Springs have the ability to provide a force and store energy. We can see this as:

We have to push/pull a spring to compress it or expand it. Big springs hold up our cars.
Kids' toy guns require that you do work to compress the spring, storing the energy as spring potential energy ( $LaTeX: E_{SP}$ or just ). Later this energy is released as the kinetic energy of the toy bullet.

Hooke’s law is named after Robert Hooke, who documented that when you pull or push on a spring, the spring stretches a displacement that is proportional to the force applied to the spring.

So, we can write: $LaTeX: F\propto\Delta x,$ where $LaTeX: \Delta x$ is the displacement of the end of the spring from the neutral position not the length of the spring.

The ratio between the force you put on a spring and the spring’s displacement depends on the spring constant (rigidity) so we can write this as an equation, called Hooke’s Law: LaTeX: F=kx, where LaTeX: k is the spring constant, and LaTeX: x is really $LaTeX: \underline{\Delta x},$ how far the spring is stretched. What are the units of ? Consider: "how much force does that spring apply to me?" If you have to pull with 5 N to get an extension of 50 cm, you conclude that you have to pull 10 N per meter of extension, or LaTeX: k = 10 N/m. Because of the linear relation between force and displacement, we could graph the displacement of the end of the spring versus the force applied.

Graph of force vs. displacement of a spring showing the linear relationship known as Hooke's Law. At right, you see Hooke’s Law on a $LaTeX: \vec{F}-\vec{x}$ graph. The solid red line represents the force you have to put on the spring to displace it a certain amount. The dotted line is the force the spring puts on you. These forces are the same interaction acting in opposite directions on your finger and the spring. The steeper the line, the more force you need for the same displacement, meaning a greater spring constant.

Exercise 2.6.1

Please prove to yourself that the slope of the (solid) line is the spring constant, LaTeX: k.

The energy stored in a spring, is $LaTeX: E_S = \frac{1}{2}kx^2.$

Again, LaTeX: x is really $LaTeX: \Delta x,$ the elongation or compression of the spring. This potential energy must be equal to the work you did stretching the spring: $LaTeX: W = \vec{F}_{ave}\cdot\Delta\vec{x}.$ The force increases as we stretch the spring. Where does the 1/2 come from? The 1/2 is explained in the exercises below.

Exercise 2.6.2

Prove to yourself that the work you do is the area under the LaTeX: F-x graph. Show using simple geometry that our formula for the stored potential energy in a spring is correct.

Exercise 2.6.3

Show using calculus that when force changes with LaTeX: x, you must change the product to an integral, adding up all the little bits of work you do as you stretch the spring:

$LaTeX: E_S = W = \sum \Delta W = \sum\vec{F}\cdot\Delta\vec{x},$

$LaTeX: E_S = W = \int dW = \int\vec{F}\cdot d\vec{x},$ integrated over the full extension, $LaTeX: \Delta x = x_f-x_i$ .

Please make the substitution of $LaTeX: \vec{F}=k\vec{x}$ into the integral and find the correct expression for spring potential energy: $LaTeX: \frac{1}{2}k(\Delta x)^2.$

Exercise 2.6.4

I compress a spring 2 cm with a force of 50 N and lock it in place. I put a 100 g mass on the spring and then let it go, straight up!

How fast is the mass moving immediately after release?
How high did it get?

Were you mindful of which lens you're using? Did you draw a good picture?

Exercise 2.6.5

You may notice that you can compress and release a spring many many times and it doesn't get hot.

What does this tell you about a spring's ability to store energy?
Was it reasonable of us to use the equation above equating the stored potential energy in a spring to the work done on the spring or do we need to consider conversion to thermal energy?

Exercise 2.6.6

Imagine that a ball has a perfectly elastic collision with a rigid wall and bounces back with its original speed in the opposite direction.

Momentum is a vector. Did its momentum change? If so, how is momentum conserved?
Energy is a scalar. Did its energy change from beginning to end? Was there an energy transition anywhere in the process (like when the ball turned around)?