4.5.3Stretching, bending and compressing
You should be able to give examples of the forces involved in stretching, bending or compressing an object, and explain why, to change the shape of an object, more than one force has to be applied. This is limited to stationary objects only.
The reason more than one force is needed: a single force applied to a stationary object would simply make it move off, rather than change its shape. To deform it, the forces must be arranged so that the object stays where it is while different parts of it are pushed or pulled in different directions.
Stretching a spring hanging from a stand: the weight pulls down while the stand pulls up. Compressing a spring: hands push inwards from both ends. Bending a ruler: the ends are pushed one way while the middle is pushed the other.
You should also be able to describe the difference between elastic deformation and inelastic deformation caused by stretching forces.
• Elastic deformation: the object returns to its original shape and length when the forces are removed. • Inelastic deformation: the object does not return to its original shape and length — it stays permanently deformed.
Common mistake
"More than one force" does not mean the forces must be equal and opposite in a simple pair — bending needs three. What matters is that the object does not simply accelerate away, so its shape changes instead.
4.5.3Force and extension
The extension of an elastic object, such as a spring, is directly proportional to the force applied, provided that the limit of proportionality is not exceeded.
force = spring constant × extension F = k e
force, F, in newtons, N; spring constant, k, in newtons per metre, N/m; extension, e, in metres, m.
You should be able to recall and apply this equation. This relationship also applies to the compression of an elastic object, where 'e' would be the compression of the object.
You should be able to describe the difference between a linear and a non-linear relationship between force and extension, and calculate a spring constant in linear cases.
• Linear: the graph of force against extension is a straight line through the origin, so extension is directly proportional to force. The spring constant is the gradient. • Non-linear: the graph curves, so equal increases in force no longer give equal increases in extension. This happens beyond the limit of proportionality.
The extension is not the length of the spring — it is the increase in length from its original unstretched length. Subtract before plotting.
Worked example
A spring of natural length 0.120 m stretches to 0.180 m when a force of 3.0 N is applied, within the limit of proportionality. Calculate the spring constant.
- Find the extension, not the final length: e = 0.180 − 0.120 = 0.060 m.
- Write the equation: F = k e.
- Rearrange: k = F ÷ e.
- Substitute: k = 3.0 ÷ 0.060 = 50.
- Give the unit: newtons per metre.
Answer: 50 N/m
Common mistake
Use the extension, in metres, not the total length and not centimetres. Forgetting to subtract the original length, or leaving the extension in cm, are the two errors that ruin most spring constant calculations.
- Extension
- the increase in length of an elastic object when a force is applied; it is directly proportional to the force provided the limit of proportionality is not exceeded
- Spring constant
- the force needed per metre of extension, in newtons per metre, given by force = spring constant × extension
- Limit of proportionality
- the point beyond which extension is no longer directly proportional to the force applied
4.5.3Energy stored in a stretched spring
A force that stretches or compresses a spring does work, and elastic potential energy is stored in the spring. Provided the spring is not inelastically deformed, the work done on the spring and the elastic potential energy stored are equal.
That equality is why the energy can be recovered — a catapult or a clockwork mechanism gives back the work that was done on it, as long as the spring was not stretched past the point where it deforms permanently.
You should be able to calculate work done in stretching or compressing a spring, up to the limit of proportionality, using the equation:
elastic potential energy = 0.5 × spring constant × extension² Ee = ½ k e²
This equation is given on the equation sheet, and it also appears under changes in energy. You should be able to calculate relevant values of stored energy and energy transfers.
Worked example
A spring with a spring constant of 50 N/m is extended by 0.060 m. Calculate the elastic potential energy stored.
- Write the equation: Ee = 0.5 × k × e².
- Square the extension first: 0.060² = 0.0036.
- Substitute: Ee = 0.5 × 50 × 0.0036.
- 0.5 × 50 = 25, and 25 × 0.0036 = 0.09.
- Give the unit: joules.
Answer: 0.09 J
Required practical 6
Investigate the relationship between force and extension for a spring
- Clamp the spring securely to a stand, with the stand weighted down or clamped to the bench so it cannot topple.
- Set a metre rule vertically beside the spring and record the position of the bottom of the unloaded spring — this is the original length.
- Hang a 1 N weight from the spring, wait for it to settle, and record the new position, reading at eye level to avoid parallax error.
- Add weights one at a time, recording the position after each.
- Calculate the extension for each load by subtracting the original length from each reading.
- Remove the weights one at a time and record the positions again, to check whether the spring returns to its original length.
You should be able to interpret data from an investigation of the relationship between force and extension — which means reading the gradient for the spring constant, and identifying where the line stops being straight.
Common mistake
Plot force on the vertical axis and extension on the horizontal, as the specification's own graph does. Plotting them the other way round makes the gradient 1 ÷ k rather than k.
- Elastic potential energy
- the energy stored in a stretched or compressed spring; provided the spring is not inelastically deformed, it equals the work done on the spring
Before you practise
- More than one force is needed to change the shape of a stationary object, or it would simply move off.
- Elastic deformation reverses when the forces are removed; inelastic deformation is permanent.
- Extension is directly proportional to force, provided the limit of proportionality is not exceeded.
- F = k e must be recalled; e is the extension, in metres, and the same equation applies to compression.
- A linear relationship gives a straight line through the origin, and the gradient is the spring constant.
- Work done on a spring equals the elastic potential energy stored, provided it is not inelastically deformed.
- Ee = ½ k e² is given on the equation sheet — square the extension before multiplying.
- Required practical 6: add weights one at a time, measure extension from the original length, and plot force against extension.