Spring Potential Energy: Definition, Equation, Units (W/ Examples)

From a taut bowstring sending an arrow flying through the air to a kid cranking a jack-in-the-box enough to make it pop out so fast you can barely see it happening, spring potential energy is all around us.

In archery, the archer draws back the bowstring, pulling it away from its equilibrium position and transferring energy from her own muscles to the string, and this stored energy is called ​spring potential energy​ (or ​elastic potential energy​). When the bowstring is released, this is released as kinetic energy in the arrow.

The concept of spring potential energy is a key step in many situations involving the conservation of energy, and learning more about it gives you insight into more than just jack-in-the-boxes and arrows.

Spring potential energy is a form of stored energy, much like gravitational potential energy or electrical potential energy, but one associated with springs and ​elastic​ objects.

Imagine a spring hanging vertically from the ceiling, with somebody pulling down on the other end. The stored energy that results from this can be quantified exactly if you know how far down the string has been pulled, and how that specific spring responds under external force.

More precisely, the potential energy of the spring depends on its distance, ​x​, that it has moved from its "equilibrium position" (the position it would rest at in the absence of external forces), and its spring constant, ​k​, which tells you how much force it takes to extend the spring by 1 meter. Because of this, ​k​ has units of newtons/meter.

The spring constant is found in Hooke's law, which describes the force required to make a spring stretch ​x​ meters from its equilibrium position, or equally, the opposite force from the spring when you do:

\(F=-kx\)

The negative sign tells you that the spring force is a restoring force, which acts to return the spring to its equilibrium position. The equation for spring potential energy is very similar, and it involves the same two quantities.

Spring potential energy ​PE​_spring is calculated using the equation:

\(PE_{spring} = \frac{1}{2}kx^2\)

The result is a value in joules (J), because spring potential is a form of energy.

In an ideal spring – one that is assumed to have no friction and no appreciable mass – this is equal to how much work you did on the spring in extending it. The equation has the same basic form as the equations for kinetic energy and rotational energy, with the ​x​ in place of the ​v​ in the kinetic energy equation and the spring constant ​k​ in place of mass ​m​ – you can use this point if you need to memorize the equation.

Calculating spring potential is simple if you know the displacement caused by the spring stretch (or compression), ​x​ and the spring constant for the spring in question. For a simple problem, imagine a spring with the constant ​k​ = 300 N/m being extended by 0.3 m: what is the potential energy stored in the spring as a result?

This problem involves the potential energy equation, and you're given the two values you need to know. You just need to plug in the values ​k​ = 300 N/m and ​x​ = 0.3 m to find the answer:

\(\begin{aligned}
PE_{spring} &= \frac{1}{2}kx^2 \
&=\frac{1}{2}×300 \;\text{N/m} × (0.3 \;\text{m})^2 \
&= 13.5 \;\text{J}
\end{aligned}\)

For a more challenging problem, imagine an archer drawing back the string on a bow preparing to fire an arrow, bringing it back up to 0.5 m from its equilibrium position and pulling the string with a maximum force of 300 N.

Here, you're given the force ​F​ and the displacement ​x​, but not the spring constant. How do you tackle a problem like this? Luckily, Hooke's law describes the relationship between, ​F​, ​x​ and the constant ​k​, so you can use the equation in the following form:

\(k=\frac{F}{x}\)

To find the value of the constant before calculating the potential energy as before. However, since ​k​ appears in the elastic potential energy equation, you can substitute this expression into it and calculate the result in a single step:

\(\begin{aligned}
PE_{spring}&=\frac{1}{2}kx^2 \
&=\frac{1}{2}\frac{F}{x}x^2 \
&=\frac{1}{2}Fx \
&= \frac{1}{2}× 300 \;\text{N} × 0.5 \;\text{m} \
&= 75 \;\text{J}
\end{aligned}\)

So, the fully taut bow has 75 J of energy. If you then need to calculate the maximum speed of the arrow, and you know its mass, you can do this by applying the conservation of energy using the kinetic energy equation.