Compare a local force derivative with a secant stiffness

Define displacement relative to equilibrium

Hooke’s linear model is F=−kx, where x is extension or compression measured from the stated equilibrium coordinate. The negative sign gives a restoring force, not a negative stiffness. The external force needed to hold a displacement is opposite to the spring’s force when no other interactions contribute; do not put both on one free-body diagram without identifying their agents and recipients.

Challenge a linear assumption

The original supplied curve is F=−6x−beta x³, with beta≥0. It reduces to Hooke’s law when beta=0 and approaches a linear relation for sufficiently small displacements. A measured curved force trace cannot be represented by a single k over every displacement simply because it passes through the origin.

Derive two different stiffness quantities

The local tangent stiffness is−dF/dx=6+3beta x². It describes an infinitesimal force change near the selected displacement. The secant ratio−F/x=6+beta x² compares the origin to a nonzero point. At zero displacement the raw ratio is 0/0 and is omitted, while its mathematical limit is 6 N/m. A defined limit does not make the original division valid at zero.

Work the original comparison

For beta=4 and x=.5, restoring force is −3−.5=−3.5 N. Secant stiffness is 7 N/m, while tangent stiffness is 9 N/m. At x=−.5 the force reverses direction but both stiffness values stay the same because the model is symmetric. For beta=0 both definitions give 6 away from zero, consistent with the linear special case.

Connect derivatives to a prediction

For a small extra displacement dx near x=.5, the first-order change is dF≈−9 dx. Using the secant 7 in place of the tangent would predict the wrong local slope. The approximation error grows as the extra step grows; retain the actual curve when comparing finite changes. Potential energy requires a separate signed force integral, which the next energy category will derive.

Transfer and material limits

For F=−5x−2x³, tangent stiffness at x=1 is 11 N/m. No single-valued polynomial tests hysteresis, plastic deformation, temperature dependence or failure load. The model is an original elastic comparison, not a validated material law or permission to stretch physical equipment.

Public numerical transfer practice

Try a new practice case

For F=−5x−2x³ N with x in metres, find local tangent stiffness at x=1 m.

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original prose, models and cases, checked 30 October 2026 against the current official AP Physics C Mechanics CED and OpenStax University Physics. Values are supplied or generated, not measured apparatus records. No physical launch, suspension, moving machinery or practical safety validation is requested. Virtual captures and public answers do not certify independent mastery or laboratory participation.