Let's imagine that we are pulling down a spring. We can imagine that the spring will extend more if we pull it with a stronger force. But exactly what relationship do the force applied on the spring and the extension of the spring have? 😕
Hooke's law explains the relationship between the force applied on a spring and the extension of the spring. According to Hooke's law, the force applied on a spring and the extension of the spring are directly proportional.
If the force applied on the spring is doubled, the extension of the spring will also double.
F = k x e
F = force applied on spring (N)
k = spring constant (N/m)
e = extension of spring (m)
Each spring has its own spring constant, and the spring constant is a measure of how stiff the spring is.
If you want to investigate whether the spring obeys Hooke's law using the above apparatus, what should you do?
- You also need a ruler to measure the length of extension of the spring.
- First, measure the original length of the spring without any weight added to the spring.
- Then, add a known weight to the spring.
- Next, measure how much the spring has extended due to the weight.
- Repeat the experiment for a range of values of the weight added to the spring.
- You can also repeat the same experiment a few times for each weight and calculate the average value of spring extension.
- When you have obtained experimental data for the weight added and corresponding extension length of spring, you can plot the force-extension graph.
If the spring obeys Hooke's law....
- The graph should be a linear graph passing through the origin.
- The force applied on the spring (weight) is proportional to the extension of the spring.
- The slope of the graph is the spring constant of the spring used.
Using Hooke's law numerically
If the spring extended 4 cm when 8 N weight was added to the spring, how much will the spring extend if 10 N of weight is added to the spring?
Step I. Calculate the spring constant.
8N = k x 4cm = k x 0.04m
k = 8N / 0.04m = 200 N/m
Step II. Find the extension of spring when 10 N of weight is added.
10N = 200N/m x e
e = 10N / (200N/m) = 0.05 m = 5 cm