Forces and Elasticity

To stretch, bend or compress an object you need to apply force, but one force alone is not enough.

To change the shape of a stationary object by stretching, bending or compressing, more than one force must act on it.

If only a single force acted, the object would simply move off in the direction of that force rather than change shape. To deform it, forces must act in more than one place. Pulling a spring means holding one end while pulling the other; squashing a sponge means pushing from both sides.

Type of deformation	What the forces do	Example
Stretching	Pull the ends apart	Stretching an elastic band
Compressing	Push the ends together	Squashing a spring in a pen
Bending	Push and pull across the object	Bending a ruler over an edge

In each case, at least two forces act together to hold and deform the object at the same time.

What happens when the force is removed tells you which kind of deformation occurred.

Elastic deformation: the object returns to its original shape once the deforming force is removed.

Inelastic (plastic) deformation: the object stays permanently changed in shape after the force is removed.

A steel spring stretched gently springs back to its original length: that is elastic. A metal paperclip bent right open stays bent: that is inelastic. The same object can behave either way depending on how far it is deformed. Stretch a spring a little and it returns; stretch it too far and it stays permanently longer, because it has passed its elastic limit.

Feature	Elastic deformation	Inelastic deformation
Shape after force removed	Returns to original	Permanently changed
Example	Spring gently stretched	Spring stretched too far
Energy stored	Elastic potential energy, recoverable	Not fully recovered

For a spring, the amount it stretches is linked to the force applied, up to a point.

Extension is directly proportional to force, provided the limit of proportionality is not exceeded.

Extension, $e$, is the increase in length from the spring's natural (unstretched) length. This relationship is written as:

$F=ke$

where $F$ is the force in newtons (N), $k$ is the spring constant in newtons per metre (N/m) and $e$ is the extension in metres (m).

You must recall and apply this equation.

The spring constant $k$ measures stiffness: a large $k$ means a stiff spring that needs a big force for a small extension. A small $k$ means a spring that stretches easily.

Worked example — a spring with a spring constant of 25 N/m is stretched by an extension of 0.08 m. Calculate the force applied.

$F=ke=25\times 0.08=2.0\text{N}$

The force applied is 2.0 N.

Rearranging $F=ke$ lets you find the spring constant or the extension.

To find the spring constant:

$k=\frac{F}{e}$

Worked example — a force of 6.0 N stretches a spring by 0.15 m, staying below the limit of proportionality. Calculate the spring constant.

$k=\frac{F}{e}=\frac{6.0}{0.15}=40\text{N/m}$

The spring constant is 40 N/m.

To find the extension:

$e=\frac{F}{k}$

Worked example — a spring with a spring constant of 50 N/m has a force of 12 N applied. Calculate the extension.

$e=\frac{F}{k}=\frac{12}{50}=0.24\text{m}$

The extension is 0.24 m. Extension must be in metres, so convert from centimetres or millimetres before substituting.

Plotting force against extension shows exactly when $F=ke$ holds.

Up to the limit of proportionality, the graph is a straight line through the origin (linear). Beyond it, the line curves (non-linear) and extension is no longer proportional to force.

Region	Shape of graph	Is $F=ke$ valid?
Below the limit of proportionality	Straight line through origin	Yes
Above the limit of proportionality	Curved	No

While the graph is linear, its gradient equals the spring constant $k$, because $F=ke$ rearranges to $k=F/e$. Once past the limit of proportionality, equal increases in force produce larger and larger extensions, so the line bends away from the straight part. Reading the extension at a given force is only reliable in the linear region.

Stretching a spring means doing work, and that work is stored in the spring.

The work done to stretch (or compress) an elastic object, up to the limit of proportionality, is stored as elastic potential energy.

Provided the spring is not stretched past the limit of proportionality, all the work done goes into elastic potential energy and is fully recovered when the spring is released. This energy is calculated with:

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

where $E_e$ is elastic potential energy in joules (J), $k$ is the spring constant in N/m and $e$ is the extension in metres (m).

This equation is given on the Physics equation sheet; you do not need to recall it.

Worked example — a spring with a spring constant of 200 N/m is stretched by 0.10 m, within the limit of proportionality. Calculate the elastic potential energy stored.

$E_e=\frac{1}{2}k{e}^2=\frac{1}{2}\times 200\times (0.10{)}^2=\frac{1}{2}\times 200\times 0.010=1.0\text{J}$

The elastic potential energy stored is 1.0 J. Note the extension is squared, so doubling the extension quadruples the stored energy.

This required practical investigates the relationship between the force applied to a spring and its extension.

Apparatus: a spring hung from a clamp stand and boss, a metre ruler (or millimetre rule) fixed vertically alongside, a set of slotted masses on a hanger, and a set square to read the ruler accurately.

Variables:

• Independent variable: the force applied (changed by adding known masses, where weight = mass × g).
• Dependent variable: the extension of the spring (length now minus original length).
• Control variables: the same spring throughout and the same starting (natural) length.

Method:

1. Hang the spring from the clamp and record its natural length with no masses attached.
2. Add a known mass, wait for the spring to settle, and measure the new length. Calculate the force from the added weight ($W=mg$) and the extension from the change in length.
3. Add masses one at a time, recording force and extension each time, then remove them one at a time to check the readings.
4. Plot a graph of force (y-axis) against extension (x-axis).

Why each step is done: the ruler is vertical and read with a set square at eye level to avoid parallax error; masses are added gradually so the spring is not taken past its limit of proportionality and permanently stretched; repeating on the way down checks the deformation was elastic.

Expected result: the graph is a straight line through the origin while extension is proportional to force. Its gradient equals the spring constant, $k$. If masses are added too far, the line curves at the top, marking the limit of proportionality.

1. Using extension in centimetres or millimetres

In $F=ke$ and $E_e=\frac{1}{2}k{e}^2$, extension must be in metres. A 5 cm extension is 0.05 m; forgetting to convert gives an answer out by a large factor.

2. Using the total length instead of the extension

Extension is the increase in length, not the full stretched length. Subtract the natural length from the new length before using $F=ke$.

3. Forgetting to square the extension in the energy equation

$E_e=\frac{1}{2}k{e}^2$ squares the extension. A common slip is to multiply by $e$ once instead of $e^2$, which gives the wrong energy.

4. Applying $F=ke$ beyond the limit of proportionality

$F=ke$ only holds while the force–extension graph is a straight line. Past the limit of proportionality the relationship no longer applies, so do not use the equation there.

5. Confusing the spring constant with the extension

The spring constant $k$ is a fixed property of the spring (its stiffness in N/m). The extension $e$ changes with the force applied. They are different quantities with different units.