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🔭 Energy & Power

Work & Power Calculator

From force and distance, find the work done (W = F·d); from work and time, find the power (P = W/t). Results in joules, watts, kilowatts and horsepower.

Work (joules)
Power (watts)
kW & horsepower
Force at an angle
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Work & power — Quick answer

Work is force times distance moved; power is that work per second. Same work done faster means more power.

W = F · d (· cos θ)  [joules]
P = W / t = F · v  [watts] · hp = W / 745.7

Worked example: F = 200 N over d = 15 m in t = 10 s. W = 3000 J; P = 3000/10 = 300 W (0.30 kW, 0.40 hp).

3000 J of work, different times

TimePower≈ hp
5 s600 W0.80
10 s300 W0.40
30 s100 W0.13

Used for: lifting, machines, motors, exercise, energy problems.

🔭 Work & Power Calculator

Enter force and distance for work. Add a time for power. An angle (force vs motion) is optional.

Work done
Power
Power (kW)
Power (hp)

⚠️ Work needs movement along the force — a force at 90° to the motion does zero work. Power is mechanical (shaft) output; for a motor's electrical input, divide by efficiency.

In physics, work has a precise meaning: a force doing work only when it moves something, with W = F·d measuring the energy transferred. Power then measures how fast that energy is delivered — the work divided by the time, in watts. The same job (lifting a load, pushing a crate) always takes the same work, but doing it quickly demands more power. That single distinction — total energy versus rate of energy — separates a strong machine from a fast one, and underlies every motor and engine rating.

Reviewed: June 19, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: the definitions of mechanical work and power.

The work & power equations

Work done
W = F × d × cos θ (θ = 0 when force is along the motion)
Power
P = W / t = F × v
Unit conversions
1 kW = 1000 W · 1 hp = 745.7 W

With force in newtons and distance in metres, work comes out in joules; divide by seconds for power in watts. The cosine term handles a force applied at an angle — only the component along the direction of motion does work, so a force perpendicular to the path (like the tension in a string whirling a ball) does none. The P = F·v form is the handy one for steady motion, linking a pushing force directly to the power at a given speed.

Worked example — pushing a crate

Scenario: Pushing a crate with 200 N over 15 m, taking 10 seconds, force aligned with the motion.

Work
W = 200 × 15 × cos 0° = 3000 J
Power
P = 3000 / 10 = 300 W = 0.30 kW ≈ 0.40 hp

The push transfers 3,000 joules of energy over the 15 m, and at a steady pace over 10 seconds that is 300 watts of power. Do the same job in 5 seconds and the work is unchanged at 3,000 J, but the power doubles to 600 W — you are working twice as fast. If the force were applied at 60° to the motion instead, only cos 60° = half of it would count, halving the work to 1,500 J.

Frequently Asked Questions

How do you calculate work done?

W = F × d (× cos θ if angled). 200 N over 15 m = 3000 J. No movement = no work.

How do you calculate power?

P = W/t (watts), or P = F·v. 3000 J in 10 s = 300 W.

Work vs power?

Work = energy transferred (J); power = how fast (W). Running upstairs = same work, more power than walking.

Watts to horsepower?

hp = W ÷ 745.7. 300 W ≈ 0.40 hp; 1 kW ≈ 1.34 hp. Metric PS uses 735.5 W.

Does holding something do work?

No — no displacement, no work. W = F×d = 0 when d = 0, even though your muscles tire.

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