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Hooke's Law: What is it & Why it Matters (w/ Equation & Examples)

Anyone who has played with a slingshot has probably noticed that, in order for the shot to go really far, the elastic must be really stretched out before it is released. Similarly, the tighter a spring is squished down, the bigger a bounce it will have when released.

While intuitive, these outcomes are also described elegantly with a physics equation known as Hooke's law.

TL;DR (Too Long; Didn't Read)

Hooke's law states that the amount of force needed to compress or extend an elastic object is proportional to the distance compressed or extended.

An example of a proportionality law, Hooke's law describes a linear relationship between restoring force F and displacement x. The only other variable in the equation is a proportionality constant, k.

British physicist Robert Hooke discovered this relationship around 1660, albeit without math. He stated it first with a Latin anagram: ut tensio, sic vis. Translated directly, this reads "as the extension, so the force."

His findings were critical during the scientific revolution, leading to the invention of many modern devices, including portable clocks and pressure gauges. It was also critical in developing such disciplines as seismology and acoustics, as well as engineering practices like the ability to calculate stress and strain on complex objects.

Elastic Limits and Permanent Deformation

Hooke's law has also been called the law of elasticity. That said, it doesn't only apply to obviously elastic material such as springs, rubber bands and other "stretchable" objects; it can also describe the relationship between the force to change the shape of an object, or elastically deform it, and the magnitude of that change. This force can come from a squeeze, push, bend or twist, but only applies if the object returns to its original shape.

For example, a water balloon hitting the ground flattens out (a deformation when its material is compressed against the ground), and then bounces upward. The more the balloon deforms, the larger the bounce will be – of course, with a limit. At some maximum value of force, the balloon breaks.

When this happens, an object is said to have reached its elastic limit, a point when permanent deformation occurs. The broken water balloon will no longer go back to its round shape. A toy spring, such as a Slinky, that has been over-stretched will stay permanently elongated with large spaces between its coils.

While examples of Hooke's law abound, not all materials obey it. For example, rubber and some plastics are sensitive to other factors, such as temperature, that affect their elasticity. Calculating their deformation under some amount of force is thus more complex.

Spring Constants

Slingshots made out of different types of rubber bands don't all act the same. Some will be harder to pull back than others. That's because each band has its own spring constant.

The spring constant is a unique value depending on the elastic properties of an object and determines how easily the length of the spring changes when a force is applied. Therefore, pulling on two springs with the same amount of force is likely to extend one further than the other unless they have the same spring constant.

Also called the proportionality constant for Hooke's law, the spring constant is a measure of an object's stiffness. The larger the value of the spring constant, the stiffer the object and the harder it will be to stretch or compress.

Equation for Hooke's Law

The equation for Hooke's law is:

F = -kx

where F is force in newtons (N), x is displacement in meters (m) and k is the spring constant unique to the object in newtons/meter (N/m).

The negative sign on the right side of the equation indicates that the displacement of the spring is in the opposite direction from the force the spring applies. In other words, a spring being pulled downwards by a hand exerts an upwards force that is opposite from the direction it is being stretched.

The measurement for x is displacement from the equilibrium position. This is where the object normally rests when no forces are applied to it. For the spring hanging downwards, then, x can be measured from the bottom of the spring at rest to the bottom of the spring when it is pulled out to its extended position.

More Real-World Scenarios

While masses on springs are commonly found in physics classes – and serve as a typical scenario for investigating Hooke's law – they are hardly the only instances of this relationship between deforming objects and force in the real world. Here are several more examples where Hooke's law applies that can be found outside the classroom:

  • Heavy loads causing a vehicle to settle, when the suspension system compresses and lowers the vehicle towards the ground.
  • A flagpole buffeting back and forth in the wind away from its fully upright equilibrium position.
  • Stepping onto the bathroom scale, which records the compression of a spring inside to calculate how much additional force your body added.
  • The recoil in a spring-loaded toy gun.
  • A door slamming into a wall-mounted doorstop.
  • Slow-motion video of a baseball hitting a bat (or a football, soccer ball, tennis ball, etc., on impact during a game).
  • A retractable pen that uses a spring to open or close.
  • Inflating a balloon.

Explore more of these scenarios with the following example problems.

Hooke's Law Problem Example #1

A jack-in-the-box with a spring constant of 15 N/m is compressed -0.2 m under the lid of the box. How much force does the spring provide?

Given the spring constant k and displacement x, solve for force F:

F = -kx

F = -15 N/m(-0.2 m)

F = 3 N

Hooke's Law Problem Example #2

An ornament hangs from a rubber band with a weight of 0.5 N. The spring constant of the band is 10 N/m. How far does the band stretch as a result of the ornament?

Remember, weight is a force – the force of gravity acting on an object (this is also evident given the units in newtons). Therefore:

F = -kx

0.5 N = -(10 N/m)x

x = -0.05 m

Hooke's Law Problem Example #3

A tennis ball hits a racket with a force of 80 N. It deforms briefly, compressing by 0.006 m. What is the spring constant of the ball?

F = -kx

80 N = -k(-0.006 m)

k = 13,333 N/m

Hooke's Law Problem Example #4

An archer uses two different bows to shoot an arrow the same distance. One of them requires more force to pull back than the other one. Which has a larger spring constant?

Using conceptual reasoning:

The spring constant is a measure of an object's stiffness, and the stiffer the bow is, the harder it will be to pull back. So, the one that requires more force to use must have a larger spring constant.

Using mathematical reasoning:

Compare both bow situations. Since both of them will have the same value for displacement x, the spring constant must change with the force for the relationship to hold. Larger values are shown here with uppercase, bold letters, and smaller values with lowercase.

F = -Kx vs. f = -kx

References

About the Author

Amy Dusto is a high school science teacher and a freelance writer. She holds a Bachelor of Arts in Natural Sciences area and a Master of Arts in Science Writing from Johns Hopkins University. She has contributed to Discovery.com, Climate.gov, Science News and Symmetry Magazine, among other outlets.