Forces, Elasticity and Hooke’s Law

Stretch an elastic band and it springs back; bend a paperclip too far and it stays bent. Pull gently on a spring and it lengthens neatly in proportion to your pull; pull too hard and it never returns to its original shape. The way materials respond when forces stretch, squash or bend them is described by elasticity and, for many springs, by a simple rule called Hooke's law. This lesson, part of Topic P2 (Forces) of OCR Gateway Science A, distinguishes elastic from inelastic deformation, sets out Hooke's law $F = ke$F=ke, introduces the limit of proportionality, walks through the required practical on stretching a spring, and (Higher tier) calculates the elastic potential energy stored.

By the end of this lesson you should be able to distinguish elastic from inelastic deformation, state and use Hooke's law $F = ke$F=ke, explain the limit of proportionality, describe the required practical for a spring, interpret force–extension graphs, and (Higher tier) calculate elastic potential energy with $E = \dfrac{1}{2}ke^2$E=21​ke2.

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Elastic and Inelastic Deformation

To deform an object means to change its shape by stretching, squashing, bending or twisting it. This requires more than one force — a single force would simply make the object accelerate; you need at least two forces (for example, pulling both ends of a spring, or pressing a ball against a table) to change its shape.

There are two kinds of deformation:

• Elastic deformation: the object returns to its original shape when the forces are removed. A stretched spring, a squashed rubber ball and a bent ruler (within limits) all spring back — they have deformed elastically.
• Inelastic (plastic) deformation: the object does not return to its original shape; it stays permanently deformed. A bent paperclip, squashed modelling clay and an over-stretched spring have deformed inelastically.

Whether a deformation is elastic or inelastic depends partly on the material and partly on how big the force is: many objects deform elastically for small forces but inelastically once the force becomes too large.

Exam Tip: Elastic deformation reverses when the force is removed (the object springs back); inelastic deformation is permanent (the object stays deformed). Remember that deforming an object needs more than one force, otherwise it would just move.

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Hooke’s Law

For many springs and elastic objects, the extension is directly proportional to the force applied — this is Hooke's law:

$F = ke$F=ke

where $F$F is the force applied (in $\text{N}$N), $k$k is the spring constant (in $\text{N/m}$N/m) and $e$e is the extension (in $\text{m}$m). The extension is the increase in length — the stretched length minus the original (natural) length — not the total length.

The spring constant $k$k measures the stiffness of the spring: a large $k$k means a stiff spring that needs a big force for a small extension, while a small $k$k means a spring that stretches easily. Its unit, $\text{N/m}$N/m, tells you the force needed per metre of extension.

Hooke's law rearranges to:

$F = ke \qquad k = \dfrac{F}{e} \qquad e = \dfrac{F}{k}$F=kek=eF​e=kF​

Worked Example 1 — Finding the force

A spring has a spring constant of $25 \text{ N/m}$25 N/m and is stretched by an extension of $0.08 \text{ m}$0.08 m. Calculate the force applied.

Step 1 — write Hooke's law: $F = ke$F=ke.

Step 2 — substitute: $F = 25 \times 0.08$F=25×0.08.

Step 3 — calculate: $F = 2 \text{ N}$F=2 N.

Answer: the force applied is $2 \text{ N}$2 N.

Worked Example 2 — Finding the spring constant

A force of $12 \text{ N}$12 N stretches a spring by $0.06 \text{ m}$0.06 m. Calculate the spring constant.

Step 1 — rearrange for $k$k: $k = \dfrac{F}{e}$k=eF​.

Step 2 — substitute: $k = \dfrac{12}{0.06}$k=0.0612​.

Step 3 — calculate: $k = 200 \text{ N/m}$k=200 N/m.

Answer: the spring constant is $200 \text{ N/m}$200 N/m.

Worked Example 3 — Finding the extension

A spring of spring constant $300 \text{ N/m}$300 N/m has a force of $30 \text{ N}$30 N applied to it. Calculate the extension.

Step 1 — rearrange for $e$e: $e = \dfrac{F}{k}$e=kF​.

Step 2 — substitute: $e = \dfrac{30}{300}$e=30030​.

Step 3 — calculate: $e = 0.1 \text{ m}$e=0.1 m.

Answer: the extension is $0.1 \text{ m}$0.1 m ($10 \text{ cm}$10 cm).

Exam Tip: In $F = ke$F=ke, the $e$e is the extension (the increase in length), not the total length — always subtract the natural length first. Keep the extension in metres so the spring constant comes out in $\text{N/m}$N/m.

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The Limit of Proportionality

Hooke's law does not hold for all forces. If you keep increasing the force, you eventually reach the limit of proportionality — the point beyond which the extension is no longer proportional to the force. Below this limit, doubling the force doubles the extension (the spring obeys Hooke's law); above it, the spring extends more for each extra newton, and the relationship breaks down.

On a force–extension graph (force on the vertical axis, extension on the horizontal axis), Hooke's law gives a straight line through the origin up to the limit of proportionality. Beyond that point the line curves, bending toward the extension axis as the spring stretches more easily.

On the straight part of the graph the gradient is the spring constant $k$k (force ÷ extension). If a spring is stretched beyond its limit of proportionality and then beyond its elastic limit, it deforms inelastically and will not return to its original length.

Exam Tip: Up to the limit of proportionality, the force–extension graph is a straight line through the origin and the spring obeys $F = ke$F=ke; beyond it, the line curves and the extension is no longer proportional to the force. On the straight part, the gradient equals the spring constant.

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Required Practical: Force and Extension for a Spring

A key P2 required practical is to investigate the relationship between the force applied to a spring and its extension, and so find its spring constant and limit of proportionality.

Method (numbered):

1. Clamp a spring vertically from a stand, with a ruler fixed alongside it. Record the original (unstretched) length of the spring (or the starting position of the pointer).
2. Hang a mass hanger from the spring and add known masses (e.g. $100 \text{ g}$100 g at a time). Each $100 \text{ g}$100 g mass adds a known weight (force) of about $0.98 \text{ N}$0.98 N (using $W = mg$W=mg).
3. After adding each mass, record the new length of the spring from the ruler, reading at eye level to avoid parallax error.
4. Calculate the extension for each load: extension = new length − original length.
5. Repeat the readings (removing the masses and re-loading) and take a mean. Continue adding masses until the spring clearly stops obeying Hooke's law.
6. Plot a graph of force (weight) against extension. The straight part shows Hooke's law; its gradient gives the spring constant; the point where it starts to curve is the limit of proportionality.