A rubber band is stretched and then released. It returns to its original shape. What type of deformation has the rubber band undergone?
Strand 1 · Mechanics and Matter
Physics Year 2 Learner Material, Section 1: Dimension, Vectors, Flotation and Deformation
Deformation is the change in the shape, size, or structure of a material when a force is applied to it. This could include stretching, compressing, bending, twisting, or any other type of change. After the force is removed, the material may either return to its original shape (elastic deformation) or remain permanently changed (plastic deformation).
Different materials react differently to the same force. For example, a 1000 N force might stretch one material, while the change in another material may be impossible to perceive under the same force. This is why, before building a bridge, materials that might be used are sent to a research laboratory for testing. These tests help engineers figure out if the materials can handle the forces they will experience.
Figure 1.17: Image showing elastic deformation
Figure 1.18: Image showing plastic deformation
Activity 1.16 Online Video Analysis of Elastic and Plastic Deformation Objective
1. To independently research and analyse examples of elastic and plastic deformation through online videos.
2. To apply the concepts of elastic and plastic deformation to real-world scenarios.
Materials needed
1. Computer or device with internet access
2. Notebook or worksheet for recording observations What to do
1. Research online videos
a. Use a search engine to find videos related to elastic and plastic deformation.
b. Look for videos that show clear examples of materials being deformed under stress.
c. Consider using keywords in your search such as “elastic deformation,” “plastic deformation,” “ properties of materials,” and “stress-strain relationship.”
2. Watch and analyse
a. Select at least three videos that you find interesting or informative.
b. Watch each video carefully and take note of the materials being deformed and the nature of the deformation.
c. Identify whether the deformation is elastic or plastic based on your observations.
3. Complete the worksheet or notebook
a. Record your observations in a worksheet or notebook.
b. Answer the following questions for each video:
i. What is the material being deformed?
ii. Does the material return to its original shape after the stress is removed?
iii. Is the deformation elastic or plastic?
iv. Can you explain using ideas about particles and intermolecular forces why the material behaves in this way?
Note
1. Look for videos that provide clear explanations and visual demonstrations.
2. Consider using educational websites or platforms that offer reliable content.
3. Take notes while watching the videos to help you remember key points.
4. Feel free to search for additional videos if you need more examples.
Activity 1.17 Exploring Deformation
Materials Needed
1. Spring or rubber band
2. Retort stand or hook
3. Small weights (50 g, 100 g, 150 g, 200 g masses)
4. Aluminium foil
5. Mould clay
6. Notebook or worksheet for recording observations What to do
1. Observe and record your observations of the behaviour of each of the following objects when they experience and force and then that force is removed; do they exhibit elastic or plastic behaviour? How can you tell?
a. A spring suspended from a retort stand, stretch using small weight on a hanger
b. A piece of aluminium foil, crumpled by hand
c. A piece of mould clay, compressed or twisted by hand Review
1. Use your observations recorded to differentiate between elastic and plastic deformations.
2. Give examples of materials that show elastic behaviour.
3. Give examples of materials that show plastic behaviour.
4. Discuss what is likely to cause elastic materials to undergo plastic deformation.
Activity 1.18 Categorising Scenarios into elastic and plastic deformation Objective
1. To differentiate between elastic and plastic deformation.
2. To categorise various scenarios into either elastic or plastic deformation.
Materials needed
1. Scenarios or case studies describing different deformations
2. Sticky notes or index cards (for grouping)
3. Markers or pens What to do
1. Group the scenarios
a. Identify a friend or friends to form a small learner group
b. Work individually or in small groups to analyse each scenario.
c. Decide whether the deformation described is elastic or plastic.
d. Use sticky notes or index cards to label each scenario as “Elastic” or “Plastic.”
2. Discuss and justify
a. Discuss your groupings with your classmates and explain your reasoning.
b. Provide evidence or examples to support your choices.
c. Address any misconceptions or disagreements that may arise.
Scenarios
1. Stretching a rubber band
2. Bending a paperclip
3. Bouncing a ball
4. Moulding clay
5. Spring compression
6. Car crash impact
7. Bending a fishing rod
8. Tennis ball on a racket
9. Wind blowing a tree branch
10. Pressing a foam cushion
11. Dent in a car door
12. Squeezing aluminium foil
13. Stretching a metal wire beyond its elastic limit
14. Forging metal
Hooke’s law is a helpful rule that explains how springs and other stretchy things behave.
Imagine a spring like a rubber band. When you pull on a rubber band, it gets longer. The more you pull, the longer it becomes. Hooke’s law says that the amount a spring stretches depends on how hard you pull it.
We can use a diagram to help us understand this better:
Figure 1.19: diagram showing a spring with different forces applied to it, resulting in different amounts of stretch In the diagram, the first spring is not being pulled by a force and so is unextended.
The second spring is being pulled with a small force, so it stretches only a little.
The third spring is being pulled with a larger force, so it stretches more.
Hooke’s law is a very important rule in science. It helps us understand how things like bridges, cars, and even our bodies work. It’s like a puzzle piece that fits into the bigger picture of how the world around us works.
Hooke’s law states that, provided the elastic limit is not exceeded, the extension in an elastic material is proportional to the load or applied force.
Figure 1.20: Diagram to demonstrate Hooke’s law Mathematically, F = ke where:
• F is the force exerted on the spring
• k is the spring constant (a measure of the stiffness of the spring)
• e is the displacement of the spring from its equilibrium position
Activity 1.19 Experimental verification of Hooke’s Law Objective To verify Hooke’s Law and investigate the relationship between the force applied to a spring and its extension. Note: in the absence of practical equipment an online simulation can be used:
HTTPS://phet.colorado.edu/en/simulations/hookes-law Materials Needed
1. Spring
2. Ruler
3. Weights or masses (of varying sizes)
4. Stand
5. Clamp
6. Graph paper
Figure 1.21: A diagram of the apparatus used to investigate Hooke’s Law.
What to do:
1. Set up the experiment
a. Attach one end of the spring to a fixed point on the stand using the clamp.
b. Ensure the spring hangs vertically.
c. Measure the initial length of the spring when it is unstretched.
2. Apply weights and measure extensions
a. Add weights to the other end of the spring one at a time.
b. After each weight is added, measure the new length of the spring.
c. Calculate the extension of the spring by subtracting the initial length from the new length.
d. Record the force applied (weight = mass × acceleration due to gravity) and the corresponding extension in a table.
3. Repeat with different weights
a. Continue adding weights and measuring extensions until you have a sufficient number of data points.
b. Ensure that you do not stretch the spring beyond its elastic limit, as this will invalidate Hooke’s Law.
4. Graph the data
a. Plot a graph of force (on the y-axis) against extension (on the x-axis).
b. Draw a line of best fit onto the graph (a single straight line, using a ruler, which is centred through as many of the point as possible)
c. Note that if the spring has exceeded its elastic limit, the graph will become shallower at the end. You could ignore this part of the graph when drawing the line of best fit.
5. Observe the graph
a. If the graph is a straight line passing through the origin, this indicates a linear relationship between force and extension.
b. This confirms Hooke’s Law, which states that the force exerted by a spring is directly proportional to its extension.
c. The slope of the line represents the spring constant.
6. Calculate the spring constant The spring constant (k) can be calculated using the formula:
k = force / extension Use values of force and extension taken from a single point on the line of best fit.
7. Extension:If you finish this task quickly, try repeating the experiment but using two springs in a) series and b) parallel. How does the effective spring constant compare the spring constant of just one spring?
Figure 1.22: Springs in Series and Parallel
Energy stored in an elastic material Think of an elastic band. When you stretch it, you’re creating tension within the material. The application of this tension increases the material’s store of elastic potential energy.
The more you stretch the elastic band, the more potential energy you store in it.
This is because you’re increasing the tension within the material, making it want to return to its original shape.
When you let go of the stretched elastic band, the tension causes it to snap back to its original shape. This movement is powered by the released potential energy, which is converted into kinetic energy (the energy of motion).
This principle applies to many other elastic materials as well, such as rubber, bungee cords, and even certain types of metal. They can all store potential energy when stretched or deformed and release it as kinetic energy when allowed to return to their original shape.
Mathematically, U = 1_ 2 Fe = 1_ 2 k e²where:
• U is the potential energy stored in the elastic material
• k is the spring constant
• e is the displacement of the elastic material from its equilibrium position
Activity 1.20 Calculating the Spring Constant and Energy Stored in a Spring Read the worked example below before attempting the example questions that follow.
1. A spring with an unstretched length of 20 cm is stretched to a length of 30 cm when a 500-gram mass is hung from it.
a. Calculate the spring constant.
b. Calculate the elastic potential energy stored in the spring when it is stretched to its maximum length. (g = 9.81 ms⁻²) Step-by-Step Solution
1. Calculating the Spring Constant
a. Convert the mass to kilograms 500 grams = 0.5 kg
b. Calculate the force exerted by the mass
i. Force = Mass × acceleration due to gravity
ii. Force = 0.5 kg × 9.81 ms⁻²= 4.905 N
c. Calculate the extension of the spring and convert to metres
i. Extension = Final length - Initial length
ii. Extension = 30 cm - 20 cm = 10 cm = 0.1 m
d. Use Hooke’s Law to calculate the spring constant
i. Hooke’s Law: Force = Spring constant × Extension
ii. Spring constant = Force_______ Extension
ii. Spring constant = 4.905 N/0.1 m = 49.05 Nm⁻¹Therefore, the spring constant is 49.05 Nm⁻¹.
2. Calculating the Elastic Potential Energy
a. Use the formula for elastic potential energy Elastic potential energy = (1/2) × Spring constant × Extension²
b. Plug in the values Elastic potential energy = (1/2) × 49.05 Nm⁻¹× (0.1 m)²
c. Calculate the energy Elastic potential energy = 0.24525 J Therefore, the elastic potential energy stored in the spring is 0.25 Joules, rounded to two decimal places.
Practice Problems
Now, using the worked example as a guide, solve the following problems individually or in groups.
1. A car’s suspension system uses springs to absorb shocks from the road. If a 1000 kg car compresses a spring 5 cm when parked,
a. What is the spring constant of the suspension?
b. How much energy is stored in the spring when it is compressed? (g = 9.81 ms⁻²)
2. A bungee jumper with a mass of 70 kg jumps from a bridge with a bungee cord that has an unstretched length of 50 meters. If the jumper stretches the cord to a maximum length of 75 meters before bouncing back,
a. What is the spring constant of the bungee cord?
b. How much energy is stored in the cord at its maximum extension? (g = 9.81 ms⁻²)
3. A spring-loaded toy gun has a spring constant of 100 N/m. If the gun stores 0.5 Joules of energy when the spring is compressed, what is the maximum extension of the spring?
4. A car’s suspension system has a spring constant of 50,000 N/m. If the car’s weight compresses the spring by 5 cm, how much energy is stored in the spring?
Imagine a rubber band. When you pull it, it stretches. The force you’re applying to stretch it, per unit of its cross-sectional area, is called stress. The percentage increase in length is called strain.
Tensile stress is the ratio of force to cross-sectional area. It is expressed in Nm⁻².
Mathematically, tensile stress = F/A Tensile strain is the ratio of extension to the original length of the material. It has no unit as it is simply a scale factor by which the length has changed.
Mathematically, tensile strain = e/lₒ Young’s Modulus Young’s modulus is a measure of how stretchy or stiff a material is. It is similar to the spring constant, k, but takes into account the dimensions of the object under stress. The Young Modulus of a particular material is always the same, whereas the spring constant will depend on the length and cross-sectional area of the object.
A rubber band has a low Young’s modulus, meaning it stretches easily. A metal rod has a high Young’s modulus, meaning it’s very stiff and doesn’t stretch much.
An alternative wording of Hooke’s law is: provided the elastic limit is not exceeded, stress is proportional to strain.
stress ∝ strain stress = E × strain Then Young’s modulus E is defined as the ratio of tensile stress to tensile strain.
It is expressed in Nm⁻².
E = stress_ strain E = F_ A ÷ e_ lₒ E = F lₒ_ Ae Where F is the force required to stretch the elastic string or spring of length lₒ and cross-sectional area A, through a length e.
Activity 1.21 Researching stress Draw diagrams to show the effect of tensile, compressional and shear stress.
Describe the effects that these different types of stresses can have on materials.
You should also research materials which have a particularly high breaking stress; what sorts of applications are these materials used in?
Activity 1.22 Calculating Tensile Stress, Tensile Strain, and Young’s Modulus Read the worked example below before attempting the example questions that follow.
Worked example
1. A metal wire of length 2 metres and a cross-sectional area of 0.001 m² is stretched by 0.005 metres when a tensile force of 1000 N is applied.
Calculate:
a. the tensile stress on the wire.
b. the tensile strain in the wire.
c. the Young’s modulus of the material.
Step-by-Step Solution
1. Calculate Tensile Stress
Identify the formula for tensile stress:
Tensile Stress = Force_________________ Cross − sectional Area = F/A Substitute the known values and compute to get the answer From the problem, we know that:
- Force F = 1000 N
- Cross-sectional area A = 0.001 m² Now, substitute these values into the formula:
Tensile Stress = 1000 N_ 0.001 m²Tensile Stress = 1,000,000 Nm⁻²2. Calculate Tensile Strain Identify the formula for tensile strain Tensile Strain = Extension___________ Original length = e/L Substitute the known values and compute to get the answer
- Extension e = 0.005 metres
- Original length L = 2 metres Now, substitute these values into the formula:
Tensile Strain = 0.005 m_ 2 m Tensile Strain = 0.0025 or 0.25%
3. Calculate Young’s Modulus
Identify the formula for Young’s modulus Young′s modulus = Tensile Stress___________ Tensile Strain Substitute the known values and compute to get the answer From the previous steps, we know that:
- Tensile stress = 1,000,000 N/m²
- Tensile strain = 0.0025 Now, substitute these values into the formula:
Young′s modulus = 1,000,000 Nm⁻²____________ 0.0025 Young′s modulus = 400,000,000 Nm⁻²Practice Problems Now, using the worked example as a guide, solve the following problems individually or in groups.
1. A steel guitar string has a length of 0.8 metres and a cross-sectional area of 0.0005 m². When tuned, the string stretches by 0.002 metres under a tension force of 600 N.
a. Calculate the tensile stress in the guitar string.
b. Determine the tensile strain in the string.
c. Find the Young’s modulus of the steel string.
2. A construction worker uses a steel rod with a cross-sectional area of 0.002 m² to lift materials. The rod is 2 metres long and stretches by 0.001 metres under a load of 5000 N.
a. Calculate the tensile stress in the steel rod.
b. Determine the tensile strain in the rod.
c. What is the Young’s modulus of the steel rod?
3. A copper wire has an original length of 1.5 metres and an extension of 0.003 metres when subjected to a load. The wire has a cross-sectional area of 0.0004 m², and the Young’s modulus of copper is 1.1×10¹¹Nm⁻². Calculate the force applied to the copper wire.
Activity 1.23 Young’s Modulus Comparison and Application through Discussion Objective
1. To compare the Young’s modulus of various materials using provided data.
2. To discuss the significance of these comparisons in material selection and engineering applications.
Data
Table 1.2: Young’s Modulus of Various Materials
Material Young’s Modulus (GPa)
Steel 200 Aluminium 70
Copper 120 Rubber 0.01
Wood (Oak) 10-12 Glass 60-70
Plastic (Nylon) 1.5-3 What to do
1. Identify a friend or friends to form a small learner group.
2. Observe the provided data, and examine the Young’s modulus values for each material. Identify the materials with the highest and lowest Young’s modulus.
3. Analyse if there are any trends or patterns in the data based on material properties (e.g., metals vs. non-metals, crystalline vs. amorphous).
4. Engage in group discussions with the following discussion prompts:
a. How does the structure of a material relate to its Young’s modulus?
b. Can you explain the trends observed in the Young’s modulus values of different materials?
c. How can the Young’s modulus of a material be related to its other mechanical properties (e.g., strength, ductility)?
d. In what types of engineering applications is a high Young’s modulus desirable?
e. Can you think of examples where a low Young’s modulus is beneficial in engineering applications?
5. Each group should appoint one member to summarise and record the answers from the discussions.
6. After the discussions, each group should share one key point from their discussion with the class or peers.
Activity 1.24 Experimental Determination of Young’s Modulus of a Wire Objective To experimentally determine the Young’s modulus of a wire by measuring its stress and strain under increasing loads. Note: in the absence of practical equipment an online simulation could be used https://www.doitpoms.ac.uk/tlplib/thermal-expansion/simulation.php You can also watch a video detailing the experimental set up and procedure here:
https://www.youtube.com/watch?v=OxV4S8pU6Co Materials needed
1. Length of wire (such as constantan), minimum 1.5m
2. Vernier callipers or micrometer screw gauge
3. Metre ruler(s)
4. Stand or clamp
5. Pulley
6. Hanging masses
7. Fiduciary marker What to do
1. Measure wire’s cross-sectional area
a. Measure the diameter of the wire using vernier callipers or a micrometer screw gauge.
b. Calculate the cross-sectional area of the wire using the formula:
Area = π d²__ 4
2. Set up the experiment
a. Attach one end of the spring to a stand or clamp.
b. Hang a pulley over the edge of the stand or clamp.
c. Attach a mass hanger to the end of the spring that passes over the pulley.
3. Measure initial length
a. Place a fiduciary marker on the wire approximately 15cm away from the point where it passes over the pulley. Leave a metre ruler in place underneath the fiduciary marker such that changes in its position will be measurable.
b. Measure the initial position of the fiduciary marker relative to the clamped end of the wire, using a metre ruler.
4. Apply loads and measure extensions
a. Gradually add masses to the hanging mass hanger.
b. For each added mass, measure the new position of the fiduciary marker and calculate the extension of the wire.
c. Record the force applied (weight of the hanging masses) and the corresponding extension.
5. Calculate stress and strain
a. Calculate the stress on the wire using the formula:
Stress = Force_________________ Cross − sectional Area
b. Strain: Calculate the strain on the wire using the formula:
Strain = Extension___________ Original length
c. Record your values in a tabular form
Table 1.3: Table to record values obtained from determining Young’s Modulus of a wire Force/N Diameter/m Area/ m²Final length/m Extension/m Stress/ Nm⁻²Strain
6. Plot the stress-strain curve
a. Plot the stress values on the y-axis and the strain values on the x-axis.
b. Determine the slope of the linear portion of the stress-strain curve which represents the Young’s modulus of the wire.
Notes
1. Ensure accurate measurements of diameter, length, and extension.
2. Choose a suitable range of loads to observe the wire’s behaviour within its elastic limit.
3. Use appropriate data analysis techniques (e.g., linear regression) to determine the Young’s modulus.
4. Handle masses and equipment carefully to avoid accidents.
Activity 1.25 Preparation of a Poster on Material Properties Objective
1. To research and understand the properties of brittle, ductile, and malleable materials.
2. To create a visual representation of the properties and provide examples of materials exhibiting each behaviour.
Materials needed
1. Poster board or paper
2. Markers, pens, or coloured pencils
3. Research materials (books, internet, scientific journals) What to do
1. Research material properties
a. Research the definitions of brittle, ductile, and malleable materials.
b. Research about the characteristics of each type of material, such as their behaviour under stress or deformation.
c. Find examples of materials that exhibit each property.
2. Organise information
a. Decide on the layout and organisation of your poster or summary sheet.
b. Outline the key points you want to include, such as definitions, examples, and visual aids.
3. Design and create the poster or summary sheet:
a. Use diagrams, illustrations, or images to represent the properties and examples.
b. Write clear and concise explanations of the terms and characteristics.
c. Include specific examples of materials for each property.
d. Use colours and fonts that are visually appealing and easy to read.
4. Review and revise:
a. Ensure that the information is accurate and up-to-date.
b. Make sure the poster or summary sheet is easy to understand and visually appealing.
c. Check that all the required information is included.
A rubber band is stretched and then released. It returns to its original shape. What type of deformation has the rubber band undergone?
A spring obeys Hooke’s law. A force of N stretches it by m. What is the spring constant of the spring?
A steel rod of original length m and cross-sectional area m is stretched by m under a tensile force of N. Calculate the Young’s modulus of the steel.
Which of the following defines tensile strain?
A student hangs masses on a spring. After adding a large mass, the spring stretches more than expected from Hooke’s law and does not return to its original length when the masses are removed. What is the best explanation?
The Ghana Highway Authority is constructing a pedestrian footbridge at Tema. Before the steel cables are used, samples are sent to a materials laboratory. A steel cable of original length and cross-sectional area is stretched by under a tensile force of . When the force is removed, the cable returns to its original length. A sample of clay from the site, however, remains permanently deformed after a similar force is applied and removed.
Distinguish between elastic deformation and plastic deformation. Use the behaviour of the steel cable and the clay sample in your answer.
State Hooke's law.
The steel cable obeys Hooke's law. Calculate: (i) the tensile stress in the cable; (ii) the tensile strain in the cable; (iii) the Young's modulus of the steel cable.
When a larger force is applied to the steel cable, it no longer returns to its original length. Explain what has happened to the cable and why engineers must not exceed the elastic limit in the footbridge cables.
A second steel cable is made of the same material, but its original length is twice that of the first cable and its cross-sectional area is the same. It is stretched by the same tensile force of . State and explain how the extension of the second cable compares with that of the first cable.