riaology
Unit 5 · Lesson 4 of 4

A varying force requires an area, not one multiplication.

Stretching a spring becomes harder as the displacement grows. Sum the little force–distance contributions.

Change the input. Follow the mathematics.

What you'll be able to do
  • Set up the model and calculate its work input.
  • Connect the equation to the changing visual.
  • Explain the assumptions and identify a misleading interpretation.
Predict before exploring

What will change if you increase stretch? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Linear spring stretched by 0.2 metersStretch: 0.2 m
Stretch (m) versus Spring force (N). Force. Selected work input: 4 J. A varying force requires an integral. Final force times full displacement overestimates the work for a linear spring by a factor of two.00200.5401Spring force (N)Stretch (m)
Force

work inputW=12(200)(0.2)2=4 JW=\frac12(200)(0.2)^2=4\,\mathrm J

A varying force requires an integral. Final force times full displacement overestimates the work for a linear spring by a factor of two.

Assumptions, units, and limits

Quasi-static linear spring, stretched from its unstressed position; no friction or kinetic-energy change.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

01

From the situation to an equation

For each small extension, incremental work is force times incremental displacement. Hooke's law gives a linearly increasing force, so its work is the triangular area under the force curve. A pumping-water problem uses slice weight times each slice's lift distance instead.

F(x)=kx,W=∫0aF(x) dx=12ka2F(x)=kx,\quad W=\int_0^aF(x)\,dx=\frac12ka^2
02

Keep the reasoning visible

dW=F(x) dxdW=F(x)\,dx

A varying force requires an integral. Final force times full displacement overestimates the work for a linear spring by a factor of two.

03

What makes this model useful?

Quasi-static linear spring, stretched from its unstressed position; no friction or kinetic-energy change.

Multiplying final force by total stretch assumes a constant force and overestimates this spring work. Define the reference position.

A worked example

Calculate the starting scenario

  1. Use stretch = 0.2 m; spring constant = 200 N/m.
  2. W=12(200)(0.2)2=4 JW=\frac12(200)(0.2)^2=4\,\mathrm J
  3. A varying force requires an integral. Final force times full displacement overestimates the work for a linear spring by a factor of two. The checked quantity is work input.
Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the work input using stretch 0.2 m, spring constant 200 N/m. State the meaning of its sign and units.

Checked quantity: work input · J. Follow the definition for finite-cutoff and numerical-model questions.

Hint

For each small extension, incremental work is force times incremental displacement. Hooke's law gives a linearly increasing force, so its work is the triangular area under the force curve. A pumping-water problem uses slice weight times each slice's lift distance instead.

Explore the visual solution
Linear spring stretched by 0.2 metersStretch: 0.2 m
Stretch (m) versus Spring force (N). Force. Selected work input: 4 J. A varying force requires an integral. Final force times full displacement overestimates the work for a linear spring by a factor of two.00200.5401Spring force (N)Stretch (m)
Force

work inputW=12(200)(0.2)2=4 JW=\frac12(200)(0.2)^2=4\,\mathrm J

A varying force requires an integral. Final force times full displacement overestimates the work for a linear spring by a factor of two.

Assumptions, units, and limits

Quasi-static linear spring, stretched from its unstressed position; no friction or kinetic-energy change.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

For each small extension, incremental work is force times incremental displacement. Hooke's law gives a linearly increasing force, so its work is the triangular area under the force curve. A pumping-water problem uses slice weight times each slice's lift distance instead.

F(x)=kx,W=∫0aF(x) dx=12ka2F(x)=kx,\quad W=\int_0^aF(x)\,dx=\frac12ka^2
Calculate and explain

Apply the idea to a new scenario

Stretch a k=300k=300 N/m spring by 0.3 m. Find work.

Checked quantity: work input · J. Follow the definition for finite-cutoff and numerical-model questions.

Hint

For each small extension, incremental work is force times incremental displacement. Hooke's law gives a linearly increasing force, so its work is the triangular area under the force curve. A pumping-water problem uses slice weight times each slice's lift distance instead.

Explore the visual solution
Linear spring stretched by 0.3 metersStretch: 0.3 m
Stretch (m) versus Spring force (N). Force. Selected work input: 13.5 J. A varying force requires an integral. Final force times full displacement overestimates the work for a linear spring by a factor of two.00450.5901Spring force (N)Stretch (m)
Force

work inputW=12(300)(0.3)2=13.5 JW=\frac12(300)(0.3)^2=13.5\,\mathrm J

A varying force requires an integral. Final force times full displacement overestimates the work for a linear spring by a factor of two.

Assumptions, units, and limits

Quasi-static linear spring, stretched from its unstressed position; no friction or kinetic-energy change.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

For each small extension, incremental work is force times incremental displacement. Hooke's law gives a linearly increasing force, so its work is the triangular area under the force curve. A pumping-water problem uses slice weight times each slice's lift distance instead.

F(x)=kx,W=∫0aF(x) dx=12ka2F(x)=kx,\quad W=\int_0^aF(x)\,dx=\frac12ka^2
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Linear spring stretched by 0.2 metersStretch: 0.2 m
Stretch (m) versus Spring force (N). Force. Selected work input: 4 J. A varying force requires an integral. Final force times full displacement overestimates the work for a linear spring by a factor of two.00200.5401Spring force (N)Stretch (m)
Force

work inputW=12(200)(0.2)2=4 JW=\frac12(200)(0.2)^2=4\,\mathrm J

A varying force requires an integral. Final force times full displacement overestimates the work for a linear spring by a factor of two.

Assumptions, units, and limits

Quasi-static linear spring, stretched from its unstressed position; no friction or kinetic-energy change.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. For each small extension, incremental work is force times incremental displacement. Hooke's law gives a linearly increasing force, so its work is the triangular area under the force curve. A pumping-water problem uses slice weight times each slice's lift distance instead.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

PreviousCalculus II

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