Hooke's Law and Elastic Potential Energy Calculations

Measuring how much a spring changes

Imagine hanging a load from a spring. The spring becomes longer, but its extension is not its complete stretched length. It is the extra length:

$e=\text{stretched length}-\text{original length}$

For example, a spring that changes from $20\text{ cm}$ to $25\text{ cm}$ has an extension of $5\text{ cm}$, or $0.05\text{ m}$. Equations in this lesson require extension or compression in metres.

The same ideas apply when an elastic object is compressed. In that case, $e$ represents how much shorter the object becomes.

Force, extension and the spring constant

Within the directly proportional region, increasing the applied force increases the extension in the same ratio. Doubling the force doubles the extension; tripling the force triples it. This relationship is commonly called Hooke's law:

$F=ke$

Here:

• $F$ is force in newtons (N);
• $k$ is the spring constant in newtons per metre (N/m);
• $e$ is extension or compression in metres (m).

The spring constant measures stiffness. A spring with a large $k$ needs a large force for each metre of extension, so it is harder to stretch or compress. A smaller $k$ means that the same force produces a greater change in length.

For a car suspension spring with $k=60\,000\text{ N/m}$ compressed by $2.0\text{ cm}=0.020\text{ m}$,

$F=ke=60\,000\times0.020=1\,200\text{ N}.$

The equation can also be rearranged:

$k=\frac{F}{e} \qquad\text{and}\qquad e=\frac{F}{k}.$

For instance, if a $200\text{ N}$ force extends a spring from $0.60\text{ m}$ to $1.00\text{ m}$, then $e=0.40\text{ m}$ and

$k=\frac{200}{0.40}=500\text{ N/m}.$

Where Hooke's law applies

The equation $F=ke$ describes the straight-line, directly proportional relationship only while the limit of proportionality has not been exceeded. Beyond this point, equal increases in force no longer produce equal increases in extension, so the relationship becomes non-linear and a single constant value of no longer describes the whole response.

The limit of proportionality and the onset of permanent, inelastic deformation are related ideas, but they should not simply be treated as the same named point. “No longer proportional” tells you about the force–extension relationship; “inelastically deformed” tells you that the object will not completely return to its original shape after the force is removed.

Work done and stored energy

A force does work as it stretches or compresses a spring. This transfers energy to the spring's elastic potential energy store. Provided that the spring is not inelastically deformed, the work done on it equals the elastic potential energy stored.

Up to the limit of proportionality, the stored energy is:

$E_e=\frac12ke^2$

where $E_e$ is in joules (J), $k$ is in N/m and $e$ is in m. The factor of $\frac12$ appears because the required force rises from zero to its final value as the spring is extended; the force is not at its final value throughout the movement.

A spring with $k=150\text{ N/m}$ extended by $6\text{ cm}=0.06\text{ m}$ stores

$E_e=\frac12\times150\times(0.06)^2=0.27\text{ J}.$

This is also the work done in stretching it, provided it has not been inelastically deformed.

Why extension has such a strong effect

Extension is squared in the energy equation. If the same spring is stretched twice as far within the proportional region, its stored energy becomes

$\frac12k(2e)^2=4\left(\frac12ke^2\right).$

It therefore stores four times as much energy, not twice as much. Tripling the extension gives nine times the energy. By contrast, doubling $k$ at the same extension doubles the stored energy.

The energy equation can be rearranged when another quantity is unknown:

$k=\frac{2E_e}{e^2}$

or

$e=\sqrt{\frac{2E_e}{k}}.$

After finding $e$, remember that the original length is obtained by subtracting the extension from the stretched length.

Force and extension

$F=ke$

• $F$: force in N
• $k$: spring constant in N/m
• : or compression in m