Physics 9702 · Paper 1

Hooke’s Law & the Spring Constant

Hooke's Law & the Spring Constant

A pair of forces deforms a solid: a tensile pair stretches it, a compressive pair squashes it.
Hooke's law: the extension is proportional to the force, up to a limit.
Use $F = kx$ , where $k$ is the spring constant in N m^-1.

The same bar under a tensile pair of forces (stretched longer than its dashed original outline) and under a compressive pair (squashed shorter). — A tensile pair of forces stretches the bar longer than its original shape (dashed); a compressive pair squashes it shorter.

Extension and Hooke's law

F = kx

Symbols

$F$ = force applied to the spring (the load) (N)
$k$ = spring constant, the stiffness (N m⁻¹)
$x$ = extension, the increase in length (m)

Extension is the increase in length (new minus original), not the new length.
Hooke's law says double the force, double the extension, up to the limit of proportionality. Past that point the straight-line rule no longer holds.
A bigger $k$ means a stiffer spring: you need more force for the same stretch.
On a force-extension graph, $k$ is the gradient of the straight part: it is a straight line through the origin while Hooke's law holds, and $k=\text{force}/\text{extension}$ . Read it off the straight section only.
A thicker sample of the same material and length gives a larger $k$ , because $k$ rises with the cross-sectional area. (9702/23/M/J/24 Q3(e))

Extension is the change in length, not the final length. Always work out new minus original first. Graphs that plot the length of the spring against force appear again and again in current papers: subtract the natural length before using any formula (9702/11/O/N/25 Q20).

Drag the load below to see the extension and the graph change.

extension x

k = 200 N m⁻¹·x = 0.050 m·F = kx = 10.00 N·E_p = ½kx² = 0.250 J

Worked example

Spring constant from one load

A spring extends 0.040 m when a 6.0 N load hangs from it.

Rearrange $F=kx$ to $k=F/x$ .
$k=6.0/0.040=150\text{ N m}^{-1}$ .

Answer

k=150\text{ N m}^{-1}

Your turn— tap to reveal the worked answer (9702/23/M/J/24 Q3)

State Hooke's law, then find the spring constant from a force-extension graph whose straight part rises to 5.4 N at an extension of 0.060 m. (9702/23/M/J/24 Q3(a), (c)(i))

Hooke's law: extension is proportional to the force, up to the limit of proportionality.
$k=\text{gradient}=5.4/0.060=90\text{ N m}^{-1}$ .

Answer

k=90\text{ N m}^{-1}

Common mistake

The word matters: extension is proportional to force, not equal to force. “Extension equals force” scores zero on a state-the-law question. And convert cm or mm to metres before using

F=kx

with N m^-1.

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Hooke’s Law & the Spring Constant