What is the relationship between force and elasticity in Physics?

In Physics, particularly at the AQA GCSE level, the relationship between force and elasticity is described by Hooke's Law. This law states that the extension of an elastic object, like a spring, is directly proportional to the force applied to it, provided the limit of proportionality is not exceeded. This means that doubling the force will double the extension, as long as the material remains elastic.

How does Hooke's Law apply to real-world scenarios?

Hooke's Law is applicable in various real-world scenarios, such as in the design of springs used in vehicles, mattresses, and measuring devices like spring scales. Understanding this law helps engineers and designers ensure that materials and structures can withstand forces without permanent deformation, which is crucial for safety and functionality.

What is meant by the 'limit of proportionality' in the context of elasticity?

The 'limit of proportionality' refers to the maximum point up to which Hooke's Law is valid. Beyond this point, the material no longer returns to its original shape after the force is removed, indicating that it has been permanently deformed. This is an important concept in AQA GCSE Physics as it helps in understanding the limits of material elasticity.

How can you calculate the spring constant in an experiment?

To calculate the spring constant (k) in an experiment, you can use the formula derived from Hooke's Law: F = kx, where F is the force applied, and x is the extension of the spring. By measuring the force applied and the corresponding extension, you can rearrange the formula to k = F/x to find the spring constant. This calculation is a key part of the Forces and Elasticity topic in AQA GCSE Physics.

What are some examples of elastic and inelastic materials?

Elastic materials, such as rubber bands and springs, return to their original shape after the force is removed, as long as they are not stretched beyond their elastic limit. In contrast, inelastic materials, like clay or putty, do not return to their original shape after deformation. Understanding these differences is crucial for AQA GCSE Physics students studying the properties of materials under the Forces and Elasticity subtopic.

Force vs extension

At a glance

$F = k \, e$ — Hooke's law (force ∝ extension within limit of proportionality).
$k$ = spring constant (N/m); stiff springs have large k.
Elastic deformation: returns to original shape when force is removed.
Inelastic deformation: permanent change in shape.
$E_e = \tfrac{1}{2}ke^2$ — energy stored in a stretched spring.
Required Practical 6: investigate F vs e for a spring.

What you should be able to do

4.5.3 — State and apply Hooke's law $F = ke$ .
4.5.3 — Distinguish elastic from inelastic deformation.
4.5.3 — Calculate energy stored in a spring using $E_e = \tfrac{1}{2}ke^2$ .
4.5.3 / RP6 — Describe the force-extension experiment.

Study notes

Hooke's law: F = ke

Force proportional to extension, until you exceed the limit of proportionality.

Hooke's law: the extension of a spring is directly proportional to the force applied, provided the spring is within its limit of proportionality.

$F = k \, e$

$F$ in N.
$k$ in N/m (spring constant).
$e$ in m (extension).

Extension = new length − natural length. (NOT the new length.)

Worked example. A spring of $k = 250$ N/m is pulled with 5 N. Find extension. $e = F/k = 5/250 = 0.020\,\text{m} = 2.0\,\text{cm}$

Beyond the limit of proportionality, the F vs e graph stops being linear — the spring stretches MORE for each additional newton. Eventually the spring may deform permanently (inelastic).

F = ke (within limit).
k = stiffness in N/m.
e is the EXTENSION (change in length).

Elastic vs inelastic deformation

Elastic = returns; inelastic = permanent change.

Elastic deformation: the spring/material returns to its original shape and length when the force is removed. F vs e graph is a STRAIGHT line through the origin.

Inelastic deformation: the material is permanently stretched/squashed. After the load is removed, it does NOT return to the original length. The F vs e graph curves AWAY from straight and may flatten.

For example, stretching a coil spring with a small load: elastic. Stretching too far: inelastic and the coils never close back together.

Straight line = elastic (Hooke's law obeyed). Curve = inelastic deformation.

Elastic returns; inelastic deforms permanently.
Straight line = elastic; curve = inelastic.
Limit of proportionality marks the boundary.

Energy stored: $E_e = \tfrac{1}{2}ke^2$

Work done stretching a spring = elastic PE stored in it.

Stretching a spring by extension $e$ requires work because the force needed grows from 0 to $F = ke$ . Average force is $\tfrac{1}{2}ke$ , so:

$E_e = \frac{1}{2}\,k\,e^2$

This is on AQA's equation sheet (since 2024).

Worked example. A spring with $k = 100$ N/m is stretched 6 cm.

$e = 0.06\,\text{m}$ .
$E_e = \tfrac{1}{2} \times 100 \times 0.06^2 = 50 \times 0.0036 = 0.18\,\text{J}$ .

This energy is released (e.g. as kinetic energy of a launched object) when the spring is released — provided the deformation was elastic.

$E_e = \tfrac{1}{2}ke^2$ — on equation sheet.
Comes from average force × extension.
Use only within elastic limit.

RP6 — investigating F vs e

Hang masses on a spring; measure extension; plot graph; find k.

Apparatus: clamp stand, spring, ruler (clamped vertically), set of masses (e.g. 100 g pieces), pointer/eye-level marker on the spring.

Method:

Measure the natural length of the spring (no load).
Hang a known mass; weight = mg.
Measure the new length.
Calculate extension = new length − natural length.
Repeat for at least 5 different masses.
Plot F (y-axis) against e (x-axis).
The straight-line region gives k as the gradient (slope = F/e = k).

Sources of error:

Parallax when reading the ruler.
Spring oscillation when adding masses → wait for it to settle.
Exceeding limit of proportionality → take care not to over-stretch.

Independent variable: force (= weight applied). Dependent: extension. Control: same spring; consistent ruler position.

F vs e graph: gradient = k.
Identify the linear (elastic) region.
Wait for oscillation to settle.

Quick recap

F = ke within limit of proportionality.
k = spring constant (N/m).
Elastic deformation: returns; inelastic: permanent.
$E_e = \tfrac{1}{2}ke^2$ (on equation sheet).
RP6: hang weights on a spring, plot F vs e, gradient = k.

Exam tips

Convert cm → m before using F = ke.
e = NEW LENGTH − NATURAL LENGTH (not just new length).
On RP6, mention 'wait for oscillation to settle' for a procedure mark.
Sketch F-e graph carefully: straight + curve after limit of proportionality.

Frequently asked questions

Sources: AQA GCSE Physics 8463 specification — section 4.5.3; AQA Required Practical 6 handbook · Last reviewed 2026-05-25 · AQA GCSE · Physics 8463 Higher & Foundation Tier

Forces and Elasticity