Deformation and the Force–Extension Investigation

Changing the shape of a stationary object requires more than one force. If you pull one end of a spring while holding the other, your pull acts at one end and your holding hand provides an opposing force at the other. In a hanging-spring experiment, the load pulls down and the support pulls up. The forces balance once the spring is stationary, but they act on different parts of it and can change its shape. A single unbalanced force would make the object accelerate rather than remain stationary.

What happens when the forces are removed?

A spring stretched gently usually returns to its original length when the load is removed. This is elastic deformation: the change in shape is reversible.

If the spring is stretched too far, it may remain longer after the load is removed. This is inelastic deformation: there is a permanent change in shape. The distinction depends on what happens after unloading, not simply on how much the object stretches.

Measuring extension and describing stiffness

Extension is the increase in length, not the whole length of the stretched object:

$$e = L - L_0$$

Here, $L$ is the loaded length and $L_0$ is the original unloaded length. For example, a spring whose length changes from 8 cm to 11 cm has an extension of 3 cm, or 0.03 m.

Within a spring's proportional region, doubling the applied force doubles its extension. This relationship is called Hooke's law:

$$F = ke$$

$F$ is force in newtons (N), $k$ is the spring constant in newtons per metre (N/m), and $e$ is extension in metres (m). A larger spring constant means a stiffer spring: more force is needed for the same extension.

The equation can be rearranged to find either unknown:

$$k = \frac{F}{e} \qquad e = \frac{F}{k}$$

For example, a spring extending by 0.03 m under a force of 6 N has $k = 6/0.03 = 200\text{ N/m}$. For this spring, a 4 N force would produce an extension of $4/200 = 0.02\text{ m}$, provided it remains in the proportional region. Conversely, an extension of 0.01 m would require $F = 200 \times 0.01 = 2\text{ N}$.

The same relationship applies to compression of an elastic object. In that case, $e$ is the amount by which its length decreases, measured in metres.

Reading a force–extension graph

A graph with force on the vertical axis and extension on the horizontal axis shows direct proportionality as a straight line through the origin. Equal increases in force produce equal increases in extension. Beyond the limit of proportionality, the graph curves: the relationship becomes non-linear and a single constant spring constant no longer describes it.

Data for A spring's proportional and non-linear regions

The graph is straight up to an extension of 0.03 m and a force of 6 N. Beyond that point, its gradient changes. This identifies the limit of proportionality in this example. A curved loading graph alone does not tell you whether the deformation is permanent; that requires checking what happens when the force is removed.

In the straight-line region, the gradient gives the spring constant:

$$k = \frac{\Delta F}{\Delta e}$$

The symbol $\Delta$ means 'change in'. Using the points (0.01 m, 2 N) and (0.03 m, 6 N), the gradient is $(6-2)/(0.03-0.01) = 200\text{ N/m}$. A steeper force-against-extension line represents a stiffer spring.

Investigating force and extension: required practical 18

The aim is to investigate how changing the applied force affects a spring's extension. A suitable method uses a spring suspended from a clamp stand, a mass hanger and slotted masses, and a vertical ruler beside a pointer attached near the spring's lower end.

1. Secure the clamp stand to the bench and position the ruler close to the pointer. With no load attached, record the pointer's initial reading.
2. Add the mass hanger and record the total hanging mass. Wait for the spring to stop moving, then read the pointer at eye level.
3. Add masses in equal increments, recording the total mass and ruler reading each time. Stay within a safe loading range.
4. Remove the masses and check whether the spring returns to its initial reading. Repeat the loading measurements, ideally three times, and calculate mean readings.
5. Calculate each extension by subtracting the original unloaded reading from the mean loaded reading. Convert extensions into metres.
6. Calculate the applied force from the total hanging mass, including the hanger, using $F = mg$. Use mass in kilograms and the stated gravitational field strength in N/kg.
7. Plot force against extension and draw a best-fit line or curve appropriate to the results. Identify any straight-line region through the origin and calculate its gradient to find $k$.

The independent variable is the applied force; the dependent variable is extension. Use the same spring throughout so that you are investigating its response rather than comparing different springs.

A pointer makes the reading position clear. Reading at eye level reduces parallax error, and waiting for oscillations to stop makes readings more consistent. Repeats help reveal unusual readings and reduce the effect of random variation. If the graph bends, restrict spring-constant calculations to its proportional region.

Wear eye protection in case the spring breaks. Keep feet clear of the hanging masses, add masses carefully, and place a soft mat beneath them to reduce damage if they fall. Securing the stand helps prevent the apparatus from toppling.

$$F = ke \qquad k = \frac{F}{e} \qquad e = \frac{F}{k}$$