What Newton's second law actually claims
Newton's second law says that the acceleration of an object is proportional to the net force acting on it and inversely proportional to its mass. Written as F = ma it looks like a definition of force, and in one sense it is — but the physical content is in the two proportionalities. Double the net force on the same object and you double its acceleration. Double the mass under the same net force and you halve it.
Three words in that sentence carry all the weight. Net: it is the resultant of every force acting, so a crate pushed with 400 N against 250 N of friction has a net force of 150 N, not 400 N. Mass: this is inertial mass in kilograms, the object's resistance to being accelerated, and it is not the same quantity as weight. Acceleration: not velocity. A car cruising at a steady 100 km/h has zero acceleration and therefore zero net force on it, even though the engine is working hard — thrust and drag are equal and opposite.
The direction matters too. Force and acceleration are vectors and they always point the same way, because mass is a positive scalar. If your working produces a force pointing one way and an acceleration pointing the other, you have made a sign error, and this calculator will tell you so.
The law also underpins the definition of the newton itself: one newton is the force that gives one kilogram an acceleration of one metre per second squared. Every force unit in the table below is anchored to that definition, which is why the conversions are exact rather than measured.
Rearranging the equation, and the version that is more general
The three arrangements are trivial algebra: F = ma, a = F/m, m = F/a. Two of them have a division, and each has a case where the answer does not exist. You cannot find mass from a zero acceleration, because a zero net force is consistent with any mass at all; and you cannot find acceleration for a zero mass, because dividing by zero is undefined. The calculator returns no answer in those cases rather than a misleading number.
Newton did not write it as F = ma. His statement was that the change of motion is proportional to the impressed force, where "motion" meant what we now call momentum. The modern form of that is ΣF = dp/dt, the rate of change of momentum. Expand the derivative of p = mv and you get m(dv/dt) + v(dm/dt). The second term vanishes whenever mass is constant, which leaves ma.
That second term is not always negligible. A rocket burning propellant, a hopper discharging onto a moving belt, a chain being lifted link by link off a table — in every one of those the mass of the system changes as it moves, and F = ma gives the wrong answer while the momentum form gives the right one. If your problem has mass entering or leaving, use momentum.
Finally, note what the law needs to be true at all: an inertial reference frame. In an accelerating frame — a braking bus, a spinning turntable — objects appear to accelerate with no force acting, and you must either move to an inertial frame or add the fictitious forces that account for the frame's own acceleration. The centripetal force calculator deals with the most common case of this.
Worked example: accelerating a 1,400 kg car
A 1,400 kg car accelerates from rest to 27.78 m/s (100 km/h) in 8.5 s on level ground. Aerodynamic drag and rolling resistance together average 420 N over that run. What force must the tyres deliver?
- Find the acceleration. Assuming it is roughly constant, a = Δv/t = 27.78 ÷ 8.5 = 3.268 m/s², which is 0.333 g.
- Find the net force. Fnet = ma = 1400 × 3.268 = 4,575 N. This is the resultant, not the tractive effort.
- Add back the resisting forces. The tyres must overcome resistance and supply the net force: 4,575 + 420 = 4,995 N at the contact patch.
- Check it against grip. The car weighs 1400 × 9.80665 = 13,729 N. A tractive force of 4,995 N needs a friction coefficient of at least 4,995 ÷ 13,729 = 0.364 at the driven wheels, which dry asphalt supplies comfortably — see the friction force calculator for the limit.
- Convert if you need to. 4,995 N ÷ 4.4482216153 = 1,123 lbf, or ÷ 9.80665 = 509 kgf.
Reverse the problem to see the second lesson. Keep the same 4,995 N of tractive effort but load the car with 400 kg of passengers and luggage. The net force is now 4,995 − 420 = 4,575 N acting on 1,800 kg, so a = 2.542 m/s² and the run to 100 km/h takes 27.78 ÷ 2.542 = 10.9 s. Take the two ratios separately, because they are not the same number: the acceleration falls by 1 − 1400/1800 = 22%, while the time, which goes as 1/a and therefore as m, rises by 1800/1400 − 1 = 29%. Both are a ∝ 1/m in action, and quoting the mass increase as the acceleration loss is the easy slip to make.
Reading the result, and the mass-versus-weight trap
Check the acceleration in g first, because that is the figure with intuition attached. Ordinary road-car acceleration is 0.2–0.4 g — the worked example below reaches 0.333 g — and hard braking on dry asphalt is limited by tyre grip to around 1 g, since the deceleration cannot exceed μg and μ for rubber on dry asphalt is near 1. Impacts run orders of magnitude higher, because the same velocity change happens in milliseconds rather than seconds. If your answer is orders of magnitude away from the class of problem you are solving, an input unit is wrong.
Then check that you used mass, not weight. This is the single most common error in the equation, and it is much easier to make in imperial units, where the pound is used for both. A 3,000 lb car does not have a mass of 3,000 in any coherent unit system: in slugs its mass is 3000 ÷ 32.174 = 93.2 slug, and in kilograms it is 1,361 kg. If you feed 3,000 into F = ma with feet and seconds, your answer is out by a factor of 32.2. The calculator offers slugs and pounds-mass as separate unit options precisely so you do not have to make that decision under pressure.
Finally, remember that the force it reports is the net force. If you want the force a person, motor or actuator has to apply, add back every resisting force: friction, drag, the component of weight along a slope, and the inertia of anything else being dragged along. Those additions are usually larger than students expect and are where real engineering estimates live.
Force units and their exact relationships
| Unit | Symbol | Value in newtons | What it is |
|---|---|---|---|
| Newton | N | 1 | The SI unit: 1 kg·m/s² |
| Kilonewton | kN | 1,000 | Convenient for structural and vehicle loads |
| Pound-force | lbf | 4.4482216153 | Weight of one pound-mass at standard gravity |
| Kilogram-force | kgf | 9.80665 | Weight of one kilogram at standard gravity |
| Dyne | dyn | 0.00001 | The CGS unit: 1 g·cm/s² |
| Poundal | pdl | 0.138254954376 | Force accelerating 1 lb at 1 ft/s² |
The pound-force and kilogram-force are defined using standard gravity of exactly 9.80665 m/s², which is why 1 lbf = 0.45359237 × 9.80665 N exactly.
The mistakes that break F = ma
- Using the applied force instead of the net force. Friction, drag and the slope component of weight all subtract. This is the most frequent error by a wide margin.
- Substituting weight for mass. Weight is a force in newtons; mass is in kilograms. Dividing a force by a force gives a dimensionless number, not an acceleration.
- Mixing pounds-mass with feet and seconds. In that system the consistent mass unit is the slug, and a factor of 32.174 goes missing if you use pounds directly.
- Applying it in a non-inertial frame. Inside an accelerating vehicle, objects move with no visible force acting. Work in the ground frame or add the frame's own acceleration explicitly.
- Using it where mass changes. Rockets, discharging hoppers and lifted chains need ΣF = dp/dt, not ma. The momentum calculator handles that form.
- Assuming zero acceleration means zero force. It means zero net force. A book on a table has two large forces on it that happen to cancel.
- Forgetting that force and acceleration are vectors. Resolve into components along sensible axes and apply the law to each axis independently.
Where this fits with the other two laws and with energy
The second law only tells you the acceleration. To get anywhere useful you usually pair it with kinematics: hand the acceleration to the final velocity calculator to find how fast the object ends up and how far it travels, or to the free fall calculator if the only force is weight. The first law is the special case a = 0, and the third law — every force has an equal and opposite reaction on another body — is what tells you which forces belong in your net-force sum and which act on something else entirely.
Energy gives you a parallel route to the same answers, often with less work. Multiply F = ma by displacement and you get the work–energy theorem, which the work calculator and the kinetic energy calculator use: net work done equals change in kinetic energy. When you know distances but not times, the energy route is shorter. When you know times but not distances, the momentum route via impulse is shorter. All three are the same physics in different clothing.
For rotational problems there is a direct analogue: τ = Iα, where torque replaces force, moment of inertia replaces mass and angular acceleration replaces linear acceleration. Everything you know about F = ma transfers, including the traps: the torque must be the net torque, and the moment of inertia must be taken about the axis you are actually rotating around.