riaology
Unit 5 · Lesson 5 of 7

Only the along-motion force component does work.

Pushing sideways can exert a force without doing work along the displacement. Rotate the force to see the signed result.

Change the input. Follow the mathematics.

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

What will change if you increase angle between force and motion? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Vector components or geometric direction diagramReferenceAlong-motion projection determines work
Angle (degrees) versus Work (J). Work. Selected signed work: 10 J. Only the force component along motion does work. Perpendicular force gives zero dot product; opposing force gives negative work.-20009020180Work (J)Angle (degrees)
Work

signed workW=10(2)cos⁡(60∘)=10 JW=10(2)\cos(60^\circ)=10\,\mathrm J

Only the force component along motion does work. Perpendicular force gives zero dot product; opposing force gives negative work.

Assumptions, units, and limits

Force magnitude 10 N and displacement magnitude 2 m. Degrees are converted to radians inside cosine.

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

01

From the situation to an equation

The dot product equals the sum of component products and also the product of magnitudes times cosine of their angle. It measures projection alignment. Orthogonal vectors have zero dot product, while opposite directions give a negative result.

W=F⋅d=∣F∣∣d∣cos⁡θW=\mathbf F\cdot\mathbf d=|\mathbf F||\mathbf d|\cos\theta
02

Keep the reasoning visible

a⋅b=∑iaibi\mathbf a\cdot\mathbf b=\sum_i a_i b_i

Only the force component along motion does work. Perpendicular force gives zero dot product; opposing force gives negative work.

03

What makes this model useful?

Force magnitude 10 N and displacement magnitude 2 m. Degrees are converted to radians inside cosine.

A zero dot product does not mean either vector has zero magnitude. Work can be negative when force opposes motion.

A worked example

Calculate the starting scenario

  1. Use angle between force and motion = 60 degrees.
  2. W=10(2)cos⁡(60∘)=10 JW=10(2)\cos(60^\circ)=10\,\mathrm J
  3. Only the force component along motion does work. Perpendicular force gives zero dot product; opposing force gives negative work. The checked quantity is signed work.
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 signed work using angle between force and motion 60 degrees. State the meaning of its sign and units.

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

Hint

The dot product equals the sum of component products and also the product of magnitudes times cosine of their angle. It measures projection alignment. Orthogonal vectors have zero dot product, while opposite directions give a negative result.

Explore the visual solution
Vector components or geometric direction diagramReferenceAlong-motion projection determines work
Angle (degrees) versus Work (J). Work. Selected signed work: 10 J. Only the force component along motion does work. Perpendicular force gives zero dot product; opposing force gives negative work.-20009020180Work (J)Angle (degrees)
Work

signed workW=10(2)cos⁡(60∘)=10 JW=10(2)\cos(60^\circ)=10\,\mathrm J

Only the force component along motion does work. Perpendicular force gives zero dot product; opposing force gives negative work.

Assumptions, units, and limits

Force magnitude 10 N and displacement magnitude 2 m. Degrees are converted to radians inside cosine.

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.

The dot product equals the sum of component products and also the product of magnitudes times cosine of their angle. It measures projection alignment. Orthogonal vectors have zero dot product, while opposite directions give a negative result.

W=F⋅d=∣F∣∣d∣cos⁡θW=\mathbf F\cdot\mathbf d=|\mathbf F||\mathbf d|\cos\theta
Calculate and explain

Apply the idea to a new scenario

For angle 120°, force magnitude 10 N and displacement 2 m, find work.

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

Hint

The dot product equals the sum of component products and also the product of magnitudes times cosine of their angle. It measures projection alignment. Orthogonal vectors have zero dot product, while opposite directions give a negative result.

Explore the visual solution
Vector components or geometric direction diagramReferenceAlong-motion projection determines work
Angle (degrees) versus Work (J). Work. Selected signed work: -10 J. Only the force component along motion does work. Perpendicular force gives zero dot product; opposing force gives negative work.-20009020180Work (J)Angle (degrees)
Work

signed workW=10(2)cos⁡(120∘)=−10 JW=10(2)\cos(120^\circ)=-10\,\mathrm J

Only the force component along motion does work. Perpendicular force gives zero dot product; opposing force gives negative work.

Assumptions, units, and limits

Force magnitude 10 N and displacement magnitude 2 m. Degrees are converted to radians inside cosine.

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.

The dot product equals the sum of component products and also the product of magnitudes times cosine of their angle. It measures projection alignment. Orthogonal vectors have zero dot product, while opposite directions give a negative result.

W=F⋅d=∣F∣∣d∣cos⁡θW=\mathbf F\cdot\mathbf d=|\mathbf F||\mathbf d|\cos\theta
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
Vector components or geometric direction diagramReferenceAlong-motion projection determines work
Angle (degrees) versus Work (J). Work. Selected signed work: 10 J. Only the force component along motion does work. Perpendicular force gives zero dot product; opposing force gives negative work.-20009020180Work (J)Angle (degrees)
Work

signed workW=10(2)cos⁡(60∘)=10 JW=10(2)\cos(60^\circ)=10\,\mathrm J

Only the force component along motion does work. Perpendicular force gives zero dot product; opposing force gives negative work.

Assumptions, units, and limits

Force magnitude 10 N and displacement magnitude 2 m. Degrees are converted to radians inside cosine.

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

The reasoning. The dot product equals the sum of component products and also the product of magnitudes times cosine of their angle. It measures projection alignment. Orthogonal vectors have zero dot product, while opposite directions give a negative result.

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.

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