Newton’s second law says the net force on an object equals its rate of change of momentum. If you know how quickly an object’s speed changes, you can work out the force that must have acted on it.
How to use the Newton’s Second Law Calculator
- Mass m
- Initial velocity v₀
- Final velocity v₁
- Time difference Δt
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Enter the mass, the initial and final velocities and the time. The acceleration is the velocity change divided by the time; the force is mass times acceleration.
Inputs explained
| Field | What to enter |
|---|---|
| Mass m | Units: kg, g, mg, t, lb, oz |
| Initial velocity v₀ | Units: m/s, km/h, mph, ft/s, knots, km/s |
| Final velocity v₁ | Units: m/s, km/h, mph, ft/s, knots, km/s |
| Time difference Δt | Units: s, ms, min, h, days |
What each result means
- Acceleration = ([Final velocity v₁] – [Initial velocity v₀])÷[Time difference Δt]
- Net force = [Mass m]×[Acceleration]
- Change in momentum = [Mass m]×([Final velocity v₁] – [Initial velocity v₀])
Example values
How the main result changes when Mass m changes and everything else stays at the example values above:
| Mass m (kg) | Net force (N) |
|---|---|
| 500 | 1,000 |
| 750 | 1,500 |
| 1,000 | 2,000 |
| 1,250 | 2,500 |
| 1,500 | 3,000 |
| 2,000 | 4,000 |
Formula
a = (v₁ − v₀)/Δt, F = m·a = m·Δv/Δt
Worked example
Inputs — Mass m: 1,000 kg; Initial velocity v₀: 0 m/s; Final velocity v₁: 20 m/s; Time difference Δt: 10 s.
Results — Acceleration = 2 m/s²; Net force = 2,000 N; Change in momentum = 20,000 kg·m/s.
Key terms
- Inertia
- Resistance of an object to changes in its motion.
Frequently asked questions
What does Newton's second law state?
F = ma: the net force equals mass times acceleration.
What if the force is zero?
The velocity stays constant (first law).
Does the law apply to rockets?
Yes, in the form F = dp/dt, accounting for changing mass.
Why is a bigger force needed for a heavier car?
More mass needs more force for the same acceleration.