Calculate kinetic energy (Ek = ½mv²), velocity, or mass in Joules, kJ, and ft-lb. Includes step-by-step formulas, momentum, and real-world impact examples.
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Kinetic energy is the work needed to accelerate a body of a given mass from rest to its stated velocity. Once gained during acceleration, the body maintains this kinetic energy unless its speed changes. According to the Work-Energy Theorem, net work done on an object equals the change in its kinetic energy: W = ΔEk.
The fundamental insight of Ek = ½mv² is the quadratic relationship with velocity. While momentum (p = mv) scales linearly with speed, kinetic energy scales with speed squared. If you drive at 100 km/h instead of 50 km/h, your momentum doubles, but your kinetic energy and necessary braking distance quadruple (4×). At 150 km/h, your kinetic energy is nine times (9×) greater than at 50 km/h.
To solve for Kinetic Energy: Ek = ½ × m × v². Multiply mass in kilograms by velocity in meters per second squared, then divide by 2.
To solve for Velocity: Multiply kinetic energy by 2, divide by mass, and take the square root: v = √(2Ek / m).
To solve for Mass: Multiply kinetic energy by 2, then divide by velocity squared: m = 2Ek / v².
1 Joule (J) is the mechanical work performed by a force of 1 Newton acting through a displacement of 1 meter. In imperial standards, 1 foot-pound force (ft·lbf) is approximately 1.35582 Joules (or 1 J ≈ 0.737562 ft·lbf).
In electrical terms, 1 Watt-hour (Wh) equals 3,600 Joules, and 1 kilowatt-hour (kWh) equals 3,600,000 Joules (3.6 Megajoules). An electric car battery storing 72 kWh contains approximately 259.2 Megajoules of chemical-electrical energy.
The reference table below compares kinetic energy and momentum across typical real-world moving bodies.
| Moving Body / Scenario | Object Mass | Velocity | Kinetic Energy (J) | Momentum (kg·m/s) |
|---|---|---|---|---|
| Major League Fastball | 145 g (0.145 kg) | 42 m/s (94 mph) | 128 J | 6.09 kg·m/s |
| 9mm Handgun Bullet | 9.0 g (0.009 kg) | 360 m/s (1,180 ft/s) | 583 J | 3.24 kg·m/s |
| Olympic Sprinter (Usain Bolt) | 94 kg | 12.4 m/s (27.8 mph) | 7,224 J (7.22 kJ) | 1,166 kg·m/s |
| Sedan at City Speed (50 km/h) | 1,500 kg | 13.89 m/s (31 mph) | 144,676 J (144.7 kJ) | 20,835 kg·m/s |
| Sedan at Highway Speed (100 km/h) | 1,500 kg | 27.78 m/s (62 mph) | 578,704 J (578.7 kJ) | 41,670 kg·m/s |
| Loaded Semi-Truck (80,000 lbs) | 36,287 kg | 26.82 m/s (60 mph) | 13,050,470 J (13.05 MJ) | 973,217 kg·m/s |
| Boeing 737 at Landing Speed | 65,000 kg | 72 m/s (140 knots) | 168,480,000 J (168.5 MJ) | 4,680,000 kg·m/s |
Deepen your mechanical physics calculations with our dedicated force and electrical power engines.
To solve for acceleration, vehicle thrust, or Newton’s Second Law relationships, check our Force Calculator (F = ma), or launch our Electrical Power Calculator to evaluate energy consumption.
Launch Force Calculator →A 1,500 kg passenger sedan travels at 108 km/h (30 m/s). Compute the kinetic energy and stopping momentum.
The vehicle possesses 675 kJ of energy. Braking systems must dissipate this entire amount as heat to bring the car to a complete stop.
A 9mm pistol bullet with a mass of 9.0 grams (0.009 kg) leaves the muzzle at a velocity of 360 m/s (1,181 ft/s).
The 9g projectile carries 583.2 Joules (430.2 ft·lb) of kinetic energy upon exiting the muzzle.
An aerospace test projectile with a mass of 500 kg must deliver 10 Megajoules (10,000,000 J) of kinetic energy upon impact. Calculate required speed.
The projectile must achieve 200 m/s (720 km/h or 447.4 mph) to deliver 10 MJ of kinetic energy.
A surveillance drone flying at 15 m/s (54 km/h) is determined to possess 450 Joules of kinetic energy. Calculate its flight mass.
The drone has an operational mass of exactly 4.0 kg (8.82 pounds).
Kinetic energy is calculated with velocity squared (v²). Because of this quadratic relationship, 2² = 4. If a vehicle increases its speed from 50 km/h to 100 km/h, its kinetic energy increases fourfold (4×), requiring roughly 4 times longer braking distance under equal friction.
Multiply the kinetic energy by 2, divide by the mass in kilograms, and take the square root of the result: v = √(2 × Ek / m). For example, if a 500 kg object has 10 MJ (10,000,000 J) of energy: v = √(20,000,000 / 500) = √40,000 = 200 m/s.
One foot-pound force equals approximately 1.355818 Joules. To convert Joules to foot-pounds, divide by 1.355818 (or multiply by approximately 0.737562). For example, 1,000 Joules equals approximately 737.56 ft·lb.
Momentum (p = m × v) is a vector quantity proportional to velocity, indicating an object’s resistance to stopping or change in direction. Kinetic energy (Ek = ½mv²) is a scalar quantity proportional to velocity squared, representing the capacity to do mechanical work or cause impact deformation.
No. Kinetic energy is a scalar quantity and has magnitude only, with no directional vector component. An object moving at 20 m/s north has the exact same kinetic energy as an object moving at 20 m/s east, provided their masses are identical.
According to the Work-Energy Theorem, stopping distance d equals kinetic energy divided by average braking friction force: d = Ek / F_braking. Because kinetic energy scales with speed squared, doubling your driving speed quadruples your vehicle’s physical braking distance.
Explore our Force Calculator (/engineering-physics/force-calculator/) to solve Newton’s Second Law (F = ma) and acceleration, or visit our Electrical Power Calculator (/engineering-physics/electrical-power-calculator/) for energy conversion and wattage analysis.