the elastic limit is not exceeded, k will be the elastic constant.

F_{1} = 3N

e_{1} = 40mm

â‡’ \( \scriptsize e_1 = \left(\normalsize \frac{40}{1000} \right) \scriptsize m \\ = \scriptsize 0.04 m\)

*F _{1} = ke*

_{1}

â‡’ \(\scriptsize k = \normalsize \frac{F_1}{e_1} \)

â‡’ \(\scriptsize k = \normalsize \frac{3}{0.04} \)

â‡’ \(\scriptsize k = 75\: Nm^{-1} \)

Let the force stretching the material be F_{2}

F_{2} = ?

e_{2} = 45mm

â‡’ \( \scriptsize e_2 = \left(\normalsize \frac{45}{1000} \right) \scriptsize m \\ = \scriptsize 0.045 m\)

k = 75 N/m

*F _{2} = ke*

_{2}

*F _{2}* = 75 x 0.045

*F _{2}* = 3.375N

âˆ´ additional force = *F _{2}* –

*F*

_{1}= 3.375 N – 3 N

= 0.375 N

### Example 4:

A string of length 2m extends by 0.02m when a force of 3N is applied. Find the new length when a force of 15N is applied.

**Solution:**

*F = ke*

F = 3N, e = 0.02, k = ?

K = \(\frac{F}{e} = \frac{3}{0.02}\)

K = 150Nm^{-1}

When F = 15N, e = ?

F = ke

15 = 150 x e

e = \( \frac{15}{150}\)

e = 0.1m

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