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Work and Power Calculator

Solve common Work and Power questions in seconds: use P = W/t, W = F·d·cos(θ), or P = F·v (optional cos(θ)). Includes quick picks, unit conversions, step-by-step, plus a mini gauge + chart/diagram.

Background

Work is energy transferred by a force acting through a distance. Power is how fast energy is transferred (work per time). Only the parallel component of a force does work—so angles matter: W = F·d·cos(θ).

Enter values

Tip: If you know work and time, use P = W/t. If you know force and distance, use W = F·d·cos(θ). If you know force and speed, use P = F·v.

Power from work & time

Uses P = W/t.

Select units to auto-convert.

Select units to auto-convert (internally uses seconds).

Select units to auto-convert.

Quick intuition

1 hp ≈ 746 W. Phone charger ~20 W. Laptop ~60 W. Kettle ~1500 W.

Options

Rounding affects display only.

Chips prefill and calculate immediately.

Result

No results yet. Enter values and click Calculate.

How to use this calculator

  • Choose a mode: P = W/t, W = F·d·cos(θ), or P = F·v.
  • Pick what you want to solve for, enter the other values, then click Calculate.
  • Use the visuals: gauge for power scale, and the chart/diagram to “see” how cos(θ) changes results.

How this calculator works

  • Unit conversions: internally converts to SI (J, s, N, m, m/s, W), then converts back for display.
  • Angles: only the component along the motion matters: F∥ = F·cos(θ).
  • Negative results: if cos(θ) is negative (force opposes motion), work/power becomes negative.

Formulas & Equations Used

Power from work & time: P = W/t

Work (constant force): W = F·d·cos(θ)

Power from force & speed: P = F·v (optional: P = F·v·cos(θ))

Example Problem & Step-by-Step Solution

Example 1 — Power from work & time

A motor does W = 600 J of work in t = 10 s. Find power.

  1. Use P = W/t.
  2. Compute: P = 600/10 = 60 W.

Example 2 — Work from force & distance (angle matters)

You pull a box with F = 50 N at θ = 30° over d = 8 m. Find the work done by the pulling force.

  1. Use W = F·d·cos(θ).
  2. Compute cos(30°) ≈ 0.866.
  3. W = 50·8·0.866 ≈ 346 J (positive → force helps the motion).

Example 3 — Power from force & speed

A cyclist pushes with an effective forward force of F = 120 N while moving at v = 5 m/s. Find the power output (assume θ = 0°).

  1. Use P = F·v.
  2. P = 120·5 = 600 W (about 0.8 hp).

Frequently Asked Questions

Q: What’s the difference between work and power?

Work is total energy transferred. Power is the rate: how fast that work happens.

Q: When is work (or power) negative?

When the force points opposite the motion (cos(θ) < 0), meaning energy is being removed from the object.

Q: Why does the angle matter?

Only the component along the motion does work: F∥ = F·cos(θ). At 90°, cos(θ)=0, so work and power are zero (for that force).

Q: Can I compute power from force and speed?

Yes. For constant force along the motion, P = F·v. If the force is at an angle, use P = F·v·cos(θ).