7.6 - Force & Extension (Hooke's Law)
The experiment
Aim
To find out how the force applied to a spring affects how far it stretches, to find the spring's spring constant, and to work out how much energy is stored in it.
Background
- The extension is the increase in length from the spring's natural length, its length with nothing hanging on it.
- Hooke's law says the extension is directly proportional to the force. While it holds, , where is the force in N, is the extension in m, and is the spring constant in N/m. A stiff spring has a large spring constant.
- Hooke's law only holds up to the limit of proportionality. Past it, the graph curves.
- Stretching a spring also stores energy in its elastic potential energy store.
- You can't measure the spring constant directly, but you can measure the spring's length as the force on it increases.
- So in this practical you find the extension at each force and plot force against extension. The gradient of the straight part is the spring constant.
Variables
- Independent variable - The force on the spring, set by the masses hung on it.
- Dependent variable - The extension of the spring.
- Control variables - The spring, the position of the ruler, and the marker read against it.
Method
- Set up - Hang the spring from a clamp stand, and clamp a millimetre ruler upright beside it. Fix a strip of tape to the bottom of the spring as a marker.
- Measure the natural length - With nothing on the spring, read the marker against the ruler with your eye level with it.
- Add the masses - Add 100 g masses one at a time, each pulling with about 1 N, and read the length once the spring stops moving. Take at least six readings, without going past the spring's rated load.
- Unload the spring - Take the masses off one at a time, and check whether the spring goes back to its natural length.
- Repeat - Run through the loads a second time, and find the mean length at each force.
- Work out the extensions - Take the natural length off each mean length.
- Plot the graph - Force on the y-axis against extension on the x-axis, with a line of best fit.
To do the same for a metal wire, clamp one end and run the wire horizontally over a pulley at the edge of the bench, with the weights hanging on the far end and a piece of tape on the wire as a marker to read against a ruler. A wire extends by very little, so the loads have to be larger and the ruler read carefully. For a rubber band, hang it in place of the spring and load it the same way. The three give different shapes of graph, which is the point of testing more than one.
Safety
- Eye protection - A spring stretched too far can snap back.
- Stop the stand tipping - Clamp a heavy weight to its base, and put a tray under the masses.
- Stand, don't sit - So you can move away if the masses fall.
Worked example - Spring constant and energy stored
A student loaded a spring 100 g at a time, ran through the loads twice, and found the mean extension at each force.
| Force (N) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Mean extension (cm) | 0.0 | 3.8 | 7.6 | 11.5 | 15.3 | 19.0 |
They then loaded a second spring further.
| Force (N) | 0 | 1.0 | 2.0 | 3.0 | 4.0 | 5.0 | 6.0 |
|---|---|---|---|---|---|---|---|
| Extension (cm) | 0 | 2.0 | 4.0 | 6.0 | 8.0 | 10.5 | 14.0 |
Show that the first spring obeys Hooke's law, find its spring constant and the energy stored in it at 4 N, then find the second spring's limit of proportionality and spring constant.
Step 1 - Check the ratio of force to extension
Divide the force by the extension for each reading of the first spring. 1 ÷ 3.8, 2 ÷ 7.6, 3 ÷ 11.5, 4 ÷ 15.3, and 5 ÷ 19.0 all give 0.26 N/cm.
The ratio is constant, so the extension is directly proportional to the force and the spring obeys Hooke's law.
Step 2 - Find the spring constant from the gradient
With force on the y-axis and extension on the x-axis, the gradient is the spring constant. Change the extension to metres first, so 19.0 cm is 0.190 m.
Step 3 - Find the energy stored at 4 N
At 4 N the extension is 0.153 m. Use .
The area under the line, a triangle, gives the same answer.
Area = 0.5 × 0.153 × 4.0 = 0.31 J
The area method also works past the limit of proportionality, where the shape under the line is no longer a triangle. There, count the squares on the graph paper instead. Count the whole squares, add the part-squares that are more than half covered, and multiply by the value of one square, worked out from the scales on the two axes.
Step 4 - Find the second spring's limit of proportionality
Each extra newton adds 2.0 cm up to 4.0 N, but 2.5 cm and then 3.5 cm after that. The limit of proportionality is 4.0 N.
Step 5 - Find the second spring's spring constant
Use only the straight section, up to 4.0 N and 0.080 m.
Common exam questions
Question 1
Why is the ruler read with your eye level with the marker?
Reading it from above or below makes the marker appear next to the wrong mark, which is a parallax error.
Question 2
How would you check whether the spring has been stretched past its elastic limit?
Take the masses off one at a time and see whether it goes back to its natural length. If it stays longer, it has been permanently stretched past its elastic limit.
Question 3
How do you find the energy stored in a stretched spring from its force-extension graph?
It's the area under the graph up to that extension. While the line is straight the area is a triangle, so it's half the base times the height. Where the line has curved, count the squares under it instead.
Question 4
A student plots extension on the y-axis and force on the x-axis instead. What does the gradient of their graph give?
1 ÷ the spring constant, because the axes are the other way round. The spring constant is then 1 ÷ gradient.
Question 5
A rubber band gives a force-extension graph that isn't a straight line at any point. What does that tell you?
It doesn't obey Hooke's law, so it has no single spring constant.