GCSE physics Experiment to investigate force applied to a spring and resulting extension describe experimental procedure, method of processing results, Hooke's Law graphs and calculations what happens if stretched beyond the elastic limit IGCSE/GCSE Physics revision notes

GCSE level physics exam revision notes on Forces 4

4.3 An experiment to investigate the force applied to a spring and resulting extension - experimental procedure, method of processing results, Hooke's Law graphs and calculations

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[KEY POINTS and learning objectives for this page, after initial notes]

Index of physics notes on FORCES section 4 on Elastic potential energy

Practice exam question on Hooke's Law and experiments

4.3a An experiment to investigate the force applied to a spring and resulting extension

How can we investigate the relationship between a spring's extension and weights added to it?

What is the relationship between the extension of a spring when increasing weights are connected to it?

If weights are attached to a firmly fixed suspended spring, the spring will elongate depending on the value of the weight attached. The greater the weight, the greater the spring is extended. The extra length the spring attains is called the spring extension and this phenomena can be systematically investigated using the simple apparatus described below.

When a weight is added and the spring is static, the weight of the mass (and the spring itself) is counterbalanced by the force of tension in the spring.

A metre rule is fixed in a vertical position using a stand and several clamps. Preferably with the linear scale pointing downwards!

The metre rule scale can be read in mm or cm.

From the top of the 1 m ruler a spring is suspended with a hook-base is added to which extra mass can be added e.g. in 50g increments (0.5 N force increase increment).

Fix a pointer onto the hook to which the weights will be attached. If you can't fix a pointer on, just use the base of the weight hook and sight it horizontally onto the scale.

You take the initial reading with no extra weight on (other than the weight of the spring itself plus hook) and take the initial reading on the scale. This first reading with no extra load applied is the crucial starting point for all the successive measurements.

You then add an extra mass and take the scale reading in mm/cm. From each successive reading you must subtract the initial reading to obtain the true extension of the spring (I haven't actually shown this in the table of results below).

Record a minimum of five observations carefully in a prepared table and convert the mm/cm scale readings to the extension in m.

Below is a typical table of results, already corrected by subtracting the 'initial' reading.

50 g load = 0.05 kg ~ force/weight of 0.5 N

Hooke's law experiment graph for different spring strengths For simplicity I've taken gravity as 10 N/kg, therefore every 50 g mass added equals an incremental weight increase of 0.5 N.

From the data table you plot a graph of total force (= tension) versus the total extension in the spring length (graph sketched on the left).

(the tension in the spring equals the force created downwards by the weight of mass).

Draw the line of best fit from the 0,0 graph origin.

If the spring is truly elastic a linear graph is obtained.

This means a simple linear equation describes the behaviour of the spring under these conditions.

The experiment is a simple proof and demonstration that the extension of a spring or any elastic material is directly proportional to the force applied (the load or weight in newtons).

This relationship is expressed with the simple equation:

force = a spring constant x extension

F = ke

where F = the applied force in newtons (N), e = the spring extension in metres (m)

and k is the spring (elastic) constant in N/m.

The stiffer (stronger) the spring (or any material being stretched) the greater the spring constant.

and you would see a steeper gradient of the graph line, or a smaller gradient for a weaker spring,

the point illustrated by the 'theoretical' purple lines on the above graph.

The strength of a spring and hence the spring constant, depends on the material and the thickness of the material wound into a spring shape.

This linear equation relationship between force applied and the extension (or compression) of an elastic material is also known as Hooke's Law of proportionality.

This can be stated as:

The extension of a spring or wire (or any elastic object) is proportional to the load (force applied)

or

If the deformation of a material is proportional to the force applied, the material is truly elastic and is said to obey Hooke's Law (a law of proportionality).

From the graph you can calculate the spring constant e.g. rearranging the equation (Hooke's Law equation)

k = F/e = gradient of the graph = 3.0/0/0.06 = 50 N/m

4.3b Applying the idea of Hooke's equation F=ke to various situations

Five important points to note:

This equation works for compression where e is the difference between the full length and compressed length

The spring constant varies with the material of the spring, the size and number of coils of the spring.

The stronger/stiffer the spring the greater the value of the spring constant k.

This spring system is the basis for simple instruments used to measure the weight of an object like a fish you have caught!

A force meter is used for experiments in a laboratory - school, college, university or in the engineering industry where it is used e.g. to test the strength of materials.

Above is an illustration of a simple instrument for weighing objects.

It is essentially a 'force meter' calibrated to read in g and kg.

(so it takes into account gravity at the Earth's surface, but it wouldn't be any good on the Moon or Mars with their different strength of gravitational fields)

Prior to taking a reading the pointer should be adjusted to read zero.

You place the object on the hook which stretches the spring and read off its weight on the calibrated scale.
Applying the Hooke's Law equation to two contrasting situations.
Applying the F=ke equation to contrasting situations with respect to the value of the spring constant, and the rearrangement k=F/e.

(a) For a sensitive spring balance to weigh a 'light' objective you need it to operate down a relatively long scale i.e. to stretch easily.

Therefore you need a spring with a small value of k, so that a small force F (the weight of the object being weighed), creates a greater spring extension.
The diameter of the spring coils will be relatively small.
You should also appreciate a linear weight scale is only accurate if the spring obeys Hooke's Law!

(b) For a car suspension spring, it needs to absorb a large force F to give a 'cushioning' effect for the comfort of the driver and passengers.

Therefore you need spring coils with a much larger diameter and a much bigger spring constant k, to give a relative small compression distance e.
See photographs in section 4.2 Work is done in stretching or compressing a material, energy stored, shock absorbers

4.3c Extending the investigation and alternative graphs

Different groups in a class can look at different springs, or if time permits each group of students can look at several springs. The class results can be pooled and graphs drawn.
Instead of plotting force versus extension (which I prefer) you can plot extension versus force.
Since F = ke, e = F / k, so the gradient will be 1 / k, the reciprocal of the spring constant.
In the 'idealised' right-hand graph of spring extension versus force, springs A and C results did not go beyond the limit of proportionality.
However, the results for spring B showed a deviation from linearity and the graph curves upwards from point L, the limit of proportionality. (in a sense the limit of obeying Hooke's Law).
BUT, the B graph does not mean beyond the limit of elasticity.

4.3d What happens if you keep on increasing the force applied to an elastic material?

In the above experiment, if you add even more weights to the spring then the resulting graph of results may not be linear for the higher weight readings.

This is because the spring is overstretched beyond its elastic limit (the limit of proportionality).

Beyond point L Hooke's Law is no longer obeyed.

In other words the non-linear section of the graph is beyond L, the limit of proportionality - the spring stretches more than you expect and the graph begins to curve over.

From zero force to L Hooke's Law is obeyed - the linear section of elasticity.

After that, between point L and point D, the stretching is greater than expected - non-linear graph, but the spring will return to its original length - the spring is still behaving elastically, but only for a relatively small further increase in the applied force.

Just because an object behaves elastically, it doesn't mean that Hooke's Law is obeyed.

An elastic band is 'elastic' but it doesn't obey Hooke's Law!

Eventually at point D, called the elastic limit, the force is too great and the spring will not return to its original length at all - permanent deformation beyond the limit of elasticity.

This happens with a repeatedly stretched elastic band - eventually it breaks!

From point D onwards the spring behaves with plastic deformation.

On the right-hand graph, an alternative representation of the graphical data, I've indicated the permanent extension showing the spring will NOT return to its original length.

Sub-index of physics notes: FORCES 4. Elastic potential energy

4.3e Key points on elastic potential energy Hooke's law and spring experiments

Information sources for Doc Brown's key points: IGCSE-GCSE physics are based on textbooks & syllabus-specifications for students taking the UK AQA, Edexcel, OCR 21st Century Science, OCR Gateway science suite, WJEC, CCEA and CIE GCSE physics 9-1 level science examinations

A structured set of summary revision notes on Hooke’s Law and its required practical, tailored to the GCSE/IGCSE Physics specifications across WJEC, CCEA, CIE, AQA, Edexcel, and OCR:

Hooke’s Law: Core Concept

Definition: Hooke’s Law states that the extension of an elastic object is directly proportional to the force applied, provided the object is within its limit of proportionality.
Equation (watch the units):
F = kx Where:
- F = force applied (N)
- k = spring constant (N/m)
- x = extension (m)

The spring constant ( k ) indicates how stiff the spring is.

larger ( k ) means a stiffer spring.

Hooke’s Law Experiment (Usually a Required Practical)

Apparatus

Clamp stand, boss, and clamp
Spring
Ruler (vertical)
Pointer (attached to spring)
Masses (e.g. 100 g/0.1 kg increments)
Safety goggles

Method

Measure the initial length of the spring.
Add masses incrementally and record the new length each time.
Calculate extension:
Extension} = New length - Original length
Plot a graph of force (weight) versus extension.
Determine the spring constant from the gradient of the straight-line portion.

Ensure measurements are taken at eye level to avoid parallax error. Remove weights after each trial to check for plastic deformation.

Typical Exam Board Specifications concerning Hooke's Law

Coverage of Hooke’s Law & Experiment
Required practical, force-extension graph, spring constant, elastic limit.
Hooke’s Law, practical skills, graph analysis, energy stored in springs.
Hooke’s Law, spring constant, limit of proportionality, experiment setup.
Force-extension graph, spring constant, safety and accuracy tips.
Prescribed practical, force-extension graph, elastic limit.
Hooke’s Law, spring behaviour, practical application, graph interpretation.

Student Tips

Memorise the equation ( F = kx ) and understand each variable.
Practice plotting and interpreting graphs - straight line through origin = obeys Hooke’s Law.
Revise safety precautions: goggles, stable clamp stand, avoid overloading spring.
Understand the limit of proportionality - beyond this, Hooke’s Law no longer applies.
Use consistent units: convert cm to m when calculating extension.
Check for anomalies: repeat measurements and calculate averages for reliability.

Applications involving Hooke's Law

Hooke’s Law isn’t just a classroom concept - it’s quietly working behind the scenes in loads of everyday tech and engineering marvels.

Here’s a breakdown of where it shows up in the real world:

Engineering & Transportation

Car Suspension Systems: Springs compress and extend to absorb shocks, keeping your ride smooth.
Shock Absorbers: Use Hooke’s Law to dampen vibrations and protect sensitive components.
Bridge Expansion Joints: Elastic materials stretch and compress to accommodate temperature changes and traffic loads.

Precision Devices

Mechanical Watches: The balance spring (hairspring) regulates timekeeping by oscillating with a restoring force.
Spring Scales: Measure weight by stretching a spring - force is proportional to extension.
Manometers: Use elastic membranes to measure pressure changes in fluids.

Toys and Everyday Gadgets

Retractable Pens: Springs inside the click mechanism obey Hooke’s Law to extend and retract the nib.
Toy Guns: Springs store elastic energy and release it to fire projectiles.
Bathroom Scales: Internal springs compress under weight, translating force into readable values.

Household Applications

Mattresses & Cushions: Springs compress under body weight and return to shape - classic elastic behavior.
Clothespins: Spring-loaded design uses Hooke’s Law to grip clothes securely.
Door Closers: Springs stretch and compress to control door movement.

Medical and Biomechanics

Prosthetics: Elastic components mimic muscle and tendon behavior, storing and releasing energy.
Tendons & Ligaments: Biological tissues stretch and recoil, following Hooke’s Law within limits.
Stents: Expand elastically to fit within arteries, then hold shape.

Industrial and Structural Uses

Valves & Switches: Springs provide tension and return force in mechanical systems.
Clutches & Brakes: Control engagement and disengagement using elastic deformation.
Energy Harvesting Devices: Convert mechanical deformation into electrical energy - the piezoelectric effect e.g. used in pressure monitoring instruments.

4.3f Keywords, phrases and learning objectives for elastic potential energy

Be able to describe an experiment to investigate the force applied to a spring and resulting extension.

Explain the experimental procedure to validate whether a stretched material obeys Hooke's Law - processing data with graphs and calculations.

Know what happens if the material (e.g. a spring) is stretched beyond the proportionality limit and the elastic limit and be able to interpret a graph of force versus spring extension to aid your explanation.

Understand the applications of Hooke's Law equation and relating it to the dimensions of a spring the relative values of k and e for a particular use.

QUESTIONS

GCSE level physics - practise exam questions on Hooke's Law spring experiments

A joint re-edited AI-doc b experiment in question design

Jot down your responses and check out the answers: ANSWERS

If you think there are any errors, please email me asap at chem55555@hotmail.com

I don't mind if students/teachers do a selected printout of these questions and answers.

Q1. A spring is tested in the lab. A series of weights are hung from it and the extension is measured. The results show that the extension doubles when the force doubles, up to a certain point. Which equation correctly describes this behaviour while the spring obeys Hooke’s Law?

A. $F = x/k$ B. $F = k x$ C. $F = k /x$ D. $F = kx 2$

Q2. In a Hooke’s Law experiment, a student plots a graph of force (vertical axis) against extension (horizontal axis) for a spring. While the spring obeys Hooke’s Law, what should the graph look like?

A. A straight line that curves upwards at high forces

B. A straight line that passes through the origin

C. A horizontal line

D. A curve that starts at the origin and becomes steeper

Q3. A spring is stretched in a Hooke’s Law experiment. At first, the force–extension graph is a straight line, but then it begins to curve. Later, when the force is removed, the spring does not return to its original length. Which statement is correct?

A. The spring has passed the limit of proportionality but not the limit of elasticity

B. The spring has passed the limit of elasticity but not the limit of proportionality

C. The spring has passed both the limit of proportionality and the limit of elasticity

D. The spring is still obeying Hooke’s Law

Q4. A student plots a force–extension graph for a spring. The straight‑line section has a gradient of $20 N/m$ . What does this gradient represent?

A. The limit of elasticity

B. The mass of the weights used

C. The maximum extension

D. The spring constant $k$

Q5. In a school experiment to investigate Hooke’s Law, which of the following is the best procedure?

A. Hang different masses from the spring, measure the extension from the unstretched length, and plot force against extension

B. Hang one mass from the spring and measure its length once

C. Hang different masses from the spring, measure the total length only, and plot mass against length

D. Hang different masses from the spring and measure the extension, but plot extension against mass

Q6. A spring has a spring constant of $50 N/m$ . It is stretched by $12$ $cm$ in the Hooke’s Law region. What is the force applied to the spring?

A. $4.2 N$ B. $6.0 N$ C. $0.24 N$ D. $60 N$

Q7. A student continues adding masses to a spring beyond the recommended maximum. The last few points on the force–extension graph lie below the straight‑line trend. What is the most likely explanation?

A. The spring constant has increased

B. The spring has been permanently stretched and no longer obeys Hooke’s Law

C. The spring is now perfectly elastic

D. The extension is still directly proportional to force

Q8. Which of the following is a good example of Hooke’s Law being used in everyday technology?

A. The heating element in an electric kettle

B. The spring in a car suspension system

C. The blades of a wind turbine

D. The lens in a camera

Q9. A spring obeys Hooke’s Law up to an extension of $0.20 m$ . As the extension increases within this region, what happens to the energy stored in the spring?

A. It stays the same because the spring constant is fixed

B. It decreases because the spring is getting longer

C. It increases as the extension increases

D. It appears only when the spring reaches the limit of elasticity

Q10. Four different graphs show how extension changes with force for four springs. Which graph represents a spring that obeys Hooke’s Law over the whole range shown?

A. A straight line that starts above the origin and slopes upwards

B. A curve that starts at the origin and becomes less steep

C. A straight line that passes through the origin

D. A horizontal line at constant extension

Jot down your responses and check out the answers: ANSWERS

If you think there are any errors, please email me asap at chem55555@hotmail.com

I don't mind if students/teachers do a selected printout of these questions and answers.

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ANSWERS

GCSE level physics - practise exam questions on Hooke's Law spring experiments

A joint re-edited AI-doc b experiment in question design

Jot down your responses and check out the answers: ANSWERS

If you think there are any errors, please email me asap at chem55555@hotmail.com

I don't mind if students/teachers do a selected printout of these questions and answers.

Q1. A spring is tested in the lab. A series of weights are hung from it and the extension is measured. The results show that the extension doubles when the force doubles, up to a certain point. Which equation correctly describes this behaviour while the spring obeys Hooke’s Law?

A. $F = x/k$ B. $F = k x$ C. $F = k /x$ D. $F = kx 2$

Answer: B

Explanation: Hooke’s Law states that, within the limit of proportionality, the force $F$ applied to a spring is directly proportional to its extension $x$ :

F = k x

where $k$ is the spring constant. Doubling the force doubles the extension, which matches a linear proportional relationship.

Common error/misconception: Thinking the relationship must involve division (like $F = k /x$ ) or a square term (like $F = kx 2$ .) Those would not give a straight‑line graph through the origin.

Q2. In a Hooke’s Law experiment, a student plots a graph of force (vertical axis) against extension (horizontal axis) for a spring. While the spring obeys Hooke’s Law, what should the graph look like?

A. A straight line that curves upwards at high forces

B. A straight line that passes through the origin

C. A horizontal line

D. A curve that starts at the origin and becomes steeper

Answer: B

Explanation: When Hooke’s Law is obeyed, force is directly proportional to extension. A directly proportional relationship gives a straight line through the origin on a force–extension graph. The gradient of this line is the spring constant $k$ .

Common error/misconception: Confusing “straight line” with “any straight line.” It must pass through the origin to show direct proportionality. A straight line not through the origin would mean the spring has some extension even with zero force.

Q3. A spring is stretched in a Hooke’s Law experiment. At first, the force–extension graph is a straight line, but then it begins to curve. Later, when the force is removed, the spring does not return to its original length. Which statement is correct?

A. The spring has passed the limit of proportionality but not the limit of elasticity

B. The spring has passed the limit of elasticity but not the limit of proportionality

C. The spring has passed both the limit of proportionality and the limit of elasticity

D. The spring is still obeying Hooke’s Law

Answer: C

Explanation: The point where the graph stops being a straight line is the limit of proportionality—beyond this, force is no longer directly proportional to extension. The point beyond which the spring does not return to its original length when the force is removed is the limit of elasticity. In this scenario, the graph has curved (past limit of proportionality) and the spring does not fully return (past limit of elasticity).

Common error/misconception: Many students think the limit of proportionality and limit of elasticity are the same point. They are related but not identical; the spring can stop obeying Hooke’s Law before it becomes permanently deformed.

Q4. A student plots a force–extension graph for a spring. The straight‑line section has a gradient of $20 N/m$ . What does this gradient represent?

A. The limit of elasticity

B. The mass of the weights used

C. The maximum extension

D. The spring constant $k$

Answer: D

Explanation: In the linear region where Hooke’s Law applies, the gradient of the force–extension graph is the spring constant $k$ . From $F = k x$ , rearranging gives $k = F x$ , which is exactly the gradient (change in force divided by change in extension).

Common error/misconception: Confusing the gradient with the limit of elasticity or maximum extension. The limit of elasticity is a point where permanent deformation begins, not the gradient.

Q5. In a school experiment to investigate Hooke’s Law, which of the following is the best procedure?

A. Hang different masses from the spring, measure the extension from the unstretched length, and plot force against extension

B. Hang one mass from the spring and measure its length once

C. Hang different masses from the spring, measure the total length only, and plot mass against length

D. Hang different masses from the spring and measure the extension, but plot extension against mass

Answer: A

Explanation: To investigate Hooke’s Law properly, you need multiple measurements of extension for different forces. Force is calculated from mass using $F = m g$ . You then plot force (vertical axis) against extension (horizontal axis) to see if the relationship is directly proportional.

Common error/misconception: Plotting mass instead of force, or plotting extension on the vertical axis. While mass is easier to measure, Hooke’s Law is defined in terms of force and extension, so the graph should be $F$ versus $x$ .

Q6. A spring has a spring constant of $50 N/m$ . It is stretched by $12$ $cm$ in the Hooke’s Law region. What is the force applied to the spring?

A. $4.2 N$ B. $6.0 N$ C. $0.24 N$ D. $60 N$

Answer: B

Explanation: Use Hooke’s Law and convert from cm to m:

F = k x = 50 N/m \times 0.12 m = 6.0 N

So the force applied is $6.0 N$ .

Common error/misconception: Forgetting to convert extension into metres or mis‑multiplying (e.g. thinking $50 \times 0.12 = 4.2$ ). Some students also mistakenly square the extension.

Q7. A student continues adding masses to a spring beyond the recommended maximum. The last few points on the force–extension graph lie below the straight‑line trend. What is the most likely explanation?

A. The spring constant has increased

B. The spring has been permanently stretched and no longer obeys Hooke’s Law

C. The spring is now perfectly elastic

D. The extension is still directly proportional to force

Answer: B

Explanation: When the spring is overstretched, it can be permanently deformed. The force–extension graph then deviates from the straight line expected from Hooke’s Law. Points lying below the line indicate that the spring is extending more than expected for a given force, showing it has been damaged and no longer behaves elastically.

Common error/misconception: Thinking the spring constant has increased. In fact, the effective stiffness has decreased—the spring is “softer” and extends more for the same force.

Q8. Which of the following is a good example of Hooke’s Law being used in everyday technology?

A. The heating element in an electric kettle

B. The spring in a car suspension system

C. The blades of a wind turbine

D. The lens in a camera

Answer: B

Explanation: Car suspension systems use springs that compress when the car goes over bumps. As long as the springs operate within their elastic limit, the force they exert is proportional to their compression, following Hooke’s Law. This helps absorb shocks and keep the ride smooth.

Common error/misconception: Choosing any device that contains metal or moves. Hooke’s Law specifically describes the relationship between force and extension (or compression) in elastic materials like springs, not general mechanical or electrical components.

Q9. A spring obeys Hooke’s Law up to an extension of $0.20 m$ . As the extension increases within this region, what happens to the energy stored in the spring?

A. It stays the same because the spring constant is fixed

B. It decreases because the spring is getting longer

C. It increases as the extension increases

D. It appears only when the spring reaches the limit of elasticity

Answer: C

Explanation: As a spring is stretched within the Hooke’s Law region, work is done on the spring and energy is stored as elastic potential energy. The more the extension (while still obeying Hooke’s Law), the more energy is stored.

Common error/misconception: Thinking energy only appears at the limit of elasticity or that it stays constant because $k$ is constant. The spring constant being fixed does not mean the energy is fixed.

Q10. Four different graphs show how extension changes with force for four springs. Which graph represents a spring that obeys Hooke’s Law over the whole range shown?

A. A straight line that starts above the origin and slopes upwards

B. A curve that starts at the origin and becomes less steep

C. A straight line that passes through the origin

D. A horizontal line at constant extension

Answer: C

Explanation: A spring obeying Hooke’s Law over the whole range will show a directly proportional relationship between force and extension: a straight line through the origin. The gradient is the spring constant. Any curve or line not passing through the origin indicates the relationship is not directly proportional over the whole range.

Common error/misconception: Picking any straight line, even if it doesn’t pass through the origin. Direct proportionality requires both a straight line and zero extension at zero force.

If you think there are any errors, please email me asap at chem55555@hotmail.com

I don't mind if students/teachers do a selected printout of these questions and answers.

Information sources for Doc Brown's key points: IGCSE-GCSE physics are based on textbooks & syllabus-specifications for students taking the UK AQA, Edexcel, OCR 21st Century Science, OCR Gateway science suite, WJEC, CCEA and CIE GCSE physics 9-1 level science examinations

Hooke’s Law: Core Concept

Hooke’s Law Experiment (Usually a Required Practical)

Apparatus

Method

Typical Exam Board Specifications concerning Hooke's Law

Student Tips

Engineering & Transportation

Precision Devices

Toys and Everyday Gadgets

Household Applications

Medical and Biomechanics

Industrial and Structural Uses

Q1. A spring is tested in the lab. A series of weights are hung from it and the extension is measured. The results show that the extension doubles when the force doubles, up to a certain point. Which equation correctly describes this behaviour while the spring obeys Hooke’s Law?

Q2. In a Hooke’s Law experiment, a student plots a graph of force (vertical axis) against extension (horizontal axis) for a spring. While the spring obeys Hooke’s Law, what should the graph look like?

Q3. A spring is stretched in a Hooke’s Law experiment. At first, the force–extension graph is a straight line, but then it begins to curve. Later, when the force is removed, the spring does not return to its original length. Which statement is correct?

Q4. A student plots a force–extension graph for a spring. The straight‑line section has a gradient of 20 N/m. What does this gradient represent?

Q5. In a school experiment to investigate Hooke’s Law, which of the following is the best procedure?

Q6. A spring has a spring constant of 50 N/m. It is stretched by 12 cm in the Hooke’s Law region. What is the force applied to the spring?

Q7. A student continues adding masses to a spring beyond the recommended maximum. The last few points on the force–extension graph lie below the straight‑line trend. What is the most likely explanation?

Q8. Which of the following is a good example of Hooke’s Law being used in everyday technology?

Q9. A spring obeys Hooke’s Law up to an extension of 0.20 m. As the extension increases within this region, what happens to the energy stored in the spring?

Q10. Four different graphs show how extension changes with force for four springs. Which graph represents a spring that obeys Hooke’s Law over the whole range shown?

Q1. A spring is tested in the lab. A series of weights are hung from it and the extension is measured. The results show that the extension doubles when the force doubles, up to a certain point. Which equation correctly describes this behaviour while the spring obeys Hooke’s Law?

Q2. In a Hooke’s Law experiment, a student plots a graph of force (vertical axis) against extension (horizontal axis) for a spring. While the spring obeys Hooke’s Law, what should the graph look like?

Q3. A spring is stretched in a Hooke’s Law experiment. At first, the force–extension graph is a straight line, but then it begins to curve. Later, when the force is removed, the spring does not return to its original length. Which statement is correct?

Q4. A student plots a force–extension graph for a spring. The straight‑line section has a gradient of 20 N/m. What does this gradient represent?

Q5. In a school experiment to investigate Hooke’s Law, which of the following is the best procedure?

Q6. A spring has a spring constant of 50 N/m. It is stretched by 12 cm in the Hooke’s Law region. What is the force applied to the spring?

Q7. A student continues adding masses to a spring beyond the recommended maximum. The last few points on the force–extension graph lie below the straight‑line trend. What is the most likely explanation?

Q8. Which of the following is a good example of Hooke’s Law being used in everyday technology?

Q9. A spring obeys Hooke’s Law up to an extension of 0.20 m. As the extension increases within this region, what happens to the energy stored in the spring?

Q10. Four different graphs show how extension changes with force for four springs. Which graph represents a spring that obeys Hooke’s Law over the whole range shown?

Q4. A student plots a force–extension graph for a spring. The straight‑line section has a gradient of $20 N/m$ . What does this gradient represent?

Q6. A spring has a spring constant of $50 N/m$ . It is stretched by $12$ $cm$ in the Hooke’s Law region. What is the force applied to the spring?

Q9. A spring obeys Hooke’s Law up to an extension of $0.20 m$ . As the extension increases within this region, what happens to the energy stored in the spring?

Q4. A student plots a force–extension graph for a spring. The straight‑line section has a gradient of $20 N/m$ . What does this gradient represent?

Q6. A spring has a spring constant of $50 N/m$ . It is stretched by $12$ $cm$ in the Hooke’s Law region. What is the force applied to the spring?

Q9. A spring obeys Hooke’s Law up to an extension of $0.20 m$ . As the extension increases within this region, what happens to the energy stored in the spring?