Kinetic and Potential Energy Calculator & Simulator

Calculate Ec, Ep and total mechanical energy with real-time free-fall simulation.

Input parameters

Calculation mode

Mass

Object mass

1.000e-41.000e+6

Height

Height above reference level

01000

Gravity preset

Gravity

System gravitational acceleration

m/s²

0.011000

Energy unit

Velocity unit (results)

Time unit (results)

Ideal model without air friction.

Free-fall simulation

g = 9.807 m/s²

Ep / Ec distributionEp 100% / Ec 0%

Results

Potential energy (Ep)

196.13 J

Kinetic energy (Ec)

0 J

Total mechanical energy (Em)

196.13 J

Impact velocity

m/s

Fall time

1.428

Height where Ep = Ec

Linear momentum

28.01 kg·m/s

Average power

137.3 W

Fundamentals & Explanation

Potential Energy

E_p = m g h

Gravitational potential energy depends on mass $m$ (kg), gravitational acceleration $g$ (m/s²), and height $h$ (m) above a chosen reference level. The reference level is arbitrary — only differences in $E_p$ have physical meaning.

Kinetic Energy

E_c = \frac{1}{2} m v^2

Kinetic energy scales with the square of velocity — doubling speed quadruples $E_c$ . It is always non-negative and equals zero only when the object is at rest.

Conservation of Energy

E_m = E_p + E_c = \text{constant}

v_{impact} = \sqrt{2 g h_0},\quad T = \sqrt{\frac{2 h_0}{g}}

In ideal free fall (no drag), total mechanical energy $E_m$ is conserved throughout the fall. Every joule lost by $E_p$ is gained by $E_c$ .

Linear Momentum

p = m v

Linear momentum $p$ (kg·m/s) measures the quantity of motion. In free fall from rest, $p$ grows linearly with time while $E_c$ grows quadratically — they are related by $E_c = p^2 / (2m)$ .

Physical interpretation

This simulator calculates gravitational potential energy ( $E_p = mgh$ ) and kinetic energy ( $E_c = \frac{1}{2}mv^2$ ) applying the principle of conservation of mechanical energy. In free fall from rest, the object starts with 100% potential energy and arrives at the ground with 100% kinetic energy.

The height where both energies are equal is always $h = h_0 / 2$ , regardless of mass or gravity.

Worked example

A 2 kg object dropped from 10 m on Earth ( $g = 9.807\ \text{m/s}^2$ ):

$E_p = 2 \times 9.807 \times 10 = 196.1\ \text{J}$
$v_{impact} = \sqrt{2 \times 9.807 \times 10} \approx 14.00\ \text{m/s}$
$T = \sqrt{2 \times 10 / 9.807} \approx 1.428\ \text{s}$
$p_{impact} = 2 \times 14.00 = 28.01\ \text{kg}{\cdot}\text{m/s}$

On the Moon ( $g = 1.62\ \text{m/s}^2$ ), the same object would take 3.51 s to fall and hit at only 5.69 m/s — same $E_p$ , but much slower impact.

Model limitations

Air resistance is not modeled. Real terminal velocity limits how fast objects actually fall.
Uniform gravity is assumed. For heights above ~10 km, $g$ decreases with altitude.
Classical mechanics only. At speeds above ~10% of the speed of light, relativistic corrections become significant.
Point mass. Rotation and deformation of the object are ignored.

References

OpenStax University Physics Vol. 1 — Work and Energy NIST — Standard Acceleration of Gravity (g = 9.80665 m/s²)