Power and Efficiency

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 Power and Efficiency 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 power and efficiency 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. Power and Efficiency 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: power and efficiency.
Capstone Studio instructional photograph

Dynamics — Power and Efficiency: 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

P	Quantity produced by "P= = F• v ε= = out" — 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.

dU Pout U
dt Pin U in
Adapted from Hibbeler, R.C., Engineering Mechanics, 10th ed., Prentice Hall, 2003.

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 — Power and Efficiency

A 5079 lb vehicle travelling 70 ft/s must stop in 247 ft. What constant resisting force is required?

Given

W = 5079 lb
v₁ = 70 ft/s
d = 247 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
$m = 5079/32.2 = 157.7 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(157.7)(70²) = 386,446 ft·lb
Work–energy — F·d = KE
Substituting
$F = 386, 446/247 = 1, 565 l b$

Answer: F ≈ 1,565 lb

Why the other options are there

50,379 lb (weight used as mass)
386,446 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

Example 2

Work–energy: force needed to stop a mass — Power and Efficiency (2)

A 3169 lb vehicle travelling 60 ft/s must stop in 245 ft. What constant resisting force is required?

Given

W = 3169 lb
v₁ = 60 ft/s
d = 245 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
$m = 3169/32.2 = 98.42 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(98.42)(60²) = 177,149 ft·lb
Work–energy — F·d = KE
Substituting
$F = 177, 149/245 = 723.1 l b$

Answer: F ≈ 723.1 lb

Why the other options are there

23,282 lb (weight used as mass)
177,149 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

Example 3

Work–energy: force needed to stop a mass — Power and Efficiency (3)

A 6283 lb vehicle travelling 48 ft/s must stop in 121 ft. What constant resisting force is required?

Given

W = 6283 lb
v₁ = 48 ft/s
d = 121 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
$m = 6283/32.2 = 195.1 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(195.1)(48²) = 224,783 ft·lb
Work–energy — F·d = KE
Substituting
$F = 224, 783/121 = 1, 858 l b$

Answer: F ≈ 1,858 lb

Why the other options are there

59,818 lb (weight used as mass)
224,783 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

Example 4

Work–energy: force needed to stop a mass — Power and Efficiency (4)

A 3535 lb vehicle travelling 65 ft/s must stop in 214 ft. What constant resisting force is required?

Given

W = 3535 lb
v₁ = 65 ft/s
d = 214 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
$m = 3535/32.2 = 109.8 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(109.8)(65²) = 231,916 ft·lb
Work–energy — F·d = KE
Substituting
$F = 231, 916/214 = 1, 084 l b$

Answer: F ≈ 1,084 lb

Why the other options are there

34,896 lb (weight used as mass)
231,916 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

Example 5

Work–energy: force needed to stop a mass — Power and Efficiency (5)

A 3564 lb vehicle travelling 53 ft/s must stop in 179 ft. What constant resisting force is required?

Given

W = 3564 lb
v₁ = 53 ft/s
d = 179 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
$m = 3564/32.2 = 110.7 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(110.7)(53²) = 155,455 ft·lb
Work–energy — F·d = KE
Substituting
$F = 155, 455/179 = 868.5 l b$

Answer: F ≈ 868.5 lb

Why the other options are there

27,964 lb (weight used as mass)
155,455 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

Example 6

Work–energy: force needed to stop a mass — Power and Efficiency (6)

A 2645 lb vehicle travelling 47 ft/s must stop in 177 ft. What constant resisting force is required?

Given

W = 2645 lb
v₁ = 47 ft/s
d = 177 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
$m = 2645/32.2 = 82.14 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(82.14)(47²) = 90,727 ft·lb
Work–energy — F·d = KE
Substituting
$F = 90, 727/177 = 512.6 l b$

Answer: F ≈ 512.6 lb

Why the other options are there

16,505 lb (weight used as mass)
90,727 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

Example 7

Work–energy: force needed to stop a mass — Power and Efficiency (7)

A 5576 lb vehicle travelling 45 ft/s must stop in 125 ft. What constant resisting force is required?

Given

W = 5576 lb
v₁ = 45 ft/s
d = 125 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
$m = 5576/32.2 = 173.2 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(173.2)(45²) = 175,332 ft·lb
Work–energy — F·d = KE
Substituting
$F = 175, 332/125 = 1, 403 l b$

Answer: F ≈ 1,403 lb

Why the other options are there

45,166 lb (weight used as mass)
175,332 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

Example 8

Work–energy: force needed to stop a mass — Power and Efficiency (8)

A 3906 lb vehicle travelling 47 ft/s must stop in 205 ft. What constant resisting force is required?

Given

W = 3906 lb
v₁ = 47 ft/s
d = 205 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
$m = 3906/32.2 = 121.3 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(121.3)(47²) = 133,981 ft·lb
Work–energy — F·d = KE
Substituting
$F = 133, 981/205 = 653.6 l b$

Answer: F ≈ 653.6 lb

Why the other options are there

21,045 lb (weight used as mass)
133,981 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

Example 9

Work–energy: force needed to stop a mass — Power and Efficiency (9)

A 6985 lb vehicle travelling 34 ft/s must stop in 105 ft. What constant resisting force is required?

Given

W = 6985 lb
v₁ = 34 ft/s
d = 105 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
$m = 6985/32.2 = 216.9 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(216.9)(34²) = 125,383 ft·lb
Work–energy — F·d = KE
Substituting
$F = 125, 383/105 = 1, 194 l b$

Answer: F ≈ 1,194 lb

Why the other options are there

38,451 lb (weight used as mass)
125,383 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

Example 10

Work–energy: force needed to stop a mass — Power and Efficiency (10)

A 4867 lb vehicle travelling 60 ft/s must stop in 210 ft. What constant resisting force is required?

Given

W = 4867 lb
v₁ = 60 ft/s
d = 210 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
$m = 4867/32.2 = 151.1 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(151.1)(60²) = 272,068 ft·lb
Work–energy — F·d = KE
Substituting
$F = 272, 068/210 = 1, 296 l b$

Answer: F ≈ 1,296 lb

Why the other options are there

41,717 lb (weight used as mass)
272,068 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Power and Efficiency

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

Power and Efficiency 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.