Whenever you stretch a material, you are doing work. This is because you have to apply a force and the material extends in the direction of the force. You will know this if you have ever used exercise equipment with springs to develop your muscles (such as in Figure 7.16). Similarly, when you push down on the end of a springboard before diving, you are doing work. You transfer energy to the springboard, and you recover the energy when it pushes you up into the air.
We call the energy in a deformed solid the elastic potential energy or strain energy. If the material has been strained elastically (the elastic limit has not been exceeded), the energy can be recovered. If the material has been plastically deformed, some of the work done has gone into moving atoms past one another and the energy cannot be recovered.
The material warms up slightly. We can determine how much elastic potential energy is involved from a force–extension graph: see Figure 7.17. We need to use the equation that defines the amount of work done by a force. That is:

$work\,done = \,force\, \times \,distance\,moved\,in\,the\,direction\,of\,the\,force$

**Figure 7.16:** Using springs to help you exercise is hard work

This equation only holds when the force is constant. When you stretch a spring the force varies; so how can you find the work done?
There are two approaches. You can:
- use the average force in the equation for work; this works well where the force–extension graph is a straight line
- add together many small extensions, in each of which the force hardly changes; adding together lots of very small extensions shows us that the work done is the area under the force–extension graph.

Force / area= work done / Extension — **Figure 7.17:** Elastic potential energy is equal to the area under the force–extension graph

First, consider the linear region of the graph where Hooke’s law is obeyed, OA. The graph in this region is a straight line through the origin. The extension x is directly proportional to the applied force F. There are two ways to find the work done.

Method 1

We can think about the average force needed to produce an extension x. The average force is half the final force F, and so we can write:

$\begin{array}{l}
elastic\,potential\,energy\, = \,work\,done\\
elastic\,potential\,energy\, = \,\frac{{final\,force}}{2} \times extension\\
E = \frac{1}{2}\,Fx
\end{array}$

Method 2

The other way to find the elastic potential energy is to recognise that we can get the same answer by finding the area under the graph. The area shaded in Figure 7.17 is a triangle whose area is given by:

$area\, = \,\frac{1}{2} \times base \times height$

This again gives:

$E = \frac{1}{2}Fx$

The work done in stretching or compressing a material is always equal to the area under the graph of force against extension.
This is true whatever the shape of the graph, provided we draw the graph with extension on the horizontal axis. If the graph is not a straight line, we cannot use the Fx relationship, so we have to resort to counting squares or some other technique to find the answer.
Note that the elastic potential energy relates to the elastic part of the graph (i.e. up to the elastic limit), so we can only consider the force–extension graph up to the elastic limit.
There is an alternative equation for elastic potential energy. We know that, according to Hooke’s law, applied force F and extension x are related by $F = kx$, where k is the force constant. Substituting for F gives:

$\begin{array}{l}
elastic\,potential\,energy\, = \,\frac{1}{2}\,Fx\\
= \frac{1}{2} \times kx \times x\\
= \frac{1}{2}k{x^2}
\end{array}$

KEY EQUATION

$\begin{array}{l}
elastic\,potential\,energy\, = \,\frac{1}{2}\,Fx\\
= \frac{1}{2}Fx = \frac{1}{2}k{x^2}
\end{array}$

20) A force of $12 N$ extends a length of rubber band by $18 cm$. Estimate the energy stored in this rubber band. Explain why your answer can only be an estimate.

21) A spring has a force constant of $4800\,N\,{m^{ - 1}}$. Calculate the elastic potential energy when it is compressed by $2.0 mm$.

WORKED EXAMPLE

4) Figure 7.18 shows a simplified version of a force–extension graph for a piece of metal. Find the elastic potential energy when the metal is stretched to its limit of proportionality, and the total work that must be done to break the metal.

Force/N - Extension / ${10^{ - 3}}m$ — **Figure 7.18:** For Worked example 4

Step 1: The elastic potential energy when the metal is stretched to its elastic limit is given by the area under the graph up to the elastic limit. We will have to take the limit of proportionality,
the point A, as the elastic limit. The graph is a straight line up to $x = 5.0 mm$, $F = 20 N$, so the elastic potential energy is the area of triangle $0AB$:
$\begin{array}{l}
elastic\,potential\,energy\, = \,\frac{1}{2}\,Fx\\
\frac{1}{2} \times 20 \times 5.0 \times {10^{ - 3}}\\
= 0.05J
\end{array}$
Step 2: To find the work done to break the metal, we need to add on the area of the rectangle
ABCD:
$\begin{array}{l}
work\,done\, = \,total\,area\,under\,graph\\
= \,0.05 + (20 \times 25 \times {10^{ - 3}})\\
= 0.05 + 0.050\\
= 0.55J
\end{array}$

22) Figure 7.19 shows force–extension graphs for two materials. For each of the following questions, make the statement required. Also explain how you deduce your answer from the graphs.
a: State which polymer has the greater stiffness.
b: State which polymer requires the greater force to break it.
c: State which polymer requires the greater amount of work to be done in order to break it.

REFLECTION

What is the most important thing that you learned personally in this chapter?
Think of examples where having materials with high or low Young modulus is useful.
Make sure you know the formulae for stress, strain and Young modulus and can write them down in terms of F, A, x and L.

SUMMARY

Density is defined as the mass per unit volume of a substance:

$density = \frac{{mass}}{{volume}}$

Pressure is defined as the normal force acting per unit cross-sectional area:

$pressure = \frac{{normal\,force}}{{cross - \sec tional\,area}}$

Pressure in a fluid increases with depth: $p = \rho gh$
Upthrust on an object in a fluid is given by $F = \rho gV$ (Archimedes’ principle).
Hooke’s law states that the extension of a material is directly proportional to the applied force,
provided the limit of proportionality is not exceeded. For a spring or a wire, $F = kx$, where k is the force constant. The force constant has units of $N\,{m^{ - 1}}$.
Stress is defined as:

$stress = \frac{{force}}{{cross - \sec tional\,area}}$ or $\sigma = \frac{F}{A}$

Strain is defined as:

$strain = \frac{{exension}}{{original\,length}}$ or $\varepsilon = \frac{x}{L}$

To describe the behaviour of a material under tensile and compressive forces, we have to draw a graph of stress against strain. The gradient of the initial linear section of the graph is equal to the Young modulus. The Young modulus is an indication of the stiffness of the material.
The Young modulus E is given by:

$E = \frac{{stress}}{{strain}} = \frac{\sigma }{\varepsilon }(unit:\,pascal\,(Pa)\,or\,N\,{m^{ - 2}})$

The area under a force–extension graph is equal to the work done by the force.
For a spring or a wire obeying Hooke’s law, the elastic potential energy E is given by:

$E = \frac{1}{2}Fx$ or $E = \frac{1}{2}k{x^2}$

1) Which force is caused by a difference in pressure? [1]
A: drag
B: friction
C: upthrust
D: weight

2) Two wires P and Q both obey Hooke’s law. They are both stretched and have the same strain. The Young modulus of P is four times larger than that of Q.
The diameter of P is twice that of Q.
What is the ratio of the tension in P to the tension in Q? [1]
A: $\frac{1}{2}$
B: 1
C: 16

3) a: i- Define density. [1]
ii- State the SI base units in which density is measured. [1]
b: i- Define pressure. [1]
ii- State the SI base units in which pressure is measured. [1]
[Total: 4]

4) Sketch a force–extension graph for a spring that has a spring constant of $20\,N\,{m^{ - 1}}$ and that obeys Hooke’s law for forces up to $5.0 N$. Your graph should cover forces between 0 and $6 N$ and show values on both axes. [3]

5) Two springs, each with a spring constant $20\,N\,{m^{ - 1}}$, are connected in series.
Draw a diagram of the two springs in series and determine the total extension if a mass, with weight $2.0 N$, is hung on the combined springs. [5]

6) A sample of fishing line is $1.0 m$ long and is of circular cross-section of radius $0.25 mm$. When a weight is hung on the line, the extension is 50 mm and the stress is $2.0 \times {10^8}\,Pa$. Calculate:
a: the cross-sectional area of the line [1]
b: the weight added [2]
c: the strain in the line [2]
d: the Young modulus [2]
e: the percentage and absolute uncertainties in the Young modulus if the uncertainty in the extension is $ \pm 1\,mm$. [2]
[Total: 9]

7) This is the force–extension graph for a metal wire of length $2.0 m$ and crosssectional area $1.5 \times {10^{ - 7}}\,{m^2}$.

a: Calculate the Young modulus. [3]
b: Determine the energy stored in the wire when the extension is $0.8 mm$. [2]
c: Calculate the work done in stretching the wire between $0.4 mm$ and $0.8mm$. [2]
[Total: 7]

8) A piece of wax is attached to a newton-meter. In air, the reading on the newtonmeter is $0.27 N$ and when submerged in water of density $1000\,kg\,{m^{ - 3}}$ the reading is $0.16 N$. Calculate:
a: the upthrust on the wax when in water [1]
b: the volume of the wax [2]
c: the reading on the newton-meter when the wax is submerged in a liquid of density $800\,kg\,{m^{ - 3}}$. [2]
[Total: 5]

9) a: These are stress–strain curves for three different materials, P, Q and R.
State and explain which material has the greatest Young modulus. [2]

b: Describe an experiment to determine the Young modulus for a material in the form of a wire. Include a labelled diagram and explain how you would make the necessary measurements. Show how you would use your measurements to calculate the Young modulus. [7]
[Total: 9]

10) a: State the meaning of tensile stress and tensile strain. [2]
b: A vertical steel wire of length 1.6 m and cross-sectional area $1.3 \times {10^{ - 6}}\,{m^2}$ carries a weight of $60 N$. The Young modulus for steel is $2.1 \times {10^{11}}\,Pa$.
Calculate:
i- the stress in the wire [2]
ii- the strain in the wire [2]
iii- the extension produced in the wire by the weight. [2]
[Total: 6]

11) To allow for expansion in the summer when temperatures rise, a steel railway line laid in cold weather is pre-stressed by applying a force of $2.6 \times {10^5}\,N$ to
the rail of cross-sectional area $5.0 \times {10^{ - 3}}\,{m^2}$.
If the railway line is not pre-stressed, a strain of $1.4 \times {10^{ - 5}}$ is caused by each degree Celsius rise in temperature. The Young modulus of the steel is $2.1 \times {10^{11}}\,Pa$
a: State and explain whether the force applied to the rail when it is laid should be tensile or compressive. [2]
b: Calculate:
i- the strain produced when the rail is laid [3]
ii- the temperature rise when the rail becomes unstressed. [2]
[Total: 7]

12) This is a stress–strain graph for a metal wire.

The wire has a diameter of $0.84 mm$ and a natural length of $3.5 m$.
Use the graph to determine:
a: the Young modulus of the wire [3]
b: the extension of the wire when the stress is $0.60 GPa$ [2]
c: the force that causes the wire to break, assuming that the cross-sectional area of the wire remains constant [3]
d: the energy stored when the wire has a stress of $0.60 GPa$. [3]
[Total: 11]

13) This is a force–extension graph for a spring.

a: State what is represented by:
i- the gradient of the graph [1]
ii- the area under the graph. [1]
b: The spring has force constant $k = 80\,N\,{m^{ - 1}}$. The spring is compressed by $0.060 m$, within the limit of proportionality, and placed between two
trolleys that run on a friction-free, horizontal track. Each trolley has a mass of $0.40 kg$. When the spring is released the trolleys fly apart with equal speeds but in opposite directions.
i- How much energy is stored in the spring when it is compressed by $0.060 m$? [2]
ii- Explain why the two trolleys must fly apart with equal speeds. [2]
iii- Calculate the speed of each trolley. [2]
[Total: 8]

14) a: Liquid of density ρ fills a cylinder of base area A and height h.
i- Using the symbols provided, state the mass of liquid in the container. [1]
ii- Using your answer to i, derive a formula for the pressure exerted on the base of the cylinder. [2]
b: A boy stands on a platform of area $0.050\,{m^2}$ and a manometer measures the pressure created in a flexible plastic container by the weight W of the
boy, as shown.

platform area $0.050\,{m^2}$ / weight W / flexible plastic containner / manometer / metre rule showing distances in cm / water — **Figure 7.24**

The density of water is $1000\,kg\,{m^{ - 3}}$. Determine:
i- the pressure difference between the inside of the plastic container and ii- the atmosphere outside [2]
the weight W of the boy. [2]
[Total: 7]
15) This diagram shows water in a container filled to a depth of $0.50 m$. The density of water is $1000\,kg\,{m^{ - 3}}$.

a: Calculate the pressure at X on the base of the container. [2]
b: Explain why the pressure at X must be equal to the pressure at Y. [1]
c: Explain why the force downwards on the base of the container is larger than the weight of the liquid in the container. [2]
[Total: 5]

16) A light spring that obeys Hooke’s law has an unstretched length of $0.250 m$.
When an object of mass $2.0 kg$ is hung from the spring the length of the spring becomes $0.280 m$. When the object is fully submerged in a liquid of density $1200\,kg\,{m^{ - 3}}$, the length of the spring becomes $0.260 m$.
Calculate:
a: the spring constant of the spring. [2]
b: the upthrust on the object. [2]
c: the volume of the object. [2]
d: the density of the material from which the object is made. [2]
[Total: 8]

After studying the chapter, complete a table like this:

I can	See topic…	Needs more work	Almost there	Ready to move on
define and use density and pressure	7.1, 7.2
derive and use the equation $\Delta p = \rho gh\Delta h$	7.2
understand that upthrust is caused by a difference in pressure and use the equation $\Delta F = \rho gh\Delta V$	7.3
understand tensile and compressive forces, load, extension, limit of proportionality, elastic deformation, plastic deformation and elastic limit	7.4
recall and use Hooke’s law and the formula for the spring constant:	7.4
define and use stress, strain and the Young modulus	7.5
describe an experiment to measure the Young modulus	7.5
determine elastic potential or strain energy.	7.6