Hooke's Law and Elasticity
This topic covers how materials stretch under force, distinguishing between elastic (spring-like) and inelastic (permanent) deformation. It involves interpreting force-extension graphs, applying Hooke's Law, and calculating the energy stored in stretched objects.
Part of the ESAT Physics syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- The gradient of a linear force-extension graph represents the spring constant, 'k'. A steeper gradient means a stiffer material.
- The area under a force-extension graph represents the work done on the material. For an elastic material, this is stored as elastic potential energy.
- Elastic deformation is temporary; the material returns to its original length when the force is removed. Inelastic (or plastic) deformation is permanent.
- Hooke's Law is only valid up to the limit of proportionality, where the force-extension graph is a straight line through the origin.F = kx
- If a material is stretched beyond its elastic limit, not all the work done to stretch it is recovered when the force is removed; some energy is lost, usually as heat.
Diagram
› Why does this happen?
Why the graph tells you everything
The two most important features of a force-extension graph are its gradient (how steep it is) and the area underneath it.
- Gradient = Stiffness: The gradient is calculated as 'rise / run', which for this graph is Force / Extension. Since the spring constant 'k' is defined by k = F / x, the gradient of the straight-line section is a direct measure of k. A steeper line means a bigger force is needed for the same extension, so the material is stiffer.
- Area = Energy Stored: Work is done when a force stretches a material. Because the force isn't constant (it increases as you stretch it), the work done is equal to the area under the force-extension graph. For the linear part of the graph, this area is a triangle. So, the elastic potential energy stored is E = Area = 1/2 × base × height = 1/2 × extension × force. We write this as E = 1/2 × F × x.
What's happening inside the material?
Imagine the atoms in a solid are connected by tiny springs (these are the chemical bonds between them).
- Elastic Stretching: When you apply a force, you are stretching these atomic bonds, pulling the atoms slightly further apart. When you remove the force, the bonds pull the atoms back to their original positions, just like a spring recoiling. This temporary change is called elastic deformation. Hooke's Law works here because the bonds stretch in a predictable, linear way.
- Permanent Stretching: If you pull too hard (beyond the elastic limit), whole layers of atoms slide over each other into new positions. When you release the force, they are now bonded to new neighbours and don't slide back. This permanent change is called inelastic (or plastic) deformation. The energy used to move these layers is converted into heat.
Formulae
F = k x To relate force (F), spring constant (k), and extension (x) for a material obeying Hooke's Law (i.e., within its limit of proportionality).
E = 1/2 F x To calculate the elastic potential energy (E) stored in a stretched material when you know the final force (F) and extension (x).
E = 1/2 k x2 To calculate the elastic potential energy (E) when you know the spring constant (k) and the extension (x). This is derived by substituting F = kx into E = 1/2 Fx.
Definitions
- Limit of Proportionality
- The point on a force-extension graph beyond which the extension is no longer directly proportional to the applied force. The graph ceases to be a straight line.
- Elastic Limit
- The maximum force that can be applied to a material without causing permanent deformation (inelastic extension). Often very close to the limit of proportionality.
- Spring Constant (k)
- A measure of the stiffness of an object. It is the force required to produce a unit extension. Its units are Newtons per metre (Nm⁻¹).
- Extension (x)
- The increase in length of an object from its original, unloaded length when a tensile force is applied.
Worked example
A spring with a spring constant of 500 Nm⁻¹ is hung vertically. An object of mass 2 kg is attached to it and gently lowered. How much elastic potential energy is stored in the spring? (Assume g = 10 N/kg)
- 1
First, calculate the force acting on the spring, which is the weight of the object:
F = m g = 2 kg × 10 N/kg = 20 N - 2
Next, use Hooke's Law to find the extension of the spring:
F = kx, so x = F / k - 3
Substitute the values:
x = 20 N / 500 Nm⁻¹ = 2 / 50 m = 0.04 m - 4
Finally, calculate the stored energy.
Using E = 1/2 Fx is simplest here:E = 1/2 × 20 N × 0.04 m - 5
Calculate the result:
E = 10 × 0.04 J = 0.4 J
Answer: 0.4 J
Common mistakes
- ×Forgetting the '1/2' in the energy formula (E = 1/2 kx²). This is a very frequent source of 'off by a factor of 2' errors.
- ×Mixing up units. The spring constant is often in Nm⁻¹, so extension must be converted from cm or mm into metres before calculating energy in Joules.
- ×Confusing the force-extension graph with a velocity-time graph. The gradient is stiffness (k), not acceleration, and the area is energy, not displacement.
- ×Applying Hooke's Law or the energy formulae beyond the limit of proportionality where they are no longer valid.
No-calculator tips
- ✓When calculating energy with E = 1/2 kx², it's often easier to square the extension 'x' first, especially if it's a simple decimal like 0.1 or 0.2, before multiplying by k.
- ✓To calculate a spring constant k = F/x where x is in cm, e.g., F=30N, x=5cm. Calculate 30/5 = 6 N/cm. Then convert to N/m by multiplying by 100: 6 × 100 = 600 Nm⁻¹.
- ✓When dealing with parallel or series springs, remember the rules are the 'opposite' of resistors. Parallel springs add up (ktotal = k1 + k2), making the system stiffer. Series springs add reciprocally (1/ktotal = 1/k1 + 1/k2), making the system less stiff.