# How to calculate extension using hooke’s law?

Hello,

I am given load(N) and length(mm).

e.g. load 1N = 58mm(length).

The question is asking to calculate extension. I know f=kl where l is extension giving an equation of l=f/k. But i have f not k in the table. What should I do?

f=force

k=spring constant or stiffness Nm-1

l=extension or length m/cm/mm

Thank you very much!

1. Load “is” the force. They are just simply two names for the same thing.

As for the k constant of stiffness, this is ultimately your goal to calculate.

You are given the total length of the member, at any given tension load. Subtract the initial length of the member, to get the extension distance. That is the x in the formula F=k*x.

If your material is somehow a perfect linear elastic material, and your measurements are also perfect, then you should be able to take any given load divided by any given extension distance, and get the spring constant.

However, most of the time, this isn’t the case. You have deviations from the linear elastic trendline, that can fool you. It is best to get a good plot on a graph, and see the overall picture. Heck, even do some trendline analysis, to get what the slope of the linear elastic trendline actually is.

Also, after reaching a certain point in the material’s loading, it will no longer be linear elastic. It will be in the realm of irreversible deformation, due to exceeding the yield strength. It still can hold the load, but on release, it will not restore to the initial length.

To avoid including the irreversible deformation in your trendline, it is important that you look at your load vs extension plot, and seek the exact point, where the graph ceases to be close to a straight line. Do not include anything more extended than that, in it.

2. THE QUESTION IS VERY GENERAL ONE.Give the specific problem.

## Relevant information

Hooke’s law states that the force required to extend or compress a spring by some distance is directly proportional to that distance. The stiffness of the spring is a constant factor characteristic. The property of elasticity states that it takes twice the much force to stretch a spring twice as long. This linear dependence of displacement on stretching is known as Hooke’s law. This law is named after 17th-century British physicist Robert Hooke.

It says that the amount of stress we apply on any object is equal to that amount of strain is observed on it, which means Stress ∝ Strain.

Hooke’s Law Formula is given as

F = -K x

Where,

• F is the amount of force applied in N,
• x is the displacement in the spring in m,
• k is the spring constant or force constant.

Hooke’s law formula can be applied to determine the force constant, displacement, and force in a stretched spring.

### Solved Examples

Example 1

A spring is stretched by 10 cm and has a force constant of 2 cm /dyne. Determine the Force applied.

Solution:

Given parameters are

Force constant k is 2 cm/dyne,

Extension x = 10 cm.

The force applied formula is given by

F = – k x

= – 2 × 10 cm

= – 20 N

Example 2

Determine the force constant if a force of 100 N is stretching a spring by 0.8 m.

Solution:

Given parameters are

Force F = 100 N,

Extension, x = 0.2 m.

The force constant formula is given by

k = – F / x

= – 100 / 0.8

k = – 125 N/m.

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