Physics · Mechanics

Newton’s second law: net force

Q: What force does $F = ma$ calculate?

The equation gives the net force on the object: $\vec F_{\mathrm{res}}=m\vec a$. For 8 kg and an acceleration of $\frac{3 m}{s^{2}}$, this is 24 N in the acceleration direction. When several forces act, net force is their vector sum. It is not an extra force to draw alongside weight, tension or friction. First identify the object you are studying and include only forces acting on that object. Then choose axes and add components with their signs. Forces on other objects do not belong in this sum. This object selection prevents a reaction force from incorrectly cancelling the force that accelerates your object.

Q: How do you calculate acceleration?

Divide net force by mass: $\vec a=\frac{\vec F_{\mathrm{res}}}{m}$, with $m>0$. Calculate components separately when forces act in several directions. For 24 N along the positive x-axis and a mass of 8 kg, $a_x=3\,\mathrm{m/s^2}$. Its sign specifies direction relative to your chosen axis; it does not by itself say whether the object speeds up or slows down. That also depends on velocity direction. A nonzero acceleration component gives mass as $m=\frac{F_{\mathrm{res},x}}{a_x}$. If $a_x=0$, this division is invalid. A negative calculated mass indicates inconsistent signs or unsuitable inputs within this model.

Q: What is the difference between force and weight?

Weight is an individual gravitational force: $F_G=mg$. At $g =\frac{ 9.81 m}{s^{2}}$, a 10 kg object has a weight of 98.1 N, or about 98 N after rounding. Net force also includes every other external force, such as the normal force from a floor. If the object rests on a horizontal support and these are its only vertical forces, their magnitudes are equal and their directions opposite. Vertical net force is then zero even though weight acts. Draw both forces on the same object. Weight is not generally identical to net force, and normal force does not equal $mg$ in every situation.

Newton’s second law relates the vector sum of external forces to mass times acceleration. An individual force equals net force only when the other forces cancel or are absent.

BasicExam-relevantLower secondaryUpper secondary

Create flashcards or a quiz I already have an account

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Formula

Calculate

\sum F_{ext} = F_{res} = m a

Force F in N; mass m in kg; acceleration a in m/s². One newton equals one kg·m/s².

Variables & units

Variables & units
Symbol	Meaning	Unit
$\sum F_{\mathrm{ext}} = F_{\mathrm{res}}$	Vector sum of all external forces on the selected system (net force)	N
$m$	Mass of the body	kg
$a$	Acceleration	m/s²

LaTeX

\sum \vec F_{\mathrm{ext}} = \vec F_{\mathrm{res}} = m \vec a

QUANTA STUDYquanta-study.de

A block on a horizontal surface. Normal force N points upward and weight mg downward; they cancel. The resultant external force F_res and acceleration a point to the right. — Schematic free-body diagram on a horizontal contact surface: normal force and weight balance. The horizontal net force summarizes the horizontal forces; it is not an additional force. The dashed arrow shows acceleration.

Force and acceleration calculator (F = ma)

Find net force, mass or acceleration along one axis, in a non-accelerating frame of reference.

Find

Mass m

Mass of the accelerated body, e.g. 1200 kg for a car.

Acceleration a

Acceleration with sign, pointing the same way as the net force, e.g. 2.5 m/s².

m/s²

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for F, substitute, evaluate

1. Rearrange for F
$F=ma$
2. Substitute values (in SI units)
$F=1200\cdot 2.5=3000$
3. Result
F = 3000 N

Result: Net force F (sum of all forces)

F = 3000 N

With m = 1200 kg and a = 2.5 m/s², F = 3000 N.

Quick tips

Formula triangle: F on top, m and a below. Cover the quantity you need and the rest is the calculation, e.g. a = F ÷ m.
Add all forces with their signs first, then divide by m: 42 N of pull and 18 N of friction give a 24 N net force.

Assumptions and limits

Inertial frame and constant mass m > 0. F is the complete signed net force along the same axis as a, not an arbitrary individual force.

Derivation & background

For a closed system in an inertial frame, net external force is the rate of change of momentum. With constant mass, $\vec p=m\vec v$ gives $\sum\vec F_{\mathrm{ext}}=m\vec a$ . Include weight, normal force and resistance whenever they act on the selected system. An open system with mass flow also needs a momentum-flux balance.

Exam blueprint

Validity range

Applies in inertial frames of classical mechanics and for constant mass; F is the vector sum of external forces on the chosen system. Relativistic speeds require a relativistic model. With mass flow, such as rocket exhaust, the balance must also account for momentum flux.

Derivation steps

Newton’s second law is an empirical law. For a closed system of constant mass, momentum balance gives $\sum\vec F=m\vec a$ .

1Select the object or closed system. Include only external forces acting on it.
2Choose an inertial frame and positive axes, and assume constant mass.
3Start with momentum balance $\sum\vec F=\frac{\mathrm d\vec p}{\mathrm dt}$ , the physical law.
4In classical mechanics, $\vec p=m\vec v$ . Constant mass gives $\frac{\mathrm d\vec p}{\mathrm dt}=m\frac{\mathrm d\vec v}{\mathrm dt}$ .
5Using $\vec a=\frac{\mathrm d\vec v}{\mathrm dt}$ gives $\sum\vec F=m\vec a$ . Along the x axis, $\sum F_x=ma_x$ .
6Check signs and units. Mass flow, rotating frames or very high speeds require an extended model.

Rearrangements

Acceleration from net force and mass

a = \frac{F_{\mathrm{res}}}{m}

F_res is the sum of all external forces; use one individual force alone only when opposing forces are absent or balanced.

Task variant

A net external force of 24 N acts on an 8 kg body. Find a.

Approach: $a=\frac{F_{\mathrm{res}}}{m}$ .
With 24 N and 8 kg, $a=3\,\mathrm{m/s^2}$ along net force.

Check: $ma=24\,\mathrm N$ .

Common mistakes

Using mass in grams or confusing force with weight.

Always use mass in kg; weight is only the special case $F_{G} = m\cdot{}g$ .

Exam context

Typical exams combine a free-body diagram, net force and acceleration.

Formula cluster

Mechanics fundamentals

Connects forces, motion and energy arguments.

Ekin = ½mv²

Kinetic Energy

F = G·m₁m₂/r²

Newton's Law of Universal Gravitation

F = k·x

Hooke's Law (Spring Force)

F_R = μ·F_N

Sliding Friction Force

Worked example

A 1,200 kg car accelerates at $\frac{3 m}{s^{2}}$ on a level road. Its net force is $F_{\mathrm{res}}=ma=3600\,\mathrm N$ . The driving force must also overcome air and rolling resistance. Only an idealized problem without resistance makes driving force equal to net force.

Applications

Vehicle dynamics, spaceflight (rocket propulsion), biomechanics (gait analysis), materials testing

How do you calculate with Newton's Second Law?

Here is how to work through a typical Newton's Second Law (ΣF_ext = F_res = ma) task step by step:

1
Task
A net external force of 24 N acts on an 8 kg body. Find a.
Solution path
1. Approach: $a=\frac{F_{\mathrm{res}}}{m}$ .
2. With 24 N and 8 kg, $a=3\,\mathrm{m/s^2}$ along net force.
Check: $ma=24\,\mathrm N$ .

Understand and apply the equation

Why is an empty shopping trolley easier to accelerate than a full one? What matters is its mass and the sum of the forces acting on it. Newton’s second law describes this relationship: net force equals mass times acceleration.

What does F mean in F = ma?

In Newton’s second law, F is the net force: the vector sum of all external forces on the object being studied. It is not an additional individual force. The form F = m·a applies to constant mass in an inertial frame. Acceleration means a change in velocity: an object speeds up, slows down or changes direction. If you pull a trolley to the right while friction acts to the left, both forces must be included.[1]

\sum_i\vec F_i=m\vec a,\qquad 1\,\mathrm N=1\,\mathrm{kg\,m/s^2}

Every force arrow must act on the same selected body or system.

At fixed mass, doubling net force doubles acceleration. At fixed net force, doubling mass halves acceleration. This does not imply doubled velocity: velocity also depends on duration and initial conditions.

The simple form applies to an object of constant mass in an inertial frame: a reference frame that neither accelerates nor rotates. For an extended object, net force determines the acceleration of its centre of mass. To describe its rotation, you also need the torques.[1]

How do you calculate force, mass and acceleration?

Along a chosen axis, $F_{\mathrm{net}}=ma$ uses signed force and acceleration components. For $m>0$ , $a=\frac{F_{\mathrm{net}}}{m}$ . If $a\ne0$ , rearranging gives $m=\frac{F_{\mathrm{net}}}{a}$ . Use the total external force component for $F_{\mathrm{net}}$ , rather than one pulling force alone. When $a=0$ and $F_{\mathrm{net}}=0$ , this equation does not determine mass.[1]

The relationship can be investigated using a trolley on a straight, horizontal track with low friction. First compare different net forces at the same mass. Then compare different masses at the same net force. Determine acceleration from the change in velocity and the corresponding time interval. Account for friction and measurement uncertainty.[1]

a=\frac{\Delta v}{\Delta t}

For constant acceleration, this ratio gives the acceleration throughout the time interval. Otherwise, it gives the average acceleration.[1]

Doubling force or mass

Baseline
Net force (N)
12
Mass (kg)
6
Acceleration (m/s²)
2
Double force
Net force (N)
24
Mass (kg)
6
Acceleration (m/s²)
4
Double mass
Net force (N)
12
Mass (kg)
12
Acceleration (m/s²)
1

Three model cases: 12 newtons and 6 kilograms give 2 metres per second squared; doubling force gives 4, doubling mass gives 1.
Case	Net force (N)	Mass (kg)	Acceleration (m/s²)
Baseline	12	6	2
Double force	24	6	4
Double mass	12	12	1

QUANTA STUDYquanta-study.de

Original model values comparing net forces. The bars show acceleration.[1]

Free with a Quanta account. Your download stays ready after registration.

The two measurement series investigate different relationships. At fixed mass, the ratio of acceleration to net force stays constant. At fixed net force, mass times acceleration stays constant. Together, these give $a\propto\frac{F_{\mathrm{res}}}{m}$ . In SI units, one newton is the force that accelerates one kilogram at one metre per second squared. The proportionality factor is therefore one, giving $F_{\mathrm{res}}=ma$ . Experiments support this relationship; algebraic rearrangement alone does not prove a law of nature.[1]

Free-body diagram and signs

Select the body first. Draw only forces exerted on it by other bodies, such as weight, normal force, tension and friction. The reaction force of an interaction acts on the other body and must not be added to this body’s balance. If two bodies are treated as one system, forces between them are internal.[3][6]

Choose the positive direction first. In the horizontal example, $F_1$ is the pulling force and $F_2$ is the opposing resistance. Then $\sum F_x=F_1-F_2$ . With no vertical acceleration, $F_N-mg=0$ : normal force and weight balance. Write a separate force balance for each direction.

\sum F_x=ma_x,\qquad \sum F_y=ma_y

Compare components using identical axes and signs on both sides.

Forces on the pulled object

QUANTA STUDYquanta-study.de

Horizontal forces from example 1. Right is positive. Vertical forces balance and are omitted here. The net force is 24 N to the right; it is not an additional third force.[3]

Free with a Quanta account. Your download stays ready after registration.

From momentum to mass times acceleration

Newton’s second law is an empirical law supported by experiments within its range of validity. Rearranging an equation does not prove it. For a closed system without mass flow, $\sum\vec F_{\mathrm{ext}}=\frac{\mathrm d\vec p}{\mathrm dt}$ . In classical mechanics, momentum is $\vec p=m\vec v$ . Constant mass gives the familiar form.[4]

\vec p=m\vec v\quad\Longrightarrow\quad \frac{\mathrm d\vec p}{\mathrm dt}=m\frac{\mathrm d\vec v}{\mathrm dt}=m\vec a

Mass can be taken outside the time derivative only when constant.

A rocket ejects exhaust and loses mass. The rocket alone is therefore an open system. Its motion also depends on the momentum carried away by the exhaust. A complete balance includes this momentum flux. For ordinary problems with an object of constant mass, the simple form of Newton’s second law is sufficient.[5]

Force balances: pulling against resistance and an accelerating lift

Pulling against resistance

Given: m = 12 kg; horizontal pull 42 N; resistance 18 N; no vertical acceleration.
Find: Find a and velocity change after 3 s with constant forces.

Approach: Choose pulling direction as positive and sum both horizontal forces with signs. Constant net force gives constant acceleration and hence velocity change.
$\begin{aligned}\sum F_x&=F_1-F_2\\{}a_x&=\frac{\sum F_x}m\\{}\Delta v_x&=a_x\Delta t\end{aligned}$
Choose positive direction along pulling force and subtract forces.
$\sum F_x=42-18=24\,\mathrm N$
Divide by mass.
$a_x=\frac{24}{12}=2.0\,\mathrm{m/s^2}$
Apply constant acceleration over 3 s.
$\Delta v_x=a_x\Delta t=2.0\cdot3=6.0\,\mathrm{m/s}$

Result: a = 2.0 m/s² along the pull; Δv = 6.0 m/s.

Check: Check: mass times acceleration plus resistance gives $24+18=42\,\mathrm N$ . Calculating $\frac{42}{12}=3.5\,\frac{\mathrm m}{\mathrm{s^2}}$ would omit resistance.

Normal force in an accelerating lift

Given: m = 68 kg; acceleration 1.2 m/s² upward; g = 9.81 m/s². Only normal force and weight act vertically.
Find: Find the force exerted by the floor on the person.

Approach: Take the positive axis upward. Floor reaction and weight act oppositely; their difference accelerates the person with the lift.
$\begin{aligned}F_N-mg&=ma\\{}F_N&=m(g+a)\end{aligned}$
Take upward as positive: normal force upward, weight downward.
$F_N-mg=ma$
Solve for $F_N$ and substitute mass and both accelerations.
$F_N=m(g+a)=68(9.81+1.2)=748.68\,\mathrm N$
Check weight and net force separately.
$\begin{aligned}mg&=667.08\,\mathrm N\\{} F_N-mg&=81.60\,\mathrm N\end{aligned}$

Result: The floor exerts about 749 N upward. The person’s mass remains 68 kg.

Check: With downward acceleration, $a=-1.2\,\mathrm{m/s^2}$ and normal force is $F_N=585.48\,\mathrm N$ . This does not determine direction of travel: a lift moving downward accelerates upward when braking.

View diagram: forces in the lift

Forces on the person in the lift

QUANTA STUDYquanta-study.de

For example 2: the upward force from the floor exceeds the downward weight. The difference of 81.60 N accelerates the person upward.[3]

Free with a Quanta account. Your download stays ready after registration.

Angled forces and changing direction

Perpendicular net-force components of 6 N and 8 N give a magnitude of 10 N. For a 2 kg object, $a_x=3\,\mathrm{m/s^2}$ , $a_y=4\,\mathrm{m/s^2}$ and $|\vec a|=5\,\mathrm{m/s^2}$ . Acceleration points along net force; its angle to the positive x axis is about 53.1°.

|\vec F|=\sqrt{F_x^2+F_y^2},\qquad |\vec a|=\frac{|\vec F|}{m}

Pythagoras applies to the perpendicular components.

Uniform circular motion has constant speed but continually changing velocity direction. Acceleration is therefore nonzero, and net force points towards the centre. Constant speed and constant velocity vector are different conditions.

Assumptions and common misinterpretations

The simple form assumes an inertial frame. In accelerating or rotating coordinates, additional inertial terms are needed to retain the same equation form. Classical momentum mv is not the appropriate model at relativistic speeds. For macroscopic bodies moving slowly, classical force balance is often sufficient.[1]

Normal force does not always equal $mg$ . It acts perpendicular to the contact surface and depends on acceleration and the other forces. Tension equals weight only under suitable conditions. Write the force balance before setting these quantities equal.[2]

When $\sum\vec F=0$ , $\vec a=0$ . The object stays at rest or continues at constant velocity. During braking, velocity and acceleration point in opposite directions. Negative $a_x$ means acceleration opposite to the positive x axis; it does not automatically mean rest or backward travel.

Five problems with solutions

Calculate · Foundation

Forces of 50 N rightward and 20 N leftward act on 7.5 kg. Find a.

Show hint

Calculate signed net force first.

Learning goal: Add opposing forces into signed net force first.

Show solution and calculation

Approach: Right is positive.
Add the forces on the selected object: $F_{\mathrm{res},x}=50\,\mathrm N-20\,\mathrm N=30\,\mathrm N$ .
Thus $a_x=\frac{30\,\mathrm N}{7.5\,\mathrm{kg}}=4.0\,\mathrm{m/s^2}$ rightward.

Check: $ma_x=30\,\mathrm N$ matches the force balance.

a_x=\frac{50-20}{7.5}=4.0\,\mathrm{m/s^2}

Calculate · Application

A 900 kg vehicle accelerates at 1.6 m/s² with total resistance 360 N. What driving force is required?

Show hint

Call driving force $F_1$ and resistance $F_2$ . Solve $F_1-F_2=ma$ for $F_1$ .

Learning goal: Distinguish driving force from net acceleration force and resistance.

Show solution and calculation

Approach: The driving direction is positive.
From $F_1-F_2=ma$ , obtain $F_1=ma+F_2$ .
Substitution gives $F_1=1800\,\mathrm N$ .
Of this, 360 N balances resistance; net force is 1440 N.

Check: $\frac{1800-360}{900}\,\mathrm{m/s^2}=1.6\,\mathrm{m/s^2}$ .

\begin{aligned}F_1&=ma+F_2\\&=900\,\mathrm{kg}\cdot1.6\,\mathrm{m/s^2}+360\,\mathrm N\\&=1800\,\mathrm N\end{aligned}

Transfer · Application

A 60 kg person experiences normal force 528.6 N in a lift. Find acceleration for g = 9.81 m/s², upward positive. Only normal force and weight act vertically.

Show hint

Write the vertical balance $F_N-mg=ma$ . Solve for $a$ , then interpret its sign.

Learning goal: Infer lift acceleration from normal force and gravity.

Show solution and calculation

Approach: Upward is positive.
Normal force and weight give $F_N-mg=ma$ .
Hence $a=\frac{F_N}{m}-g=-1.0\,\mathrm{m/s^2}$ , downward.

Check: Weight is 588.6 N; net force is $528.6-588.6=-60.0\,\mathrm N$ , matching $ma=-60.0\,\mathrm N$ . Direction of travel is unknown without initial velocity.

a=\frac{F_N}{m}-g

Interpret · Application

A 0.20 kg ball reaches the highest point of a vertical throw without air resistance. Its instantaneous velocity there is zero; g=10 m/s². A graph annotation therefore claims “net force zero”. Find force and acceleration in an inertial frame with upward positive, and explain subsequent motion.

Show hint

Newton’s law relates force to the change of velocity, not to instantaneous velocity itself.

Learning goal: Recognize that acceleration can be nonzero when instantaneous velocity is zero.

Show solution and calculation

Approach: only gravity acts after release.
With upward positive, force is −2.0 N and acceleration −10 m/s².
Velocity subsequently becomes negative and the ball falls.

Check: zero velocity is only an instantaneous turning point. If acceleration also stayed zero, the ideal model would leave the ball there, which gravity rules out.

\begin{aligned}v(t_{\mathrm{top}})&=0\\F_y&=-mg=-(0.20)(10)\,\mathrm N=-2.0\,\mathrm N\\a_y&=\frac{F_y}{m}=-10\,\mathrm{m/s^2}\end{aligned}

[1][2]

Error analysis · Application

A hand pushes a 4 kg cart horizontally with +12 N on an ideal frictionless track. The cart pushes back on the hand with −12 N. A solution adds both forces on the cart and obtains $a=0$ . Correct the free-body diagram and calculate cart acceleration. Vertical forces balance.

Show hint

Label the body acted on by each force. The cart’s reaction acts on the hand, not back on the cart itself.

Learning goal: Avoid canceling interaction pairs in one body’s force balance.

Show solution and calculation

Approach: choose the cart alone as the body under consideration.
The hand’s horizontal force is +12 N; the reaction on the hand does not belong in that balance.
Cart acceleration is therefore +3.0 m/s².

Check: the interaction pair is equal and opposite but acts on different bodies. Only a balance of their combined system would treat both as internal forces.

\begin{aligned}F_{\mathrm{cart}}&=+12\,\mathrm N\\a_{\mathrm{cart}}&=\frac{F_{\mathrm{cart}}}{m}=\frac{12\,\mathrm N}{4\,\mathrm{kg}}=+3.0\,\mathrm{m/s^2}\\\vec F_{\mathrm{cart\to hand}}&=-\vec F_{\mathrm{hand\to cart}}\end{aligned}

[1][2][6]

How to check a solution

Do all force arrows refer to the same body? Have all forces on the selected object been included or reasonably neglected? Are axes and units consistent? Substitute ma back into the force balance. The original diagram above illustrates the principle schematically; multiple forces still require vector addition.

Further motion calculations need initial conditions. v = v₀ + at applies only to constant acceleration over the chosen interval. If resistance increases with speed, acceleration is not automatically constant. Force balance remains the starting point; determine the time evolution consistently with it.

Key points

Net force determines acceleration. Add individual forces with their directions first.
More force means more acceleration at fixed mass. More mass means less acceleration at fixed force.
Acceleration and velocity need not point in the same direction.
The simple equation assumes constant mass and an inertial frame in classical mechanics.

[1]

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Clarified the direct answer on net force and rearrangements for axis components; made the existing examples easier to find through the section heading and added contextual links to friction and momentum.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible introduction, experimental context, three explanatory figures and a summary; revised wording, notation and model limits.

Cite this reference page

Quanta. Newton's Second Law. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/newton-zweites-gesetz

When citing a specific scientific statement, also cite the specialist source linked to that statement.

Suggest a correction with a reference

Frequently asked questions about Newton's Second Law

What force does

F = ma

calculate?+

The equation gives the net force on the object: $\vec F_{\mathrm{res}}=m\vec a$ . For 8 kg and an acceleration of $\frac{3 m}{s^{2}}$ , this is 24 N in the acceleration direction. When several forces act, net force is their vector sum. It is not an extra force to draw alongside weight, tension or friction. First identify the object you are studying and include only forces acting on that object. Then choose axes and add components with their signs. Forces on other objects do not belong in this sum. This object selection prevents a reaction force from incorrectly cancelling the force that accelerates your object.

How do you calculate acceleration?+

Divide net force by mass: $\vec a=\frac{\vec F_{\mathrm{res}}}{m}$ , with $m>0$ . Calculate components separately when forces act in several directions. For 24 N along the positive x-axis and a mass of 8 kg, $a_x=3\,\mathrm{m/s^2}$ . Its sign specifies direction relative to your chosen axis; it does not by itself say whether the object speeds up or slows down. That also depends on velocity direction. A nonzero acceleration component gives mass as $m=\frac{F_{\mathrm{res},x}}{a_x}$ . If $a_x=0$ , this division is invalid. A negative calculated mass indicates inconsistent signs or unsuitable inputs within this model.

What is the difference between force and weight?+

Weight is an individual gravitational force: $F_G=mg$ . At $g =\frac{ 9.81 m}{s^{2}}$ , a 10 kg object has a weight of 98.1 N, or about 98 N after rounding. Net force also includes every other external force, such as the normal force from a floor. If the object rests on a horizontal support and these are its only vertical forces, their magnitudes are equal and their directions opposite. Vertical net force is then zero even though weight acts. Draw both forces on the same object. Weight is not generally identical to net force, and normal force does not equal $mg$ in every situation.

What assumptions does

F = ma

require?+

The simple form assumes constant mass and an inertial frame in classical mechanics. A rocket requires accounting for the momentum flux carried by its exhaust; inserting changing rocket mass into a closed-object force balance is insufficient. Speeds near the speed of light require a relativistic model. Accelerating or rotating reference frames need additional inertial forces if you want to use the same form of the motion equation there. Before calculating, check the object boundary, mass exchange and reference frame. At ordinary low speeds, the classical model is a very good approximation. Acceleration then describes the rate of change of velocity, rather than directly specifying distance travelled.

Why must all forces be included?+

Forces can reinforce or oppose one another. Omitting a force can make the calculated acceleration too large, too small or even point in the wrong direction. Draw all forces on the same object first and add them with appropriate signs. Check components along each axis separately instead of simply adding all force magnitudes. Zero net force means constant velocity in an inertial frame; the object need not be at rest. Conversely, a nonzero force opposed to velocity can initially slow the object down. As a check, identify the interaction responsible for each drawn force and ask whether that force actually acts on the object you selected.

Quanta exam set

Curated exam set for "Newton's Second Law":

Question (front)

Which formula describes Newton's Second Law?

Answer in your set

Question (front)

How do you rearrange Newton's Second Law for Acceleration from net force and mass?

Answer in your set

Question (front)

Which common mistake happens with Newton's Second Law?

Answer in your set

+ 7 more cards: units, variables, derivation, example, exam task

The formula set contains 10 cards. After importing, review them and start practicing; your ratings determine later review intervals.

Create flashcards or a quiz I already have an account

Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §5.3 Newton’s Second Law

More formulas: Mechanics

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

s = \frac{1}{2} a t^2 + v_0 t

Uniformly Accelerated Motion (Position–Time Equation)

v = a \cdot t + v_0

Velocity-Time Law

h = \frac{1}{2} g t^2

Free Fall

Spaced-repetition apps

Compare scheduling workflows

Study evidence

Read methods, results and limits

Compare flashcard apps

FSRS support and study features

Create a formula exam set

From Essential in Quanta

Retain Newton's Second Law for exams

Create a curated FSRS exam set for ΣF_ext = F_res = ma: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Create flashcards or a quiz Already registered? Continue directly

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Quanta Study editorial team

Newton’s second law: net force

Newton’s second law relates the vector sum of external forces to mass times acceleration. An individual force equals net force only when the other forces cancel or are absent.

BasicExam-relevantLower secondaryUpper secondary

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

\sum F_{\mathrm{ext}} = F_{\mathrm{res}}

Vector sum of all external forces on the selected system (net force)

m

Mass of the body

a

Acceleration

m/s²

Derivation & background

Exam blueprint

Validity range

Derivation steps

Newton’s second law is an empirical law. For a closed system of constant mass, momentum balance gives $\sum\vec F=m\vec a$ .

1Select the object or closed system. Include only external forces acting on it.
2Choose an inertial frame and positive axes, and assume constant mass.
3Start with momentum balance $\sum\vec F=\frac{\mathrm d\vec p}{\mathrm dt}$ , the physical law.
4In classical mechanics, $\vec p=m\vec v$ . Constant mass gives $\frac{\mathrm d\vec p}{\mathrm dt}=m\frac{\mathrm d\vec v}{\mathrm dt}$ .
5Using $\vec a=\frac{\mathrm d\vec v}{\mathrm dt}$ gives $\sum\vec F=m\vec a$ . Along the x axis, $\sum F_x=ma_x$ .
6Check signs and units. Mass flow, rotating frames or very high speeds require an extended model.

How do you calculate with Newton's Second Law?

Here is how to work through a typical Newton's Second Law (ΣF_ext = F_res = ma) task step by step:

Task

A net external force of 24 N acts on an 8 kg body. Find a.

Solution path

Approach: $a=\frac{F_{\mathrm{res}}}{m}$ .
With 24 N and 8 kg, $a=3\,\mathrm{m/s^2}$ along net force.

Check: $ma=24\,\mathrm N$ .

Case

Net force (N)

Mass (kg)

Acceleration (m/s²)

Baseline

Double force

Double mass

Key points

Net force determines acceleration. Add individual forces with their directions first.
More force means more acceleration at fixed mass. More mass means less acceleration at fixed force.
Acceleration and velocity need not point in the same direction.
The simple equation assumes constant mass and an inertial frame in classical mechanics.

[1]

Frequently asked questions about Newton's Second Law

What force does

F = ma

calculate?+

How do you calculate acceleration?+

What is the difference between force and weight?+

What assumptions does

F = ma

require?+

Why must all forces be included?+

Retain Newton's Second Law for exams

Create a curated FSRS exam set for ΣF_ext = F_res = ma: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Newton’s second law: net force

Variables & units

Force and acceleration calculator (F = ma)

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Mechanics fundamentals

Worked example

Applications

How do you calculate with Newton's Second Law?

What does F mean in F = ma?

How do you calculate force, mass and acceleration?

Doubling force or mass

Baseline

Double force

Double mass

Free-body diagram and signs

Forces on the pulled object

From momentum to mass times acceleration

Force balances: pulling against resistance and an accelerating lift

Pulling against resistance

Normal force in an accelerating lift

Forces on the person in the lift

Angled forces and changing direction

Assumptions and common misinterpretations

Five problems with solutions

Forces of 50 N rightward and 20 N leftward act on 7.5 kg. Find a.

A 900 kg vehicle accelerates at 1.6 m/s2 with total resistance 360 N. What driving force is required?

A 60 kg person experiences normal force 528.6 N in a lift. Find acceleration for g = 9.81 m/s2, upward positive. Only normal force and weight act vertically.

A 0.20 kg ball reaches the highest point of a vertical throw without air resistance. Its instantaneous velocity there is zero; g=10 m/s2. A graph annotation therefore claims “net force zero”. Find force and acceleration in an inertial frame with upward positive, and explain subsequent motion.

A hand pushes a 4 kg cart horizontally with +12 N on an ideal frictionless track. The cart pushes back on the hand with −12 N. A solution adds both forces on the cart and obtains a=0a=0a=0. Correct the free-body diagram and calculate cart acceleration. Vertical forces balance.

How to check a solution

Key points

Sources for this explanation

Frequently asked questions about Newton's Second Law

Subject references and further reading

More formulas: Mechanics

Retain Newton's Second Law for exams

Newton’s second law: net force

Variables & units

Force and acceleration calculator (F = ma)

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Mechanics fundamentals

Worked example

Applications

How do you calculate with Newton's Second Law?

What does F mean in F = ma?

How do you calculate force, mass and acceleration?

Doubling force or mass

Baseline

Double force

Double mass

Free-body diagram and signs

Forces on the pulled object

From momentum to mass times acceleration

Force balances: pulling against resistance and an accelerating lift

Pulling against resistance

Normal force in an accelerating lift

Forces on the person in the lift

Angled forces and changing direction

Assumptions and common misinterpretations

Five problems with solutions

Forces of 50 N rightward and 20 N leftward act on 7.5 kg. Find a.

A 900 kg vehicle accelerates at 1.6 m/s2 with total resistance 360 N. What driving force is required?

A 60 kg person experiences normal force 528.6 N in a lift. Find acceleration for g = 9.81 m/s2, upward positive. Only normal force and weight act vertically.

A 0.20 kg ball reaches the highest point of a vertical throw without air resistance. Its instantaneous velocity there is zero; g=10 m/s2. A graph annotation therefore claims “net force zero”. Find force and acceleration in an inertial frame with upward positive, and explain subsequent motion.

A 900 kg vehicle accelerates at 1.6 m/s² with total resistance 360 N. What driving force is required?

A 60 kg person experiences normal force 528.6 N in a lift. Find acceleration for g = 9.81 m/s², upward positive. Only normal force and weight act vertically.

A 0.20 kg ball reaches the highest point of a vertical throw without air resistance. Its instantaneous velocity there is zero; g=10 m/s². A graph annotation therefore claims “net force zero”. Find force and acceleration in an inertial frame with upward positive, and explain subsequent motion.

A hand pushes a 4 kg cart horizontally with +12 N on an ideal frictionless track. The cart pushes back on the hand with −12 N. A solution adds both forces on the cart and obtains $a=0$ . Correct the free-body diagram and calculate cart acceleration. Vertical forces balance.

A 900 kg vehicle accelerates at 1.6 m/s² with total resistance 360 N. What driving force is required?

A 60 kg person experiences normal force 528.6 N in a lift. Find acceleration for g = 9.81 m/s², upward positive. Only normal force and weight act vertically.

A 0.20 kg ball reaches the highest point of a vertical throw without air resistance. Its instantaneous velocity there is zero; g=10 m/s². A graph annotation therefore claims “net force zero”. Find force and acceleration in an inertial frame with upward positive, and explain subsequent motion.

A hand pushes a 4 kg cart horizontally with +12 N on an ideal frictionless track. The cart pushes back on the hand with −12 N. A solution adds both forces on the cart and obtains $a=0$ . Correct the free-body diagram and calculate cart acceleration. Vertical forces balance.