Potential Energy in Gravity Field

Dynamics · FE Reference Handbook section

Dynamics

2 formulas

10 exam-style examples

~49 min

All Dynamics lectures

Learning objectives

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

This chapter section covers Potential Energy in Gravity Field 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 potential energy in gravity field describes physically and when it applies.
State every one of the 2 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. Potential Energy in Gravity Field 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: potential energy in gravity field.
Capstone Studio instructional photograph

Dynamics — Potential Energy in Gravity Field: 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 2 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

Vg	Quantity produced by "Vg = mgh" — read its definition and unit from the handbook line directly above the equation.
where h	Quantity produced by "where h = the elevation above some specified datum." — read its definition and unit from the handbook line directly above the equation.

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 — Potential Energy in Gravity Field

A 7718 lb vehicle travelling 67 ft/s must stop in 110 ft. What constant resisting force is required?

Given

W = 7718 lb
v₁ = 67 ft/s
d = 110 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
$m = 7718/32.2 = 239.7 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(239.7)(67²) = 537,983 ft·lb
Work–energy — F·d = KE
Substituting
$F = 537, 983/110 = 4, 891 l b$

Answer: F ≈ 4,891 lb

Why the other options are there

157,482 lb (weight used as mass)
537,983 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

Example 2

Work–energy: force needed to stop a mass — Potential Energy in Gravity Field (2)

A 4032 lb vehicle travelling 36 ft/s must stop in 141 ft. What constant resisting force is required?

Given

W = 4032 lb
v₁ = 36 ft/s
d = 141 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
$m = 4032/32.2 = 125.2 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(125.2)(36²) = 81,141 ft·lb
Work–energy — F·d = KE
Substituting
$F = 81, 141/141 = 575.5 l b$

Answer: F ≈ 575.5 lb

Why the other options are there

18,530 lb (weight used as mass)
81,141 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

Example 3

Work–energy: force needed to stop a mass — Potential Energy in Gravity Field (3)

A 4560 lb vehicle travelling 70 ft/s must stop in 248 ft. What constant resisting force is required?

Given

W = 4560 lb
v₁ = 70 ft/s
d = 248 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
$m = 4560/32.2 = 141.6 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(141.6)(70²) = 346,957 ft·lb
Work–energy — F·d = KE
Substituting
$F = 346, 957/248 = 1, 399 l b$

Answer: F ≈ 1,399 lb

Why the other options are there

45,048 lb (weight used as mass)
346,957 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

Example 4

Work–energy: force needed to stop a mass — Potential Energy in Gravity Field (4)

A 7763 lb vehicle travelling 49 ft/s must stop in 246 ft. What constant resisting force is required?

Given

W = 7763 lb
v₁ = 49 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
$m = 7763/32.2 = 241.1 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(241.1)(49²) = 289,425 ft·lb
Work–energy — F·d = KE
Substituting
$F = 289, 425/246 = 1, 177 l b$

Answer: F ≈ 1,177 lb

Why the other options are there

37,884 lb (weight used as mass)
289,425 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

Example 5

Work–energy: force needed to stop a mass — Potential Energy in Gravity Field (5)

A 2779 lb vehicle travelling 59 ft/s must stop in 70 ft. What constant resisting force is required?

Given

W = 2779 lb
v₁ = 59 ft/s
d = 70 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
$m = 2779/32.2 = 86.30 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(86.30)(59²) = 150,213 ft·lb
Work–energy — F·d = KE
Substituting
$F = 150, 213/70 = 2, 146 l b$

Answer: F ≈ 2,146 lb

Why the other options are there

69,098 lb (weight used as mass)
150,213 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

Example 6

Work–energy: force needed to stop a mass — Potential Energy in Gravity Field (6)

A 6699 lb vehicle travelling 66 ft/s must stop in 209 ft. What constant resisting force is required?

Given

W = 6699 lb
v₁ = 66 ft/s
d = 209 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
$m = 6699/32.2 = 208.0 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(208.0)(66²) = 453,119 ft·lb
Work–energy — F·d = KE
Substituting
$F = 453, 119/209 = 2, 168 l b$

Answer: F ≈ 2,168 lb

Why the other options are there

69,811 lb (weight used as mass)
453,119 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

Example 7

Work–energy: force needed to stop a mass — Potential Energy in Gravity Field (7)

A 3974 lb vehicle travelling 49 ft/s must stop in 151 ft. What constant resisting force is required?

Given

W = 3974 lb
v₁ = 49 ft/s
d = 151 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
$m = 3974/32.2 = 123.4 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(123.4)(49²) = 148,161 ft·lb
Work–energy — F·d = KE
Substituting
$F = 148, 161/151 = 981.2 l b$

Answer: F ≈ 981.2 lb

Why the other options are there

31,595 lb (weight used as mass)
148,161 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

Example 8

Work–energy: force needed to stop a mass — Potential Energy in Gravity Field (8)

A 4141 lb vehicle travelling 39 ft/s must stop in 154 ft. What constant resisting force is required?

Given

W = 4141 lb
v₁ = 39 ft/s
d = 154 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
$m = 4141/32.2 = 128.6 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(128.6)(39²) = 97,802 ft·lb
Work–energy — F·d = KE
Substituting
$F = 97, 802/154 = 635.1 l b$

Answer: F ≈ 635.1 lb

Why the other options are there

20,450 lb (weight used as mass)
97,802 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

Example 9

Work–energy: force needed to stop a mass — Potential Energy in Gravity Field (9)

A 6613 lb vehicle travelling 49 ft/s must stop in 189 ft. What constant resisting force is required?

Given

W = 6613 lb
v₁ = 49 ft/s
d = 189 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
$m = 6613/32.2 = 205.4 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(205.4)(49²) = 246,550 ft·lb
Work–energy — F·d = KE
Substituting
$F = 246, 550/189 = 1, 304 l b$

Answer: F ≈ 1,304 lb

Why the other options are there

42,005 lb (weight used as mass)
246,550 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

Example 10

Work–energy: force needed to stop a mass — Potential Energy in Gravity Field (10)

A 7776 lb vehicle travelling 49 ft/s must stop in 62 ft. What constant resisting force is required?

Given

W = 7776 lb
v₁ = 49 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
$m = 7776/32.2 = 241.5 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(241.5)(49²) = 289,910 ft·lb
Work–energy — F·d = KE
Substituting
$F = 289, 910/62 = 4, 676 l b$

Answer: F ≈ 4,676 lb

Why the other options are there

150,566 lb (weight used as mass)
289,910 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy in Gravity Field

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

Potential Energy in Gravity Field contains 2 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.