1000 kg car at 20 m/s
- Quantity
- Kinetic energy ½mv²
- Solve for
- Energy
- Mass
- 1000 kg
- Speed
- 20 m/s
- Energy or work
- 200,000 J
- Energy or work
- 0.055556 kWh
Checked against: ½ × 1000 × 20² = 200000 J = 0.0555… kWh
Kinetic energy (½mv²), potential energy (mgh), spring energy, work (Fd cos θ) and power (W/t), solved for any variable, plus the speed after a drop.
By AdminDrafted with AI writing toolsUpdated Checked examples: 8
1,000 kg moving at 20 m/s carries 200,000 J of kinetic energy; doubling the speed would quadruple it.
Six formulas cover the common mechanical-energy problems: kinetic energy ½mv², gravitational potential energy mgh, spring energy ½kx², work Fd cos θ, average power W ÷ t, and the speed after a frictionless drop, √(v₀² + 2gh). Pick one, choose the variable to solve for, and the others become inputs.
Energy comes out in joules, with the kilowatt-hour equivalent beside it from 1 Wh (3,600 J) up; 1 kWh = 3.6 MJ. The default, a 1,000 kg car at 20 m/s (72 km/h), carries 200,000 J, or 0.0556 kWh; at 40 m/s it would carry four times as much.
Potential energy is measured from whichever level you call h = 0, so only differences in height matter. The drop mode ignores friction and air resistance, which makes its speeds an upper limit for real falls.
Kinetic energy is half the mass times the speed squared: KE = ½mv², in joules when mass is in kilograms and speed in meters per second. A 2 kg ball moving at 3 m/s carries ½ × 2 × 3² = 9 J. This is the energy of the whole object moving, valid at speeds far below the speed of light.
Because speed is squared, doubling it quadruples the energy. OpenStax's kinetic energy lesson works through more examples.
| Mass (kg) | Speed (m/s) | Kinetic energy (J) |
|---|---|---|
| 1000 | 20 | 200000 |
| 1000 | 40 | 800000 |
| 2 | 3 | 9 |
| 1 | 0 | 0 |
A 1,000 kg car at 20 m/s (72 km/h) carries 200,000 J; at 40 m/s it carries 800,000 J. Convert km/h to m/s by dividing by 3.6 before squaring.
Rearranged, v = √(2KE/m) and m = 2KE/v². A 70 kg runner carrying 3,500 J of kinetic energy is moving at √(2 × 3,500 ÷ 70) = √100 = 10 m/s.
An object at rest has zero kinetic energy whatever its mass, so mass cannot be found from KE when the speed is zero.
Lifting a mass m through a height h stores PE = mgh. With standard gravity, g = 9.80665 m/s², raising 50 kg by 10 m stores 50 × 9.80665 × 10 = 4,903.325 J, about 4.9 kJ.
Height is measured from a reference level you choose, such as the floor. Only differences in potential energy matter, so use the same reference for every position in a problem. OpenStax's potential energy lesson explains why.
A spring with stiffness k, stretched or compressed by x, stores E = ½kx². A 200 N/m spring compressed 10 cm (0.1 m) stores ½ × 200 × 0.1² = 1 J; compressed 20 cm, it stores 4 J.
The formula assumes Hooke's law, F = kx, which holds only while the spring stays in its elastic range and returns to its original length.
A constant force F that moves an object a distance d does work W = Fd cos θ, where θ is the angle between the force and the direction of motion. For 100 N over 5 m, the work is 500 J at 0°, 250 J at 60°, zero at 90° and −500 J at 180°.
Negative work means the force opposes the motion, as friction and brakes do. The work-energy theorem says the net work on an object equals its change in kinetic energy.
Stopping a 1,000 kg car from 20 m/s therefore takes −200,000 J of net work. If the brakes supply a constant 8,000 N, the car stops in 200,000 ÷ 8,000 = 25 m.
Power is the rate of doing work: P = W/t, in watts, where 1 W is 1 J per second. Delivering 3,600 J in 60 s is 3,600 ÷ 60 = 60 W. Rearranged, W = Pt and t = W/P.
A 100 W device running for one hour transfers 100 × 3,600 = 360,000 J, which is 0.1 kWh. One kilowatt-hour is 3,600,000 J. OpenStax's power lesson separates this average power from the instantaneous rate.
Without air resistance, the potential energy lost in a fall becomes kinetic energy: mgh = ½mv², so the mass cancels. From rest, v = √(2gh). A 20 m drop ends at √(2 × 9.80665 × 20) ≈ 19.81 m/s.
With a starting speed v₀, the speed at the bottom is √(v₀² + 2gh). Impact speed grows with the square root of height: from 80 m, four times higher, it is about 39.61 m/s, only twice as fast. Real falls are slower because of drag.
For signed motion and time, use the kinematics calculator; for collisions, the momentum and collision calculator. Electrical power uses P = VI in the Ohm's law calculator, and engine power from torque and rpm comes from the horsepower calculator.
Checked against: ½ × 1000 × 20² = 200000 J = 0.0555… kWh
Checked against: v = √(2E/m) = √(200/0.145) = 37.1390676…
Checked against: 50 × 9.80665 × 10
Checked against: ½ × 200 × 0.1² = 1 J
Kinetic energy is KE = ½mv²; with mass in kilograms and speed in metres per second the result is in joules. Because speed is squared, doubling it quadruples the energy: a 1,000 kg car has 200 kJ at 20 m/s and 800 kJ at 40 m/s. Rearranged for speed, v = √(2KE ÷ m), so a 0.145 kg ball carrying 100 J moves at 37.1 m/s.
Work is energy transferred by a force, W = Fd cos θ, measured in joules; power is how fast that transfer happens, P = W ÷ t, measured in watts, where 1 W = 1 J/s. Lifting 50 kg through 10 m takes 4,903 J however it is done; doing it in 5 s needs 981 W, and spreading it over 60 s needs 81.7 W.
One kilowatt-hour is 3,600,000 J, because it is 1,000 W sustained for 3,600 s. One kilocalorie, the food Calorie, is 4,184 J, using the thermochemical calorie of exactly 4.184 J listed in NIST SP 811. The 200,000 J carried by a 1,000 kg car at 20 m/s is therefore 0.0556 kWh, or about 47.8 kcal.
Without air resistance the impact speed is v = √(2gh), whatever the mass. A drop from 20 m ends at 19.81 m/s (71.3 km/h); a drop from 5 m ends at 9.90 m/s. Speed grows with the square root of height, so four times the height only doubles the speed. Air drag makes real speeds lower, most of all for light objects and long falls.
A spring stores E = ½kx², where k is the spring constant in N/m and x the extension or compression in metres. A 200 N/m spring compressed by 10 cm stores 1 J; compressed by 20 cm it stores 4 J. The formula holds only within the elastic range, where Hooke's law F = kx applies and the spring returns to its original length.
This calculator includes 8 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.