P18: Forces and Elasticity
Hooke's law, springs and elastic deformation
📋 Key Definitions
Elastic deformation: When a force is applied to an object and it changes shape, but returns to its original shape when the force is removed. Example: a spring stretching and then returning to its original length.
Plastic deformation: When a force is applied to an object and it changes shape permanently — it does NOT return to its original shape when the force is removed. Example: overstretching a spring so it stays stretched.
Hooke's Law: The extension of a spring is directly proportional to the force applied, provided the limit of proportionality is not exceeded.
Limit of proportionality: The maximum force at which extension is still proportional to force. Beyond this point, Hooke's Law no longer applies.
Spring constant (k): A measure of the stiffness of a spring. A higher spring constant means a stiffer spring (more force needed for the same extension). Measured in N/m.
📐 Hooke's Law — F = ke
F = ke
F = force (N), k = spring constant (N/m), e = extension (m)
Extension is the increase in length from the original, unstretched length: extension = stretched length − original length.
Hooke's Law means: If you double the force, the extension doubles. If you triple the force, the extension triples. This only works up to the limit of proportionality.
Worked Example 1: Calculating force
A spring has a spring constant of 200 N/m. Calculate the force needed to extend it by 0.05 m (5 cm).
F = ke = 200 × 0.05 = 10 N
Worked Example 2: Calculating extension
A force of 15 N is applied to a spring with spring constant 300 N/m. Calculate the extension.
e = F / k = 15 / 300 = 0.05 m (or 5 cm)
Worked Example 3: Calculating spring constant
A force of 8 N causes a spring to extend by 0.04 m. Calculate the spring constant.
k = F / e = 8 / 0.04 = 200 N/m
📊 Force-Extension Graphs
When you plot force (y-axis) against extension (x-axis) for a spring:
- The graph is a straight line through the origin — this shows Hooke's Law (F is proportional to e)
- The gradient of the straight line = spring constant k
- The graph curves at the limit of proportionality — beyond this, extension is no longer proportional to force
- If the force is removed before the limit of proportionality, the spring returns to its original length (elastic deformation)
- If the force exceeds the elastic limit, the spring is permanently deformed (plastic deformation)
| Point on graph | Meaning |
|---|---|
| Limit of proportionality | Maximum force where F ∝ e. Beyond this, the graph curves. |
| Elastic limit | Maximum force for elastic deformation. Beyond this, plastic deformation occurs. The spring will not return to its original length. |
⚡ Elastic Potential Energy
When a spring is stretched or compressed, it stores elastic potential energy.
Ee = ½ke²
Ee = elastic potential energy (J), k = spring constant (N/m), e = extension (m)
Ee = ½Fe
(This is the area under a force-extension graph up to extension e)
Worked Example 4: Calculating elastic potential energy
A spring with spring constant 400 N/m is extended by 0.1 m. Calculate the elastic potential energy stored.
Ee = ½ke² = ½ × 400 × 0.1² = ½ × 400 × 0.01 = 2 J
Worked Example 5: Energy using force and extension
A force of 20 N extends a spring by 0.05 m. Calculate the elastic potential energy stored.
Ee = ½Fe = ½ × 20 × 0.05 = 0.5 J
🧪 Required Practical: Investigating Springs
Aim
To investigate the relationship between force and extension for a spring, and determine the spring constant.
Method
- Set up a clamp stand with a spring hanging from it and a ruler alongside
- Measure the original length of the spring (with no mass attached)
- Hang a 1 N (100 g) mass on the spring and measure the new length
- Calculate the extension: extension = new length − original length
- Add masses in 1 N increments, measuring the new length each time
- Calculate the extension for each force
- Plot a graph of force (y-axis) against extension (x-axis)
Analysis
- The graph should be a straight line through the origin (Hooke's Law region)
- The gradient of the straight-line section = spring constant k
- The point where the line starts to curve = limit of proportionality
Safety: Wear eye protection. Do not exceed the elastic limit of the spring (it could snap). Stand up so you can move away quickly if the spring breaks.
Improving accuracy: Take readings at eye level to avoid parallax error. Measure the original length carefully before adding any mass. Repeat the experiment and calculate a mean.
❓ Practice Questions
Q1: Foundation State Hooke's Law and explain what the limit of proportionality is.
Q2: Foundation A spring has a spring constant of 250 N/m. Calculate the force needed to extend it by 0.08 m.
Q3: Higher A force of 12 N extends a spring by 0.06 m. Calculate the spring constant and the elastic potential energy stored.
Q4: Higher Describe how to find the spring constant from a force-extension graph.
Q5: Foundation Explain the difference between elastic deformation and plastic deformation.
✅ Answers
- Hooke's Law states that extension is directly proportional to the force applied, provided the limit of proportionality is not exceeded. The limit of proportionality is the maximum force at which this proportional relationship still holds. Beyond this point, extension increases more rapidly than force.
- F = ke = 250 × 0.08 = 20 N
- k = F / e = 12 / 0.06 = 200 N/m. Ee = ½ke² = ½ × 200 × 0.06² = ½ × 200 × 0.0036 = 0.36 J
- Plot force on the y-axis and extension on the x-axis. In the Hooke's Law region, the graph is a straight line through the origin. The spring constant k equals the gradient of this straight line (gradient = F/e = k).
- Elastic deformation: the object returns to its original shape when the force is removed. Plastic deformation: the object does NOT return to its original shape — the deformation is permanent.
🎯 Exam Tips
- Remember: extension is not the same as length. Extension = new length − original length
- The gradient of the force-extension graph = spring constant k (not the other way round)
- Hooke's Law only applies up to the limit of proportionality — always mention this condition
- For elastic potential energy, use E = ½ke² or E = ½Fe — note the ½ in both
- In the practical, describe measuring extension (not length), plotting the graph, and finding k from the gradient
- The elastic limit is different from the limit of proportionality — the elastic limit is where permanent deformation begins
🔬 Required Practical
Required Practical: Investigating Hooke's Law (Spring Extension)
Aim: To investigate the relationship between force and extension for a spring, and to determine the spring constant.
Method: 1. Set up a clamp stand with a spring hanging from it and a ruler aligned vertically alongside. 2. Measure and record the original length of the spring with no load. 3. Hang a 1 N (100 g) mass on the spring and measure the new length. 4. Calculate the extension = new length − original length. 5. Add masses in 1 N increments up to about 10 N, measuring the new length each time. 6. Calculate the extension for each force. 7. Plot a force-extension graph. 8. The gradient of the straight-line section = spring constant k.
Variables: IV: force applied (N), DV: extension (m), Control: same spring, same ruler, same ambient temperature, eye level for readings.
🔢 Maths Skills
Mathematical Skills
Use F = ke to calculate force, spring constant or extension. Use Ee = ½ke² or Ee = ½Fe for elastic potential energy. On a force-extension graph, gradient = spring constant k, and area under the line = elastic potential energy stored. Remember extension = stretched length − original length (not the total length).
Maths Example
A spring has k = 250 N/m and is stretched by 0.08 m. F = ke = 250 × 0.08 = 20 N. Energy stored: E = ½ke² = ½ × 250 × 0.08² = ½ × 250 × 0.0064 = 0.8 J. A second spring (k = 500 N/m) is stretched by the same force: e = F/k = 20/500 = 0.04 m. The stiffer spring extends half as much.
⚠️ Common Misconceptions
Watch Out!
1. Wrong: Extension is the same as the stretched length of the spring. Correct: Extension = stretched length MINUS the original length. It is the INCREASE in length, not the total length.
2. Wrong: Hooke's Law applies for all forces applied to a spring. Correct: Hooke's Law only applies up to the limit of proportionality. Beyond this point, extension is no longer proportional to force and the law breaks down.
3. Wrong: The limit of proportionality and the elastic limit are the same point. Correct: They are different. The limit of proportionality is where F ∝ e stops being true. The elastic limit is where permanent (plastic) deformation begins. The elastic limit is slightly beyond the limit of proportionality.
✍️ 6-Mark Question
Extended Answer
6 marks: Describe how you would investigate the relationship between force and extension for a spring. Explain how your results would show whether Hooke's Law is obeyed and how you would determine the spring constant.
Set up a clamp stand with a spring hanging from it and a ruler alongside. Measure the original length of the unstretched spring. Add a 1 N mass hanger and measure the new length. Calculate the extension by subtracting the original length from the new length. Add further 1 N masses one at a time, measuring the new length and calculating the extension each time, up to about 10 N. Record all results in a table. Plot a graph of force (y-axis) against extension (x-axis). If Hooke's Law is obeyed, the graph will be a straight line through the origin, showing that extension is directly proportional to force. The spring constant k equals the gradient of the straight-line section (gradient = ΔF/Δe = k). The point where the line starts to curve is the limit of proportionality, beyond which Hooke's Law no longer applies. To improve accuracy, take readings at eye level to avoid parallax error and repeat the experiment to calculate a mean.
Mark scheme: 1 mark for correct apparatus setup; 1 mark for method of adding masses and measuring extension; 1 mark for plotting force-extension graph; 1 mark for straight line through origin confirms Hooke's Law; 1 mark for gradient = spring constant k; 1 mark for identifying limit of proportionality where line curves. (6 marks total)
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
A student investigates two springs, A and B. Their force-extension data is shown below.
| Force (N) | Extension A (cm) | Extension B (cm) |
|---|---|---|
| 0 | 0 | 0 |
| 2 | 1.0 | 0.5 |
| 4 | 2.0 | 1.0 |
| 6 | 3.0 | 1.5 |
| 8 | 4.0 | 2.0 |
| 10 | 5.0 | 2.5 |
| 12 | 6.5 | 3.0 |
(a) Calculate the spring constant for each spring using the straight-line data.
(b) At what force does Spring A exceed its limit of proportionality? Explain your answer.
(c) Both springs are used in a suspension system that must not extend more than 4 cm under an 8 N load. Evaluate which spring is more suitable and explain your reasoning.
Answers: (a) Spring A: k = F/e = 2 N / 0.01 m = 200 N/m (using linear data up to 10 N). Spring B: k = 2 N / 0.005 m = 400 N/m. (b) Between 10 N and 12 N, Spring A's extension increases from 5.0 cm to 6.5 cm — a 1.5 cm increase compared to the previous 1.0 cm increase per 2 N. This shows the extension is no longer proportional to force, so the limit of proportionality is exceeded at 10 N. (c) At 8 N: Spring A extends 4.0 cm (exactly at the limit), Spring B extends 2.0 cm (well within limit). Spring B is more suitable because it stays within the 4 cm limit with a good safety margin. Spring A reaches exactly 4 cm, leaving no margin for additional loads or the non-linear region just beyond.
📝 Exam Questions by Topic
🎬 Video Resources
Ready to ace your GCSE Combined Science exams?
Get the best revision books and guides to boost your grades.