Edexcel A Level (IAL) Physics-1.27 Hooke’s Law- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -1.27 Hooke’s Law- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -1.27 Hooke’s Law- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 1.27 be able to use the Hooke’s law equation, ∆F = k∆x, where k is the stiffness of the object
Hooke’s Law \( \Delta F = k \Delta x \)
Hooke’s Law
Hooke’s Law describes how the extension of an elastic object (like a spring) depends on the force applied to it.
The relationship is:
\( \Delta F = k \Delta x \)
- \( \Delta F \) = applied force (N)
- \( \Delta x \) = extension (m)
- \( k \) = stiffness (spring constant), measured in \( \mathrm{N\,m^{-1}} \)
Understanding Hooke’s Law
- Extension is directly proportional to the force applied.
- The constant of proportionality is the stiffness \( k \).
- A larger value of \( k \) means a stiffer object (harder to stretch).
- Hooke’s Law only applies within the material’s elastic limit.
- Beyond the elastic limit, permanent deformation occurs and the law no longer holds.
Graph of Force vs Extension
- A straight line through the origin indicates Hooke’s Law is obeyed.
- The slope of the line = stiffness \( k \).
- Non-linear region indicates the spring is near or beyond its elastic limit.
Rearranging the Formula
- Stiffness:
\( k = \dfrac{\Delta F}{\Delta x} \)
- Extension:
\( \Delta x = \dfrac{\Delta F}{k} \)
Example (Easy)
A spring extends by \( 0.05\, \mathrm{m} \) when a force of \( 2.0\, \mathrm{N} \) is applied. Find its stiffness.
▶️ Answer / Explanation
\( k = \dfrac{\Delta F}{\Delta x} = \dfrac{2.0}{0.05} = 40\, \mathrm{N\,m^{-1}} \)
Example (Medium)
A spring with stiffness \( 120\, \mathrm{N\,m^{-1}} \) is stretched by \( 0.08\, \mathrm{m} \). Calculate the force required.
▶️ Answer / Explanation
\( \Delta F = k\Delta x = 120 \times 0.08 = 9.6\, \mathrm{N} \)
Example (Hard)
A spring obeys Hooke’s Law up to a maximum force of \( 15\, \mathrm{N} \). Its stiffness is \( 75\, \mathrm{N\,m^{-1}} \). What is the maximum extension before the spring exceeds its elastic limit?
▶️ Answer / Explanation
Rearrange to find extension:
\( \Delta x = \dfrac{\Delta F}{k} = \dfrac{15}{75} = 0.20\, \mathrm{m} \)
The spring can extend a maximum of \( 0.20\, \mathrm{m} \) before permanent deformation begins.