Forces and ElasticityAQA GCSE Physics: Revision notes
Section 1
What is the difference between elastic and inelastic deformation?
Elastic deformation occurs when an object returns to its original shape and size after the deforming force is removed. The material has not been permanently changed, and the deformation is temporary.
Inelastic deformation occurs when an object does not return to its original shape and size after the deforming force is removed. The material is permanently changed and will not recover its original form.
| Type | Returns to original shape? | Example | Energy outcome |
|---|---|---|---|
| Elastic | Yes | Stretching a rubber band (within limits) | Stored as elastic potential energy |
| Inelastic | No | Bending a copper wire | Converted to heat and sound |
Most materials exhibit elastic behaviour up to a certain point (called the limit of proportionality), beyond which they begin to show inelastic behaviour.
Think of elastic deformation like a sponge – it squashes and returns to its original shape. Inelastic deformation is like plasticine – it stays deformed even after you stop pressing.
Section 2
What is Hooke's Law and how do we apply it?
Hooke's Law states that the force applied to a spring is directly proportional to its extension, provided the limit of proportionality is not exceeded.
This relationship is expressed by the equation:
F = ke
Where:
- F = force applied (in Newtons, N)
- k = spring constant (in N/m)
- e = extension (in metres, m)
Key points:
- The extension is the increase in length from the original length
- The spring constant k is a measure of the stiffness of the spring – a larger k value means a stiffer spring that requires more force to stretch it
- This law only applies within the elastic limit – once exceeded, the relationship is no longer linear
- The equation can also be rearranged to find k = F/e
Examiners expect you to state Hooke's Law as a word equation AND give the symbolic form F = ke. When using the equation, always substitute units to show you understand what each variable represents.
A spring has a spring constant of 50 N/m. If a force of 10 N is applied, the extension is: e = F/k = 10/50 = 0.2 m (or 20 cm). This shows the extension is proportional to the force applied.
Section 3
How do we determine spring constant from a force–extension graph?
A force–extension graph is plotted with force (N) on the vertical axis and extension (m) on the horizontal axis.
Reading the spring constant from the graph:
- The graph shows a straight-line relationship through the origin up to the limit of proportionality
- The gradient (slope) of the straight-line section equals the spring constant k
- To calculate the gradient, use: k = ΔF / Δe (change in force divided by change in extension)
- Choose two points on the straight-line section that are far apart for accuracy
- The spring constant is measured in N/m
Beyond the limit of proportionality:
- The graph curves upwards (becomes non-linear)
- The relationship is no longer directly proportional
- Hooke's Law no longer applies
- The material is beginning to undergo inelastic deformation
Many students read the spring constant from a single pair of coordinates (e.g., F = 5 N, e = 0.1 m) without calculating the gradient. Always use two well-separated points and calculate ΔF/Δe to avoid errors.
If a force–extension graph shows (0, 0) and (10 N, 0.2 m) on the straight-line section, the spring constant is: k = 10/0.2 = 50 N/m. This means 50 N is required to extend the spring by 1 m.
Section 4
What is the experiment to investigate force and extension?
Aim: To investigate the relationship between force applied to a spring and its extension, and to determine the spring constant.
Equipment:
- Spring
- Ruler (or measuring scale)
- Masses and mass hanger
- Clamp and stand (to hold the spring vertically)
- Pointer or fixed reference marker
Method:
- Clamp the spring vertically and attach a pointer at its bottom to mark the initial length
- Record the initial length of the spring (e = 0)
- Add a mass to the hanger and allow the system to reach equilibrium
- Measure the new length of the spring and calculate the extension (new length – original length)
- Record the force applied (F = mg, where g = 10 m/s²)
- Repeat by adding further masses in equal increments (e.g., 50 g each time)
- Record extension and corresponding force for each mass addition
- Continue until the spring reaches its limit of proportionality or deforms visibly
- Plot a force–extension graph with force on the vertical axis and extension on the horizontal axis
- Identify the straight-line region and calculate the spring constant from the gradient
Safety and accuracy considerations:
- Ensure the spring is not overloaded beyond its elastic limit
- Use a reference point (pointer) to ensure accurate extension measurements
- Allow time for equilibrium after each mass addition
- Repeat measurements for reliability
Examiners want to see that you understand why measurements must be taken from a fixed reference point (the initial length) and that you know F = mg is the force calculation. State that the spring must reach equilibrium before reading the extension.
Section 5
What is the significance of the limit of proportionality on a force–extension graph?
The limit of proportionality is a crucial point on a force–extension graph where the linear relationship between force and extension breaks down.
Significance:
-
Below the limit of proportionality: The material obeys Hooke's Law; force and extension are directly proportional; the graph is a straight line through the origin; the material will fully recover its original shape when the force is removed (elastic behaviour)
-
At the limit of proportionality: This is the last point where F = ke applies accurately; beyond this point, the relationship is no longer linear
-
Beyond the limit of proportionality: The graph curves upwards; Hooke's Law no longer applies; for the same increase in force, the extension increases by a larger amount; the material begins to undergo permanent (inelastic) deformation; when the force is removed, the material will not fully return to its original length
Key distinction:
- The limit of proportionality is where the straight-line section ends
- The elastic limit is slightly beyond this point – the maximum stress beyond which permanent deformation occurs
- In GCSE-level questions, these terms are often used interchangeably, but the limit of proportionality is the point where the graph deviates from linearity
When describing the significance, examiners expect you to explain what happens to the graph (it stops being straight) and the material (it starts to show inelastic deformation). Always link the graph appearance to the physical behaviour of the material.
The limit of proportionality is like the point where a rubber band goes from being 'recoverable' to 'permanently stretched' – once you pass it, the rubber band won't snap back properly.
Section 6
How do we calculate elastic potential energy stored in a spring? (Higher Tier)
When a spring is stretched or compressed, it stores elastic potential energy. This energy is released when the spring returns to its original length.
Elastic Potential Energy Equation:
Ee = ½ke²
Where:
- Ee = elastic potential energy (in Joules, J)
- k = spring constant (in N/m)
- e = extension (in metres, m)
Key points:
- This equation is only valid within the elastic limit where Hooke's Law applies
- The energy stored is proportional to the square of the extension – doubling the extension stores four times the energy
- The energy is stored as elastic potential energy in the bonds between atoms in the material
- When the spring is released, this potential energy is converted to kinetic energy (if the spring is free to move) or dissipated as heat and sound (if the spring is held fixed)
Alternative form (using Hooke's Law): Since F = ke, we can also express elastic potential energy as:
Ee = ½Fe (where F is the force at maximum extension)
This shows that energy depends on both the force applied and the extension achieved.
A spring with spring constant k = 200 N/m is stretched by e = 0.15 m. The elastic potential energy stored is: Ee = ½ × 200 × (0.15)² = ½ × 200 × 0.0225 = 2.25 J. Note: the extension must be squared, making the energy very sensitive to changes in extension.
Students often forget to square the extension in the equation. Ee = ½ke² means k multiplied by e², not (½ke) × e. Always use brackets or calculate e² first to avoid this error.
Must Know
- Elastic deformation is reversible (object returns to original shape); inelastic deformation is permanent (object does not recover original shape)
- Hooke's Law: F = ke – force is directly proportional to extension, provided the limit of proportionality is not exceeded
- Spring constant (k) is the gradient of the force–extension graph within the straight-line section; k is measured in N/m and indicates how stiff the spring is
- The limit of proportionality is where the force–extension graph stops being straight; beyond this point, Hooke's Law no longer applies and the material begins inelastic deformation
- Elastic potential energy: Ee = ½ke² – energy stored in a stretched spring is proportional to the square of the extension and is only valid within the elastic limit (Higher Tier)
- In the experiment: measure extension from a fixed reference point, use F = mg to calculate force, and plot force (vertical) against extension (horizontal) to find k from the gradient
That's the notes covered.
Carry on to the next subtopic.