Store energy in an ideal spring

Begin with a changing pull

Imagine a block attached to a horizontal spring. The spring is relaxed when the block is at x = 0. Pulling the block to the right stretches the spring; pushing it left compresses it. In either case an ideal spring pushes or pulls back toward the relaxed position. Positive x means rightward displacement, measured from equilibrium, not the spring’s full length.

The lab assumes a massless spring and a 1 kg block on a frictionless horizontal surface. Its fixed support does not move. Gravity and the support force balance vertically. The spring can compress and extend across the entire displayed range without coil contact or permanent deformation. These are model assumptions, not guarantees about a real spring.

Choose stiffness 20 N/m and displacement +20 cm. Convert centimetres to metres before calculating: 20 cm = 0.20 m. The restoring force is −4 N, leftward. At −20 cm it becomes +4 N, rightward. The equal magnitudes and opposite signs are consequences of the same restoring rule.

Interpret Hooke’s law

The rule is F = −kx. F is the signed spring force on the block in newtons, k is stiffness in newtons per metre, and x is displacement from the relaxed position in metres. The minus sign makes force point opposite displacement. It does not mean that spring force is always negative.

For k = 20 N/m and x = +0.20 m, F = −20 × 0.20 = −4 N. For x = −0.20 m, F = −20 × (−0.20) = +4 N. At x = 0, F = 0. Increasing k means more restoring force at the same displacement. It does not mean the block has a different mass.

Predict the force at +40 cm before changing the control. It is −8 N: twice the displacement gives twice the force magnitude. Save the two states and explain which variable was held fixed. A real force–displacement measurement could test whether this proportional relationship holds over the tested range; the synthetic widget assumes it.

Why the stored energy includes one half

To prepare a stretched spring slowly, an external hand must oppose its increasing restoring force. From x = 0 to x = 0.20 m, the required applied force rises from 0 to 4 N. It is not 4 N throughout the preparation. The area under the applied-force graph is a triangle: half its base, 0.20 m, multiplied by its height, 4 N. The work supplied is 0.4 J.

For an ideal spring this becomes elastic potential energy U = ½kx², with U measured in joules and zero chosen at the relaxed state. In the example, U = ½ × 20 × 0.20² = 0.4 J. Multiplying final force by distance gives 0.8 J and incorrectly treats the maximum force as constant.

Compression also stores energy: the preparation force and displacement are both leftward, so the external work is positive. Squaring x produces the same U at +0.20 m and −0.20 m. Equal stored energies do not imply equal force directions. The graph is an energy-versus-position curve, not a force diagram or a picture of the spring’s shape.

Read the landscape and its energy gap

The solid curve uses fixed axes: −0.4 to +0.4 m horizontally and 0 to 4 J vertically. It reaches its minimum at equilibrium. Moving farther from equilibrium on either side raises U quadratically. Doubling displacement magnitude from 0.20 to 0.40 m quadruples energy from 0.4 to 1.6 J, although force magnitude only doubles.

The lab prepares every release at +0.40 m from rest. The dashed line is that release’s total mechanical energy, E = ½kA², where A = 0.40 m is amplitude. At a selected position, the vertical gap between E and U represents kinetic energy K. It is not an additional energy store to add to E. At either endpoint U equals E, leaving zero K.

Change stiffness from 20 to 40 N/m while keeping x = 0.20 m. U rises from 0.4 to 0.8 J and total release energy rises from 1.6 to 3.2 J. This compares two separately prepared systems. A physical device that changes stiffness during motion can exchange energy with the system; that process is not simulated here.

Test a new case

A different ideal spring has stiffness 30 N/m and is compressed 0.10 m. Predict force, stored energy, and the work the spring does as it returns to equilibrium. Write units and signs before opening the answer.

Check the elastic-energy transfer

Force is +3 N. Stored energy is ½ × 30 × 0.10² = 0.15 J. During release toward equilibrium the spring does +0.15 J of work on the block, while elastic potential falls by 0.15 J. Force is not constant during this return. If no other force does work, the block gains 0.15 J of kinetic energy.

The spring model omits hysteresis, heating, internal spring motion and permanent deformation. A curved or different loading/unloading force graph would require revising the energy calculation. A successful virtual prediction establishes use of this ideal model, not physical measurement skill.

Background: OpenStax simple harmonic motion. The comparisons, numerical examples and investigation are original teaching scenarios.