4.2 Force-extension graphs and stored energy — Practise

Work through these on your own. Decide whether each graph is a straight line before you choose a formula, and write the unit of every answer. In the last two questions, explain your reasoning in full sentences.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A spring extends by $0.050\,\text{m}$ under a force of $10\,\text{N}$ . Find the spring constant.
A spring of constant $250\,\text{N}\,\text{m}^{-1}$ is stretched by $0.080\,\text{m}$ . Find the force.
A spring of constant $250\,\text{N}\,\text{m}^{-1}$ is stretched by $0.080\,\text{m}$ by a force of $20\,\text{N}$ . Find the energy stored.
A force-extension graph is a straight line through $(0, 0)$ and $(0.040\,\text{m}, 8.0\,\text{N})$ . Find the gradient, which is the spring constant.
The same straight-line graph reaches $8.0\,\text{N}$ at an extension of $0.040\,\text{m}$ . Find the energy stored, from the area under the graph.
A wire of length $1.50\,\text{m}$ and area $2.0 \times 10^{-7}\,\text{m}^2$ has a force-extension graph of gradient $2.0 \times 10^{4}\,\text{N}\,\text{m}^{-1}$ . Find the Young modulus.
A spring of constant $400\,\text{N}\,\text{m}^{-1}$ is stretched by $0.10\,\text{m}$ . Find the energy stored.
A wire carries a force of $60\,\text{N}$ and extends by $25\,\text{mm}$ . Find the energy stored in the wire.
A force-extension graph for a rubber cord gives the points $(0, 0)$ , $(0.020, 3.5)$ , $(0.040, 7.5)$ , $(0.060, 13)$ and $(0.080, 20)$ . Find the energy stored at an extension of $0.080\,\text{m}$ .
A spring stores $0.36\,\text{J}$ when it is extended by $0.060\,\text{m}$ . The graph is a straight line through the origin. Find the final force.
Explain why $E = \tfrac{1}{2}F\Delta L$ cannot be used to find the energy stored when the force-extension graph is not a straight line.
A spring obeys Hooke's law up to a force of $15\,\text{N}$ . A student loads it to $25\,\text{N}$ and finds that the extension is larger than $F = k\Delta L$ predicts. Explain why.

Answers and concise solutions

1. The spring constant

Answer: From Hooke's law, $k = \dfrac{F}{\Delta L} = \dfrac{10}{0.050} = 200\,\text{N}\,\text{m}^{-1}$ .

2. The force

Answer: The force is $F = k\Delta L = 250 \times 0.080 = 20\,\text{N}$ .

3. The energy stored

Answer: The graph is a straight line through the origin, so the energy is $\tfrac{1}{2}F\Delta L = \tfrac{1}{2} \times 20 \times 0.080 = 0.80\,\text{J}$ .

4. The gradient

Answer: The gradient is $\dfrac{8.0}{0.040} = 200\,\text{N}\,\text{m}^{-1}$ .

5. The energy stored

Answer: The area is a triangle, so the energy stored is $\tfrac{1}{2} \times 8.0 \times 0.040 = 0.16\,\text{J}$ .

6. The Young modulus

Answer: The gradient of a wire's force-extension graph is $k = \dfrac{EA}{L}$ , so $E = \dfrac{kL}{A} = \dfrac{2.0 \times 10^{4} \times 1.50}{2.0 \times 10^{-7}} = 1.5 \times 10^{11}\,\text{Pa}$ .

7. The energy stored

Answer: The energy stored is $\tfrac{1}{2}k(\Delta L)^2 = \tfrac{1}{2} \times 400 \times 0.10^2 = 2.0\,\text{J}$ .

8. The energy stored

Answer: The extension is $25\,\text{mm} = 0.025\,\text{m}$ , so the energy stored is $\tfrac{1}{2}F\Delta L = \tfrac{1}{2} \times 60 \times 0.025 = 0.75\,\text{J}$ .

9. The energy stored

Answer: The graph is not a straight line, so use strips. The trapezium areas are $0.035$ , $0.11$ , $0.205$ and $0.33\,\text{J}$ , and their sum is $0.68\,\text{J}$ .

10. The final force

Answer: For a straight-line graph, $E = \tfrac{1}{2}Fx$ , so $F = \dfrac{2E}{x} = \dfrac{2 \times 0.36}{0.060} = 12\,\text{N}$ .

11. Why the triangle formula fails

The formula half F times extension is the area of a triangle. It is only correct when the graph is a straight line through the origin, because then the area under the graph is a triangle.
When the graph bends, the area under it is not a triangle. Find the real area by counting squares or by splitting it into strips and adding their areas.

Answer: The formula gives the area of a triangle, which is only right for a straight line through the origin. When the graph bends, find the area under it by counting squares or by adding strips.

12. Why the extension is larger

The force of 25 N is past the limit of proportionality, so the extension is no longer proportional to the force. Hooke's law only describes the straight part of the graph.
On the graph the line bends away from the straight line, so the gradient is no longer constant. At 25 N the extension is larger than the straight line would give, and the spring may have been permanently stretched. That happens past the elastic limit.

Answer: The 25 N load is above the limit of proportionality, so the extension stops being proportional to the force and the graph bends away from the straight line. At that load the spring may also have passed its elastic limit and be permanently stretched.