Work Calculator

Professional physics Work Calculator. Compute work done by a force using W = F · d · cos θ, with unit conversions (N, lbf, m, ft) and detailed step-by-step breakdown. WCAG-accessible and mobile-first.

Work Calculator

This professional-grade physics Work Calculator computes the mechanical work done by a constant force along a straight path using W = F · d · cos(θ). It’s designed for students, engineers, and educators who need fast, precise results with rigorous unit conversions and a clear step-by-step breakdown. This tool focuses on physics work (energy transfer), not “work hours” or timecards.

Results

—
Computed quantity—
Signed value—
Equivalent (kJ)—
Equivalent (ft·lbf)—

Step-by-step

Using $W = F \\cdot d \\cdot \\cos(\\theta)$
—

Data Source and Methodology

Authoritative Data Source: OpenStax College Physics, Chapter 7 “Work, Energy, and Energy Resources” (latest edition, 2016) — openstax.org/details/books/college-physics. Unit conversion constants are taken from the NIST Guide to the SI (2019) — physics.nist.gov/cuu/Units/.

All calculations are rigorously based on the formulas and data provided by these sources. Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.

The Formula Explained

$W = \\vec{F} \\cdot \\vec{d} = F\\,d\\,\\cos(\\theta)$

Solving for each variable:

$F = \\dfrac{W}{d\\,\\cos(\\theta)}\\,,\\quad d = \\dfrac{W}{F\\,\\cos(\\theta)}\\,,\\quad \\theta = \\arccos\\!\\left(\\dfrac{W}{F\\,d}\\right)$

Glossary of Variables

    - Work (W): Energy transferred by a force along a displacement; SI unit: joule (J). - Force (F): Constant force magnitude; SI unit: newton (N). 1 lbf = 4.4482216152605 N. - Displacement (d): Straight-line distance moved; SI unit: meter (m). 1 ft = 0.3048 m. - Angle (θ): Angle between force direction and displacement; measured in degrees. - ft·lbf: Foot-pound-force; 1 ft·lbf = 1.3558179483314 J. - BTU: British thermal unit; 1 BTU = 1055.05585262 J.

How It Works: A Step-by-Step Example

Suppose a 50 N force acts through a displacement of 12 m at an angle of 30° to the direction of motion. The work done is:

$W = F\\,d\\,\\cos(\\theta) = 50\\,\\text{N} \\times 12\\,\\text{m} \\times \\cos(30^\\circ)$

$\\cos(30^\\circ) = \\sqrt{3}/2 \\approx 0.8660$

$W \\approx 600 \\times 0.8660 = 519.6\\,\\text{J}$

The sign of W indicates whether the force aids motion (positive) or opposes it (negative).

Frequently Asked Questions (FAQ)

Is this calculator for physics work or work hours?

It is exclusively for physics work (energy transfer), not for timecards or payroll.

What if the force is perpendicular to the motion?

At θ = 90°, cos(θ) = 0, so W = 0. No energy is transferred along the path.

Can I calculate the angle from known work, force, and distance?

Yes. The calculator uses θ = arccos(W / (F d)) if the ratio is between −1 and 1.

How are units converted?

Using NIST constants: 1 lbf = 4.4482216152605 N; 1 ft = 0.3048 m; 1 ft·lbf = 1.3558179483314 J; 1 BTU = 1055.05585262 J; 1 Wh = 3600 J; 1 kWh = 3.6×10^6 J.

Why is my result negative?

If θ > 90°, the force opposes the motion, leading to negative work.

Does this handle variable forces or curved paths?

No. It assumes constant force and straight-line displacement. For variable forces, use the line integral W = ∫ F · ds.

How many significant figures are shown?

Results are formatted for readability; use the raw value if you need more precision via the copy button.

Strumento sviluppato da Ugo Candido,. Contenuti verificati da,.

Ultima revisione per l'accuratezza in data: .

Audit: Complete
Formula (LaTeX) + variables + units
This section shows the formulas used by the calculator engine, plus variable definitions and units.
Formula (extracted LaTeX)
\[','\]
','
Formula (extracted text)
Using $W = F \\cdot d \\cdot \\cos(\\theta)$ —
Formula (extracted text)
$W = \\vec{F} \\cdot \\vec{d} = F\\,d\\,\\cos(\\theta)$ Solving for each variable: $F = \\dfrac{W}{d\\,\\cos(\\theta)}\\,,\\quad d = \\dfrac{W}{F\\,\\cos(\\theta)}\\,,\\quad \\theta = \\arccos\\!\\left(\\dfrac{W}{F\\,d}\\right)$
Formula (extracted text)
$W = F\\,d\\,\\cos(\\theta) = 50\\,\\text{N} \\times 12\\,\\text{m} \\times \\cos(30^\\circ)$ $\\cos(30^\\circ) = \\sqrt{3}/2 \\approx 0.8660$ $W \\approx 600 \\times 0.8660 = 519.6\\,\\text{J}$
Variables and units
  • T = property tax (annual or monthly depending on input) (currency)
Sources (authoritative):
Changelog
Version: 0.1.0-draft
Last code update: 2026-01-19
0.1.0-draft · 2026-01-19
  • Initial audit spec draft generated from HTML extraction (review required).
  • Verify formulas match the calculator engine and convert any text-only formulas to LaTeX.
  • Confirm sources are authoritative and relevant to the calculator methodology.
Verified by Ugo Candido on 2026-01-19
Profile · LinkedIn

Full original guide (expanded)

Work Calculator

This professional-grade physics Work Calculator computes the mechanical work done by a constant force along a straight path using W = F · d · cos(θ). It’s designed for students, engineers, and educators who need fast, precise results with rigorous unit conversions and a clear step-by-step breakdown. This tool focuses on physics work (energy transfer), not “work hours” or timecards.

Results

—
Computed quantity—
Signed value—
Equivalent (kJ)—
Equivalent (ft·lbf)—

Step-by-step

Using $W = F \\cdot d \\cdot \\cos(\\theta)$
—

Data Source and Methodology

Authoritative Data Source: OpenStax College Physics, Chapter 7 “Work, Energy, and Energy Resources” (latest edition, 2016) — openstax.org/details/books/college-physics. Unit conversion constants are taken from the NIST Guide to the SI (2019) — physics.nist.gov/cuu/Units/.

All calculations are rigorously based on the formulas and data provided by these sources. Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.

The Formula Explained

$W = \\vec{F} \\cdot \\vec{d} = F\\,d\\,\\cos(\\theta)$

Solving for each variable:

$F = \\dfrac{W}{d\\,\\cos(\\theta)}\\,,\\quad d = \\dfrac{W}{F\\,\\cos(\\theta)}\\,,\\quad \\theta = \\arccos\\!\\left(\\dfrac{W}{F\\,d}\\right)$

Glossary of Variables

    - Work (W): Energy transferred by a force along a displacement; SI unit: joule (J). - Force (F): Constant force magnitude; SI unit: newton (N). 1 lbf = 4.4482216152605 N. - Displacement (d): Straight-line distance moved; SI unit: meter (m). 1 ft = 0.3048 m. - Angle (θ): Angle between force direction and displacement; measured in degrees. - ft·lbf: Foot-pound-force; 1 ft·lbf = 1.3558179483314 J. - BTU: British thermal unit; 1 BTU = 1055.05585262 J.

How It Works: A Step-by-Step Example

Suppose a 50 N force acts through a displacement of 12 m at an angle of 30° to the direction of motion. The work done is:

$W = F\\,d\\,\\cos(\\theta) = 50\\,\\text{N} \\times 12\\,\\text{m} \\times \\cos(30^\\circ)$

$\\cos(30^\\circ) = \\sqrt{3}/2 \\approx 0.8660$

$W \\approx 600 \\times 0.8660 = 519.6\\,\\text{J}$

The sign of W indicates whether the force aids motion (positive) or opposes it (negative).

Frequently Asked Questions (FAQ)

Is this calculator for physics work or work hours?

It is exclusively for physics work (energy transfer), not for timecards or payroll.

What if the force is perpendicular to the motion?

At θ = 90°, cos(θ) = 0, so W = 0. No energy is transferred along the path.

Can I calculate the angle from known work, force, and distance?

Yes. The calculator uses θ = arccos(W / (F d)) if the ratio is between −1 and 1.

How are units converted?

Using NIST constants: 1 lbf = 4.4482216152605 N; 1 ft = 0.3048 m; 1 ft·lbf = 1.3558179483314 J; 1 BTU = 1055.05585262 J; 1 Wh = 3600 J; 1 kWh = 3.6×10^6 J.

Why is my result negative?

If θ > 90°, the force opposes the motion, leading to negative work.

Does this handle variable forces or curved paths?

No. It assumes constant force and straight-line displacement. For variable forces, use the line integral W = ∫ F · ds.

How many significant figures are shown?

Results are formatted for readability; use the raw value if you need more precision via the copy button.

Strumento sviluppato da Ugo Candido,. Contenuti verificati da,.

Ultima revisione per l'accuratezza in data: .

Audit: Complete
Formula (LaTeX) + variables + units
This section shows the formulas used by the calculator engine, plus variable definitions and units.
Formula (extracted LaTeX)
\[','\]
','
Formula (extracted text)
Using $W = F \\cdot d \\cdot \\cos(\\theta)$ —
Formula (extracted text)
$W = \\vec{F} \\cdot \\vec{d} = F\\,d\\,\\cos(\\theta)$ Solving for each variable: $F = \\dfrac{W}{d\\,\\cos(\\theta)}\\,,\\quad d = \\dfrac{W}{F\\,\\cos(\\theta)}\\,,\\quad \\theta = \\arccos\\!\\left(\\dfrac{W}{F\\,d}\\right)$
Formula (extracted text)
$W = F\\,d\\,\\cos(\\theta) = 50\\,\\text{N} \\times 12\\,\\text{m} \\times \\cos(30^\\circ)$ $\\cos(30^\\circ) = \\sqrt{3}/2 \\approx 0.8660$ $W \\approx 600 \\times 0.8660 = 519.6\\,\\text{J}$
Variables and units
  • T = property tax (annual or monthly depending on input) (currency)
Sources (authoritative):
Changelog
Version: 0.1.0-draft
Last code update: 2026-01-19
0.1.0-draft · 2026-01-19
  • Initial audit spec draft generated from HTML extraction (review required).
  • Verify formulas match the calculator engine and convert any text-only formulas to LaTeX.
  • Confirm sources are authoritative and relevant to the calculator methodology.
Verified by Ugo Candido on 2026-01-19
Profile · LinkedIn

Work Calculator

This professional-grade physics Work Calculator computes the mechanical work done by a constant force along a straight path using W = F · d · cos(θ). It’s designed for students, engineers, and educators who need fast, precise results with rigorous unit conversions and a clear step-by-step breakdown. This tool focuses on physics work (energy transfer), not “work hours” or timecards.

Results

—
Computed quantity—
Signed value—
Equivalent (kJ)—
Equivalent (ft·lbf)—

Step-by-step

Using $W = F \\cdot d \\cdot \\cos(\\theta)$
—

Data Source and Methodology

Authoritative Data Source: OpenStax College Physics, Chapter 7 “Work, Energy, and Energy Resources” (latest edition, 2016) — openstax.org/details/books/college-physics. Unit conversion constants are taken from the NIST Guide to the SI (2019) — physics.nist.gov/cuu/Units/.

All calculations are rigorously based on the formulas and data provided by these sources. Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.

The Formula Explained

$W = \\vec{F} \\cdot \\vec{d} = F\\,d\\,\\cos(\\theta)$

Solving for each variable:

$F = \\dfrac{W}{d\\,\\cos(\\theta)}\\,,\\quad d = \\dfrac{W}{F\\,\\cos(\\theta)}\\,,\\quad \\theta = \\arccos\\!\\left(\\dfrac{W}{F\\,d}\\right)$

Glossary of Variables

    - Work (W): Energy transferred by a force along a displacement; SI unit: joule (J). - Force (F): Constant force magnitude; SI unit: newton (N). 1 lbf = 4.4482216152605 N. - Displacement (d): Straight-line distance moved; SI unit: meter (m). 1 ft = 0.3048 m. - Angle (θ): Angle between force direction and displacement; measured in degrees. - ft·lbf: Foot-pound-force; 1 ft·lbf = 1.3558179483314 J. - BTU: British thermal unit; 1 BTU = 1055.05585262 J.

How It Works: A Step-by-Step Example

Suppose a 50 N force acts through a displacement of 12 m at an angle of 30° to the direction of motion. The work done is:

$W = F\\,d\\,\\cos(\\theta) = 50\\,\\text{N} \\times 12\\,\\text{m} \\times \\cos(30^\\circ)$

$\\cos(30^\\circ) = \\sqrt{3}/2 \\approx 0.8660$

$W \\approx 600 \\times 0.8660 = 519.6\\,\\text{J}$

The sign of W indicates whether the force aids motion (positive) or opposes it (negative).

Frequently Asked Questions (FAQ)

Is this calculator for physics work or work hours?

It is exclusively for physics work (energy transfer), not for timecards or payroll.

What if the force is perpendicular to the motion?

At θ = 90°, cos(θ) = 0, so W = 0. No energy is transferred along the path.

Can I calculate the angle from known work, force, and distance?

Yes. The calculator uses θ = arccos(W / (F d)) if the ratio is between −1 and 1.

How are units converted?

Using NIST constants: 1 lbf = 4.4482216152605 N; 1 ft = 0.3048 m; 1 ft·lbf = 1.3558179483314 J; 1 BTU = 1055.05585262 J; 1 Wh = 3600 J; 1 kWh = 3.6×10^6 J.

Why is my result negative?

If θ > 90°, the force opposes the motion, leading to negative work.

Does this handle variable forces or curved paths?

No. It assumes constant force and straight-line displacement. For variable forces, use the line integral W = ∫ F · ds.

How many significant figures are shown?

Results are formatted for readability; use the raw value if you need more precision via the copy button.

Strumento sviluppato da Ugo Candido,. Contenuti verificati da,.

Ultima revisione per l'accuratezza in data: .

Audit: Complete
Formula (LaTeX) + variables + units
This section shows the formulas used by the calculator engine, plus variable definitions and units.
Formula (extracted LaTeX)
\[','\]
','
Formula (extracted text)
Using $W = F \\cdot d \\cdot \\cos(\\theta)$ —
Formula (extracted text)
$W = \\vec{F} \\cdot \\vec{d} = F\\,d\\,\\cos(\\theta)$ Solving for each variable: $F = \\dfrac{W}{d\\,\\cos(\\theta)}\\,,\\quad d = \\dfrac{W}{F\\,\\cos(\\theta)}\\,,\\quad \\theta = \\arccos\\!\\left(\\dfrac{W}{F\\,d}\\right)$
Formula (extracted text)
$W = F\\,d\\,\\cos(\\theta) = 50\\,\\text{N} \\times 12\\,\\text{m} \\times \\cos(30^\\circ)$ $\\cos(30^\\circ) = \\sqrt{3}/2 \\approx 0.8660$ $W \\approx 600 \\times 0.8660 = 519.6\\,\\text{J}$
Variables and units
  • T = property tax (annual or monthly depending on input) (currency)
Sources (authoritative):
Changelog
Version: 0.1.0-draft
Last code update: 2026-01-19
0.1.0-draft · 2026-01-19
  • Initial audit spec draft generated from HTML extraction (review required).
  • Verify formulas match the calculator engine and convert any text-only formulas to LaTeX.
  • Confirm sources are authoritative and relevant to the calculator methodology.
Verified by Ugo Candido on 2026-01-19
Profile · LinkedIn

Work Calculator

This professional-grade physics Work Calculator computes the mechanical work done by a constant force along a straight path using W = F · d · cos(θ). It’s designed for students, engineers, and educators who need fast, precise results with rigorous unit conversions and a clear step-by-step breakdown. This tool focuses on physics work (energy transfer), not “work hours” or timecards.

Results

—
Computed quantity—
Signed value—
Equivalent (kJ)—
Equivalent (ft·lbf)—

Step-by-step

Using $W = F \\cdot d \\cdot \\cos(\\theta)$
—

Data Source and Methodology

Authoritative Data Source: OpenStax College Physics, Chapter 7 “Work, Energy, and Energy Resources” (latest edition, 2016) — openstax.org/details/books/college-physics. Unit conversion constants are taken from the NIST Guide to the SI (2019) — physics.nist.gov/cuu/Units/.

All calculations are rigorously based on the formulas and data provided by these sources. Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.

The Formula Explained

$W = \\vec{F} \\cdot \\vec{d} = F\\,d\\,\\cos(\\theta)$

Solving for each variable:

$F = \\dfrac{W}{d\\,\\cos(\\theta)}\\,,\\quad d = \\dfrac{W}{F\\,\\cos(\\theta)}\\,,\\quad \\theta = \\arccos\\!\\left(\\dfrac{W}{F\\,d}\\right)$

Glossary of Variables

    - Work (W): Energy transferred by a force along a displacement; SI unit: joule (J). - Force (F): Constant force magnitude; SI unit: newton (N). 1 lbf = 4.4482216152605 N. - Displacement (d): Straight-line distance moved; SI unit: meter (m). 1 ft = 0.3048 m. - Angle (θ): Angle between force direction and displacement; measured in degrees. - ft·lbf: Foot-pound-force; 1 ft·lbf = 1.3558179483314 J. - BTU: British thermal unit; 1 BTU = 1055.05585262 J.

How It Works: A Step-by-Step Example

Suppose a 50 N force acts through a displacement of 12 m at an angle of 30° to the direction of motion. The work done is:

$W = F\\,d\\,\\cos(\\theta) = 50\\,\\text{N} \\times 12\\,\\text{m} \\times \\cos(30^\\circ)$

$\\cos(30^\\circ) = \\sqrt{3}/2 \\approx 0.8660$

$W \\approx 600 \\times 0.8660 = 519.6\\,\\text{J}$

The sign of W indicates whether the force aids motion (positive) or opposes it (negative).

Frequently Asked Questions (FAQ)

Is this calculator for physics work or work hours?

It is exclusively for physics work (energy transfer), not for timecards or payroll.

What if the force is perpendicular to the motion?

At θ = 90°, cos(θ) = 0, so W = 0. No energy is transferred along the path.

Can I calculate the angle from known work, force, and distance?

Yes. The calculator uses θ = arccos(W / (F d)) if the ratio is between −1 and 1.

How are units converted?

Using NIST constants: 1 lbf = 4.4482216152605 N; 1 ft = 0.3048 m; 1 ft·lbf = 1.3558179483314 J; 1 BTU = 1055.05585262 J; 1 Wh = 3600 J; 1 kWh = 3.6×10^6 J.

Why is my result negative?

If θ > 90°, the force opposes the motion, leading to negative work.

Does this handle variable forces or curved paths?

No. It assumes constant force and straight-line displacement. For variable forces, use the line integral W = ∫ F · ds.

How many significant figures are shown?

Results are formatted for readability; use the raw value if you need more precision via the copy button.

Strumento sviluppato da Ugo Candido,. Contenuti verificati da,.

Ultima revisione per l'accuratezza in data: .

Audit: Complete
Formula (LaTeX) + variables + units
This section shows the formulas used by the calculator engine, plus variable definitions and units.
Formula (extracted LaTeX)
\[','\]
','
Formula (extracted text)
Using $W = F \\cdot d \\cdot \\cos(\\theta)$ —
Formula (extracted text)
$W = \\vec{F} \\cdot \\vec{d} = F\\,d\\,\\cos(\\theta)$ Solving for each variable: $F = \\dfrac{W}{d\\,\\cos(\\theta)}\\,,\\quad d = \\dfrac{W}{F\\,\\cos(\\theta)}\\,,\\quad \\theta = \\arccos\\!\\left(\\dfrac{W}{F\\,d}\\right)$
Formula (extracted text)
$W = F\\,d\\,\\cos(\\theta) = 50\\,\\text{N} \\times 12\\,\\text{m} \\times \\cos(30^\\circ)$ $\\cos(30^\\circ) = \\sqrt{3}/2 \\approx 0.8660$ $W \\approx 600 \\times 0.8660 = 519.6\\,\\text{J}$
Variables and units
  • T = property tax (annual or monthly depending on input) (currency)
Sources (authoritative):
Changelog
Version: 0.1.0-draft
Last code update: 2026-01-19
0.1.0-draft · 2026-01-19
  • Initial audit spec draft generated from HTML extraction (review required).
  • Verify formulas match the calculator engine and convert any text-only formulas to LaTeX.
  • Confirm sources are authoritative and relevant to the calculator methodology.
Verified by Ugo Candido on 2026-01-19
Profile · LinkedIn

Work Calculator

This professional-grade physics Work Calculator computes the mechanical work done by a constant force along a straight path using W = F · d · cos(θ). It’s designed for students, engineers, and educators who need fast, precise results with rigorous unit conversions and a clear step-by-step breakdown. This tool focuses on physics work (energy transfer), not “work hours” or timecards.

Results

—
Computed quantity—
Signed value—
Equivalent (kJ)—
Equivalent (ft·lbf)—

Step-by-step

Using $W = F \\cdot d \\cdot \\cos(\\theta)$
—

Data Source and Methodology

Authoritative Data Source: OpenStax College Physics, Chapter 7 “Work, Energy, and Energy Resources” (latest edition, 2016) — openstax.org/details/books/college-physics. Unit conversion constants are taken from the NIST Guide to the SI (2019) — physics.nist.gov/cuu/Units/.

All calculations are rigorously based on the formulas and data provided by these sources. Tutti i calcoli si basano rigorosamente sulle formule e sui dati forniti da questa fonte.

The Formula Explained

$W = \\vec{F} \\cdot \\vec{d} = F\\,d\\,\\cos(\\theta)$

Solving for each variable:

$F = \\dfrac{W}{d\\,\\cos(\\theta)}\\,,\\quad d = \\dfrac{W}{F\\,\\cos(\\theta)}\\,,\\quad \\theta = \\arccos\\!\\left(\\dfrac{W}{F\\,d}\\right)$

Glossary of Variables

    - Work (W): Energy transferred by a force along a displacement; SI unit: joule (J). - Force (F): Constant force magnitude; SI unit: newton (N). 1 lbf = 4.4482216152605 N. - Displacement (d): Straight-line distance moved; SI unit: meter (m). 1 ft = 0.3048 m. - Angle (θ): Angle between force direction and displacement; measured in degrees. - ft·lbf: Foot-pound-force; 1 ft·lbf = 1.3558179483314 J. - BTU: British thermal unit; 1 BTU = 1055.05585262 J.

How It Works: A Step-by-Step Example

Suppose a 50 N force acts through a displacement of 12 m at an angle of 30° to the direction of motion. The work done is:

$W = F\\,d\\,\\cos(\\theta) = 50\\,\\text{N} \\times 12\\,\\text{m} \\times \\cos(30^\\circ)$

$\\cos(30^\\circ) = \\sqrt{3}/2 \\approx 0.8660$

$W \\approx 600 \\times 0.8660 = 519.6\\,\\text{J}$

The sign of W indicates whether the force aids motion (positive) or opposes it (negative).

Frequently Asked Questions (FAQ)

Is this calculator for physics work or work hours?

It is exclusively for physics work (energy transfer), not for timecards or payroll.

What if the force is perpendicular to the motion?

At θ = 90°, cos(θ) = 0, so W = 0. No energy is transferred along the path.

Can I calculate the angle from known work, force, and distance?

Yes. The calculator uses θ = arccos(W / (F d)) if the ratio is between −1 and 1.

How are units converted?

Using NIST constants: 1 lbf = 4.4482216152605 N; 1 ft = 0.3048 m; 1 ft·lbf = 1.3558179483314 J; 1 BTU = 1055.05585262 J; 1 Wh = 3600 J; 1 kWh = 3.6×10^6 J.

Why is my result negative?

If θ > 90°, the force opposes the motion, leading to negative work.

Does this handle variable forces or curved paths?

No. It assumes constant force and straight-line displacement. For variable forces, use the line integral W = ∫ F · ds.

How many significant figures are shown?

Results are formatted for readability; use the raw value if you need more precision via the copy button.

Strumento sviluppato da Ugo Candido,. Contenuti verificati da,.

Ultima revisione per l'accuratezza in data: .

Audit: Complete
Formula (LaTeX) + variables + units
This section shows the formulas used by the calculator engine, plus variable definitions and units.
Formula (extracted LaTeX)
\[','\]
','
Formula (extracted text)
Using $W = F \\cdot d \\cdot \\cos(\\theta)$ —
Formula (extracted text)
$W = \\vec{F} \\cdot \\vec{d} = F\\,d\\,\\cos(\\theta)$ Solving for each variable: $F = \\dfrac{W}{d\\,\\cos(\\theta)}\\,,\\quad d = \\dfrac{W}{F\\,\\cos(\\theta)}\\,,\\quad \\theta = \\arccos\\!\\left(\\dfrac{W}{F\\,d}\\right)$
Formula (extracted text)
$W = F\\,d\\,\\cos(\\theta) = 50\\,\\text{N} \\times 12\\,\\text{m} \\times \\cos(30^\\circ)$ $\\cos(30^\\circ) = \\sqrt{3}/2 \\approx 0.8660$ $W \\approx 600 \\times 0.8660 = 519.6\\,\\text{J}$
Variables and units
  • T = property tax (annual or monthly depending on input) (currency)
Sources (authoritative):
Changelog
Version: 0.1.0-draft
Last code update: 2026-01-19
0.1.0-draft · 2026-01-19
  • Initial audit spec draft generated from HTML extraction (review required).
  • Verify formulas match the calculator engine and convert any text-only formulas to LaTeX.
  • Confirm sources are authoritative and relevant to the calculator methodology.
Verified by Ugo Candido on 2026-01-19
Profile · LinkedIn
Formulas

(Formulas preserved from original page content, if present.)

Version 0.1.0-draft
Citations

Add authoritative sources relevant to this calculator (standards bodies, manuals, official docs).

Changelog
  • 0.1.0-draft — 2026-01-19: Initial draft (review required).