Kinetic Energy

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 Kinetic 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 kinetic energy 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. Kinetic 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: kinetic energy.
Capstone Studio instructional photograph

Dynamics — Kinetic 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 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

T	Quantity produced by "T= mv" — 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.

Particle 2
Rigid Body
1 2 1
2 2
subscript c represents the center of mass

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

Stopping distance from kinematics

A truck travelling at 60 mph decelerates at 11.2 ft/s². What distance does it need to stop?

Given

v₀ = 60 mph
a = −11.2 ft/s²
v = 0

Find

Stopping distance d

Start with the thinking

Convert mph to ft/s before anything else: 60 mph = 88 ft/s.
Use the velocity-position relation to avoid solving for time.

Step-by-step solution

Convert
$v_{0} = 60 (1.467) = 88.0 f t / s$
Kinematics
$v^{2} = v_{0}^{2} + 2 a d$
Set v = 0
$0 = 88. 0^{2} - 2 (11.2) d$
Rearrange
$d = 7, 744/22.4$
Result
$d = 346 f t$

Answer: d ≈ 346 ft

Why the other options are there

160 ft (mph used directly)
692 ft (factor of 2 dropped)

Reference: FE Reference Handbook — Dynamics — Rectilinear motion

Example 2

Kinetic energy delivered by a drop hammer

A 3,500 lb pile hammer falls freely 4.0 ft. What is its kinetic energy and impact velocity?

Given

W = 3,500 lb
h = 4.0 ft
g = 32.2 ft/s²

Find

KE at impact and velocity

Start with the thinking

Energy conservation: potential energy converts fully to kinetic in free fall.
Mass in slugs equals W/g.

Step-by-step solution

Energy — KE = W h = 3,500(4.0) = 14,000 ft·lb
Mass
$m = W / g = 3, 500/32.2 = 108.7 s l ug$
Velocity relation
$K E = ½ m v^{2}$
Rearrange
$v = (2 K E / m) = (28, 000/108.7)$
Result
$v = 16.0 f t / s$

Answer: KE = 14,000 ft·lb, v = 16.0 ft/s

Why the other options are there

v = 8.0 ft/s (factor of 2 omitted)
KE = 435 ft·lb (weight divided by g twice)

Reference: FE Reference Handbook — Dynamics — Work and energy

Example 3

Work–energy: force needed to stop a mass — Kinetic Energy

A 3852 lb vehicle travelling 34 ft/s must stop in 101 ft. What constant resisting force is required?

Given

W = 3852 lb
v₁ = 34 ft/s
d = 101 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
$m = 3852/32.2 = 119.6 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(119.6)(34²) = 69,145 ft·lb
Work–energy — F·d = KE
Substituting
$F = 69, 145/101 = 684.6 l b$

Answer: F ≈ 684.6 lb

Why the other options are there

22,044 lb (weight used as mass)
69,145 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

Example 4

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

A 7061 lb vehicle travelling 69 ft/s must stop in 106 ft. What constant resisting force is required?

Given

W = 7061 lb
v₁ = 69 ft/s
d = 106 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
$m = 7061/32.2 = 219.3 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(219.3)(69²) = 522,010 ft·lb
Work–energy — F·d = KE
Substituting
$F = 522, 010/106 = 4, 925 l b$

Answer: F ≈ 4,925 lb

Why the other options are there

158,573 lb (weight used as mass)
522,010 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

Example 5

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

A 7854 lb vehicle travelling 47 ft/s must stop in 156 ft. What constant resisting force is required?

Given

W = 7854 lb
v₁ = 47 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
$m = 7854/32.2 = 243.9 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(243.9)(47²) = 269,402 ft·lb
Work–energy — F·d = KE
Substituting
$F = 269, 402/156 = 1, 727 l b$

Answer: F ≈ 1,727 lb

Why the other options are there

55,607 lb (weight used as mass)
269,402 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

Example 6

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

A 7390 lb vehicle travelling 64 ft/s must stop in 193 ft. What constant resisting force is required?

Given

W = 7390 lb
v₁ = 64 ft/s
d = 193 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
$m = 7390/32.2 = 229.5 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(229.5)(64²) = 470,022 ft·lb
Work–energy — F·d = KE
Substituting
$F = 470, 022/193 = 2, 435 l b$

Answer: F ≈ 2,435 lb

Why the other options are there

78,418 lb (weight used as mass)
470,022 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

Example 7

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

A 4477 lb vehicle travelling 64 ft/s must stop in 227 ft. What constant resisting force is required?

Given

W = 4477 lb
v₁ = 64 ft/s
d = 227 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
$m = 4477/32.2 = 139.0 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(139.0)(64²) = 284,748 ft·lb
Work–energy — F·d = KE
Substituting
$F = 284, 748/227 = 1, 254 l b$

Answer: F ≈ 1,254 lb

Why the other options are there

40,392 lb (weight used as mass)
284,748 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

Example 8

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

A 7457 lb vehicle travelling 42 ft/s must stop in 107 ft. What constant resisting force is required?

Given

W = 7457 lb
v₁ = 42 ft/s
d = 107 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
$m = 7457/32.2 = 231.6 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(231.6)(42²) = 204,257 ft·lb
Work–energy — F·d = KE
Substituting
$F = 204, 257/107 = 1, 909 l b$

Answer: F ≈ 1,909 lb

Why the other options are there

61,468 lb (weight used as mass)
204,257 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

Example 9

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

A 2797 lb vehicle travelling 53 ft/s must stop in 60 ft. What constant resisting force is required?

Given

W = 2797 lb
v₁ = 53 ft/s
d = 60 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
$m = 2797/32.2 = 86.86 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(86.86)(53²) = 122,000 ft·lb
Work–energy — F·d = KE
Substituting
$F = 122, 000/60 = 2, 033 l b$

Answer: F ≈ 2,033 lb

Why the other options are there

65,473 lb (weight used as mass)
122,000 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetic Energy

Example 10

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

A 7299 lb vehicle travelling 70 ft/s must stop in 91 ft. What constant resisting force is required?

Given

W = 7299 lb
v₁ = 70 ft/s
d = 91 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
$m = 7299/32.2 = 226.7 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(226.7)(70²) = 555,359 ft·lb
Work–energy — F·d = KE
Substituting
$F = 555, 359/91 = 6, 103 l b$

Answer: F ≈ 6,103 lb

Why the other options are there

196,512 lb (weight used as mass)
555,359 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetic 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

Kinetic Energy 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.

Kinetic Energy

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

T=mv

(PlaneMotion)T=mvc+Icω2

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section