AS Level Physics 9702: Force-Extension Graph, Elastic Limit and Elastic PE

A student fixes one end of a thin metal wire and hangs increasing loads from the free end, recording the extension produced at each load. A graph of load $F$ (vertical axis) against extension $x$ (horizontal axis) is plotted from the results, with the following features.

From the origin O, the graph is a straight line up to a point labelled A, where $F_A = 8.0\text{ N}$ and $x_A = 2.0\text{ mm}$ .
Beyond A, the graph curves very slightly, but a short sample loaded to any point up to a further point labelled B (only slightly beyond A) is found to return exactly to its original length once the load is removed.
The wire is then loaded further, well beyond B, to a point labelled C, where $F_C = 12.0\text{ N}$ and $x_C = 4.0\text{ mm}$ . The load is then removed completely. Once fully unloaded, the wire is measured again and is found to be $1.0\text{ mm}$ longer than its original, unstretched length.

(a) State the name given to point A, and state the feature of the graph between O and A that identifies it. Calculate the spring constant of the wire for the region between O and A. [2]

(b) State what is meant by the elastic limit. A short sample of the wire is stretched only as far as a point between A and B, and the load is then removed. State and explain whether this sample returns to its original length, and hence state the name given to point B. [3]

(c) Calculate the elastic potential energy stored in the wire when it has been loaded, within the region obeying Hooke's law, up to point A. [2]

(d) State the term used to describe the type of deformation that has occurred in the wire between B and C, given that a permanent extension of $1.0\text{ mm}$ remains after the load is fully removed. Calculate the extension of the wire, measured at C (before unloading), that was recovered elastically once the load was removed. [2]

Part (a): Point A and the spring constant

Point A is the limit of proportionality. Between O and A, the graph is a straight line passing through the origin, which shows that the load $F$ is directly proportional to the extension $x$ . This is the region in which the wire obeys Hooke’s law.

The spring constant $k$ is given by the gradient of the graph in this straight-line region: $k = \frac{F_A}{x_A}$

Converting the extension to metres, $x_A = 2.0\text{ mm} = 2.0\times10^{-3}\text{ m}$ . Substituting: $k = \frac{8.0}{2.0\times10^{-3}} = 4.0\times10^{3}\text{ N m}^{-1}$

Part (b): The elastic limit and point B

The elastic limit is the point up to which a deformed material will return exactly to its original length once the load causing the deformation is removed; beyond it, some permanent extension remains.

The sample in this question is stretched only to a point between A and B. That is, it has not yet reached or passed the elastic limit. It therefore does return to its original length once unloaded: the deformation is still fully elastic, even though the graph is curved (not straight) in this small region beyond A, so $F$ and $x$ are no longer in direct proportion.

Point B is therefore the elastic limit: the point beyond which some permanent (plastic) extension would remain after the load is removed.

Part (c): Elastic potential energy up to A

The elastic potential energy (strain energy) stored is equal to the area under the force–extension graph. Since the region up to A obeys Hooke’s law, this area is a triangle, and: $E_p = \frac{1}{2}F_A x_A$

Substituting $F_A = 8.0\text{ N}$ and $x_A = 2.0\times10^{-3}\text{ m}$ : $E_p = \frac{1}{2}\times8.0\times2.0\times10^{-3} = 8.0\times10^{-3}\text{ J}$

So the elastic potential energy stored is $8.0\times10^{-3}\text{ J}$ , i.e. $8.0\text{ mJ}$ .

Part (d): Between B and C

The wire is loaded well beyond the elastic limit B to point C, and once the load is fully removed, a permanent extension of $1.0\text{ mm}$ remains. Because some extension is left behind after unloading, the wire has undergone plastic deformation between B and C.

Not all of the extension at C is permanent, however: the extension that disappears when the load is removed was recovered elastically. This elastic (recovered) part is the total extension at C minus the permanent extension that remains: $x_{\text{elastic}} = x_C - x_{\text{permanent}} = 4.0\text{ mm} - 1.0\text{ mm} = 3.0\text{ mm}$

So $3.0\text{ mm}$ of the extension at C was recovered elastically, while $1.0\text{ mm}$ remained as permanent (plastic) extension.

Final answers

(a) A $=$ limit of proportionality; $F \propto x$ up to this point. Spring constant $k = \boxed{4.0\times10^{3}}\text{ N m}^{-1}$ .
(b) The sample returns to its original length (deformation is elastic); B $=$ elastic limit.
(c) Elastic potential energy up to A $= \boxed{8.0\times10^{-3}}\text{ J}$ (i.e. $8.0\text{ mJ}$ ).
(d) Deformation between B and C is plastic; recovered elastic extension at C $= \boxed{3.0}\text{ mm}$ .