Work Done Calculator
Calculates work done using W=F×d×cosθ — solving for work, force, distance, or the angle between them — and includes the Work-Energy Theorem calculator (W_net=ΔKE), net work from multiple forces, and a power calculator. The complete work formula W=Fd·cosθ with full step-by-step working.
✅ POSITIVE WORK
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Force
F
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Displacement
d
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Angle
θ
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Work-Energy Theorem
W_net = ΔKE = ½mv₂² − ½mv₁²
Net work done equals change in kinetic energy
Add up to 6 forces. W_net = W₁ + W₂ + ... + Wₙ (work done is scalar — values add algebraically).
Power Formulas
P = W/t | P = F × v × cos(θ)
1 hp = 745.7 W | 1 kW = 1000 W
| θ | cos(θ) | Work | Situation | Sign |
|---|---|---|---|---|
| 0° | 1.000 | W = Fd | Horizontal push, falling object | Positive ↑ |
| 30° | 0.866 | W = 0.866 Fd | Lawnmower handle at 30° | Positive ↑ |
| 45° | 0.707 | W = 0.707 Fd | Handle at 45° | Positive ↑ |
| 60° | 0.500 | W = 0.500 Fd | Handle at 60° | Positive ↑ |
| 90° | 0.000 | W = 0 | Carrying bag horizontally, normal force | Zero ⊥ |
| 120° | −0.500 | W = −0.500 Fd | Braking / retarding force | Negative ↓ |
| 180° | −1.000 | W = −Fd | Friction opposing motion | Negative ↓ |
| Situation | Approx. Work Done |
|---|---|
| Lifting 1 kg by 1 m | 9.81 J |
| Climbing one flight of stairs (70 kg person) | ~2,000 J |
| Pushing a car 1 m (500N force) | 500 J |
| Burning 1 food calorie | 4,184 J |
| 1 kWh of electricity | 3,600,000 J |
| Braking a car from 100 km/h (1500 kg) | ~578,700 J |
| Human heart beating for 1 minute | ~1 J |
| Angle Range | Work Sign | Example |
|---|---|---|
| 0° ≤ θ < 90° | W > 0 (Positive) | Pushing a box forward, applied force |
| θ = 90° | W = 0 (Zero) | Carrying a bag horizontally, normal force |
| 90° < θ ≤ 180° | W < 0 (Negative) | Friction, braking, retarding force |
| θ = 180° | W most negative | Force directly opposing motion (friction) |
Work Done Formula — W = F × d × cos(θ)
This work done calculator uses the complete work formula W = F × d × cos(θ) to calculate the work done by any force at any angle to the direction of motion. The work done formula is the foundation of mechanics energy calculations.
Each variable in the work equation physics W = F × d × cos(θ) has a precise meaning:
- W — work done, measured in Joules (J = N·m = kg·m²/s²)
- F — magnitude of the applied force (always positive, in Newtons)
- d — displacement of the object (net change in position, not total distance traveled)
- θ — the angle between the force vector and the displacement vector
- cos(θ) — the geometric factor: only the component of force along displacement does work
The key insight of the work done physics formula: only the component of force in the direction of motion does work. That component is F×cos(θ). This is why a force perpendicular to motion (θ=90°) does zero work — it has no component along the displacement.
Displacement vs Distance: Work depends on displacement (NET change in position), not total distance traveled. If you walk in a complete circle and return to your starting point, the displacement is zero, so the net work done against gravity is zero — regardless of how far you walked. This is the definition of a conservative force.
How to Calculate Work Done — Step-by-Step
Use this five-step method to calculate work done for any force-displacement problem:
- Step 1 — Identify the force magnitude F: Use the given force in Newtons (convert if needed)
- Step 2 — Measure displacement d: Find the NET displacement in meters (not path length)
- Step 3 — Determine angle θ: Find the angle between the force VECTOR and displacement VECTOR
- Step 4 — Compute cos(θ): Use a calculator or the special angle table above
- Step 5 — Multiply W = F × d × cos(θ): The result is the work done in Joules
Example 1 — Horizontal Push (θ=0°): F=200N, d=5m
- Force and displacement are parallel: θ = 0°
- cos(0°) = 1.000
- W = 200 × 5 × 1.000 = 1000 J
- Maximum possible work done — force fully aligned with motion
Example 2 — Angled Push / Lawnmower (θ=30°): F=150N, d=8m
- Handle at 30° below horizontal while moving horizontally: θ = 30°
- cos(30°) = √3/2 = 0.8660
- W = 150 × 8 × 0.8660 = 1200 × 0.8660 = 1039.2 J
- Effective horizontal force = 150 × cos(30°) = 129.9 N
Example 3 — Carrying a Bag (θ=90°): F=50N upward, d=10m horizontal
- Support force is vertical (upward); displacement is horizontal: θ = 90°
- cos(90°) = 0.000
- W = 50 × 10 × 0 = 0 J — no work done by the carrying force!
- This is the most counterintuitive case: large force, zero work done
Example 4 — Friction Opposing Motion (θ=180°): F=40N, d=6m
- Friction acts opposite to displacement: θ = 180°
- cos(180°) = −1.000
- W = 40 × 6 × (−1) = −240 J — negative work done
- Friction removes 240 J of kinetic energy from the object
Positive, Negative, and Zero Work — The Three Cases
Positive Work (W > 0)
θ < 90° — Force has component in direction of motion. Energy added TO object. Kinetic energy increases.
Zero Work (W = 0)
θ = 90° — Force perpendicular to motion. No energy transferred. cos(90°) = 0 always.
Negative Work (W < 0)
θ > 90° — Force opposes motion. Energy removed FROM object. Kinetic energy decreases.
Zero work done is the most counterintuitive case in the work done physics formula. You can exert a very large force and yet do zero work if that force is perpendicular to motion:
- Carrying a heavy bag horizontally: your arm exerts an upward force; motion is horizontal → θ=90° → zero work done by you
- Satellite in circular orbit: gravitational force points toward Earth's center; velocity is tangential → perpendicular → zero work by gravity
- Normal force from surface: perpendicular to motion along surface → always zero work
- Magnetic force: always perpendicular to velocity → always zero work done
The sign of work indicates energy direction in the work done formula: positive work means energy is transferred TO the object (it gains kinetic energy); negative work means the object does work ON something else (it loses kinetic energy). Friction always does negative work because it always opposes displacement (θ=180° always).
The Angle Between Force and Displacement
The most common source of errors in work done calculations is incorrectly identifying θ in the work formula with angle. The angle θ is the angle between the FORCE VECTOR and the DISPLACEMENT VECTOR — not the angle between force and horizontal, not the incline angle.
Case 1: Lawnmower pushed at 35° below horizontal
Force direction: 35° below horizontal. Displacement direction: horizontal (along ground).
Angle between force vector and displacement vector = θ = 35°
W = F × d × cos(35°) = F × d × 0.819
Case 2: Gravity on object sliding down 40° incline
Gravity direction: straight down (vertical). Displacement direction: along incline (40° below horizontal).
Angle between gravity vector (down) and displacement vector (along incline) = θ = 40°
W_gravity = mg × d × cos(40°) = mg × d × 0.766
The component approach makes the work formula with angle geometrically obvious: F×cos(θ) is the projection of the force vector onto the displacement direction. This is the "effective force" that actually drives the motion. The remaining component F×sin(θ) is perpendicular to motion — it does zero work done and only pushes the object into (or away from) the surface.
Work-Energy Theorem — W_net = ΔKE
The Work-Energy Theorem is one of the most powerful equations in mechanics. It states that the net work done on an object equals the change in its kinetic energy:
The Work-Energy Theorem is powerful because it lets you find final velocity without tracking acceleration step by step. If you know all the forces and the displacement, you can sum all the work done and find the velocity change directly.
Example 1 — Accelerating Object: m=2kg, v₁=3 m/s, v₂=7 m/s
- KE₁ = ½ × 2 × 3² = ½ × 2 × 9 = 9 J
- KE₂ = ½ × 2 × 7² = ½ × 2 × 49 = 49 J
- W_net = ΔKE = 49 − 9 = 40 J of net work done on the object
Example 2 — Braking Car: m=1000kg, v₁=30 m/s, v₂=0
- KE₁ = ½ × 1000 × 900 = 450,000 J
- KE₂ = 0 J
- W_net = 0 − 450,000 = −450,000 J (friction does this negative work done)
Example 3 (Work-Energy Theorem Mode B) — Find v₂: m=2kg, v₁=0, W=40J
- W_net = ½mv₂² − ½mv₁² → 40 = ½×2×v₂² − 0
- v₂² = 2×40/2 = 40 → v₂ = √40 = 6.32 m/s
Work Done by Gravity and Friction
Two forces — gravity and friction — appear in almost every work done physics problem. Their work follows directly from the work done formula W = F × d × cos(θ).
Work Done by Gravity:
- Object falling height h: W_gravity = mgh × cos(0°) = mgh (positive — gravity aids downward motion)
- Object moving horizontally: W_gravity = 0 (gravity perpendicular to horizontal displacement, zero work done)
- Object rising height h: W_gravity = mgh × cos(180°) = −mgh (negative — gravity opposes upward motion)
Gravity Example: 5 kg object falls 3 m
- F_gravity = mg = 5 × 9.81 = 49.05 N (downward)
- Displacement = 3 m (downward) → θ = 0°
- W_gravity = 49.05 × 3 × cos(0°) = 147.15 J
Work Done by Friction:
Friction always does negative work because it always opposes motion (θ=180° always). W_friction = −μmg × d (for kinetic friction on a horizontal surface). This is always negative, removing kinetic energy from the sliding object.
Friction Example: μ=0.3, m=10kg, d=4m
- Friction force = μmg = 0.3 × 10 × 9.81 = 29.43 N
- θ = 180° (friction opposes motion), cos(180°) = −1
- W_friction = 29.43 × 4 × (−1) = −117.72 J
Power — Rate of Doing Work
Power is the rate at which work is done: P = W/t. A powerful machine does the same work done in less time. The instantaneous power when a force F acts on an object moving at velocity v is P = F×v×cos(θ).
Example 1 — Car Engine: W=1000J done in 5s
- P = W/t = 1000/5 = 200 W = 0.200 kW = 0.268 hp
Example 2 — Person Climbing Stairs: 70kg, 3m height, 6 seconds
- W_gravity = mgh = 70 × 9.81 × 3 = 2060.1 J
- P = 2060.1 / 6 = 343.4 W = 0.461 hp
Example 3 — Electric Motor: F=100N, v=5 m/s, θ=0°
- P = F × v × cos(θ) = 100 × 5 × 1 = 500 W = 0.5 kW = 0.671 hp
Common Mistakes in Work Done Calculations
Mistake 1 — Using total distance instead of displacement
- ❌ Wrong: Object walks 5m forward and 5m back — total distance=10m, W = F×10
- ✅ Correct: Displacement = 0 m → W = F × 0 × cos(θ) = 0 J for any constant force
Mistake 2 — Wrong angle identification
- ❌ Wrong: Lawnmower handle at 35° — using θ = 90° − 35° = 55°
- ✅ Correct: θ is between the force vector (along handle, pointing down-forward) and the displacement vector (horizontal) → θ = 35°
Mistake 3 — Forgetting cos(θ) when force is not parallel to motion
- ❌ Wrong: F=50N, d=10m, θ=30° → W = 50 × 10 = 500 J
- ✅ Correct: W = 50 × 10 × cos(30°) = 500 × 0.866 = 433 J
Mistake 4 — Thinking zero work means zero force
- ❌ Wrong: "No work done means no force is applied"
- ✅ Correct: A large force at θ=90° does zero work done. Normal force, centripetal force, and magnetic force all do zero work — not because they are zero, but because they are perpendicular to motion.
Mistake 5 — Treating friction as doing positive work
- ❌ Wrong: Friction force = 30N, d = 5m → W_friction = +150 J
- ✅ Correct: Friction always opposes motion → θ=180° → W_friction = 30 × 5 × cos(180°) = −150 J (negative work)
Worked Examples — 8 Complete Problems
1. Pushing box horizontally: F=200N, d=5m, θ=0°
- cos(0°) = 1.000
- W = 200 × 5 × 1.000 = 1000 J (positive work)
2. Lawnmower: F=150N, d=8m, θ=35°
- cos(35°) = 0.8192
- W = 150 × 8 × 0.8192 = 1200 × 0.8192 = 983.0 J
3. Carrying bag: F=50N upward, d=10m horizontal, θ=90°
- cos(90°) = 0.000
- W = 50 × 10 × 0 = 0 J (zero work done)
4. Friction opposing motion: F=40N, d=6m, θ=180°
- cos(180°) = −1.000
- W = 40 × 6 × (−1) = −240 J (negative work)
5. Find force: W=500J, d=10m, θ=45°
- F = W/(d×cos(θ)) = 500/(10×0.7071) = 500/7.071 = 70.71 N
6. Find distance: W=1200J, F=80N, θ=30°
- d = W/(F×cos(θ)) = 1200/(80×0.8660) = 1200/69.28 = 17.32 m
7. Find angle: W=350J, F=100N, d=5m
- cos(θ) = W/(F×d) = 350/(100×5) = 350/500 = 0.700
- θ = arccos(0.700) = 45.57°
8. Work-Energy Theorem: m=3kg, v₁=4 m/s, v₂=10 m/s
- KE₁ = ½ × 3 × 16 = 24 J
- KE₂ = ½ × 3 × 100 = 150 J
- W_net = 150 − 24 = 126 J
Frequently Asked Questions — Work Done
Related Calculators
Force aids motion
Energy → object ↑
Force perpendicular
cos(90°)=0 always
Force opposes motion
Energy ← object ↓