Newton's Second Law

Equivalent forms

\vec{F}_{\text{net}} = m\vec{a}

F = m \frac{dv}{dt}

F = \frac{dp}{dt}

The simplest bridge between cause (force) and effect (acceleration).

Symbol	Name	SI unit	Dimension	Range
$F$	Forceoutput Net force applied to the object	N	MLT^-2	0 – 1000
$m$	Mass Mass of the object	kg	M	0.1 – 100
$a$	Acceleration Acceleration of the object	m/s²	LT^-2	0 – 50

Unit systems

Symbol	SI	CGS	Imperial
$F$	N	dyn	lbf
$m$	kg	g	slug
$a$	m/s²	cm/s²	ft/s²

Where it holds

Valid for non-relativistic speeds (v << 299792458 m/s) and inertial reference frames. Breaks down at quantum scales.

Dimensional analysis

F \to [MLT^{-2}]

m\cdot a \to [M]\cdot [LT^{-2}] = [MLT^{-2}]

Both sides are

[MLT^{-2}]

= Newton.

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Discovery

Isaac Newton · 1687

Published in Principia Mathematica, unifying terrestrial and celestial mechanics under three laws of motion.

Try this

How hard must you push a stalled 1200 kg car to hit 60 mph in 6 seconds?

Calculate the constant force needed to accelerate a 1200 kg car from rest to 26.8 m/s in 6 s. Apply F = ma with a = Δv/Δt.

Research status: stable

Real-world applications

Vehicle crash testing
Rocket propulsion (with variable mass: $F = dp/dt)$
Sports biomechanics
Elevator design

Common misconceptions

Force is not required to maintain constant velocity — only to change it
Heavier objects do not fall faster in vacuum
The net force, not any single force, determines acceleration

Experimental verification

Verified by Atwood machine experiments, air-track gliders, and modern force sensors with sub-Newton precision.

Derivation

From Newton's axiom: the rate of change of momentum equals the applied force.

For constant mass:

F = dp/dt = m(dv/dt) = ma

.

Limiting cases

m \to 0

⟶

F \to 0

No mass means no force needed for any acceleration.

a \to 0

⟶

F \to 0

Zero acceleration means the object is in equilibrium.

m \to \infty

⟶

F \to \infty

Infinite mass requires infinite force to accelerate.

What if…

What if mass doubles?

Force doubles for the same acceleration. A 2400 kg truck needs 10728 N instead of 5364 N.

What if on the Moon

(g = 1.62)

?

The formula still holds — $F = ma$ is universal. Only gravitational weight changes, not the law itself.

What if velocity approaches c?

Relativistic mass increases: $F = \gamma ^{3}ma$ for longitudinal acceleration. Classical $F = ma$ underestimates the required force.

1

Pushing a stalled car

Given ·

m:: 1200
a:: 4.47

Find · F

Steps

Convert 60 mph to m/s: $60 \times 0.4470 = 26.8\,\mathrm{m/s}$
Find acceleration: $a = \Delta v/\Delta t = 26.8/6 = 4.47\,\mathrm{m/s}^{2}$
Apply $F = ma$ : $F = 1200 \times 4.47 = 5364\,\mathrm{N}$

Result ·

F = ma = 1200 \times 4.47 = 5364\,\mathrm{N} \approx 5.4\,\mathrm{kN}

2

Elevator acceleration

Given ·

F net:: 800
m:: 80

Find · a

Steps

Person mass: 80 kg, scale reads $80 \times 9.8 + 800 = 1584\,\mathrm{N}$
Net upward force beyond weight: $F_net = 800\,\mathrm{N}$
$a = F$ _net/ $m = 800/80 = 10\,\mathrm{m/s}^{2}$

Result ·

a = F/m = 800/80 = 10\,\mathrm{m/s}^{2}

(upward beyond gravity)

Newton's Law of Universal Gravitation→Hooke's Law→Work-Energy Theorem→