Elasticity and Hooke's law: WASSCE Physics

A spring scale in the market, the springs under a motorbike seat, a catapult and the rubber in a sandal all stretch and spring back. How far they stretch, and when they stop springing back, is the subject of elasticity.

Learning objectives

By the end of this lesson you should be able to:

define elasticity, elastic limit and plastic deformation
state Hooke's law and use $F = ke$
interpret a load–extension graph for a spring and a metal wire
describe an experiment to verify Hooke's law

Elasticity

Definition

Elasticity is the property of a material that enables it to return to its original shape and size after the force deforming it is removed.

A material that does not return to its original shape is plastic (for example plasticine, wet clay, chewing gum). Many materials are elastic for small forces and plastic for large ones: a spring stretched too far stays permanently longer.

Hooke's law

Definition

Hooke's law: the extension of an elastic material is directly proportional to the force (load) applied, provided the elastic limit (more precisely, the limit of proportionality) is not exceeded.

F = ke

where $F$ is the force (N), $e$ the extension (m) and $k$ the spring constant (or force constant) in N m⁻¹.

The extension is the increase in length: $e = l - l_0$ , where $l_0$ is the original (natural) length.

The spring constant measures stiffness: a large $k$ means a stiff spring that stretches little.

Worked example

A spring of natural length 20 cm stretches to 24 cm when a 6 N weight hangs on it. Find (a) the spring constant, (b) the length when the load is 9 N (within the elastic limit).

(a) Extension $e = 24 - 20 = 4\ \text{cm} = 0.04\ \text{m}$ .

k = \dfrac{F}{e} = \dfrac{6}{0.04} = 150\ \text{N m}^{-1}

\begin{aligned}k &= \dfrac{F}{e} \\ &= \dfrac{6}{0.04} \\ &= 150\ \text{N m}^{-1}\end{aligned}

(b) $e = \dfrac{9}{150} = 0.06\ \text{m} = 6\ \text{cm}$ , so the length is $20 + 6 = 26$ cm.

Common mistake

Use the extension, not the stretched length. In part (b) above, many candidates answer 6 cm instead of 26 cm, or use 24 cm as if it were the extension.

The load–extension graph

Load against extension for a metal wire: a straight line from the origin up to the limit of proportionality P, then a curve through the elastic limit E, the yield point Y and finally the breaking point B

Figure 1. Load–extension graph for a metal wire. WASSCE Tutor diagram.

Point	Meaning
O to P	straight line through the origin: Hooke's law is obeyed; gradient $= k$ (when load is on the y-axis)
P, limit of proportionality	beyond this, extension is no longer proportional to load
E, elastic limit	the greatest load from which the material still returns to its original length
Y, yield point	the material suddenly stretches a lot for little extra load: it starts to flow (plastic)
B, breaking point	the wire snaps

Beyond the elastic limit the deformation is permanent: removing the load leaves a permanent set (extension).

Experiment: verifying Hooke's law

Clamp a spiral spring vertically with a pointer at its lower end against a vertical metre rule. Hang a light scale pan.
Record the pointer reading with no load ( $l_0$ ).
Add loads in equal steps (e.g. 0.5 N), recording the pointer reading each time.
Remove the loads one at a time and record the readings again (they should match, showing the elastic limit was not exceeded).
Calculate the extension for each load. Plot load (y) against extension (x).
A straight line through the origin verifies Hooke's law; its gradient is $k$ .

Precautions: avoid parallax when reading the pointer (eye level); allow the spring to come to rest before reading; do not exceed the elastic limit; take readings while loading and unloading.

Work done in stretching (elastic potential energy)

The work done in stretching a spring is the area under the force–extension graph:

W = \tfrac{1}{2}Fe = \tfrac{1}{2}ke^2

\begin{aligned}W &= \tfrac{1}{2}Fe \\ &= \tfrac{1}{2}ke^2\end{aligned}

Worked example

How much energy is stored in the spring above (k = 150 N m⁻¹) when it is stretched by 4 cm?

W = \tfrac{1}{2} \times 150 \times 0.04^2 = 0.12\ \text{J}

\begin{aligned}W &= \tfrac{1}{2} \times 150 \times 0.04^2 \\ &= 0.12\ \text{J}\end{aligned}

Key points

Elasticity: returning to the original shape when the force is removed.
Hooke's law: $F = ke$ , valid up to the limit of proportionality.
Extension = new length − natural length.
Beyond the elastic limit, deformation is permanent; then come the yield point and breaking point.
Energy stored $= \tfrac{1}{2}Fe = \tfrac{1}{2}ke^2$ .

WASSCE examination focus

WAEC tests this

State Hooke's law (include "provided the elastic limit is not exceeded"), solve spring problems with $F = ke$ , interpret the load–extension graph (label the elastic limit, yield point, breaking point), and describe the experiment to verify Hooke's law with precautions. Spring-balance practicals are common.

Quick check

A spring of $k = 200$ N m⁻¹ is pulled by 10 N. Find its extension.
What does the gradient of a load–extension graph give?
What happens if a spring is loaded beyond its elastic limit?

Answers: 1. $e = 10 \div 200 = 0.05$ m (5 cm). 2. The spring constant $k$ (with load on the vertical axis). 3. It is permanently stretched and does not return to its original length.

Elasticity and Hooke's law