Library / Fundamentals / FUN-02
Fundamentals
Work and kinetic energy
Net work needed to change the speed of a mass.
- Ref
- FUN-02
- Rev
- v1
- Field
- Fundamentals
- Result
- W [kJ]
W in joules (J), m in kilograms, u and v the initial and final speed in metres per second. Negative work means the body slows down.
The net work done on a body equals the change in its kinetic energy. Speeding up takes positive work from the net force; slowing down means the net force does negative work.
Calculator
- Kinetic energy at the start75 kJ
- Kinetic energy at the end468.8 kJ
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Energy depends on speed squared. Going from 20 to 30 m/s costs far more than going from 10 to 20 m/s.
- Reversing the formula gives stopping work: braking from v to rest removes ½mv² of energy, which becomes heat.
- Only the component of force along the motion does work. A force perpendicular to the motion does none.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (kJ) | Local step | Change in result |
|---|---|---|---|---|---|---|
| m Mass | 200 to 20,000 kg | 52.5 | 5,250 | 5,198 | +10% | +10 % |
| v Final speed | 25 to 40 m/s | 393.8 | 1,125 | 731.3 | +10% | +25 % |
| u Initial speed | 0 to 20 m/s | 168.8 | 468.8 | 300 | +10% | −4 % |
Sweep of Final speed
| Final speed (m/s) | Net work (kJ) |
|---|---|
| 25 | 393.8 |
| 26.88 | 466.7 |
| 28.75 | 544.9 |
| 30.63 | 628.4 |
| 32.5 | 717.2 |
| 34.38 | 811.2 |
| 36.25 | 910.5 |
| 38.13 | 1,015 |
| 40 | 1,125 |
Sweep one variable
Variables and how each one contributes
Net work
What the equation gives you. Everything else feeds it.
Mass
Mass of the body (constant). Directly proportional: twice the mass needs twice the work for the same speed change.
Initial speed
Speed at the start of the interval. A higher starting speed means less work is needed to reach the same final speed, and the effect grows with u².
Final speed
Speed at the end of the interval. The most powerful variable: work grows with the square of the final speed, so +10% speed gives about +21% energy.
Worked example with the values above
- Kinetic energy before: ½ × 1,500 × 10² = 75,000 J. After: ½ × 1,500 × 25² = 468,800 J.
- Net work is the difference: 468,800 − 75,000 = 393,800 J.
- 393.8 kJ of net work is needed.
Assumptions and limits
- Constant mass particle or rigid body.
- W is the net work of all forces, not of one force.
- Speeds in an inertial frame.
Common mistakes
- Squaring after subtracting: it is v² − u², not (v − u)².
- Leaving speeds in km/h or knots instead of m/s.
- Forgetting that work is a scalar and can be negative.
Statistics and uncertainty
The result is a difference of two squares, so errors in the speeds are amplified: δ(v²)/v² ≈ 2 δv/v. A 3% error in speed gives about a 6% error in kinetic energy. When u and v are close, the difference becomes small and its relative error grows.
Where it is used
- Energy needed to accelerate a vehicle or a projectile.
- Brake energy and heat load.
- Impact energy of a moving mass.
Maths you need
- Newton’s second law: work is force acting through a distance
- Squares: energy depends on speed squared