Why a Spring’s Force May Not Be Linear
A simple spring model assumes that force is proportional to extension: doubling the extension doubles the force. Real springs may depart from this ideal behavior, especially over a larger range of motion. A nonlinear force-extension law lets us describe that departure and calculate the work needed to stretch the spring.
Consider the model \(F(x)=kx+cx^3\), where \(x\) is the extension from the spring’s natural length. The first term is Hooke’s-law force; the cubic term adjusts the force as extension grows. For \(k>0\) and \(c>0\), both terms increase the force for positive extension.
We will calculate the work done by an external force as the spring is extended quasistatically from \(x=a\) to \(x=b\), with \(0\le a\le b\). Quasistatic extension means the spring is moved slowly enough that the applied force balances the spring’s force at each position. The work is the integral of that applied force over the extension interval. The spring’s own force points opposite the extension, so its work has the opposite sign.
Deriving the Work Formula
For a small increase in extension, the work is approximately the force at that extension multiplied by the small change in position. Adding these contributions over the interval leads to the definite integral of the force. Substituting the nonlinear model and integrating each term gives a formula for the total work.
For external work during extension from \(a\) to \(b\), use \(W=\int_a^b F(x)\,dx\).
With \(F(x)=kx+cx^3\), the integral is \(W=\int_a^b(kx+cx^3)\,dx\).
An antiderivative is \(\frac{k}{2}x^2+\frac{c}{4}x^4\). Evaluate it at \(b\) and subtract its value at \(a\).
This formula applies when the stated force law describes the force throughout the interval and the motion is extension in the positive \(x\)-direction. If the spring is compressed or the applied force does not balance the spring force, the force direction and the work sign must be handled accordingly. Here \(W\ge 0\) for \(k,c\ge 0\) and \(0\le a\le b\).
The formula also gives a direct comparison with Hooke’s law. If the same spring is modeled by \(F_H(x)=kx\), the ideal Hooke’s-law work over the same interval is \(W_H=\frac{k}{2}(b^2-a^2)\). Therefore, the cubic term contributes the difference shown below. When \(c>0\) and \(b>a\ge0\), this contribution is positive, so the nonlinear model requires more work than the Hooke’s-law model over that interval.
Worked Examples
Worked Example: Work from a Nonzero Extension
A spring has force law \(F(x)=200x+8000x^3\), with \(x\) in metres and force in newtons. Find the external work required to extend it from \(0.01\ \mathrm{m}\) to \(0.03\ \mathrm{m}\), and compare with the Hooke’s-law model \(F_H(x)=200x\).
Here \(k=200\ \mathrm{N/m}\), \(c=8000\ \mathrm{N/m^3}\), \(a=0.01\ \mathrm{m}\), and \(b=0.03\ \mathrm{m}\). The linear contribution is
The cubic contribution is
Thus the nonlinear model gives \(W=0.08+0.0016=0.0816\ \mathrm{J}\). The Hooke’s-law result is \(W_H=0.08\ \mathrm{J}\), so the cubic term adds \(0.0016\ \mathrm{J}\). As a check, differentiating the antiderivative \(\frac{200}{2}x^2+\frac{8000}{4}x^4\) gives \(200x+8000x^3\), the specified force. Substituting the endpoints into that antiderivative gives the same difference, \(0.0816\ \mathrm{J}\).
Worked Example: A Ten-Percent Increase Over Hooke’s Law
Use \(F(x)=100x+10000x^3\) to calculate work from \(0.02\ \mathrm{m}\) to \(0.04\ \mathrm{m}\). Determine how much larger this is than the Hooke’s-law work, as a percentage of the Hooke’s-law result.
The Hooke’s-law contribution is
The cubic contribution is
The total nonlinear work is \(W=0.06+0.006=0.066\ \mathrm{J}\). Relative to Hooke’s-law work, the increase is
Therefore the nonlinear model requires \(0.066\ \mathrm{J}\), which is 10% more than the ideal linear model over this interval. To check, the antiderivative is \(50x^2+2500x^4\); its derivative is \(100x+10000x^3\). Its endpoint difference is \(50(0.0016-0.0004)+2500(0.00000256-0.00000016)=0.06+0.006=0.066\ \mathrm{J}\).
Worked Example: Comparing Two Nonlinear Contributions
A spring is modeled by \(F(x)=120x+6000x^3\). Find the work to extend it from \(0.02\ \mathrm{m}\) to \(0.05\ \mathrm{m}\), then state the cubic contribution as a percentage of the Hooke’s-law work.
For \(k=120\ \mathrm{N/m}\), the linear contribution is
For \(c=6000\ \mathrm{N/m^3}\), the cubic contribution is
The total work is \(0.126+0.009135=0.135135\ \mathrm{J}\). The cubic contribution relative to the Hooke’s-law work is
So the spring requires \(0.135135\ \mathrm{J}\), and its cubic term adds 7.25% of the Hooke’s-law work on this interval. The check is that \(\frac{d}{dx}(60x^2+1500x^4)=120x+6000x^3\), and evaluating \(60x^2+1500x^4\) at \(0.05\) and \(0.02\) gives \(0.135135\) and \(0\!+\!0.000? \) respectively; explicitly, \(60(0.02)^2+1500(0.02)^4=0.024+0.00024=0.02424\), while \(60(0.05)^2+1500(0.05)^4=0.15+0.009375=0.159375\). Their difference is \(0.159375-0.02424=0.135135\ \mathrm{J}\).
A Common Shortcut That Fails
For an ideal Hooke’s-law spring extended from its natural length, work is often written as one-half the final force multiplied by the final extension. That shortcut depends on the force increasing linearly from zero to its final value. A nonlinear force-extension graph is not a straight line, so its area under the graph is generally not the area of that triangle.
For example, with \(F(x)=200x+8000x^3\), extending from zero to \(0.03\ \mathrm{m}\) gives
The final force is \(F(0.03)=200(0.03)+8000(0.03)^3=6+0.216=6.216\ \mathrm{N}\). The incorrect shortcut gives \(\frac12(6.216)(0.03)=0.09324\ \mathrm{J}\), not \(0.09162\ \mathrm{J}\). The correction is to integrate the actual force law over the interval; the one-half shortcut is not valid for this nonlinear model.
Interpreting and Checking the Result
The cubic term becomes more influential at larger extensions because its force contribution grows as \(x^3\), while its work contribution from the natural length grows as \(x^4\). This does not mean that the cubic term always dominates: its size depends on both \(c\) and the extension range. Comparing the two work contributions is a direct way to assess the importance of the nonlinear correction for a particular interval.
Dimensional checks help catch errors. Since \(kx\) and \(cx^3\) must both have units of force, their coefficients have units \(\mathrm{N/m}\) and \(\mathrm{N/m^3}\). Multiplying force by distance gives work in \(\mathrm{N\,m}=\mathrm{J}\). In the integrated expression, \(kb^2\) and \(cb^4\) each have units of joules.
Check Your Understanding
Use the work formula and show enough steps to make your answers checkable.
- For \(F(x)=80x+4000x^3\), find the work to extend from \(0\) to \(0.02\ \mathrm{m}\).
- For a force law \(F(x)=kx+cx^3\), write the Hooke’s-law work and the additional cubic contribution over \(a\le x\le b\).
- Why does the cubic term contribute more work over an interval at larger extensions, when \(c>0\)?
- For \(F(x)=50x+2000x^3\), find the work from \(0.01\ \mathrm{m}\) to \(0.03\ \mathrm{m}\), and compare it with Hooke’s-law work.
- In what special situation is one-half the final force times the final extension a valid work calculation for stretching from the natural length?
