Energy in Hooke’s Law of Deformation

In order to produce a deformation, work must be done. That is, a force must be exerted through a distance, whether you pluck a guitar string or compress a car spring. If the only result is deformation, and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy. The potential energy stored in a spring is \({\text{PE}}_{\text{el}}=\cfrac{1}{2}{\mathrm{kx}}^{2}\). Here, we generalize the idea to elastic potential energy for a deformation of any system that can be described by Hooke’s law. Hence,

\({\text{PE}}_{\text{el}}=\cfrac{1}{2}{\mathrm{kx}}^{2},\)

where \({\text{PE}}_{\text{el}}\) is the elastic potential energy stored in any deformed system that obeys Hooke’s law and has a displacement \(x\) from equilibrium and a force constant \(k\).

It is possible to find the work done in deforming a system in order to find the energy stored. This work is performed by an applied force \({F}_{\text{app}}\). The applied force is exactly opposite to the restoring force (action-reaction), and so \({F}_{\text{app}}=\text{kx}\). This figure shows a graph of the applied force versus deformation \(x\) for a system that can be described by Hooke’s law. Work done on the system is force multiplied by distance, which equals the area under the curve or \((1/2){\mathrm{kx}}^{2}\)(Method A in the figure). Another way to determine the work is to note that the force increases linearly from 0 to \(\mathrm{kx}\), so that the average force is \((1/2)\mathrm{kx}\), the distance moved is \(x\), and thus \(W={F}_{\text{app}}d=[(1/2)\text{kx}](x)=(1/2){\mathrm{kx}}^{2}\) (Method B in the figure).

The graph here represents applied force, given along y-axis, versus deformation or displacement, given along x axis. The slope is linear slanting and the slope area is covered between x axis and the slope, given by F is equal to k multiplied by x, where k is constant and x is displacement. The force applied along y-axis is given by half of k multiplied by x. Along with the graph, two methods are provided to calculate weight, W. The first method gives the solution by multiplying half of b multiplied by h, whereas in the second we can get the solution by multiplying f with x.

A graph of applied force versus distance for the deformation of a system that can be described by Hooke’s law is displayed. The work done on the system equals the area under the graph or the area of the triangle, which is half its base multiplied by its height, or \(W=(1/2){\mathrm{kx}}^{2}\).

Example: Calculating Stored Energy: A Tranquilizer Gun Spring

We can use a toy gun’s spring mechanism to ask and answer two simple questions: (a) How much energy is stored in the spring of a tranquilizer gun that has a force constant of 50.0 N/m and is compressed 0.150 m? (b) If you neglect friction and the mass of the spring, at what speed will a 2.00-g projectile be ejected from the gun?

The figure a shows an artistic impression of a tranquilizer gun, which shows the inside of it revealing the gun spring and a panel just below it, in the outside area, attached to the spring. This stage shows the gun before it is cocked, and the spring is uncompressed covering the entire inside area. The figure b shows the gun with the spring in the compressed mode. The spring has been compressed to a distance x, where x distance shows the vacant area inside the gun through which the spring has been compressed. The panel is also moving along the spring. And a bullet of mass m is shown at the front of the compressed spring. The spring here has elastic potential energy, represented by P E sub e l. The figure c is the third stage of the above two stages of the gun. The spring here is released from the compressed stage releasing the bullet in the outer forward direction with velocity V and the spring’s potential energy is converted into kinetic energy, represented here by K E.

(a) In this image of the gun, the spring is uncompressed before being cocked. (b) The spring has been compressed a distance \(x\), and the projectile is in place. (c) When released, the spring converts elastic potential energy \({\text{PE}}_{\text{el}}\) into kinetic energy.

Strategy for a

(a): The energy stored in the spring can be found directly from elastic potential energy equation, because \(k\) and \(x\) are given.

Solution for a

Entering the given values for \(k\) and \(x\) yields

\(\begin{array}{lll}{\text{PE}}_{\text{el}}& =& \cfrac{1}{2}{\mathrm{kx}}^{2}=\cfrac{1}{2}(\text{50}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{N/m}){(0\text{.}\text{150}\phantom{\rule{0.25em}{0ex}}\text{m})}^{2}=0\text{.}\text{563}\phantom{\rule{0.25em}{0ex}}\text{N}\cdot \text{m}\\ & =& 0\text{.}\text{563}\phantom{\rule{0.25em}{0ex}}\text{J}\end{array}\)

Strategy for b

Because there is no friction, the potential energy is converted entirely into kinetic energy. The expression for kinetic energy can be solved for the projectile’s speed.

Solution for b

Identify known quantities:
\({\text{KE}}_{\text{f}}={\text{PE}}_{\text{el}}\phantom{\rule{0.25em}{0ex}}or\phantom{\rule{0.25em}{0ex}}1/2{\mathit{mv}}^{2}=(1/2){\mathit{kx}}^{2}={\text{PE}}_{\mathrm{el}}=0\text{.}\text{563}\phantom{\rule{0.25em}{0ex}}\text{J}\)
Solve for \(v\):
\(v={[\cfrac{2{\text{PE}}_{\text{el}}}{m}]}^{1/2}={[\cfrac{2(0\text{.}\text{563}\phantom{\rule{0.25em}{0ex}}\text{J})}{0\text{.}\text{002}\phantom{\rule{0.25em}{0ex}}\text{kg}}]}^{1/2}=\text{23}\text{.}7{(\text{J/kg})}^{1/2}\)
Convert units: \(23.7\phantom{\rule{0.25em}{0ex}}\text{m}/\text{s}\)

Discussion

(a) and (b): This projectile speed is impressive for a tranquilizer gun (more than 80 km/h). The numbers in this problem seem reasonable. The force needed to compress the spring is small enough for an adult to manage, and the energy imparted to the dart is small enough to limit the damage it might do. Yet, the speed of the dart is great enough for it to travel an acceptable distance.

Check your Understanding

Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations of the ruler. In what way could you modify this simple experiment to increase the rigidity of the system?

Solution

You could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment.

Check your Understanding

If you apply a deforming force on an object and let it come to equilibrium, what happened to the work you did on the system?

Solution

It was stored in the object as potential energy.

Energy in Hooke’s Law of Deformation