Principle of Work and Energy

Dynamics · FE Reference Handbook section

Dynamics

3 formulas

10 exam-style examples

~51 min

All Dynamics lectures

Learning objectives

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

This chapter section covers Principle of Work and Energy 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 principle of work and energy describes physically and when it applies.
State every one of the 3 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. Principle of Work and Energy 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: principle of work and energy.
Capstone Studio instructional photograph

Dynamics — Principle of Work and Energy: 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 3 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

T2 + V2	Quantity produced by "T2 + V2 = T1 + V1" — read its definition and unit from the handbook line directly above the equation.
U1→2	Quantity produced by "U1→2 = the work done by the nonconservative forces in moving between state 1 and state 2. Care must be exercised during" — 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.

If Ti and Vi are, respectively, the kinetic and potential energy of a particle at state i, then for conservative systems (no energy
dissipation or gain), the law of conservation of energy is
If nonconservative forces are present, then the work done by these forces must be accounted for. Hence
computations to correctly compute the algebraic sign of the work term. If the forces serve to increase the energy of the system,
U1→2 is positive. If the forces, such as friction, serve to dissipate energy, U1→2 is negative.

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

Work–energy: force needed to stop a mass — Principle of Work and Energy

A 6782 lb vehicle travelling 48 ft/s must stop in 64 ft. What constant resisting force is required?

Given

W = 6782 lb
v₁ = 48 ft/s
d = 64 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 6782/32.2 = 210.6 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(210.6)(48²) = 242,636 ft·lb
Work–energy — F·d = KE
Substituting
$F = 242, 636/64 = 3, 791 l b$

Answer: F ≈ 3,791 lb

Why the other options are there

122,076 lb (weight used as mass)
242,636 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 2

Work–energy: force needed to stop a mass — Principle of Work and Energy (2)

A 6840 lb vehicle travelling 67 ft/s must stop in 111 ft. What constant resisting force is required?

Given

W = 6840 lb
v₁ = 67 ft/s
d = 111 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 6840/32.2 = 212.4 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(212.4)(67²) = 476,782 ft·lb
Work–energy — F·d = KE
Substituting
$F = 476, 782/111 = 4, 295 l b$

Answer: F ≈ 4,295 lb

Why the other options are there

138,310 lb (weight used as mass)
476,782 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 3

Work–energy: force needed to stop a mass — Principle of Work and Energy (3)

A 2804 lb vehicle travelling 52 ft/s must stop in 202 ft. What constant resisting force is required?

Given

W = 2804 lb
v₁ = 52 ft/s
d = 202 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 2804/32.2 = 87.08 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(87.08)(52²) = 117,733 ft·lb
Work–energy — F·d = KE
Substituting
$F = 117, 733/202 = 582.8 l b$

Answer: F ≈ 582.8 lb

Why the other options are there

18,767 lb (weight used as mass)
117,733 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 4

Work–energy: force needed to stop a mass — Principle of Work and Energy (4)

A 4101 lb vehicle travelling 45 ft/s must stop in 230 ft. What constant resisting force is required?

Given

W = 4101 lb
v₁ = 45 ft/s
d = 230 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 4101/32.2 = 127.4 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(127.4)(45²) = 128,952 ft·lb
Work–energy — F·d = KE
Substituting
$F = 128, 952/230 = 560.7 l b$

Answer: F ≈ 560.7 lb

Why the other options are there

18,053 lb (weight used as mass)
128,952 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 5

Work–energy: force needed to stop a mass — Principle of Work and Energy (5)

A 4992 lb vehicle travelling 39 ft/s must stop in 246 ft. What constant resisting force is required?

Given

W = 4992 lb
v₁ = 39 ft/s
d = 246 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 4992/32.2 = 155.0 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(155.0)(39²) = 117,901 ft·lb
Work–energy — F·d = KE
Substituting
$F = 117, 901/246 = 479.3 l b$

Answer: F ≈ 479.3 lb

Why the other options are there

15,433 lb (weight used as mass)
117,901 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 6

Work–energy: force needed to stop a mass — Principle of Work and Energy (6)

A 4243 lb vehicle travelling 30 ft/s must stop in 130 ft. What constant resisting force is required?

Given

W = 4243 lb
v₁ = 30 ft/s
d = 130 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 4243/32.2 = 131.8 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(131.8)(30²) = 59,297 ft·lb
Work–energy — F·d = KE
Substituting
$F = 59, 297/130 = 456.1 l b$

Answer: F ≈ 456.1 lb

Why the other options are there

14,687 lb (weight used as mass)
59,297 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 7

Work–energy: force needed to stop a mass — Principle of Work and Energy (7)

A 5364 lb vehicle travelling 36 ft/s must stop in 63 ft. What constant resisting force is required?

Given

W = 5364 lb
v₁ = 36 ft/s
d = 63 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 5364/32.2 = 166.6 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(166.6)(36²) = 107,946 ft·lb
Work–energy — F·d = KE
Substituting
$F = 107, 946/63 = 1, 713 l b$

Answer: F ≈ 1,713 lb

Why the other options are there

55,173 lb (weight used as mass)
107,946 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 8

Work–energy: force needed to stop a mass — Principle of Work and Energy (8)

A 7087 lb vehicle travelling 49 ft/s must stop in 236 ft. What constant resisting force is required?

Given

W = 7087 lb
v₁ = 49 ft/s
d = 236 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 7087/32.2 = 220.1 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(220.1)(49²) = 264,222 ft·lb
Work–energy — F·d = KE
Substituting
$F = 264, 222/236 = 1, 120 l b$

Answer: F ≈ 1,120 lb

Why the other options are there

36,051 lb (weight used as mass)
264,222 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 9

Work–energy: force needed to stop a mass — Principle of Work and Energy (9)

A 3066 lb vehicle travelling 60 ft/s must stop in 112 ft. What constant resisting force is required?

Given

W = 3066 lb
v₁ = 60 ft/s
d = 112 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 3066/32.2 = 95.22 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(95.22)(60²) = 171,391 ft·lb
Work–energy — F·d = KE
Substituting
$F = 171, 391/112 = 1, 530 l b$

Answer: F ≈ 1,530 lb

Why the other options are there

49,275 lb (weight used as mass)
171,391 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

Example 10

Work–energy: force needed to stop a mass — Principle of Work and Energy (10)

A 2320 lb vehicle travelling 64 ft/s must stop in 156 ft. What constant resisting force is required?

Given

W = 2320 lb
v₁ = 64 ft/s
d = 156 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 2320/32.2 = 72.05 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(72.05)(64²) = 147,558 ft·lb
Work–energy — F·d = KE
Substituting
$F = 147, 558/156 = 945.9 l b$

Answer: F ≈ 945.9 lb

Why the other options are there

30,457 lb (weight used as mass)
147,558 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Principle of Work and Energy

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

Principle of Work and Energy contains 3 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.