Elastic Potential 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 Elastic 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 elastic potential 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. Elastic 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: elastic potential energy.
Capstone Studio instructional photograph

Dynamics — Elastic 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 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

Fs	Quantity produced by "Fs = k s" — read its definition and unit from the handbook line directly above the equation.
where s	Quantity produced by "where s = the change in length of the spring from the undeformed length of the spring." — read its definition and unit from the handbook line directly above the equation.
V2 - V1	Quantity produced by "V2 - V1 = k ` s 22 - s12 j /2" — 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.

For a linear elastic spring with modulus, stiffness, or spring constant, k, the force in the spring is
In changing the deformation in the spring from position s1 to s2, the change in the potential energy stored in the spring is

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 — Elastic Potential Energy

A 4804 lb vehicle travelling 48 ft/s must stop in 138 ft. What constant resisting force is required?

Given

W = 4804 lb
v₁ = 48 ft/s
d = 138 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
$m = 4804/32.2 = 149.2 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(149.2)(48²) = 171,870 ft·lb
Work–energy — F·d = KE
Substituting
$F = 171, 870/138 = 1, 245 l b$

Answer: F ≈ 1,245 lb

Why the other options are there

40,103 lb (weight used as mass)
171,870 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

Example 2

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

A 4537 lb vehicle travelling 46 ft/s must stop in 67 ft. What constant resisting force is required?

Given

W = 4537 lb
v₁ = 46 ft/s
d = 67 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
$m = 4537/32.2 = 140.9 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(140.9)(46²) = 149,073 ft·lb
Work–energy — F·d = KE
Substituting
$F = 149, 073/67 = 2, 225 l b$

Answer: F ≈ 2,225 lb

Why the other options are there

71,644 lb (weight used as mass)
149,073 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

Example 3

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

A 5669 lb vehicle travelling 70 ft/s must stop in 151 ft. What constant resisting force is required?

Given

W = 5669 lb
v₁ = 70 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
$m = 5669/32.2 = 176.1 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(176.1)(70²) = 431,337 ft·lb
Work–energy — F·d = KE
Substituting
$F = 431, 337/151 = 2, 857 l b$

Answer: F ≈ 2,857 lb

Why the other options are there

91,980 lb (weight used as mass)
431,337 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

Example 4

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

A 4330 lb vehicle travelling 61 ft/s must stop in 204 ft. What constant resisting force is required?

Given

W = 4330 lb
v₁ = 61 ft/s
d = 204 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
$m = 4330/32.2 = 134.5 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(134.5)(61²) = 250,185 ft·lb
Work–energy — F·d = KE
Substituting
$F = 250, 185/204 = 1, 226 l b$

Answer: F ≈ 1,226 lb

Why the other options are there

39,490 lb (weight used as mass)
250,185 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

Example 5

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

A 5058 lb vehicle travelling 60 ft/s must stop in 185 ft. What constant resisting force is required?

Given

W = 5058 lb
v₁ = 60 ft/s
d = 185 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
$m = 5058/32.2 = 157.1 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(157.1)(60²) = 282,745 ft·lb
Work–energy — F·d = KE
Substituting
$F = 282, 745/185 = 1, 528 l b$

Answer: F ≈ 1,528 lb

Why the other options are there

49,213 lb (weight used as mass)
282,745 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

Example 6

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

A 4936 lb vehicle travelling 60 ft/s must stop in 240 ft. What constant resisting force is required?

Given

W = 4936 lb
v₁ = 60 ft/s
d = 240 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
$m = 4936/32.2 = 153.3 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(153.3)(60²) = 275,925 ft·lb
Work–energy — F·d = KE
Substituting
$F = 275, 925/240 = 1, 150 l b$

Answer: F ≈ 1,150 lb

Why the other options are there

37,020 lb (weight used as mass)
275,925 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

Example 7

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

A 5863 lb vehicle travelling 39 ft/s must stop in 72 ft. What constant resisting force is required?

Given

W = 5863 lb
v₁ = 39 ft/s
d = 72 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
$m = 5863/32.2 = 182.1 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(182.1)(39²) = 138,472 ft·lb
Work–energy — F·d = KE
Substituting
$F = 138, 472/72 = 1, 923 l b$

Answer: F ≈ 1,923 lb

Why the other options are there

61,928 lb (weight used as mass)
138,472 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

Example 8

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

A 4526 lb vehicle travelling 61 ft/s must stop in 79 ft. What constant resisting force is required?

Given

W = 4526 lb
v₁ = 61 ft/s
d = 79 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
$m = 4526/32.2 = 140.6 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(140.6)(61²) = 261,510 ft·lb
Work–energy — F·d = KE
Substituting
$F = 261, 510/79 = 3, 310 l b$

Answer: F ≈ 3,310 lb

Why the other options are there

106,590 lb (weight used as mass)
261,510 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

Example 9

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

A 6155 lb vehicle travelling 54 ft/s must stop in 110 ft. What constant resisting force is required?

Given

W = 6155 lb
v₁ = 54 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
$m = 6155/32.2 = 191.1 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(191.1)(54²) = 278,695 ft·lb
Work–energy — F·d = KE
Substituting
$F = 278, 695/110 = 2, 534 l b$

Answer: F ≈ 2,534 lb

Why the other options are there

81,582 lb (weight used as mass)
278,695 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Elastic Potential Energy

Example 10

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

A 3057 lb vehicle travelling 67 ft/s must stop in 185 ft. What constant resisting force is required?

Given

W = 3057 lb
v₁ = 67 ft/s
d = 185 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
$m = 3057/32.2 = 94.94 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(94.94)(67²) = 213,088 ft·lb
Work–energy — F·d = KE
Substituting
$F = 213, 088/185 = 1, 152 l b$

Answer: F ≈ 1,152 lb

Why the other options are there

37,089 lb (weight used as mass)
213,088 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Elastic 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

Elastic Potential 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.