Work U is defined as

Dynamics · FE Reference Handbook section

Dynamics

6 formulas

10 exam-style examples

~57 min

All Dynamics lectures

Learning objectives

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

This chapter section covers Work U is defined as 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 work u is defined as describes physically and when it applies.
State every one of the 6 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. Work U is defined as 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: work u is defined as.
Capstone Studio instructional photograph

Dynamics — Work U is defined as: 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 6 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

U	Quantity produced by "U = ∫F • dr" — read its definition and unit from the handbook line directly above the equation.
Spring Us	Quantity produced by "Spring Us = −  k s 2 − k s 1" — read its definition and unit from the handbook line directly above the equation.
Couple moment U M	Quantity produced by "Couple moment U M = M" — 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.

1 2 1 2
2 2

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 — Work U is defined as

A 4844 lb vehicle travelling 36 ft/s must stop in 237 ft. What constant resisting force is required?

Given

W = 4844 lb
v₁ = 36 ft/s
d = 237 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
$m = 4844/32.2 = 150.4 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(150.4)(36²) = 97,482 ft·lb
Work–energy — F·d = KE
Substituting
$F = 97, 482/237 = 411.3 l b$

Answer: F ≈ 411.3 lb

Why the other options are there

13,244 lb (weight used as mass)
97,482 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Work U is defined as

Example 2

Work–energy: force needed to stop a mass — Work U is defined as (2)

A 3043 lb vehicle travelling 52 ft/s must stop in 117 ft. What constant resisting force is required?

Given

W = 3043 lb
v₁ = 52 ft/s
d = 117 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
$m = 3043/32.2 = 94.50 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(94.50)(52²) = 127,768 ft·lb
Work–energy — F·d = KE
Substituting
$F = 127, 768/117 = 1, 092 l b$

Answer: F ≈ 1,092 lb

Why the other options are there

35,164 lb (weight used as mass)
127,768 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Work U is defined as

Example 3

Work–energy: force needed to stop a mass — Work U is defined as (3)

A 6801 lb vehicle travelling 57 ft/s must stop in 103 ft. What constant resisting force is required?

Given

W = 6801 lb
v₁ = 57 ft/s
d = 103 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
$m = 6801/32.2 = 211.2 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(211.2)(57²) = 343,113 ft·lb
Work–energy — F·d = KE
Substituting
$F = 343, 113/103 = 3, 331 l b$

Answer: F ≈ 3,331 lb

Why the other options are there

107,264 lb (weight used as mass)
343,113 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Work U is defined as

Example 4

Work–energy: force needed to stop a mass — Work U is defined as (4)

A 6393 lb vehicle travelling 67 ft/s must stop in 203 ft. What constant resisting force is required?

Given

W = 6393 lb
v₁ = 67 ft/s
d = 203 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
$m = 6393/32.2 = 198.5 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(198.5)(67²) = 445,624 ft·lb
Work–energy — F·d = KE
Substituting
$F = 445, 624/203 = 2, 195 l b$

Answer: F ≈ 2,195 lb

Why the other options are there

70,685 lb (weight used as mass)
445,624 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Work U is defined as

Example 5

Work–energy: force needed to stop a mass — Work U is defined as (5)

A 2017 lb vehicle travelling 51 ft/s must stop in 115 ft. What constant resisting force is required?

Given

W = 2017 lb
v₁ = 51 ft/s
d = 115 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
$m = 2017/32.2 = 62.64 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(62.64)(51²) = 81,463 ft·lb
Work–energy — F·d = KE
Substituting
$F = 81, 463/115 = 708.4 l b$

Answer: F ≈ 708.4 lb

Why the other options are there

22,810 lb (weight used as mass)
81,463 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Work U is defined as

Example 6

Work–energy: force needed to stop a mass — Work U is defined as (6)

A 5369 lb vehicle travelling 50 ft/s must stop in 198 ft. What constant resisting force is required?

Given

W = 5369 lb
v₁ = 50 ft/s
d = 198 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
$m = 5369/32.2 = 166.7 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(166.7)(50²) = 208,424 ft·lb
Work–energy — F·d = KE
Substituting
$F = 208, 424/198 = 1, 053 l b$

Answer: F ≈ 1,053 lb

Why the other options are there

33,895 lb (weight used as mass)
208,424 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Work U is defined as

Example 7

Work–energy: force needed to stop a mass — Work U is defined as (7)

A 6096 lb vehicle travelling 55 ft/s must stop in 103 ft. What constant resisting force is required?

Given

W = 6096 lb
v₁ = 55 ft/s
d = 103 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
$m = 6096/32.2 = 189.3 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(189.3)(55²) = 286,342 ft·lb
Work–energy — F·d = KE
Substituting
$F = 286, 342/103 = 2, 780 l b$

Answer: F ≈ 2,780 lb

Why the other options are there

89,517 lb (weight used as mass)
286,342 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Work U is defined as

Example 8

Work–energy: force needed to stop a mass — Work U is defined as (8)

A 2848 lb vehicle travelling 52 ft/s must stop in 61 ft. What constant resisting force is required?

Given

W = 2848 lb
v₁ = 52 ft/s
d = 61 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
$m = 2848/32.2 = 88.45 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(88.45)(52²) = 119,581 ft·lb
Work–energy — F·d = KE
Substituting
$F = 119, 581/61 = 1, 960 l b$

Answer: F ≈ 1,960 lb

Why the other options are there

63,123 lb (weight used as mass)
119,581 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Work U is defined as

Example 9

Work–energy: force needed to stop a mass — Work U is defined as (9)

A 5571 lb vehicle travelling 53 ft/s must stop in 232 ft. What constant resisting force is required?

Given

W = 5571 lb
v₁ = 53 ft/s
d = 232 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
$m = 5571/32.2 = 173.0 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(173.0)(53²) = 242,996 ft·lb
Work–energy — F·d = KE
Substituting
$F = 242, 996/232 = 1, 047 l b$

Answer: F ≈ 1,047 lb

Why the other options are there

33,726 lb (weight used as mass)
242,996 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Work U is defined as

Example 10

Work–energy: force needed to stop a mass — Work U is defined as (10)

A 4116 lb vehicle travelling 47 ft/s must stop in 62 ft. What constant resisting force is required?

Given

W = 4116 lb
v₁ = 47 ft/s
d = 62 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
$m = 4116/32.2 = 127.8 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(127.8)(47²) = 141,184 ft·lb
Work–energy — F·d = KE
Substituting
$F = 141, 184/62 = 2, 277 l b$

Answer: F ≈ 2,277 lb

Why the other options are there

73,325 lb (weight used as mass)
141,184 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Work U is defined as

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

Work U is defined as contains 6 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.