Spring Constant Revision Notes
Subject: Physics | Level: GCSE | Exam Board: OCR
This guide provides a comprehensive, exam-focused breakdown of the OCR GCSE Physics topic on the Spring Constant (2.7). It covers Hooke's Law, elastic potential energy, and the required practical (PAG P1), equipping candidates with the knowledge to secure maximum marks.
Revision Notes & Key Concepts
Revision Podcast Transcript
Hello, and welcome to your GCSE Physics revision podcast. I'm your tutor for today, and we're diving into one of the most rewarding topics in the Forces unit — the Spring Constant, Hooke's Law, and elastic deformation. Whether you're sitting Foundation or Higher tier, this topic is absolutely worth mastering, because it combines clear definitions, straightforward calculations, and a required practical all in one neat package. So let's get started. --- SECTION ONE: CORE CONCEPTS --- Let's begin with the big idea. When you stretch or compress a spring, it pushes or pulls back against you. The stiffer the spring, the harder it resists being deformed. Scientists measure this stiffness using a quantity called the spring constant, which we give the symbol k. The spring constant tells you exactly how many newtons of force are needed to stretch a spring by one metre. Its unit is newtons per metre, written N/m or N per m. Now, the relationship between force and extension was first described by Robert Hooke in 1676, and we still call it Hooke's Law today. The law states that the extension of a spring is directly proportional to the force applied to it, provided the force does not exceed the limit of proportionality. In equation form, we write this as F equals k times x. Let's unpack that. F is the force applied, measured in newtons. k is the spring constant, measured in newtons per metre. And x is the extension — that is, how much the spring has stretched beyond its natural, unloaded length — measured in metres. This is a critical point that catches many candidates out in the exam. The extension is NOT the total length of the spring. If your spring has a natural length of 10 centimetres and you stretch it to 15 centimetres, the extension is only 5 centimetres — the difference. Always subtract the original length. And always, always convert your extension into metres before substituting into any formula. Let me give you a feel for what the spring constant means physically. Imagine two springs. Spring A has a spring constant of 50 newtons per metre — it's quite soft, like a slinky. Spring B has a spring constant of 500 newtons per metre — it's very stiff, like the spring in a car suspension. If you apply the same force to both, Spring A will extend much further than Spring B. Higher k means stiffer spring, smaller extension for the same force. Now let's talk about the force-extension graph, because this is where OCR really likes to test you. When you plot force on the y-axis and extension on the x-axis, Hooke's Law gives you a straight line through the origin. The gradient of that straight line equals the spring constant k. This is beautiful because it means you can determine k experimentally just by measuring the gradient — and that's exactly what you do in PAG P1, the required practical. But here's where it gets interesting. The straight line doesn't go on forever. At a certain point, the line begins to curve. The point where it first deviates from being straight is called the limit of proportionality. Beyond this point, the relationship between force and extension is no longer proportional — the graph curves upward. OCR specifically requires you to use the term "limit of proportionality" to describe this point, not "elastic limit," which is a slightly different concept that comes just after it. Let's be precise about the difference. The limit of proportionality is the point beyond which force and extension are no longer proportional — the graph is no longer straight. The elastic limit is the point beyond which the spring will not return to its original shape when the force is removed. Between these two points, the spring is still elastic but no longer obeys Hooke's Law. Beyond the elastic limit, we enter plastic deformation, where the spring is permanently deformed. So to summarise the two types of deformation. Elastic deformation: the spring returns to its original shape when the force is removed. This happens below the elastic limit. Plastic deformation: the spring is permanently deformed and does not return to its original shape. This happens beyond the elastic limit. --- SECTION TWO: ELASTIC POTENTIAL ENERGY --- Now let's talk about energy. When you stretch a spring elastically, you do work on it, and that energy is stored as elastic potential energy. The formula for this is E equals one half times k times x squared. E is the elastic potential energy in joules, k is the spring constant in newtons per metre, and x is the extension in metres. There's a beautiful graphical interpretation of this formula. The elastic potential energy stored is equal to the area under the force-extension graph in the linear region. For a straight line through the origin, that area is a triangle, and the area of a triangle is one half times base times height — which is exactly one half times x times F, and since F equals kx, this gives us one half kx squared. So the formula and the graph are telling you the same thing. When answering exam questions about energy, always explicitly state that you are calculating the area under the line. Examiners award a mark for this statement, separate from the calculation itself. One more critical unit conversion to drill in. If extension is given in centimetres, divide by 100 to get metres. If given in millimetres, divide by 1000. Because x is squared in the energy formula, getting the unit wrong doesn't just shift your answer by a factor of 100 — it shifts it by a factor of 10,000. That's the difference between 0.05 joules and 500 joules. Always convert first. --- SECTION THREE: THE REQUIRED PRACTICAL — PAG P1 --- OCR's Practical Activity Group P1 requires you to investigate the relationship between force and extension for a spring. Let me walk you through it. The apparatus you need is: a retort stand with a clamp, a spring, a metre rule fixed vertically alongside the spring, a mass hanger, and a set of slotted masses. You also need a pointer attached to the bottom of the spring to make readings more precise. The method is as follows. First, measure and record the natural length of the spring with no masses attached — this is your zero-extension reference. Then add masses one at a time, typically 100 gram increments, and record the new total length of the spring each time. Calculate the extension by subtracting the natural length from each total length. Calculate the force by multiplying the mass in kilograms by 9.8, or 10 if your exam board allows it, to get the weight in newtons. Plot force on the y-axis and extension on the x-axis. Draw a line of best fit through the linear section. Calculate the gradient of the linear section — this is your spring constant k. Common errors in this practical include: using the total length instead of the extension; parallax error when reading the ruler — always read at eye level; not repeating measurements and taking averages; and continuing to add masses beyond the elastic limit, which permanently deforms the spring. Examiners testing this practical often ask you to describe how to improve the accuracy of the method, identify anomalous results on a graph, or explain why the line curves at high forces. --- SECTION FOUR: EXAM TIPS AND COMMON MISTAKES --- Let me now give you the examiner's perspective on the most common mistakes I see candidates make, and how to avoid them. Mistake number one: using total length instead of extension. I cannot stress this enough. Extension equals final length minus original length. Write this formula at the top of your working every single time. Mistake number two: failing to convert units. The formula F equals kx requires extension in metres. The formula E equals half kx squared requires extension in metres. If the question gives you centimetres, convert immediately. Write "x equals 15 cm equals 0.15 m" as your first line of working. Mistake number three: confusing limit of proportionality with elastic limit. OCR mark schemes are specific. If a question asks you to identify the limit of proportionality on a graph, you must identify the point where the line first becomes non-linear — where it stops being straight. The elastic limit is a separate point, slightly beyond this, where permanent deformation begins. Mistake number four: inverting the gradient. If the graph has force on the y-axis and extension on the x-axis, the gradient equals k. But if the axes are swapped — extension on y, force on x — the gradient equals one over k. Always check the axis labels before calculating the gradient. Mistake number five: forgetting to square x in the energy formula. E equals half k x squared. The x is squared. If x is 0.2 metres, then x squared is 0.04, not 0.2. For command word guidance: if the question says "state," give a brief factual answer — one or two sentences maximum. If it says "explain," you must give a reason using the word "because" — for example, "the extension increases because the force increases proportionally." If it says "calculate," show every step of your working, write the formula first, substitute values, then state the answer with correct units. --- SECTION FIVE: QUICK-FIRE RECALL QUIZ --- Right, let's test your recall. I'll ask a question, pause, then give the answer. Ready? Question one: What is the unit of the spring constant? ... The answer is newtons per metre, N/m. Question two: What does the gradient of a force-extension graph represent? ... The gradient equals the spring constant k. Question three: What is the formula for elastic potential energy stored in a spring? ... E equals one half k x squared. Question four: What is the difference between elastic and plastic deformation? ... Elastic deformation is reversible — the spring returns to its original shape. Plastic deformation is permanent — the spring does not return to its original shape. Question five: A spring has a natural length of 12 cm. Under a load, it measures 20 cm. What is the extension in metres? ... The extension is 20 minus 12 equals 8 cm, which equals 0.08 metres. Question six: What does the area under a force-extension graph represent? ... The area represents the work done on the spring, or the elastic potential energy stored. --- SECTION SIX: SUMMARY AND SIGN-OFF --- Let's bring it all together. The spring constant k measures the stiffness of a spring in newtons per metre. Hooke's Law tells us that F equals kx — force equals spring constant times extension — and this holds true up to the limit of proportionality. Beyond that point, the spring may still be elastic up to the elastic limit, but beyond the elastic limit it undergoes permanent plastic deformation. The elastic potential energy stored is E equals half kx squared, and this equals the area under the force-extension graph. In PAG P1, you determine k by measuring the gradient of the linear section of your graph. And the three golden rules for the exam are: always calculate extension not total length, always convert to metres, and always check which axis is which before calculating a gradient. You've got this. The Spring Constant topic is one of the most predictable in the entire OCR specification — the same types of questions come up year after year. Master the formula, understand the graph, and nail those unit conversions, and you will be picking up marks with confidence. Good luck with your revision, and I'll see you in the next episode. Keep going — every session brings you closer to that grade.
Key Terms & Definitions
- Spring Constant (k)
- A measure of the stiffness of a spring, defined as the force per unit extension.
- Hooke's Law
- The extension of an elastic object is directly proportional to the force applied, provided that the limit of proportionality is not exceeded.
- Extension (x)
- The increase in length of a spring from its original, unloaded length.
- Limit of Proportionality
- The point beyond which the force applied to an object is no longer directly proportional to its extension.
- Elastic Deformation
- A temporary deformation where the object returns to its original shape after the deforming force is removed.
- Plastic Deformation
- A permanent deformation where the object does not return to its original shape after the deforming force is removed.
- Elastic Potential Energy (Eₑ)
- The energy stored in an elastic object when it is stretched or compressed.
Worked Examples
Worked Example
Question: A spring has a spring constant of 40 N/m. Calculate the force that must be applied to the spring to cause it to extend by 25 cm. (3 marks)
Solution: Step 1: State the correct formula. F = kx [1 mark] Step 2: Convert the extension from cm to m. x = 25 cm = 0.25 m [1 mark] Step 3: Substitute the values and calculate the force. F = 40 N/m * 0.25 m = 10 N [1 mark]
Worked Example
Question: A student's data for stretching a spring is plotted on a graph of Force (y-axis) vs Extension (x-axis). The straight-line portion of the graph passes through the origin and the point (0.20 m, 5.0 N). Calculate the elastic potential energy stored in the spring when it is stretched by 0.20 m. (4 marks)
Solution: Step 1: Calculate the spring constant (k) from the graph's gradient. k = gradient = ΔF / Δx = 5.0 N / 0.20 m = 25 N/m [2 marks for method and correct calculation of k] Step 2: State the formula for elastic potential energy. E = 0.5 * k * x² [1 mark] Step 3: Substitute the values and calculate the energy. E = 0.5 * 25 N/m * (0.20 m)² = 0.5 * 25 * 0.04 = 0.5 J [1 mark]
Worked Example
Question: Describe the difference between elastic and plastic deformation, referring to the forces involved and the effect on the spring. (4 marks)
Solution: Elastic deformation occurs when a spring is stretched but returns to its original length when the force is removed [1 mark]. This happens as long as the applied force is below the elastic limit [1 mark]. Plastic deformation is when a spring is stretched by a force beyond its elastic limit [1 mark]. In this case, the spring is permanently deformed and does not return to its original length when the force is removed [1 mark].
Practice Questions
Question: A spring has a natural length of 15 cm. When a 400 g mass is hung from it, its length becomes 23 cm. Calculate the spring constant of the spring. (g = 9.8 N/kg)
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Question: State two ways a student could improve the accuracy of the experiment to determine a spring constant.
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Question: A spring with a spring constant of 200 N/m is stretched by 10 cm. How much elastic potential energy is stored in the spring?
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Question: A force-extension graph for a different spring is plotted with Extension (m) on the y-axis and Force (N) on the x-axis. The gradient of the straight line section is 0.025. What is the spring constant of this spring?
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Question: Explain why a spring feels stiffer the more it is stretched, even within its elastic limit.
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