P19: Forces and Elasticity
Covering Hooke's law, the spring constant, elastic and plastic deformation, the limit of proportionality, elastic potential energy, and the required practical on spring extension.
Elastic and Plastic Deformation
Elastic deformation occurs when an object returns to its original shape after the force is removed. Plastic deformation occurs when an object does not return to its original shape after the force is removed - it is permanently deformed.
Examples of elastic deformation:
- Stretching a rubber band and releasing it
- Compressing a spring and letting it return
- Bending a ruler slightly and letting go
Examples of plastic deformation:
- Bending a metal paperclip so it stays bent
- Stretching a piece of Plasticine or clay
- Crushing a drinks can
A material that can return to its original shape after deformation is called elastic. A material that stays deformed is called plastic. Most materials behave elastically up to a certain force and then deform plastically.
Hooke's Law
Force = spring constant × extension
F = ke
F = force (N), k = spring constant (N/m), e = extension (m)
Hooke's law states that the extension of a spring is directly proportional to the force applied, provided the limit of proportionality is not exceeded.
The extension is the increase in length from the original (unstretched) length:
Extension = stretched length - original length
e = L - L₀
Worked Example
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
A force of 15 N extends a spring by 0.03 m. Calculate the spring constant.
k = F / e = 15 / 0.03 = 500 N/m
Worked Example
A spring with spring constant 400 N/m has a force of 24 N applied. Calculate the extension.
e = F / k = 24 / 400 = 0.06 m = 6 cm
Limit of Proportionality
The limit of proportionality is the maximum force that can be applied to a spring while still obeying Hooke's law. Beyond this point, the extension is no longer proportional to the force and the graph begins to curve.
On a force-extension graph:
- The straight-line section through the origin shows Hooke's law being obeyed (F ∝ e)
- The gradient of the straight section equals the spring constant k
- At the limit of proportionality, the line starts to curve
- Beyond the elastic limit, the spring deforms plastically and will not return to its original length
The elastic limit is usually slightly beyond the limit of proportionality. Between these two points, the spring no longer obeys Hooke's law but can still return to its original shape when the force is removed. Beyond the elastic limit, permanent deformation occurs.
On a force-extension graph, you can find the spring constant from the gradient of the straight section: k = F/e = gradient. Only use data from the straight-line section to calculate k.
Spring Constant
The spring constant (k) is a measure of the stiffness of a spring. A higher spring constant means the spring is stiffer and requires more force to produce the same extension.
| Spring type | Spring constant | Behaviour |
|---|---|---|
| Stiff spring | High k | Small extension for a given force |
| Soft spring | Low k | Large extension for a given force |
Worked Example
Spring A has k = 100 N/m and spring B has k = 300 N/m. A force of 6 N is applied to each. Compare the extensions.
Spring A: e = F/k = 6/100 = 0.06 m = 6 cm
Spring B: e = F/k = 6/300 = 0.02 m = 2 cm
Spring B is three times stiffer, so it extends one-third as much.
Elastic Potential Energy
Elastic potential energy = ½ × spring constant × extension²
Eₑ = ½ke²
Eₑ = elastic potential energy (J), k = spring constant (N/m), e = extension (m)
Elastic potential energy is the energy stored in a stretched or compressed spring. The work done in stretching a spring (within the limit of proportionality) is equal to the elastic potential energy stored.
Worked Example
A spring with k = 500 N/m is extended by 0.1 m. Calculate the elastic potential energy stored.
Eₑ = ½ke² = 0.5 × 500 × 0.1² = 0.5 × 500 × 0.01 = 2.5 J
Worked Example
A spring is compressed by 0.04 m. The elastic potential energy stored is 0.32 J. Calculate the spring constant.
Eₑ = ½ke² → 0.32 = 0.5 × k × 0.04²
0.32 = 0.5 × k × 0.0016
0.32 = 0.0008k
k = 0.32 / 0.0008 = 400 N/m
On a force-extension graph, the area under the line (between the origin and a given extension) represents the elastic potential energy stored. For a spring obeying Hooke's law, this area is a triangle: Eₑ = ½ × F × e = ½ke².
Remember that elastic potential energy is proportional to extension². Doubling the extension quadruples the energy stored. This is why the formula has e², not just e.
Multiple Springs
When springs are arranged in series (end to end), the combined spring constant is less than either individual spring. The total extension is the sum of each spring's extension.
When springs are arranged in parallel (side by side), the combined spring constant is the sum of the individual spring constants. The arrangement is stiffer overall.
Worked Example
Two identical springs, each with k = 300 N/m, are arranged in series. A force of 6 N is applied.
Each spring extends by: e = F/k = 6/300 = 0.02 m
Total extension = 0.02 + 0.02 = 0.04 m
Combined spring constant = F/e = 6/0.04 = 150 N/m
In series, the combined k is half of the individual k.
Worked Example
Two identical springs, each with k = 300 N/m, are arranged in parallel. A force of 6 N is applied.
Combined k = 300 + 300 = 600 N/m
Extension = F/k = 6/600 = 0.01 m
In parallel, the combined k is double the individual k.
Required Practical: Investigating Spring Extension
This practical investigates the relationship between force and extension for a spring, verifying Hooke's law.
Method:
- Set up a clamp stand with a spring attached to a boss and clamp. Hang a pointer and a mass hanger from the spring.
- Place a metre ruler vertically beside the spring. Record the position of the pointer (this is the original length position).
- Add a 100 g mass (0.98 N force) to the hanger and record the new pointer position.
- Repeat by adding 100 g masses one at a time, recording the pointer position each time. Add masses until the spring is clearly beyond the limit of proportionality.
- Calculate the extension for each force: extension = new position - original position.
- Plot a force-extension graph.
Expected results:
- The graph should be a straight line through the origin at first, showing Hooke's law is obeyed
- The gradient of the straight section equals the spring constant k
- The graph curves at higher forces, showing the limit of proportionality has been exceeded
Sources of error: the ruler may not be exactly vertical; parallax error when reading the ruler; the spring may already be slightly stretched; masses may not be exactly 100 g. To reduce parallax error, read the ruler at eye level.
You may be asked to describe how to investigate multiple springs in series or parallel using the same method. The key difference is the setup - springs in series are attached end to end, while springs in parallel are attached side by side to the same support.
Work Done and Elasticity
When a force stretches a spring, work is done on the spring. This work is stored as elastic potential energy in the spring. Provided the spring is not stretched beyond the limit of proportionality, the work done equals the elastic potential energy stored.
Work done = area under force-extension graph
For a linear spring: W = ½Fe = ½ke²
Worked Example
A spring is stretched by 0.2 m by a force of 50 N (within the limit of proportionality). Calculate the work done on the spring.
W = ½Fe = 0.5 × 50 × 0.2 = 5 J
This 5 J is stored as elastic potential energy.
If the spring is stretched beyond the elastic limit, some of the work done is transferred to thermal energy rather than being stored as elastic potential energy. The spring does not fully return to its original length when the force is removed.
Practice Questions
1. A force of 8 N extends a spring by 4 cm. Calculate the spring constant in N/m.
e = 4 cm = 0.04 m. k = F/e = 8/0.04 = 200 N/m.
2. A spring has a spring constant of 250 N/m. Calculate the elastic potential energy stored when it is extended by 0.08 m.
Eₑ = ½ke² = 0.5 × 250 × 0.08² = 0.5 × 250 × 0.0064 = 0.8 J.
3. Explain the difference between the limit of proportionality and the elastic limit.
The limit of proportionality is the point beyond which extension is no longer proportional to force (Hooke's law no longer applies). The elastic limit is the point beyond which the spring is permanently deformed and cannot return to its original length. The elastic limit is usually slightly beyond the limit of proportionality.
4. Two identical springs (k = 400 N/m each) are arranged in series. A force of 4 N is applied. Calculate the total extension.
Each spring extends by e = 4/400 = 0.01 m. Total extension = 0.01 + 0.01 = 0.02 m. Combined k = 4/0.02 = 200 N/m.
5. Describe how you would investigate Hooke's law in the laboratory, including the measurements you would take and the graph you would plot.
Hang a spring from a clamp stand with a pointer attached. Place a metre ruler beside the spring. Add masses one at a time and record the pointer position each time. Calculate extension = new position - original position. Calculate force = mass × g. Plot a force-extension graph. The straight-line section through the origin confirms Hooke's law; the gradient equals the spring constant.
Required Practical: Investigating Extension and Force (Hooke's Law)
Aim
To investigate the relationship between force and extension for a spring, and to identify the limit of proportionality.
Equipment
Clamp stand, boss and clamp, spring, metre ruler, pointer (or fiducial marker), mass hanger, slotted masses (100 g each), balance for checking masses.
Method
- Clamp the spring to the stand and attach the mass hanger with a pointer. Position the metre ruler vertically beside the spring.
- Record the position of the pointer on the ruler with just the mass hanger (this is the reference position for zero additional force). Ensure the ruler is vertical and read at eye level to avoid parallax error.
- Add a 100 g mass (0.98 N) to the hanger. Record the new pointer position. Calculate extension = new position − reference position.
- Repeat, adding one 100 g mass at a time, recording the pointer position and calculating the extension for each force. Continue until the spring is well beyond the limit of proportionality (the extension will start to increase by more than expected).
- Plot a force-extension graph. The straight-line section through the origin confirms Hooke's law.
Safety
Wear eye protection in case the spring snaps. Stand clear below the masses to avoid injury if they fall. Do not overload the spring beyond its elastic limit if you need to reuse it. Place a cushion or box below the masses to catch them.
Sources of Error and Improvements
Parallax error when reading the ruler: read at eye level or use a mirror behind the ruler. The spring may not return to its original length between readings: if this occurs, the elastic limit has been exceeded and a new spring is needed. Masses may not be exactly 100 g: use a balance to check. The ruler may not be vertical: use a set square to check alignment. Zero error: always measure from the unstretched position, not from zero on the ruler.
Maths Skills
F = ke Calculations
Rearrange F = ke to find any unknown: k = F/e, e = F/k. Remember to convert units: extension in metres (not cm), force in newtons, spring constant in N/m. A stiffer spring has a larger k value.
Worked Example
Two springs are connected in series. Spring A has k = 200 N/m and spring B has k = 300 N/m. A force of 6 N is applied. Calculate the total extension.
Extension of A = F/kA = 6/200 = 0.030 m. Extension of B = F/kB = 6/300 = 0.020 m. Total extension = 0.030 + 0.020 = 0.050 m.
Elastic Potential Energy: Ee = ½ke²
This formula gives the energy stored in a stretched or compressed spring within the limit of proportionality. Note that energy is proportional to extension squared: doubling the extension quadruples the stored energy. This is why overstretching a spring stores much more energy than expected and can be dangerous.
Calculating Spring Constant from a Graph
On a force-extension graph, the spring constant equals the gradient of the straight-line section: k = ΔF/Δe. Only use data points from the linear region. The area under the line (triangle) gives the elastic potential energy: E = ½Fe = ½ke².
Common Misconceptions
Hooke's Law and All Materials
All materials obey Hooke's law. Hooke's law only applies to materials up to their limit of proportionality. Beyond this point, extension is no longer proportional to force. Rubber bands do not obey Hooke's law at all — their force-extension graph is curved from the start. Metals typically obey Hooke's law over a small range, and polymers often do not.
Extension vs Stretched Length
Extension and stretched length are the same thing. Extension is the increase in length from the original unstretched length. Stretched length is the total current length. Extension = stretched length − original length. In Hooke's law, F = ke, the variable e is the extension, not the total stretched length. Using the stretched length instead of extension will give an incorrect spring constant.
6-Mark Extended Question
Describe how you would investigate the relationship between force and extension for a spring. Explain how to identify the limit of proportionality. [6 marks]
Set up a clamp stand with a spring attached by a clamp. Hang a mass hanger from the spring and position a metre ruler vertically beside it (1). Record the position of a pointer attached to the hanger with no additional masses — this is the reference position (1). Add 100 g masses one at a time, recording the pointer position each time and calculating the extension (new position minus reference position) for each force (1). Convert mass to force using F = mg where each 100 g mass provides approximately 0.98 N (1). Plot a force-extension graph. The initial section will be a straight line through the origin, confirming Hooke's law — extension is proportional to force (1). The limit of proportionality is the point where the graph starts to curve, indicating that extension is no longer proportional to force. Beyond this point, Hooke's law is no longer valid (1).
AO3: Analyse and Evaluate
A student measures the following force-extension data for a spring:
| Force (N) | Extension (cm) |
|---|---|
| 0 | 0 |
| 1.0 | 2.0 |
| 2.0 | 4.0 |
| 3.0 | 6.0 |
| 4.0 | 8.2 |
| 5.0 | 11.0 |
| 6.0 | 15.0 |
(a) Is Hooke's law being obeyed throughout? Justify your answer using data. (b) Calculate the spring constant from the data that obeys Hooke's law. (c) At which force does the limit of proportionality occur? (d) Estimate the elastic potential energy stored at 3.0 N extension.
Evaluation
(a) Hooke's law is NOT obeyed throughout. For 0–3.0 N, extension increases by 2.0 cm per 1.0 N (constant ratio), confirming proportionality. At 4.0 N the extension is 8.2 cm instead of the expected 8.0 cm, and the deviation grows (11.0 cm instead of 10.0 cm at 5.0 N; 15.0 cm instead of 12.0 cm at 6.0 N).
(b) k = F/e = 1.0/0.020 = 50 N/m (using data from the linear region).
(c) The limit of proportionality occurs at approximately 3.0 N, the last force where extension is exactly proportional.
(d) E = ½ke² = 0.5 × 50 × 0.06² = 0.5 × 50 × 0.0036 = 0.09 J.
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