Newton's second law for a particle is

Dynamics · FE Reference Handbook section

Dynamics

13 formulas

10 exam-style examples

~60 min

All Dynamics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Newton's second law for a particle is within Dynamics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what newton's second law for a particle is describes physically and when it applies.
State every one of the 13 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight.

Lecture

Why this section exists. Newton's second law for a particle is is the part of Dynamics that lets you connect a particle or rigid body moving under known forces to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as kinematics first, then a work-energy or impulse-momentum shortcut. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: newton's second law for a particle is.
Capstone Studio instructional photograph

Dynamics — Newton's second law for a particle is: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a particle or rigid body moving under known forces. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 13 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

ΣF	Quantity produced by "ΣF = d(mv)/dt" — read its definition and unit from the handbook line directly above the equation.
m	Quantity produced by "m = mass of the particle" — read its definition and unit from the handbook line directly above the equation.
v	Quantity produced by "v = velocity of the particle" — read its definition and unit from the handbook line directly above the equation.
ax	Quantity produced by "ax = Fx /m" — read its definition and unit from the handbook line directly above the equation.
where Fx	Quantity produced by "where Fx = the resultant of the applied forces, which in general can depend on t, x, and vx." — read its definition and unit from the handbook line directly above the equation.
ax ^ t h	Quantity produced by "ax ^ t h = Fx ^ t h /m" — read its definition and unit from the handbook line directly above the equation.
vx	Quantity produced by "vx = ax _t − t0 j + vxt0" — read its definition and unit from the handbook line directly above the equation.
x	Quantity produced by "x = ax _t − t0 j /2 + vxt0 _t − t0 j + xt0" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

where
For constant mass,
One-Dimensional Motion of a Particle (Constant Mass)
When motion exists only in a single dimension then, without loss of generality, it may be assumed to be in the x direction, and
If Fx only depends on t, then
where τ is the variable of integration.
If the force is constant (i.e., independent of time, displacement, and velocity) then

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Acceleration from Newton's second law — Newton's second law for a particle is

A 3617 lb cart is pulled horizontally by a 1036 lb force against rolling resistance μ = 0.20. What is its acceleration?

Given

W = 3617 lb
F = 1036 lb
μ = 0.20
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 3617/32.2 = 112.3 s l ug s$
Resistance — F_r = μW = 0.20(3617) = 723.4 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 1036 − 723.4 = 312.6 lb
Acceleration — a = ΣF/m = 312.6/112.3 = 2.783 ft/s²

Answer: a ≈ 2.78 ft/s²

Why the other options are there

0.086 ft/s² (weight used as mass)
9.22 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Newton's second law for a particle is

Example 2

Acceleration from Newton's second law — Newton's second law for a particle is (2)

A 1287 lb cart is pulled horizontally by a 807 lb force against rolling resistance μ = 0.25. What is its acceleration?

Given

W = 1287 lb
F = 807 lb
μ = 0.25
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 1287/32.2 = 39.97 s l ug s$
Resistance — F_r = μW = 0.25(1287) = 321.8 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 807 − 321.8 = 485.3 lb
Acceleration — a = ΣF/m = 485.3/39.97 = 12.141 ft/s²

Answer: a ≈ 12.14 ft/s²

Why the other options are there

0.377 ft/s² (weight used as mass)
20.19 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Newton's second law for a particle is

Example 3

Acceleration from Newton's second law — Newton's second law for a particle is (3)

A 5358 lb cart is pulled horizontally by a 795 lb force against rolling resistance μ = 0.25. What is its acceleration?

Given

W = 5358 lb
F = 795 lb
μ = 0.25
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 5358/32.2 = 166.4 s l ug s$
Resistance — F_r = μW = 0.25(5358) = 1,340 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 795 − 1,340 = -544.5 lb
Acceleration — a = ΣF/m = -544.5/166.4 = -3.272 ft/s²

Answer: a ≈ -3.27 ft/s²

Why the other options are there

-0.102 ft/s² (weight used as mass)
4.78 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Newton's second law for a particle is

Example 4

Acceleration from Newton's second law — Newton's second law for a particle is (4)

A 3190 lb cart is pulled horizontally by a 1136 lb force against rolling resistance μ = 0.15. What is its acceleration?

Given

W = 3190 lb
F = 1136 lb
μ = 0.15
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 3190/32.2 = 99.07 s l ug s$
Resistance — F_r = μW = 0.15(3190) = 478.5 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 1136 − 478.5 = 657.5 lb
Acceleration — a = ΣF/m = 657.5/99.07 = 6.637 ft/s²

Answer: a ≈ 6.64 ft/s²

Why the other options are there

0.206 ft/s² (weight used as mass)
11.47 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Newton's second law for a particle is

Example 5

Acceleration from Newton's second law — Newton's second law for a particle is (5)

A 1992 lb cart is pulled horizontally by a 724 lb force against rolling resistance μ = 0.20. What is its acceleration?

Given

W = 1992 lb
F = 724 lb
μ = 0.20
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 1992/32.2 = 61.86 s l ug s$
Resistance — F_r = μW = 0.20(1992) = 398.4 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 724 − 398.4 = 325.6 lb
Acceleration — a = ΣF/m = 325.6/61.86 = 5.263 ft/s²

Answer: a ≈ 5.26 ft/s²

Why the other options are there

0.163 ft/s² (weight used as mass)
11.70 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Newton's second law for a particle is

Example 6

Acceleration from Newton's second law — Newton's second law for a particle is (6)

A 1348 lb cart is pulled horizontally by a 471 lb force against rolling resistance μ = 0.25. What is its acceleration?

Given

W = 1348 lb
F = 471 lb
μ = 0.25
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 1348/32.2 = 41.86 s l ug s$
Resistance — F_r = μW = 0.25(1348) = 337.0 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 471 − 337.0 = 134.0 lb
Acceleration — a = ΣF/m = 134.0/41.86 = 3.201 ft/s²

Answer: a ≈ 3.20 ft/s²

Why the other options are there

0.099 ft/s² (weight used as mass)
11.25 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Newton's second law for a particle is

Example 7

Acceleration from Newton's second law — Newton's second law for a particle is (7)

A 4916 lb cart is pulled horizontally by a 815 lb force against rolling resistance μ = 0.10. What is its acceleration?

Given

W = 4916 lb
F = 815 lb
μ = 0.10
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 4916/32.2 = 152.7 s l ug s$
Resistance — F_r = μW = 0.10(4916) = 491.6 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 815 − 491.6 = 323.4 lb
Acceleration — a = ΣF/m = 323.4/152.7 = 2.118 ft/s²

Answer: a ≈ 2.12 ft/s²

Why the other options are there

0.066 ft/s² (weight used as mass)
5.34 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Newton's second law for a particle is

Example 8

Acceleration from Newton's second law — Newton's second law for a particle is (8)

A 1746 lb cart is pulled horizontally by a 908 lb force against rolling resistance μ = 0.30. What is its acceleration?

Given

W = 1746 lb
F = 908 lb
μ = 0.30
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 1746/32.2 = 54.22 s l ug s$
Resistance — F_r = μW = 0.30(1746) = 523.8 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 908 − 523.8 = 384.2 lb
Acceleration — a = ΣF/m = 384.2/54.22 = 7.085 ft/s²

Answer: a ≈ 7.09 ft/s²

Why the other options are there

0.220 ft/s² (weight used as mass)
16.75 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Newton's second law for a particle is

Example 9

Acceleration from Newton's second law — Newton's second law for a particle is (9)

A 5578 lb cart is pulled horizontally by a 922 lb force against rolling resistance μ = 0.10. What is its acceleration?

Given

W = 5578 lb
F = 922 lb
μ = 0.10
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 5578/32.2 = 173.2 s l ug s$
Resistance — F_r = μW = 0.10(5578) = 557.8 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 922 − 557.8 = 364.2 lb
Acceleration — a = ΣF/m = 364.2/173.2 = 2.102 ft/s²

Answer: a ≈ 2.10 ft/s²

Why the other options are there

0.065 ft/s² (weight used as mass)
5.32 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Newton's second law for a particle is

Example 10

Acceleration from Newton's second law — Newton's second law for a particle is (10)

A 4879 lb cart is pulled horizontally by a 580 lb force against rolling resistance μ = 0.20. What is its acceleration?

Given

W = 4879 lb
F = 580 lb
μ = 0.20
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 4879/32.2 = 151.5 s l ug s$
Resistance — F_r = μW = 0.20(4879) = 975.8 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 580 − 975.8 = -395.8 lb
Acceleration — a = ΣF/m = -395.8/151.5 = -2.612 ft/s²

Answer: a ≈ -2.61 ft/s²

Why the other options are there

-0.081 ft/s² (weight used as mass)
3.83 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Newton's second law for a particle is

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a particle or rigid body moving under known forces, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Newton's second law for a particle is contains 13 relations; you must be able to find this page in under 15 seconds.
Exam style: kinematics first, then a work-energy or impulse-momentum shortcut.
Unit rule: g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.