Potential Energy

Dynamics · FE Reference Handbook section

Dynamics

1 formulas

10 exam-style examples

~47 min

All Dynamics lectures

Learning objectives

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

This chapter section covers Potential 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 potential energy describes physically and when it applies.
State every one of the 1 relation 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 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.
Capstone Studio instructional photograph

Dynamics — Potential 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 1 relation 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

V	Quantity produced by "V = Vg + Ve , where Vg = Wy , Ve = 1/2 ks2" — 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.

The work done by an external agent in the presence of a conservative field is termed the change in potential energy.

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

A 5300 lb vehicle travelling 49 ft/s must stop in 225 ft. What constant resisting force is required?

Given

W = 5300 lb
v₁ = 49 ft/s
d = 225 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
$m = 5300/32.2 = 164.6 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(164.6)(49²) = 197,598 ft·lb
Work–energy — F·d = KE
Substituting
$F = 197, 598/225 = 878.2 l b$

Answer: F ≈ 878.2 lb

Why the other options are there

28,278 lb (weight used as mass)
197,598 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy

Example 2

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

A 5789 lb vehicle travelling 48 ft/s must stop in 71 ft. What constant resisting force is required?

Given

W = 5789 lb
v₁ = 48 ft/s
d = 71 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
$m = 5789/32.2 = 179.8 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(179.8)(48²) = 207,110 ft·lb
Work–energy — F·d = KE
Substituting
$F = 207, 110/71 = 2, 917 l b$

Answer: F ≈ 2,917 lb

Why the other options are there

93,929 lb (weight used as mass)
207,110 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy

Example 3

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

A 4842 lb vehicle travelling 44 ft/s must stop in 167 ft. What constant resisting force is required?

Given

W = 4842 lb
v₁ = 44 ft/s
d = 167 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
$m = 4842/32.2 = 150.4 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(150.4)(44²) = 145,561 ft·lb
Work–energy — F·d = KE
Substituting
$F = 145, 561/167 = 871.6 l b$

Answer: F ≈ 871.6 lb

Why the other options are there

28,066 lb (weight used as mass)
145,561 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy

Example 4

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

A 6260 lb vehicle travelling 37 ft/s must stop in 117 ft. What constant resisting force is required?

Given

W = 6260 lb
v₁ = 37 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
$m = 6260/32.2 = 194.4 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(194.4)(37²) = 133,074 ft·lb
Work–energy — F·d = KE
Substituting
$F = 133, 074/117 = 1, 137 l b$

Answer: F ≈ 1,137 lb

Why the other options are there

36,624 lb (weight used as mass)
133,074 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy

Example 5

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

A 3244 lb vehicle travelling 55 ft/s must stop in 249 ft. What constant resisting force is required?

Given

W = 3244 lb
v₁ = 55 ft/s
d = 249 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
$m = 3244/32.2 = 100.7 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(100.7)(55²) = 152,377 ft·lb
Work–energy — F·d = KE
Substituting
$F = 152, 377/249 = 612.0 l b$

Answer: F ≈ 612.0 lb

Why the other options are there

19,705 lb (weight used as mass)
152,377 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy

Example 6

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

A 7432 lb vehicle travelling 44 ft/s must stop in 78 ft. What constant resisting force is required?

Given

W = 7432 lb
v₁ = 44 ft/s
d = 78 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
$m = 7432/32.2 = 230.8 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(230.8)(44²) = 223,422 ft·lb
Work–energy — F·d = KE
Substituting
$F = 223, 422/78 = 2, 864 l b$

Answer: F ≈ 2,864 lb

Why the other options are there

92,233 lb (weight used as mass)
223,422 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy

Example 7

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

A 6559 lb vehicle travelling 69 ft/s must stop in 120 ft. What constant resisting force is required?

Given

W = 6559 lb
v₁ = 69 ft/s
d = 120 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
$m = 6559/32.2 = 203.7 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(203.7)(69²) = 484,898 ft·lb
Work–energy — F·d = KE
Substituting
$F = 484, 898/120 = 4, 041 l b$

Answer: F ≈ 4,041 lb

Why the other options are there

130,114 lb (weight used as mass)
484,898 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy

Example 8

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

A 3385 lb vehicle travelling 65 ft/s must stop in 89 ft. What constant resisting force is required?

Given

W = 3385 lb
v₁ = 65 ft/s
d = 89 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
$m = 3385/32.2 = 105.1 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(105.1)(65²) = 222,075 ft·lb
Work–energy — F·d = KE
Substituting
$F = 222, 075/89 = 2, 495 l b$

Answer: F ≈ 2,495 lb

Why the other options are there

80,346 lb (weight used as mass)
222,075 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy

Example 9

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

A 5185 lb vehicle travelling 61 ft/s must stop in 200 ft. What constant resisting force is required?

Given

W = 5185 lb
v₁ = 61 ft/s
d = 200 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
$m = 5185/32.2 = 161.0 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(161.0)(61²) = 299,587 ft·lb
Work–energy — F·d = KE
Substituting
$F = 299, 587/200 = 1, 498 l b$

Answer: F ≈ 1,498 lb

Why the other options are there

48,233 lb (weight used as mass)
299,587 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential Energy

Example 10

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

A 7349 lb vehicle travelling 39 ft/s must stop in 220 ft. What constant resisting force is required?

Given

W = 7349 lb
v₁ = 39 ft/s
d = 220 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
$m = 7349/32.2 = 228.2 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(228.2)(39²) = 173,569 ft·lb
Work–energy — F·d = KE
Substituting
$F = 173, 569/220 = 788.9 l b$

Answer: F ≈ 788.9 lb

Why the other options are there

25,404 lb (weight used as mass)
173,569 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Potential 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

Potential Energy contains 1 relation; 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.