How To Calculate Spring Force

If you have ever toyed in isolation with the sort of spring encountered in everyday objects and tools – say, the small kind inside the bottom of a "clickable" ballpoint pen – you may have noticed that it has certain general properties that set it apart from most other objects.

One of these is that it tends to return to the same size after you either stretch or compress it. Another, perhaps less obvious property is that the more you stretch or compress it, the harder it is to stretch or compress it even more.

These properties apply wholly to an ideal spring, and – to some extent – to springs used for all manner of purposes in the real world. Most other objects behave differently; those that rigidly resist deformation usually break when an applied force becomes strong enough, while others may stretch or be compressed but not return fully or at all to their original shape and size.

The unusual properties of springs, combined with a then-new conceptual framework about force and motion – advanced chiefly by Galileo Galilei and Issac Newton, led to the discovery of Hooke's law, a simple but elegant relationship that has applied to countless engineering and industrial processes in the modern world.

A spring is an ‌elastic‌ object, which means it has the various characteristics described in the previous section. That means that it resists being deformed (stretching and compression being two types of deformation), and it returns to its original dimensions provided the force remains within the spring's elastic limits.

Before the publication of Newton's laws, Robert Hooke (1635-1703) , an English scientist, discovered through some simple experimentation that the amount of deformation of objects was proportional to the forces applied to deform that object, as long as they had the property he termed "elasticity." Hooke, in fact, was a prolific scientist across almost all imaginable disciplines, even if he is not a household name today, in large part because of the sheer number of accomplished scientists operating throughout Europe in his time.

Hooke's law is very easy to work with – a luxury not often bestowed on physics work. It says that the force an elastic object exerts against deformation increases proportionally to the displacement of the spring (or elastic object) from its equilibrium position. Using less scientific jargon, this means that it becomes harder to stretch/compress/deform an elastic object the further it is from its original position at rest.

The equation describing this relationship describes the restoring force ‌F‌ that the spring exerts against the external force causing deformation. The restoring force is customarily negative to indicate it is the opposite direction of the force acting on the object, as it resists the deformation from rest position. There is a linear relationship between the opposite force of a spring and how much it has been deformed:

\(F = −kx\)

The constant ‌k‌ is called the spring constant, and ‌x‌ is the displacement from equilibrium (often the difference between the deformed and original length of the spring). The spring constant is different for different springs, as you would anticipate. Hooke's law, which you can think of as a "spring force formula," is in play in a variety of different tools and aspects of life, such as archery bows and the shock absorbers and bumpers on automobiles.

For simple examples, you can use your own head as spring force calculator. For example, if you are told that a spring exerts a force of 1,000 N when stretched by 2 m, you can divide to get the spring constant: 1,000/2 = 500 N/m.

Bear in mind that although people may think of springs more as "stretchable" than "compressible," if a spring is properly constructed (that is, has enough room between successive coils), it can be significantly compressed as well as stretched, and Hooke's law applies in both directions of deformation.

Imagine a system with a block sitting on a frictionless surface and connected to a wall by a spring that is at equilibrium, meaning that it is being neither compressed nor stretched. If you pull the block away from the wall and let it go, what do you think will happen?

At the moment you release the block, a force ‌F‌, in accordance with Newton's second law (F = ma), acts to accelerate the block toward its starting point. Thus for Hooke's law in this situation:

\(F = -kx = ma\)

From here it is possible, using ‌k‌ and ‌m‌, to predict the mathematical behavior of the oscillation, which is wavelike in nature. The block is at is fastest at the times it passes through its starting point in either direction and, more evidently, at its slowest (0) when it reverses direction.

Theory vs. Reality

What happens in this imaginary situation is that the block passes its starting point and oscillates back and forth across its starting point, being compressed by the same distance it was first stretched in each trip toward the wall and then zooming back out to where you pulled it, in a never-ending cycle. In the real world, the spring would not be ideal and its material would eventually lose its elasticity, but more importantly, friction in reality is unavoidable; its net force soon reduces the magnitude of the oscillations, and the block returns to rest.

You have seen that a spring has inherent, or built-in, properties that can be leveraged to do work in a way that, say, bubble gum or a ball bearing cannot. As a result, springs can be described in terms of not just force but energy. (Work has the same fundamental unit as energy: the newton-meter or N⋅m),

To deform the spring, you or something else must do work on it. ‌The energy you impart using your arm is "transferred" into elastic potential energy‌ when the spring is held stretched. This is analogous to an object above the ground having gravitational potential energy, and its value is:

\(E_P = \frac{1}{2}kx^2\)

Say you use a compressed spring to launch an object along a frictionless surface. The energy in this ideal situation has been "converted" entirely into kinetic energy at the instant the object leaves the spring, where:

\(E_K = \frac{1}{2}mv^2\)

Thus if you know the mass of the object, you can use algebra to solve for the velocity ‌v‌ by setting ‌_E_P‌ (initial) to ‌E_K_‌ at "launch." We can describe the total mechanical energy of such a situation as the sum of kinetic and potential energy.