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

Published by Patrick Mutisya · 14 days ago

IGCSE Physics 0625 – Effects of Forces: Spring Constant

1.5.1 Effects of Forces

Objective

Define the spring constant as force per unit extension and recall and use the equation \$k = \frac{F}{x}\$.

Definition of Spring Constant

The spring constant, denoted by \$k\$, quantifies the stiffness of a spring. It is defined as the ratio of the applied force \$F\$ to the resulting extension \$x\$ of the spring:

\$k = \frac{F}{x}\$

where:

  • \$F\$ = force applied to the spring (in newtons, N)

  • \$x\$ = extension produced (in metres, m)

  • \$k\$ = spring constant (in newtons per metre, N·m⁻¹)

Units and Dimensions

QuantitySymbolSI UnitDimension
Force\$F\$newton (N)\$\mathrm{MLT^{-2}}\$
Extension\$x\$metre (m)\$\mathrm{L}\$
Spring constant\$k\$newton per metre (N·m⁻¹)\$\mathrm{MT^{-2}}\$

Using the Equation \$k = \frac{F}{x}\$

To find any one of the three variables, rearrange the equation accordingly:

  1. Find the spring constant: \$k = \dfrac{F}{x}\$

  2. Find the force required for a given extension: \$F = k\,x\$

  3. Find the extension produced by a known force: \$x = \dfrac{F}{k}\$

Worked Example

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

Solution:

  1. Calculate \$k\$:

    \$k = \frac{F}{x} = \frac{5\ \text{N}}{0.025\ \text{m}} = 200\ \text{N·m}^{-1}\$

  2. Use \$k\$ to find the required force for \$x = 0.10\ \text{m}\$:

    \$F = k\,x = 200\ \text{N·m}^{-1} \times 0.10\ \text{m} = 20\ \text{N}\$

Thus, the spring constant is \$200\ \text{N·m}^{-1}\$ and a \$20\ \text{N}\$ force will stretch the spring by \$0.10\ \text{m}\$.

Common Mistakes to Avoid

  • Confusing extension \$x\$ with the total length of the spring. \$x\$ is the change in length, not the original length.

  • Using inconsistent units (e.g., cm for \$x\$ and N for \$F\$). Convert all lengths to metres before calculation.

  • For a compressed spring, treat \$x\$ as a positive magnitude; the direction of the force is opposite to the extension.

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?

Suggested Diagram

Suggested diagram: A spring fixed at the top with a weight hanging from the bottom, showing the original length \$L0\$, the extended length \$L\$, and the extension \$x = L - L0\$. Label the applied force \$F\$ acting downward.

Summary

The spring constant \$k\$ characterises how stiff a spring is. It is calculated using the simple ratio \$k = F/x\$, where \$F\$ is the applied force and \$x\$ is the resulting extension. Mastery of this relationship enables you to solve a wide range of problems involving springs in the IGCSE Physics syllabus.