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Restoring force

For an ideal elastic element within its proportional range,

$\vec F_H=-k\vec x$

The minus sign means the force acts opposite the displacement from equilibrium.

Use extension correctly

For a spring, $x$ is extension or compression relative to its natural length. In a vertical equilibrium, the spring tension can balance weight, but the extension is not the total spring length.

Respect the model boundary

Hooke’s law is a linear approximation. Beyond the limit of proportionality, the force–extension graph is no longer linear and the same $k$ cannot be used.

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A spring of negligible mass and length $l_{0}$ hangs from a fixed point. When a mass m is attached to the free end of the spring, the length of the spring increases to l. The tension in the spring is equal to $k \Delta x$ , where k is a constant and $\Delta x$ is the extension of the spring. What is k ?

Build the force model

Choose the system, draw a labelled free-body diagram, classify the interactions and resolve components. Apply Newton’s laws with the correct boundary: contact forces, field forces, friction, tension, buoyancy and restoring forces each have their own direction and conditions.

Track momentum

Use $ec p=m ec v$ , $ec J=\Delta ec p$ and momentum conservation only after checking external impulse. Distinguish elastic and inelastic collisions, explosions and energy transfer.

Track circular motion

The inward resultant provides $a_c=v^2/r=\omega^2r$ . It may come from tension, gravity, normal, friction or a field force. Angular and linear descriptions are linked by $v=\omega r=2\pi r/T$ .

Final checks

Ask: Which body is the system? Which forces are external? Is mass constant? Is acceleration uniform or radial? Is kinetic energy conserved, transferred or increased?

A.2.9—Hooke’s law restoring force

Apply Hooke’s Law to Elastic Restoring Force

Retrieve the A.2 Forces and Momentum Model

A.2.9 Exam Analysis