MYP Physics · topic MYP-06.04
Hooke's law and elastic response
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
The spring that stops obeying the first slope
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A wire of original length 1.0 m and cross-sectional area 1.0 mm² is loaded gradually. Its force–extension data are supplied below; segments between points are straight. The first 4 mm is reversible and proportional. After the 6 mm reading, unloading leaves 1 mm permanent extension. Supplied model: Use F = kx, stress = F/A, strain = ΔL/L and Young modulus = stress/strain in the proportional region. Supplied data (Extension / mm; Force / N): (0, 0), (2, 40), (4, 80), (6, 100). (a) Find the initial spring constant and Young modulus, converting the units explicitly. [3 marks] (b) Calculate the work done loading to 6 mm. Explain why ½Fmax xmax is invalid and whether all the work is recoverable. [3 marks]
Wire loading curve
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Force / N
Horizontal axis (x): Extension / mm
Read the exact data as a table
| Extension / mm | Force / N |
|---|---|
| 0 | 0 |
| 2 | 40 |
| 4 | 80 |
| 6 | 100 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find the initial spring constant and Young modulus, converting the units explicitly.
Reveal the worked solution
- Initial stiffness is 40/(0.002) = 2.0×10⁴ N m⁻¹. A millimetre is 10⁻³ m, while 1 mm² is 10⁻⁶ m².
- At 40 N, stress is F/A = 4.0×10⁷ Pa and strain is ΔL/L = 0.002.
- Young modulus = stress/strain = 2.0×10¹⁰ Pa. This material property requires the original geometry; k alone describes this particular wire.
- From 0 to 4 mm the area is ½(80)(0.004) = 0.16 J.
- From 4 to 6 mm it is ½(80+100)(0.002) = 0.18 J, so total loading work is 0.34 J.
- The full graph is not one straight proportional line, so ½Fmax xmax would give an incorrect 0.30 J. Permanent extension shows plastic deformation; without an unloading curve the recoverable energy cannot be quantified.
Answer (a) k = 2.0×10⁴ N m⁻¹; Young modulus = 2.0×10¹⁰ Pa. (b) Loading work = 0.34 J; not all is recoverable after permanent extension.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.