2.2 Forces and elasticity — Practise

Work through these on your own. Keep your working neat: the method matters as much as the answer.

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

A spring has a spring constant of 200 N/m. It is stretched by 0.16 m. Calculate the force applied to the spring.
A force of 15 N stretches a spring by 0.060 m. Calculate the spring constant of the spring.
A spring has a spring constant of 150 N/m. A force of 45 N is applied. Calculate the extension of the spring.
A spring has a spring constant of 250 N/m. It is stretched by 0.20 m, up to the limit of proportionality. Calculate the energy stored in the elastic potential store.
A spring has a spring constant of 400 N/m. It is compressed by 0.050 m. Calculate the energy stored in the elastic potential store.
A spring has a spring constant of 60 N/m. It is stretched by 0.35 m. Calculate the force applied to the spring.
A force is added to a spring and the extension is measured. The results are shown in the table. Use the gradient of the force-extension graph to find the spring constant, in N/m.
A spring stretches by 0.30 m when a force of 30 N is applied. The graph of force against extension is a straight line through the origin. Use the area under the graph to find the energy stored in the spring.
Describe the difference between elastic deformation and inelastic deformation. Give one example of each.
Describe the relationship between force and extension for a spring, up to the limit of proportionality and beyond it.
A spring has a spring constant of 500 N/m. It is stretched by 0.040 m. Calculate the force applied and the energy stored in the elastic potential store.

Answers and concise solutions

1. Force applied to the spring

Answer: $F = ke = 200 \times 0.16 = 32\,\text{N}$ .

2. Spring constant of the spring

Answer: $k = \dfrac{F}{e} = \dfrac{15}{0.060} = 250\,\text{N/m}$ .

3. Extension of the spring

Answer: $e = \dfrac{F}{k} = \dfrac{45}{150} = 0.30\,\text{m}$ .

4. Energy stored in the spring

Answer: $E_e = \tfrac{1}{2}ke^2 = 0.5 \times 250 \times 0.20^2 = 5.0\,\text{J}$ .

5. Energy stored in the spring

Answer: $E_e = \tfrac{1}{2}ke^2 = 0.5 \times 400 \times 0.050^2 = 0.50\,\text{J}$ .

6. Force applied to the spring

Answer: $F = ke = 60 \times 0.35 = 21\,\text{N}$ .

7. Spring constant from the graph

Answer: The graph is a straight line through the origin, so the gradient is the spring constant. Gradient $= \dfrac{12.0}{0.060} = 200\,\text{N/m}$ .

8. Energy stored, from the area under the graph

Answer: The area under a force-extension graph is the energy stored. Area $= \tfrac{1}{2} \times 0.30 \times 30 = 4.5\,\text{J}$ , which is the same as $E_e = \tfrac{1}{2}ke^2$ .

9. Elastic and inelastic deformation

In elastic deformation the object returns to its original shape and size when the forces are removed.
In inelastic deformation the object does not return to its original shape and size when the forces are removed.
A rubber band stretched a little is elastic. A spring stretched past its limit of proportionality, or a bent paper clip, is inelastically deformed.

10. Force and extension for a spring

Up to the limit of proportionality, the extension is directly proportional to the force, so the graph is a straight line through the origin.
Beyond the limit of proportionality the line curves, so the extension is no longer proportional to the force. The spring is inelastically deformed.

11. Force and energy stored

Answer: $F = ke = 500 \times 0.040 = 20\,\text{N}$ . $E_e = \tfrac{1}{2}ke^2 = 0.5 \times 500 \times 0.040^2 = 0.40\,\text{J}$ .