Newton's Second Law: F = ma

Every machine that moves — every car, rocket, robot arm and cricket ball — obeys one three-symbol sentence: $F = ma$ . When engineers land a probe on Mars or tune a metro train's braking curve, the computation under the hood is this page's recipe run millions of times. Newton's second law is the engine room of mechanics, and this module is about driving it: give me the forces on a body and its mass, and I will tell you its acceleration — and from acceleration, the entire future of its motion. The law's power is in what it connects: a cause you can list (gravity, tension, friction, thrust) to an effect you can measure (acceleration), with mass sitting between them as the exchange rate. Double the net force and the acceleration doubles; double the mass and it halves. Simple proportions — with two traps built in. First, $F$ means the net force, the vector sum of everything acting; feeding a single force into the formula while others still stand is the most common lost mark in this chapter. Second, the acceleration tracks the direction of that net force, not the direction the body happens to be moving. Every board makes this module load-bearing. CBSE and ISC want the statement, the unit, and clean two-force numericals. JEE-Main and Advanced push the vector character — perpendicular forces, equilibrants, component ledgers. NEET drills ratio templates ("same force, two masses…"). AP Physics 1, IB and A-Level test the proportionalities through graphs and design-an-experiment questions, and IGCSE wants confident $F = ma$ arithmetic. One law, ten syllabi, and after the Free-Body Diagrams module it becomes a full problem-solving machine — the one the Atwood Machine archetype runs at full throttle.

Key takeaways

Newton's second law: the net force on a body equals mass times acceleration, $\vec F_{\text{net}} = m\vec a$ — a vector equation, true component by component.
$\vec F_{\text{net}}$ is the vector sum of all forces acting on the body (superposition principle: each force acts as if the others were absent; the body answers only to the total).
The acceleration points along the net force — not necessarily along the velocity. A braking car accelerates backward while moving forward.
Proportions to own: $a \propto F$ at fixed mass; $a \propto 1/m$ at fixed force. The unit: $1\ \mathrm{N} = 1\ \mathrm{kg\,m/s^2}$ — about the weight of an apple.
The law responds instantly: the acceleration at this moment is set by the net force at this moment. Remove the force, and the acceleration — not the velocity — vanishes.
Set $\vec F_{\text{net}} = 0$ and the second law hands back the first: $a = 0$ , velocity frozen.
Key idea: F = ma is a recipe — isolate the body, add every force as a vector, divide by the mass, and the motion follows.

The core insight

Net force is not what makes bodies move — it is what makes their velocity change: $\vec F_{\text{net}} = m\vec a$ converts the vector sum of all forces on a body into its acceleration, with mass as the exchange rate, instant by instant.

Where this lives today

Every physics engine — in games, film effects and robotics simulators — is a second-law integrator: at each tick it sums the forces on every body, divides by mass, and steps velocity and position forward. When a game character's cape swirls convincingly, thousands of $F = ma$ evaluations per frame are doing the tailoring. The same loop, hardened and certified, flies real hardware: a quadcopter's flight controller estimates the forces on the airframe hundreds of times a second and solves for the motor thrusts that produce the commanded acceleration — equation (2) in three dimensions, running on a chip the size of a stamp. The frontier version is spacecraft navigation. Mission teams model every force on a probe — gravity from each body, solar radiation pressure, thruster pulses, even the recoil of escaping heat — because integrated through $F = ma$ over months, micronewtons decide whether you skim Titan's atmosphere or miss it. Three centuries on, Newton's proportion between force and change of motion is still the sentence the solar system is flown with.

The physical picture

Push a stalled car and an empty shopping trolley with the same effort. The trolley leaps; the car barely creeps. Same cause, wildly different effect — the difference is mass, matter's built-in resistance to having its velocity changed (the inheritance from the Force, Inertia & Newton's First Law module). The second law makes the comparison quantitative: acceleration is force per unit of reluctance, $a = F_{\text{net}}/m$ . And it is a steering relation as much as a speeding one — the acceleration vector always points where the net force points, growing the velocity if aligned with it, eating it if opposed, bending it if aslant. The phrase that earns its keep is net force. A block pulled across a rough floor feels four forces at once: the pull, friction, weight, and the floor's normal push. The law never asks how many forces act or how large each is alone; it asks for one thing — their vector sum — and repays you with the acceleration. Watch the whole pipeline in one frame:

Figure

The second-law pipeline — four forces in, one acceleration out

The trolley-versus-car comparison is worth graphing, because the two proportionalities inside $a = F_{\text{net}}/m$ have opposite shapes — and both are things AP, IB and A-Level exams ask you to extract from data:

Figure

Two proportionalities — a against F is a straight line, a against m is a reciprocal curve

The math

Start with the bookkeeping question: which force does the body obey when several act at once? Experiment answers with the superposition principle — each force acts as though the others were absent, and the body responds only to their vector total. That total then buys acceleration at a rate set by the mass:

\vec F_{\text{net}} = \sum_i \vec F_i = m\,\vec a \qquad (1)

where: $\vec F_i$ — each individual force on the body (weight, normal, tension, friction, thrust…), in newtons; $m$ — the mass in kilograms; $\vec a$ — the acceleration in $\mathrm{m/s^2}$ , always parallel to the net force. Read it as a machine: forces in, acceleration out, mass as the conversion rate. Nothing else about the body — its speed, its history, its direction of travel — enters at all. That omission is the trap the warning box below springs. And note the special case it contains: forces cancelling ( $\vec F_{\text{net}} = \vec 0$ ) means $\vec a = \vec 0$ — the first law, recovered as the second law's quiet corner. Because (1) is a vector equation, it is really two (or three) independent scalar equations riding under one roof — the $x$ -forces must account for the $x$ -acceleration all by themselves, and likewise $y$ :

F_{\text{net},x} = m\,a_x, \qquad F_{\text{net},y} = m\,a_y \qquad (2)

where: the subscripts pick out components along axes you choose. This is a superpower: pick axes wisely (along and perpendicular to the motion) and a messy two-dimensional problem splits into two clean one-dimensional ones. In the pipeline figure, the vertical ledger quietly reports "cancelled" and the whole problem collapses onto the horizontal line. What exactly is a newton? With mass in kilograms and acceleration in $\mathrm{m/s^2}$ , equation (1) leaves no freedom — the law itself defines the unit:

1\ \mathrm{N} = 1\ \mathrm{kg \cdot m/s^2} \qquad (3)

where: one newton gives one kilogram an acceleration of $1\ \mathrm{m/s^2}$ . For feel: an apple weighs about 1 N; your school bag perhaps 40 N; a small car's weight is around 12,000 N. Board papers in India also quote the kilogram-force: $1\ \mathrm{kgf} = 9.8\ \mathrm{N}$ , the weight of one kilogram — a unit to convert immediately on sight. The most common entry in any force ledger gets its own line. Near Earth's surface, gravity pulls on a body in proportion to its mass:

\vec F_{\text{grav}} = m\vec g \qquad (4)

where: $g \approx 9.8\ \mathrm{m/s^2}$ , pointing down. On a force diagram a 70 kg person appears as $686\ \mathrm{N}$ — never as "70 kg"; kilograms measure mass, newtons measure force. The full settlement of mass versus weight (and what a bathroom scale actually reads) is the business of the Weight, Normal Force & Apparent Weight module.

you are given	reach for	to get
all forces and the mass	$a = F_{\text{net}}/m$	the acceleration, along $\vec F_{\text{net}}$
forces at angles	equation (2), axis by axis	two clean 1-D problems
the motion ( $v$ – $t$ data, kinematics)	$a$ first, then $F_{\text{net}} = ma$	the hidden net force
a body in equilibrium	$F_{\text{net}} = 0$	one unknown force from the others
force in kgf	multiply by $9.8$	newtons, before anything else

Derivation AFrom two proportionalities to the law and its unit

We build $F = ma$ the way a laboratory would — the construction CBSE and A-Level practical questions echo. The strategy: pin down how acceleration responds to force at fixed mass, then to mass at fixed force, combine the two proportionalities, and use the freedom in the force unit to make the constant exactly one. The thing to watch is the last step — the "missing constant" is not discovered to be 1; it is chosen to be 1, and that choice is what defines the newton.

Step 1. Fix the mass. On a low-friction track, pull one trolley with one, two, three identical stretched springs in parallel. The measured accelerations come out in the ratio $1 : 2 : 3$ — acceleration is proportional to the applied force: $a \propto F$ at fixed $m$ .

Step 2. Fix the force. Pull one, two, three stacked trolleys with the same single spring. The accelerations now come out in ratio $1 : \tfrac{1}{2} : \tfrac{1}{3}$ — acceleration is inversely proportional to mass: $a \propto 1/m$ at fixed $F$ .

Step 3. Combine the proportionalities:

a \propto \frac{F}{m} \quad\Longrightarrow\quad F = k\,m\,a

Why a single constant: any quantity proportional to each of two independent variables is proportional to their product; $k$ absorbs the units.

Step 4. Choose the unit to kill the constant. Define the newton as the force giving $1$ kg exactly $1\ \mathrm{m/s^2}$ ; then $k = 1$ and

F_{\text{net}} = ma

Why this is legal: nature fixed the proportionality; humans were free to pick the force unit, and picked the one that makes the law cleanest — equation (3) is that decision written down. *Verified: the construction reproduces itself — 6 N on 2 kg gives 3 m/s², and 6 N on 4 kg gives 1.5 m/s², halving as Step 2 demands. Standard treatment: NCERT Ch. 5; Kleppner and Kolenkow section 2.4.*

Go deeper · Derivation B · 1 minWhy one vector law is two scalar laws — and why you may pick the axes

Derivation BWhy one vector law is two scalar laws — and why you may pick the axes

We justify the component recipe of equation (2), because everything in the Free-Body Diagrams module leans on it. The claim: the single vector statement $\vec F_{\text{net}} = m\vec a$ is exactly equivalent to one scalar equation per perpendicular axis, with the axes chosen any way you like. The strategy is to resolve every vector along two perpendicular directions and use the fact that perpendicular components never feed each other. The payoff to watch for: the freedom of choice in Step 3 is a solving tool, not a technicality.

Step 1. Resolve. Any force $\vec F_i$ splits uniquely into components along two chosen perpendicular axes: $F_{i,x} = F_i\cos\theta_i$ and $F_{i,y} = F_i\sin\theta_i$ , where $\theta_i$ is the angle from the $x$ -axis. Vector addition adds components independently, so the net force's components are the sums $\sum_i F_{i,x}$ and $\sum_i F_{i,y}$ .

Step 2. Match components. Two vectors are equal exactly when their components match pair by pair. So $\vec F_{\text{net}} = m\vec a$ says precisely: $\sum F_x = ma_x$ and $\sum F_y = ma_y$ — each axis keeps its own ledger, and money never moves between ledgers. Why: an x-force cannot produce y-acceleration; that is what "perpendicular" means physically — this independence is the same one that lets projectile problems treat horizontal and vertical motion separately.

Step 3. Exploit the freedom. Nothing fixed the axes; rotate them to suit the problem. Choose one axis along the acceleration and the perpendicular ledger reads $\sum F_\perp = 0$ — a free equation, usually handing you the normal force. On an incline, axes along and across the slope tame gravity's components. The physics never changes; the algebra shrinks.

Step 4. Recombine when asked for magnitudes: $F_{\text{net}} = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$ , direction $\tan\phi = \sum F_y / \sum F_x$ , and the acceleration inherits both from division by $m$ . Verified: forces of 8 N and 6 N at right angles give a net of exactly 10 N (the 6-8-10 triangle), direction 36.9 degrees from the 8 N force — matching Worked Example 3 below. Standard treatment: Halliday-Resnick-Walker Ch. 5.

Limiting cases

case	result	meaning
$F_{\text{net}} = 0$	$a = 0$ , velocity frozen	balanced forces: the first law recovered
only gravity acts	$a = g$ , whatever the mass	$mg/m = g$ : all bodies fall alike — mass cancels
$m$ very large	$a \to 0$	same push, no visible response — why you cannot shove a ship
force applied for an instant, then removed	acceleration stops; velocity stays	force controls the change, never the velocity itself
speeds near light speed	$F = ma$ fails	relativity takes over; the momentum form survives — next module
mass changing (rocket, conveyor)	$F = ma$ unreliable	retreat to $F = \Delta p/\Delta t$ — the Momentum & the General Second Law module

Common trap“The body moves in the direction of the net force”

No — the body accelerates in the direction of the net force; the velocity can point anywhere. A ball thrown straight up moves upward while its net force (gravity) and its acceleration point straight down for the entire flight — including the very top, where $v = 0$ but $a = g$ still. A braking car moves forward while its net force points backward. A satellite in circular orbit moves sideways while accelerating toward Earth's centre. Exams weaponise this: "a body always moves along the applied force" is a stock false statement, and the top-of-flight question ("what is the acceleration when the ball is momentarily at rest?") catches thousands every year — the answer is $9.8\ \mathrm{m/s^2}$ downward, unchanged. If you catch yourself matching force direction to velocity direction, stop and redraw: the law couples force to the change in velocity, never to the velocity itself.

Worked examples

1 · Board level — the merging car (CBSE/IGCSE pattern, 3 marks)

A 1500 kg car accelerates uniformly from rest to $27\ \mathrm{m/s}$ (about 100 km/h) in 9.0 s. Find the net force on it, and compare that force with the car's weight. (Take $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Acceleration from kinematics: $a = \Delta v/\Delta t = 27/9.0 = 3.0\ \mathrm{m/s^2}$ .

Step 2. Apply equation (1): $F_{\text{net}} = ma = 1500 \times 3.0 = \boxed{4500\ \mathrm{N}}$ .

Step 3. The weight for comparison: $mg = 1500 \times 9.8 = 14{,}700\ \mathrm{N}$ . The net forward force is about $0.31\,mg$ — spirited but ordinary driving.

Step 4. Read the answer: 4500 N is the net of tyre thrust minus drag and rolling resistance — the engine actually delivers more, and the road disposes of the rest.

2 · Board level — pull versus friction (ISC/A-Level pattern)

A 7.0 kg crate on a rough floor is pulled horizontally with 40 N while friction opposes with 12 N. (a) Find the acceleration. (b) Starting from rest, how far does the crate travel in 3.0 s? (c) The pull is suddenly removed at $t = 3.0$ s — what happens next?

Step 1. Vertical ledger: normal force and weight cancel (no vertical acceleration). Horizontal ledger: $F_{\text{net}} = 40 - 12 = 28\ \mathrm{N}$ .

Step 2. Equation (1): $a = 28/7.0 = \boxed{4.0\ \mathrm{m/s^2}}$ . Only after the vector sum did mass enter — never feed a single force into $F = ma$ while others are still standing.

Step 3. Distance from rest: $x = \tfrac{1}{2}at^2 = \tfrac{1}{2}(4.0)(9.0) = \boxed{18\ \mathrm{m}}$ , moving at $v = at = 12\ \mathrm{m/s}$ by then.

Step 4. Remove the pull: the net force becomes $-12\ \mathrm{N}$ (friction alone), so $a = -12/7.0 = -1.7\ \mathrm{m/s^2}$ . The crate does not stop dead — it keeps moving and decelerates. Force controls the change of velocity, never the velocity.

3 · Competitive level — perpendicular forces and the equilibrant (JEE-Main pattern, ~2.5 minutes)

Two forces of 8.0 N and 6.0 N act at right angles on a 5.0 kg block resting on a smooth horizontal table. (a) Find the block's acceleration, magnitude and direction. (b) What third force would hold the block in equilibrium?

Step 1. Net force by components (take the 8.0 N along $x$ , the 6.0 N along $y$ ): $F_{\text{net}} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\ \mathrm{N}$ — the 6-8-10 triangle.

Step 2. Acceleration: $a = 10/5.0 = \boxed{2.0\ \mathrm{m/s^2}}$ , along the net force.

Step 3. Direction: $\tan\phi = 6/8 = 0.75 \Rightarrow \phi = \boxed{36.9°\ \text{from the 8 N force}}$ — the acceleration shares this direction exactly.

Step 4. Equilibrium needs the ledger to close: the third force must cancel the resultant, so it is $\boxed{10\ \mathrm{N}}$ , opposite to the net-force direction (i.e., at $180° + 36.9°$ from the 8 N force). Adding it makes $F_{\text{net}} = 0$ , and by the first law the block's velocity — whatever it is at that instant — freezes. Timing note: budget ~2.5 min. The 3-4-5 family (6-8-10, 9-12-15) is deliberate — spot it and skip the surd arithmetic. Distractor on offer: $8 + 6 = 14\ \mathrm{N}$ ; perpendicular forces never add as plain numbers.

4 · Competitive level — one force, two masses, then both (NEET pattern, ~1.5 minutes)

A force $F$ gives a body of mass $m_1$ an acceleration of $8.0\ \mathrm{m/s^2}$ , and a body of mass $m_2$ an acceleration of $24\ \mathrm{m/s^2}$ . What acceleration does the same force give the two bodies tied together?

Step 1. Extract the masses in terms of $F$ : $m_1 = F/8$ and $m_2 = F/24$ .

Step 2. Tie them together: $m_1 + m_2 = F\left(\tfrac{1}{8} + \tfrac{1}{24}\right) = F \cdot \tfrac{3 + 1}{24} = \tfrac{F}{6}$ .

Step 3. Same force on the combination: $a = \dfrac{F}{m_1 + m_2} = \boxed{6.0\ \mathrm{m/s^2}}$ .

Step 4. Spot the template: $\dfrac{1}{a} = \dfrac{1}{a_1} + \dfrac{1}{a_2}$ — accelerations under a shared force combine like parallel resistors, because masses (the reluctances) add. Timing note: ~1.5 min with the reciprocal template; ~4 min without it. Check: the combined answer must be smaller than both individual accelerations ( $6 < 8 < 24$ ✓) — more mass, less response, always.

Exam traps

Feeding one force into $F = ma$ . The law wants the net force. If a 40 N pull fights 12 N of friction, the mass divides 28 N — not 40. Symptom: your answer ignores a force the problem took trouble to mention.

Kilograms on a force diagram. Weight is $mg$ in newtons; "70 kg" is not a force. Boards that use kgf expect an immediate conversion: $1\ \mathrm{kgf} = 9.8\ \mathrm{N}$ , so "a 5 kgf force" is $49\ \mathrm{N}$ before anything else happens.

Direction of $a$ versus direction of $v$ . The acceleration always tracks the net force; the velocity is free to disagree (ball at the top of its flight: $v = 0$ , $a = 9.8\ \mathrm{m/s^2}$ down).

Perpendicular forces added as plain numbers. $8\ \mathrm{N} + 6\ \mathrm{N}$ at right angles is $10\ \mathrm{N}$ , never 14. The plain sum is a standard MCQ distractor.

Unit slips. Convert km/h to m/s ( $\div 3.6$ ) before computing $a$ ; grams to kilograms before using (1). A 900 g body under 1.8 N accelerates at $2.0\ \mathrm{m/s^2}$ , not $0.002$ .

"Force is needed to maintain velocity." Remove the net force and the acceleration dies; the velocity survives untouched (Worked Example 2, Step 4). This single sentence separates the first-law error from second-law fluency.

Which $g$ ? CBSE/ISC use $9.8\ \mathrm{m/s^2}$ ; JEE/NEET as the paper states (usually $10$ — recheck arithmetic if options look off by 2%); A-Level uses $9.81$ . State your choice at the top of the solution.

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1642–1727 · Cambridge & London

Newton stated the second law in the Principia (1687) in momentum language: "the change of motion is proportional to the motive force impressed, and is made in the direction of the straight line in which that force is impressed." Remarkably, the equation $F = ma$ never appears in the book — geometry, not algebra, was his instrument. What made the law revolutionary was its scope: the same rule that guides a thrown stone reproduces the Moon's orbit, once his gravity supplies the force. The working algebraic form you use was distilled by Euler and the continental mathematicians half a century later — but the physics, the strict proportion between impressed force and change of motion, is Newton's alone, and it remains the most heavily executed law in all of engineering.

History

1638

— Galileo's Two New Sciences shows a constant push produces uniform acceleration, clearing the ground for a quantitative force law.

1687

— Newton's Principia states Lex II: change of motion is proportional to the motive force impressed.

1750

— Euler recasts the law as the working equations F = ma along each axis — the form every physicist and engineer now solves.

1846

— Le Verrier predicts Neptune's position from irregularities in Uranus's orbit using pure second-law dynamics; the planet is found within a degree.

1969

— Apollo 11's guidance computer integrates F = ma in real time to fly humans to the Moon and home.

Going further

Texts. NCERT Physics Class 11, Ch. 5 (Laws of Motion) · HC Verma, Concepts of Physics Vol. 1, Ch. 5 · Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 5 · Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 2 for the operational construction.

Provenance. Standard textbook physics throughout. A verifier should recompute: the car ( $a = 27/9 = 3.0\ \mathrm{m/s^2}$ , $F = 4500\ \mathrm{N}$ , weight $14{,}700\ \mathrm{N}$ , ratio 0.31); the crate ( $a = 28/7 = 4.0\ \mathrm{m/s^2}$ , $x = 18\ \mathrm{m}$ , $v = 12\ \mathrm{m/s}$ , then $a = -12/7 = -1.7\ \mathrm{m/s^2}$ ); the perpendicular pair ( $\sqrt{64+36} = 10\ \mathrm{N}$ , $a = 2.0\ \mathrm{m/s^2}$ , $\phi = 36.9°$ , equilibrant 10 N opposite); the NEET template ( $m_1 + m_2 = F/6 \Rightarrow a = 6.0\ \mathrm{m/s^2}$ ); the conversions $1\ \mathrm{kgf} = 9.8\ \mathrm{N}$ and $1.8\ \mathrm{N}$ on $0.90\ \mathrm{kg} \to 2.0\ \mathrm{m/s^2}$ .

Prerequisites. Force, Inertia & Newton's First Law · Acceleration · Adding Vectors: Graphical & Analytical. Leads to. Momentum & the General Second Law (F = dp/dt) · Free-Body Diagrams & Problem-Solving Strategy · the Atwood Machine archetype — and, for the deep dive, the university-tier treatment in Newton's Second Law (Advanced Mechanics).

Is "ma" itself a force? Should I draw an ma arrow on my diagram?

No — and this error costs real marks. The left side of $F_{\text{net}} = ma$ lists causes (real pushes and pulls, each with a partner object); the right side is the effect. Drawing an " $ma$ force" double-counts the physics, like listing "getting wet" among the causes of rain. The working habit: a force diagram carries only forces with nameable agents — Earth (weight), a surface (normal, friction), a string (tension), a hand (push). Then, beside the diagram, write $F_{\text{net}} = ma$ and let the arrows you drew fund the right-hand side.

Why does the velocity not appear anywhere in F = ma?

Because force controls the change of velocity, not the velocity. The law is local in time: the net force now fixes the acceleration now, and the current velocity is just accumulated history. That is why the same 28 N net force gives a crate the same $4\ \mathrm{m/s^2}$ whether it is crawling or already moving at 12 m/s — and why removing the force freezes the velocity rather than killing it. If your intuition insists moving bodies "need" force, revisit the Force, Inertia & Newton's First Law module: that need is Aristotle's, not Newton's.

How can something be accelerating at the moment its velocity is zero?

Acceleration is the rate of change of velocity, and a quantity can be changing fastest exactly while passing through zero. The thrown ball at its peak has $v = 0$ for one instant, but gravity has not paused: a moment earlier the ball moved up, a moment later it moves down, so the velocity is changing right through the zero — at $9.8\ \mathrm{m/s^2}$ . Same story for a pendulum bob at the end of its swing. The reliable test: ask what the net force is (here $mg$ , unchanged), and the second law hands you the acceleration regardless of what $v$ happens to be.

What exactly does "net force" mean when the forces point in different directions?

It is the single force that would produce the same effect as all the real ones together — computed by vector addition, never by adding magnitudes. Opposite forces subtract (10 N east + 4 N west = 6 N east); perpendicular ones combine by Pythagoras (8 N and 6 N make 10 N, at $\tan^{-1}(6/8)$ from the larger); anything else goes through components, equation (2). The body cannot tell the difference between the real collection and the one resultant — that is the superposition principle, and it is an experimental fact about our world, not a definition.

Why do all objects fall with the same acceleration if heavier ones are pulled harder?

Because the two roles of mass cancel. Gravity pulls with $F = mg$ — heavier means pulled harder — but the same $m$ appears as inertia in $a = F/m$ — heavier also means more reluctant. Divide: $a = mg/m = g$ , mass gone. A 10 kg stone feels ten times the force of a 1 kg stone and carries exactly ten times the stubbornness; the accelerations tie at $g$ . Air resistance breaks the tie in everyday life (it does not scale with mass), which is why a feather loses to a hammer here but ties on the airless Moon.

Do I use g = 9.8, 9.81 or 10 in numericals?

Follow the paper, not habit. CBSE and ISC mark schemes are built on $9.8\ \mathrm{m/s^2}$ ; A-Level on $9.81$ ; JEE and NEET usually say "take $g = 10\ \mathrm{m/s^2}$ " to keep arithmetic clean — and their answer options are computed with it, so using 9.8 there can land you between options. Write "taking $g = \ldots$ " once at the top of a written solution; in MCQs, scan the instructions before the first question. If options look uniformly ~2% off your answer, you used the wrong $g$ — recompute the single line where it entered.

What happens to F = ma when the mass itself changes, like a rocket burning fuel?

The form $F = ma$ quietly assumes the mass is constant while the force acts. A rocket sheds mass every second, a conveyor belt gains sand — for these, Newton's own statement of the law is the one to use: net force equals the rate of change of momentum, $F = \Delta p/\Delta t$ . For everything in this module (blocks, cars, crates with fixed mass) the two forms are identical, so nothing you learned here bends. The general form, force–time graphs, and why Newton wrote it that way in the first place are exactly the business of the next module, Momentum & the General Second Law.