2.2 Forces and elasticity — Learn

Prerequisites: Section 2.1, especially forces, newtonmeters and the unit newton; substituting numbers into a formula.

By the end of this section you can: explain why more than one force is needed to change the shape of an object; describe elastic and inelastic deformation; use the force-extension relationship for a spring; and calculate the energy stored in a stretched spring.

1. Changing the shape of an object

To stretch, bend or squeeze an object, you must apply more than one force. For a stationary object the forces are equal and opposite, so they squash or stretch the object instead of moving it. A spring stretched by your hands has a pull at each end. A sponge squeezed by your fingers has a push at each side.

2. Elastic and inelastic deformation

When the forces are removed, the object may or may not go back to its original shape.

Elastic deformation happens when the object returns to its original shape and size.
Inelastic deformation happens when the object stays changed. It does not return to its original shape.

A rubber band stretched a little springs back, so the change is elastic. A paper clip bent out of shape stays bent, so the change is inelastic.

3. Force and extension

The extension of a spring is how much longer it becomes. It is measured in metres. Push a spring instead and it gets shorter; the compression is how much shorter it becomes.

For a spring, the extension is directly proportional to the force applied, as long as the spring is not stretched too far. Directly proportional means that doubling the force doubles the extension.

F = k e

where $F$ is the force in newtons, $k$ is the spring constant in newtons per metre, and $e$ is the extension in metres. The spring constant tells you how stiff the spring is. A larger spring constant means a stiffer spring.

Worked example 1. A spring has a spring constant of 320 N/m and is stretched by 0.15 m. Calculate the force applied.

\begin{aligned} F &= k e \\ &= 320 \times 0.15 \\ &= 48\,\text{N} \end{aligned}

Worked example 2. A force of 20 N stretches a spring by 0.080 m. Calculate the spring constant.

\begin{aligned} k &= \frac{F}{e} \\ &= \frac{20}{0.080} \\ &= 250\,\text{N/m} \end{aligned}

Worked example 3. A spring has a spring constant of 120 N/m. A force of 36 N is applied. Calculate the extension.

\begin{aligned} e &= \frac{F}{k} \\ &= \frac{36}{120} \\ &= 0.30\,\text{m} \end{aligned}

4. The limit of proportionality

The straight-line relationship between force and extension holds only up to the limit of proportionality. Beyond that point, the line curves. The extension is then no longer proportional to the force, and the spring is inelastically deformed.

A graph of force against extension shows the pattern. Up to the limit of proportionality it is a straight line through the origin. Beyond the limit the line curves upwards, because each extra newton gives a larger extra extension.

Watch out. The extension is the increase in length, not the total length. A spring that is 12 cm long and stretches to 15 cm has an extension of 3 cm, which is 0.030 m. Put every length in metres before you use $F = ke$ , and do not add the natural length to the extension.

Try it: force and extension

The figure shows a spring and the graph of force against extension. The reader sets the force and the spring constant.

Predict first. Do not move the sliders yet. Write down your answers.

If you double the force on a spring, what happens to the extension?
What happens to the extension if you use a stiffer spring, with a larger spring constant?

Test. Set the spring constant to 200 N/m and the force to 10 N. Read the extension. Now double the force to 20 N and read it again. Then set the force past the limit of proportionality and watch the graph.

Explain. Why is the graph a straight line at first and a curve after the limit of proportionality? Why does a stiffer spring stretch less for the same force? Write your answer in two or three sentences.

5. Energy in a stretched spring

A force that stretches a spring does work on it. If the spring is not inelastically deformed, all of that work is stored in the elastic potential store. The energy stored is

E_e = \tfrac{1}{2} k e^2

where $k$ is the spring constant in newtons per metre and $e$ is the extension in metres. The extension is squared, so doubling the extension gives four times the energy.

Worked example 4. A spring has a spring constant of 200 N/m and is stretched by 0.20 m. Calculate the energy stored in the elastic potential store.

\begin{aligned} E_e &= \tfrac{1}{2} k e^2 \\ &= 0.5 \times 200 \times 0.20^2 \\ &= 4.0\,\text{J} \end{aligned}

The same expression gives the work done in stretching the spring, as long as the spring is not stretched past its limit of proportionality.

6. Progress check

Return to the chapter problem. The front spring has a force of 600 N and an extension of 0.020 m. Work out the spring constant. Then decide whether doubling the force to 1200 N would double the extension, and explain your answer.

You should now be able to

explain why more than one force is needed to change the shape of an object;
describe the difference between elastic deformation and inelastic deformation;
use $F = ke$ to find a force, a spring constant or an extension;
describe the force-extension graph up to and beyond the limit of proportionality;
calculate the energy stored in a stretched spring, using $E_e = \tfrac{1}{2}ke^2$ .