Define the spring constant as force per unit extension; recall and use the equation k = F / x

1.5.1 Effects of Forces – Spring Constant

Learning Objective

Define the spring constant as force per unit extension, recall and use the equation k = F ⁄ x, recognise the **limit of proportionality**, and apply the concept in calculations, practical investigations and related topics (work, power, energy stores, vectors and waves).

1. Definition, Hooke’s Law & Limit of Proportionality

• Spring constant (k) – a measure of a spring’s stiffness.
• Within the **elastic (linear) region** of a load‑extension graph the force and extension are directly proportional (Hooke’s law): \[ F = kx \] where F is the applied force and x the extension (or compression) from the natural length.
• Limit of proportionality (elastic limit) – the greatest extension for which the relationship F ∝ x remains valid.
  • Graphically it is the point up to which the load‑extension plot is a straight line through the origin.
  • Quantitatively, if the measured gradient changes by more than **5 %** between successive points, the limit has been exceeded.

2. Equation and Rearrangements

The basic definition is

From this you can solve for any of the three quantities:

1. Spring constant: \(k = \dfrac{F}{x}\)
2. Force for a given extension: \(F = k\,x\)
3. Extension for a known force: \(x = \dfrac{F}{k}\)

3. Units and Dimensions

4. Practical Investigation – Determining k

Apparatus

• Spring (unknown k) fixed at the upper end
• Set of masses (e.g. 0.1 kg, 0.2 kg, …)
• Hook or clamp
• Ruler or metre scale
• Stopwatch (optional, for later oscillation work)

Method

1. Measure the natural length \(L_0\) of the spring (no load).
2. Hang a mass \(m\) from the spring, record the new length \(L\). The extension is \(x = L - L_0\).
3. Calculate the force \(F = mg\) (use \(g = 9.8\ \text{m s}^{-2}\)).
4. Repeat steps 2–3 for at least five different masses.
5. Plot a graph of Force (N) (vertical) against Extension (m) (horizontal).
6. The gradient of the straight‑line portion equals the spring constant \(k\).
   • If the gradient varies by more than 5 % between points, you have passed the limit of proportionality.

Safety & Evaluation Checklist

• Secure the spring to avoid snapping.
• Do not exceed the limit of proportionality – watch for curvature in the graph.
• Possible errors: parallax when reading the ruler, friction at the hook, mass of the hook itself.
• Evaluate linearity (e.g., R² from spreadsheet), compare experimental \(k\) with a single‑point calculation, and note where the limit is reached.

5. Vector Addition of Perpendicular Spring Forces

6. Work, Power and Energy Stored in a Spring

• Work done in stretching/compressing: \[ W = \int_0^{x} kx'\,dx' = \frac{1}{2}kx^{2}. \] This follows directly from the definition of work \(W = Fd\) with \(F = kx\).
• Average power when the spring releases the stored energy over a time \(t\): \[ P = \frac{W}{t} = \frac{\tfrac12 kx^{2}}{t}. \] (Only a brief extension is needed for the syllabus.)
• Energy‑store comparison: Elastic (strain) energy \(\tfrac12 kx^{2}\) is another form of stored energy alongside thermal, chemical, gravitational, etc.

7. Connection to Waves

A mass–spring system performs simple harmonic motion, a classic example of a mechanical wave source. It demonstrates energy transfer without net matter transfer, satisfying the IGCSE wave requirement.

8. Worked Example (Single‑step)

Problem: A spring stretches \(0.025\ \text{m}\) when a force of \(5\ \text{N}\) is applied. Find the spring constant and the force needed to stretch the same spring by \(0.10\ \text{m}\).

1. Spring constant: \[ k = \frac{F}{x} = \frac{5\ \text{N}}{0.025\ \text{m}} = 200\ \text{N·m}^{-1} \]
2. Force for a \(0.10\ \text{m}\) extension: \[ F = kx = 200\ \text{N·m}^{-1}\times0.10\ \text{m}=20\ \text{N} \]

Result: \(k = 200\ \text{N·m}^{-1}\); a \(20\ \text{N}\) force will produce a \(0.10\ \text{m}\) stretch.

9. Multi‑step Challenge – Spring Constant, Energy & Dynamics

Question: A spring with \(k = 250\ \text{N·m}^{-1}\) is compressed by \(0.04\ \text{m}\) and then released. The compressed spring lifts a 0.5 kg block vertically.

1. (i) Calculate the maximum height the block reaches.
2. (ii) The block is attached to a light string that passes over a frictionless pulley and then to a 0.2 kg mass hanging on the other side. Determine the acceleration of the system just after release (assume the spring remains in its elastic region).

Solution Sketch

1. Elastic potential energy stored: \[ E_s = \frac12 kx^{2}= \frac12(250)(0.04)^{2}=0.20\ \text{J} \]
2. All this energy becomes gravitational potential energy of the block: \[ E_g = mgh \;\Rightarrow\; h = \frac{E_s}{mg}= \frac{0.20}{0.5\times9.8}=0.041\ \text{m} \]
3. Spring force at release: \(F_s = kx = 250\times0.04 = 10\ \text{N}\). Net upward force on the combined system: \[ F_{\text{net}} = F_s - (0.5g + 0.2g) = 10 - (4.9+1.96)=3.14\ \text{N} \] Total mass \(=0.5+0.2=0.7\ \text{kg}\). Acceleration: \[ a = \frac{F_{\text{net}}}{m_{\text{total}}}= \frac{3.14}{0.7}=4.5\ \text{m s}^{-2} \] \]

10. Common Mistakes to Avoid

• Using the total length of the spring as the extension; remember \(x = L - L_0\).
• Mixing units – always convert cm or mm to metres before using the formula.
• Applying the equation beyond the limit of proportionality; check the gradient change (≤5 %).
• For a compressed spring, keep \(x\) positive in the formula; the restoring force acts opposite to the direction of compression.
• When adding perpendicular spring forces, forget to use the Pythagorean result.

11. Practice Questions

1. A spring has a constant \(k = 150\ \text{N·m}^{-1}\). What force is required to stretch it by \(0.08\ \text{m}\)?
2. A force of \(12\ \text{N}\) stretches a spring \(0.04\ \text{m}\). Determine the spring constant.
3. If a spring with \(k = 250\ \text{N·m}^{-1}\) is compressed by \(0.015\ \text{m}\), what is the magnitude of the restoring force?
4. Multi‑step: A spring of unknown \(k\) is used in a load‑extension experiment. The data points (F / N, x / m) are (2, 0.010), (5, 0.025), (8, 0.040).
   1. Plot the points and find the gradient (use a ruler on graph paper).
   2. State the value of \(k\).
   3. Using this \(k\), calculate the force needed to compress the spring by \(0.060\ \text{m}\).
   4. Comment on whether the data indicate that the limit of proportionality has been reached.

12. Suggested Diagram

13. Summary

The spring constant \(k\) quantifies stiffness via the linear relationship \(F = kx\) (Hooke’s law) within the elastic region. It is obtained experimentally from the gradient of a load‑extension graph, and it enables calculation of forces, extensions, stored elastic energy, work, power, and dynamics of systems involving springs. Awareness of the **limit of proportionality**, the ability to combine perpendicular spring forces, and the connections to work, energy stores, power and wave phenomena ensure full mastery of the Cambridge IGCSE 0625 syllabus requirements.