Newton’s second law of motion states that the net force on a body equals the rate of change of its momentum, F = dp/dt. For a body of constant mass this becomes F = ma: acceleration is proportional to the net force, inversely proportional to mass, and points in the direction of the net force. A 1 N net force gives a 1 kg mass an acceleration of 1 m/s².
The first law says a body keeps its state of motion unless a net force acts. The second law says by how much that motion changes when a force does act, which makes it the working equation of classical mechanics, from a lift cable to a vehicle’s brakes.
What does Newton’s second law say?
Newton published the law in the Principia (1687) in terms of “change of motion”, which we now call momentum, p = mv. In modern form:
- Momentum form: Fnet = dp/dt = d(mv)/dt
- Constant-mass form: Fnet = m(dv/dt) = ma
Three points are packed into F = ma:
- F is the net (resultant) force, the vector sum of every force on the body, not any single force.
- It is a vector equation. The acceleration always points along the net force. In practice you write it in components: ΣFx = max, ΣFy = may.
- It holds in an inertial frame, one that is not itself accelerating. Inside an accelerating lift or a braking bus, you must either work from the ground frame or add a pseudo-force.
Units of force
| Unit | Definition | In newtons |
|---|---|---|
| newton (N), SI | force giving 1 kg an acceleration of 1 m/s² | 1 N = 1 kg·m/s² |
| dyne, CGS | force giving 1 g an acceleration of 1 cm/s² | 10⁻⁵ N |
| kilogram-force (kgf) | weight of 1 kg under standard gravity | 9.80665 N (≈ 9.81 N) |
Dimensional formula of force: [M L T⁻²].
How to apply F = ma: the free-body diagram method
- Pick the body (or each body separately in a connected system).
- Draw a free-body diagram: the body as a dot or box, with every external force as an arrow. Typical forces are weight mg, normal reaction N, tension T, friction f and applied force P.
- Choose axes, ideally one along the expected acceleration (along an incline, for example).
- Resolve the forces into components along those axes. The parallelogram law of forces is the rule for combining two forces into one resultant.
- Write ΣF = ma for each axis and solve. Along an axis with no motion, ΣF = 0.
Solved examples of Newton’s second law
Example 1: apparent weight in a lift
A 60 kg person stands on a weighing scale in a lift. Find the scale reading when the lift accelerates upward at 2 m/s², and when it accelerates downward at 2 m/s². Take g = 9.81 m/s².
Forces on the person: weight mg down and normal reaction N (the scale reading) up. Take up as positive.
- Accelerating up: N − mg = ma, so N = m(g + a) = 60 × (9.81 + 2) = 60 × 11.81 = 708.6 N. The scale shows 708.6 ÷ 9.81 ≈ 72.2 kg.
- Accelerating down: mg − N = ma, so N = m(g − a) = 60 × 7.81 = 468.6 N, about 47.8 kg on the scale.
- At constant speed: a = 0, so N = mg = 588.6 N (60 kg).
- If the cable snapped: a = g, so N = 0 and the person feels weightless.
Your mass never changes. The scale measures N, the “apparent weight”.
Example 2: block on a frictionless incline
A 5 kg block slides down a smooth plane inclined at 30°. Find its acceleration and the normal reaction.
- Resolve weight mg = 5 × 9.81 = 49.05 N along and perpendicular to the slope.
- Along the slope: mg sin 30° = ma, so a = g sin 30° = 9.81 × 0.5 = 4.905 m/s². The net force is 5 × 4.905 = 24.53 N.
- Perpendicular: N − mg cos 30° = 0, so N = 49.05 × 0.866 = 42.48 N.
The acceleration does not depend on the mass: a 50 kg block on the same smooth slope also accelerates at 4.905 m/s².
Example 3: force needed to stop a vehicle
A 1200 kg car travelling at 72 km/h must stop in 40 m. What average braking force is needed?
- First the motion: u = 20 m/s, v = 0, s = 40 m. From v² = u² + 2as, a = −400 ÷ 80 = −5 m/s². (This step uses the kinematic equations.)
- Then the force: F = ma = 1200 × (−5) = −6000 N, i.e. 6 kN opposing the motion.
- Momentum check: the car loses p = 1200 × 20 = 24,000 kg·m/s. Stopping takes t = 20 ÷ 5 = 4 s, and F × t = 6000 × 4 = 24,000 N·s. Same answer from F = Δp/Δt.
Double the speed to 144 km/h and, for the same 40 m, the force rises to 24 kN because a depends on u².
Variable mass: why rockets need the momentum form
F = ma assumes constant mass. A rocket burns hundreds of kilograms of propellant per second, so its mass falls continuously. Writing F = m(dv/dt) + v(dm/dt) looks natural but gives wrong answers, because the expelled gas leaves at a relative speed, not at the rocket’s speed. The correct equation for the rocket body is
m(dv/dt) = Fext + vrel × ṁ
where ṁ is the mass burn rate (kg/s) and vrel is the exhaust speed relative to the rocket. The term vrel × ṁ is the thrust. For example, an exhaust speed of 3000 m/s with a burn rate of 100 kg/s gives a thrust of 300,000 N (300 kN). The same care applies to conveyor belts being loaded with sand and to chains falling onto a table.
Common mistakes with Newton’s second law
- Mass vs weight. Mass (kg) is the amount of matter and is the same everywhere. Weight is a force, W = mg, in newtons. A 60 kg person weighs about 589 N on Earth and about 97 N on the Moon (g ≈ 1.62 m/s²).
- Using one force instead of the net force. A crate pushed with 100 N against 40 N of friction accelerates under 60 N, not 100 N.
- Adding ma as a force on the diagram. ma is the result of the forces, not another force. Only draw real pushes and pulls (in an inertial frame).
- Treating the normal reaction as always equal to mg. Example 1 and Example 2 both show N ≠ mg.
- Applying F = ma to rotation. For turning effects, the rotational analogue τ = Iα is used, where τ is the moment of a force.
How it links to the first and third laws
Put Fnet = 0 into F = ma and a = 0: the body keeps a constant velocity, which is the first law. The third law says forces come in equal and opposite pairs acting on different bodies, which is why those pairs never cancel in a single free-body diagram. Combined with the third law, the second law leads to conservation of momentum for an isolated system.
FAQs
What is the formula of Newton’s second law?
F = ma for constant mass, where F is the net force in newtons, m the mass in kilograms and a the acceleration in m/s². The general form is F = dp/dt, the rate of change of momentum.
What is 1 newton?
One newton is the net force that gives a 1 kg mass an acceleration of 1 m/s², so 1 N = 1 kg·m/s². It equals 10⁵ dyne.
Why is F = ma not valid for a rocket?
F = ma assumes the mass stays constant. A rocket loses mass as it burns fuel, so you use the momentum form, which gives m(dv/dt) = Fext + vrelṁ, with thrust equal to exhaust speed times burn rate.
Why do you feel heavier in a lift going up?
While the lift accelerates upward, the floor must push with N = m(g + a), more than your weight mg. Scales read N, so they show a higher value. Once the lift moves at constant speed, the reading returns to normal.
Is Newton’s second law valid in all frames?
It holds directly only in inertial (non-accelerating) frames. In an accelerating frame you must add a pseudo-force of −maframe to each body for the equation to work.