The **Hooke’s law** points out that the deformation experienced by an elastic object is directly proportional to the force applied to it. The constant of proportionality depends on the nature of the object, its geometry and the material from which it is made.

All materials have elastic properties to a greater or lesser extent, so they obey Hooke’s law as long as they return to their original dimensions once the force is removed. Springs and rubber bands are good examples of objects that satisfy Hooke’s Law, but so are the steel rods that are part of a bridge.

Taking a spring as an example, to keep it stretched or compressed, it is necessary to apply a force whose magnitude is F. According to Hooke’s law, the spring will experience a deformation x:

F ∝ x

The constant of proportionality, which since it is a spring is called *spring stiffness constant*is denoted as k, therefore:

F = k⋅x

In SI units, force is in newtons (N) and deformation is in meters (m). Therefore, the spring constant has units of N/m. The spring constant represents the force that must be applied to deform it by 1 m of length.

If, after stretching or compressing, the spring is released, it will move in the opposite direction to the applied force. This means that if we stretch it, it is compressed and vice versa. Therefore the force FR that **the spring exerts **is:

RF = -k⋅x

The negative sign indicates what was said before: that the force opposes the displacement, therefore this force is known as *restorative force*.

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**formula and equations**

The relationship between force and displacement in a spring was discovered by Robert Hooke (1635-1703), a noted English physicist known for his rivalry with Isaac Newton. Hooke was a versatile scientist who successfully ventured into different fields of science: mechanics, biology, astronomy and architecture.

Hooke realized that if the force applied to a spring is not very large, the spring deforms proportionally to the force, and once the force is removed, the spring returns to its natural length.

Thus, Hooke’s law in graphical form has the form of a straight line, the slope of which is the spring constant. The following image shows the force exerted on the spring to stretch it –or compress it- depending on the position x. Also note that the force does not depend on the natural length of the spring, but on its displacement.

The average force is indicated on the graph by the bar F and is equal to ½ kxf, where xf is the final position of the spring.

Both the force exerted on the spring and the force it exerts on an object attached to it are variable forces. The more you want to stretch or compress the spring, the more force you have to apply to do so.

**Work done to stretch or compress a spring**

When a force is applied that deforms the spring, work is done that is stored in the spring and can be used later.

Mechanical work is defined as the area under the graph of force F as a function of position x. To calculate the work W that a variable force F(x) does when moving an object from position x1 to position x2, we must calculate the definite integral:

In the case of the work necessary to bring a spring from its equilibrium position to the xf position, it is very simple, since the area to be calculated is that of the triangle shaded in gray in figure 4, whose formula is known:

Triangle area = ½ base. height

Therefore the work required is:

W = ½ x f . (kxf) = ½ k (xf)2

And if you want to calculate the work necessary to bring the spring from position x to position xf, it would be equivalent to calculating the area of the hatched trapezoid in figure 5:

W = ½k(xf)2 – ½kx2

**Examples of springs**

Depending on the application for which they are intended, the springs can be helical, cylindrical, conical, spiral, with a circular cross section (the most common), square or rectangular in section.

A widely used classification is according to the type of effort to which they are going to be subjected: there are torsion, flexion, compression and extension springs. The latter are used extensively and there are those that work equally well for tension and compression.

**Compression spring**

An example of a compression spring is the one used in the toy called *pogo *either* jumping stick* These springs store a lot of potential energy when they are compressed and gradually release it as they return to equilibrium. In this way the rebounds are not too abrupt.

**Extension and torsion springs**

Trampoline springs are the type of extension springs and are made with tightly wound coils, with two hooks at the ends. They are capable of retaining quite a bit of potential energy, which they then release when someone steps up and starts jumping on the mat, which also has its own elastic response, like all materials.

Torsion springs are very common, because they are used to make clothespins. Instead of hooks at the ends, these are bent at an angle, to resist the forces that tend to exert torsion.

**Materials to make springs**

The most suitable materials for manufacturing springs are those with a *last stand* (final resistance) high, that is, they withstand a great effort before breaking. It is also desirable for the material to have a high yield point, so that it does not lose its elastic qualities with little stress.

Industrial use springs are made from alloys including high carbon steel, copper, nickel and bronze.

**Hooke’s Law Applications**

Because springs have the ability to store potential energy when stretched or compressed, they are capable of doing work by moving things like mechanisms.

In this way, springs have a multitude of applications, from small and everyday objects, through automobiles, to machinery of all kinds. The springs are used to:

-Dampen vibrations.

-Manufacture retractable mechanisms: pens, clothespins, hair clips.

-Make spring scales or dynamometers

And they are also part of the mechanism of:

-Clocks.

-Trampolines.

-Locks.

-Toys.

-Weapons.

-Needle meters, for example the galvanometer, used to measure currents, voltages and resistances.

**solved exercises**

**– Exercise 1**

A force of magnitude 5.0 N is applied to a spring, causing it to stretch a length of 3.5 cm from its natural length.

a) How far is it stretched when the applied force is 7 N?

b) Find the work done by the applied force in stretching the spring 3.5 cm from its natural length.

**Solution to**

Knowing that the spring is stretched 3.5 cm by applying 5.0 N we can calculate its constant:

k = F / x = 5.0 N / 3.5 cm = 1.43 N / cm.

When a force of 7 N is applied, the following stretch is obtained:

x = F / k = 7.0 N / 1.43 N/m = 4.9 cm

**solution b**

The work required to deform a spring is given by:

W = ½ kx2 = 0.5 x 1.43 N/cm x (3.5 cm)2 = 8.76 N . cm = 8.76N. 1 x10 -2 m = 0.0876 J.

**– Exercise 2**

A spring of negligible mass and 10 cm long hangs from a support. If a 2 kg mass is attached to it, the spring stretches to 15 cm. Calculate:

a) The spring constant

b) The size of the spring when a mass of 3 kg is suspended.

**Solution to**

The stretch of the spring is x = 15 – 10 cm = 5 cm

Since the system is in static equilibrium, the force exerted by the spring when it is stretched is directed vertically upwards, to compensate for the weight, which is directed downwards, then:

FR = W → kx = mg

k = 2 x 9.8 N / 5 x10 -2 m = 392 N/m

**solution b**

When a 3 kg weight is suspended, the new force is W = 3 x 9.8 N = 29.4 N

In such a case the stretch is:

x = mg /k = 29.4 N / 392 N/m = 0.075 m = 7.5 cm

**References**

Bauer, W. 2011. Physics for Engineering and Science. Volume 1. Mc Graw Hill.

Creative Mechanisms Blog. Four different types of springs. Recovered from: creativemechanisms.com.

Figueroa, D. (2005). Series: Physics for Science and Engineering. Volume 2. Dynamics. Edited by Douglas Figueroa (USB).

Giancoli, D. 2006. Physics: Principles with Applications. 6th. Ed. Prentice Hall.

Knight, R. 2017. Physics for Scientists and Engineering: a Strategy Approach. pearson.