Kinetic Energy
Calculate KE = ½mv² (or rotational ½Iω²). Free kinetic energy calculator with unit conversions and step-by-step working.
Key formula
KE = (1/2)mv^2
How do you calculate kinetic energy?
For a moving mass treated as a particle, KE = ½ m v². Use mass in kilograms and speed in metres per second; the result is in joules. Rearrange to m = 2 KE / v² or v = √(2 KE / m) when energy is given.
About this tool
Kinetic energy measures the energy of motion, and it grows quickly as speed increases. This page keeps KE = ½mv² visible so users can solve for energy, mass, or speed without losing the unit checks that usually cost marks.
How to use this calculator
- Choose the variable you want to solve for and confirm which values are already known.
- Enter inputs in consistent units before calculating so the formula is applied correctly.
- Check whether the result is physically reasonable by comparing it with the size of the inputs.
- Use the step-by-step breakdown to see how the equation was rearranged and evaluated.
Worked examples
Example 1: 2.0 kg at 3.0 m/s
KE = ½ (2.0)(3.0)² = ½ × 2.0 × 9.0 = 9.0 J. Quick check: ½ × 2 × 9 = 9.
Example 2: find speed from 50 J
A 4.0 kg object has KE = 50 J. v = √(2 × 50 / 4.0) = √25 = 5.0 m/s. Check: ½ (4)(25) = 50 J.
Example 3: convert 36 km/h first
A 0.50 kg ball at 36 km/h → v = 36 / 3.6 = 10 m/s. KE = ½ (0.50)(10)² = 25 J. Skipping the conversion wrongly uses 36².
Frequently asked questions
What is the formula for kinetic energy?
For linear motion of a mass, KE = ½ m v². Mass m is in kg, speed v in m/s, and KE in joules. Rearrangements are m = 2 KE / v² and v = √(2 KE / m).
Why is velocity squared in kinetic energy?
The work–energy derivation for accelerating a mass from rest leads to an energy proportional to v². Doubling speed therefore roughly quadruples KE when mass is fixed.
How do I convert km/h for KE problems?
Divide km/h by 3.6 to get m/s before substituting into ½mv². Leaving km/h in the formula produces a hugely inflated joule answer.
Is kinetic energy the same as potential energy?
No. Kinetic energy depends on motion; near-surface gravitational potential energy depends on height via PE = mgh. Falling objects can convert PE to KE, but the formulas use different inputs.
When should I use ½ I ω² instead?
Use rotational kinetic energy when the problem is about spinning about an axis with moment of inertia I and angular speed ω. Otherwise stay with ½ m v² for translation.