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Engineering Sciences

4% of exam

Basic dynamics (friction, force, mass, acceleration, momentum), work, energy, and power for particles and rigid bodies, and electricity fundamentals (charge, current, voltage, power, Ohm's law, and Kirchhoff's laws).

2 concepts

A. Basic dynamics

Newtonian Dynamics, Work, Energy and Power

Newton's second law with friction, the work-energy theorem, and impulse-momentum for particles — the three bookkeeping methods that solve almost every FE dynamics question.

Dynamics rewards the engineer who picks the right bookkeeping tool before touching a number. Three laws cover the entire FE Engineering Sciences dynamics set: Newton's second law

\sum \mathbf{F}=m\mathbf{a}

when you want an instantaneous acceleration or force, the work-energy theorem when a problem links force and distance to a change in speed, and the impulse-momentum theorem when force and time govern a change in velocity — especially across impacts. The NCEES FE Reference Handbook collects all three under Dynamics (Particle Kinetics), and the points are lost not in the physics but in the setup: a missing friction term, a velocity squared that should not be squared, or a USCS mass that forgot the

g_c

factor. This concept builds the reflex of naming the method first, drawing one clean free-body diagram, and carrying units the whole way.

Newton's second law and the free-body diagram

For a particle of constant mass, Newton's second law is the vector statement that the net force equals mass times acceleration. Everything starts with a free-body diagram: isolate the body, draw every external force (weight, normal, applied, friction, tension), choose axes, and write one scalar equation per axis. On an incline it almost always pays to align one axis with the surface so the normal direction has zero acceleration. Acceleration is shared by all forces on the body — it is not a force you add to the diagram.

\sum \mathbf{F} = m\,\mathbf{a} = m\frac{d\mathbf{v}}{dt}

Friction: the force that is an inequality

Dry (Coulomb) friction acts along the contact surface, opposing relative sliding or its tendency. The handbook states it as

F \le \mu_s N

: below impending motion the friction force is whatever static value the other equations require, capped at

\mu_s N

; once sliding begins it takes the fixed kinetic value

F = \mu_k N

, usually with

\mu_k < \mu_s

. The normal force

N

is whatever the surface actually pushes with — on a flat floor

N = mg

, but on an incline

N = mg\cos\theta

, and any vertical component of an applied force changes it. Friction is never

\mu m g

by reflex; it is

\mu N

with

N

from the perpendicular equilibrium.

F_{f} = \mu_k N \quad (\text{sliding}), \qquad F_{f} \le \mu_s N \quad (\text{static})

The mass-force relation and the gc factor

In SI,

\sum F = ma

with force in newtons, mass in kilograms, and acceleration in

\text{m/s}^2

— no conversion constant appears. USCS is the trap: with mass in

\text{lbm}

, force in

\text{lbf}

, and acceleration in

\text{ft/s}^2

, the handbook writes

F = ma/g_c

where

g_c = 32.174\ \text{lbm}\cdot\text{ft}/(\text{lbf}\cdot\text{s}^2)

. The constant exists only to reconcile the pound-mass and pound-force, which were defined so that one pound-mass weighs one pound-force at standard gravity. A body of

m

lbm weighs

W = mg/g_c

lbf; at standard

g

that is numerically

W = m

, which is exactly why people forget

g_c

when an acceleration is not

g

.

F = \frac{m\,a}{g_c}, \qquad g_c = 32.174\ \frac{\text{lbm}\cdot\text{ft}}{\text{lbf}\cdot\text{s}^2}

Work, kinetic energy, and the work-energy theorem

Work is the line integral of force along the path,

U = \int \mathbf{F}\cdot d\mathbf{r}

; for a constant force at angle

\theta

to the displacement it is

U = Fs\cos\theta

. The principle of work and energy states that the net work done on a particle equals its change in kinetic energy,

T = \tfrac{1}{2}mv^2

. This is the fastest route whenever a problem gives forces and distances and asks for a speed, because time never enters. Forces perpendicular to motion (a normal force, the tension on a body moving along a level path) do no work, which is what makes the method clean.

U_{1\to 2} = T_2 - T_1 = \tfrac{1}{2}mv_2^{2} - \tfrac{1}{2}mv_1^{2}

Potential energy and conservation

When only conservative forces act — gravity and ideal springs — mechanical energy is conserved:

T_2 + V_2 = T_1 + V_1

. Gravitational potential energy is

V_g = mgh

relative to a chosen datum, and a linear spring stores

V_e = \\tfrac{1}{2}ks^2

where

s

is the deflection from the free length. When a nonconservative force such as friction is present, add its work as a separate term,

T_2 + V_2 = T_1 + V_1 + U_{1\\to 2}^{nc}

, where

U_{1\\to 2}^{nc}

is the work of the nonconservative forces only (negative for a dissipative force) and explicitly excludes the gravity and spring work already carried in

V

. Note this differs from the work-energy theorem, where

U_{1\\to 2}

is the net work of ALL forces; mixing the two double-counts gravity or spring work. Choosing the right datum and a consistent sign convention removes most arithmetic errors.

T_1 + V_1 + U_{1\\to 2}^{nc} = T_2 + V_2, \\qquad V = mgh + \\tfrac{1}{2}ks^2

Impulse, momentum, and impact

Linear momentum is

\mathbf{p} = m\mathbf{v}

, and integrating Newton's law over time gives the impulse-momentum theorem: the impulse of the net force equals the change in momentum. For a collision with no external impulse, total momentum is conserved,

m_1\mathbf{v}_1 + m_2\mathbf{v}_2 = m_1\mathbf{v}_1' + m_2\mathbf{v}_2'

. Kinetic energy is conserved only in a perfectly elastic impact (

e=1

); a perfectly plastic impact (

e=0

) has the bodies move off together and dissipates the most energy. The coefficient of restitution

e

relates the relative separation and approach speeds along the line of impact.

m\mathbf{v}_2 = m\mathbf{v}_1 + \int_{t_1}^{t_2}\mathbf{F}\,dt, \qquad e = \frac{(v_2'-v_1')_n}{(v_1-v_2)_n}

Power and efficiency

Power is the time rate of doing work,

P = dU/dt

, and for a force moving a body at velocity

v

it is the dot product

P = \mathbf{F}\cdot\mathbf{v}

. Mechanical efficiency is the ratio of useful output power to input power,

\eta = P_{out}/P_{in}

. These appear in FE problems as pump, motor, and hoist questions — watch the unit families (

1\ \text{hp} = 550\ \text{ft·lbf/s} = 746\ \text{W}

) and keep force and velocity collinear before multiplying.

P = \frac{dU}{dt} = \mathbf{F}\cdot\mathbf{v}, \qquad \eta = \frac{P_{out}}{P_{in}}

Exam strategy

Name the method before you compute. If the question pairs force with distance and asks for speed, use work-energy and avoid solving for acceleration. If it pairs force with time, or involves a collision, use impulse-momentum or conservation of momentum. If it asks for an instantaneous acceleration or a force at one instant, use

\sum F = ma

with a free-body diagram. Always resolve forces on the incline before computing

N

, write friction as

\mu N

(never

\mu mg

blindly), and in USCS insert

g_c

the moment you see lbm with a non-

g

acceleration. A quick units check on the final number — N, J, or W — catches most slips.

Key equations

Newton's second law (particle)

\sum \mathbf{F} = m\mathbf{a} = m\dfrac{d\mathbf{v}}{dt}

Net external force equals mass times acceleration. SI: $F$ in N, $m$ in kg, $a$ in m/s $^2$ . Apply per axis from a free-body diagram.

Newton's law in USCS (gc form)

F = \dfrac{m\,a}{g_c}, \quad g_c = 32.174\ \tfrac{\text{lbm}\cdot\text{ft}}{\text{lbf}\cdot\text{s}^2}

Coulomb friction

F_f \le \mu_s N, \qquad F_f = \mu_k N \ (\text{sliding})

Constant-acceleration kinematics

v = v_0 + at, \quad v^2 = v_0^2 + 2a(x-x_0)

Kinetic energy

T = \tfrac{1}{2}mv^{2}

Energy of motion of a particle. $v$ = speed (m/s), result in J. Note $v$

Work of a force

U = \int \mathbf{F}\cdot d\mathbf{r} = Fs\cos\theta\ (\text{constant } F)

$\theta$ = angle between force and displacement; forces perpendicular to motion do zero work.

Work-energy theorem

U_{1\to 2} = \tfrac{1}{2}mv_2^{2} - \tfrac{1}{2}mv_1^{2}

Conservation of energy

T_1 + V_1 + U_{1\\to 2}^{nc} = T_2 + V_2

Linear impulse-momentum

m\mathbf{v}_2 = m\mathbf{v}_1 + \int_{t_1}^{t_2}\mathbf{F}\,dt

Conservation of momentum (impact)

m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'

Coefficient of restitution

e = \dfrac{(v_2' - v_1')_n}{(v_1 - v_2)_n}, \ 0\le e\le 1

Power and efficiency

P = \mathbf{F}\cdot\mathbf{v}, \qquad \eta = \dfrac{P_{out}}{P_{in}}

Worked examples

Acceleration of a dragged box (Newton's second law with friction)

Problem. A $25\ \text{kg}$ box rests on a level floor with kinetic friction coefficient $\mu_k = 0.30$ . A horizontal force of $120\ \text{N}$ is applied. Find the acceleration. Use $g = 9.81\ \text{m/s}^2$ .

Solution. Vertical equilibrium gives the normal force:

N = mg = 25 \times 9.81 = 245.25\ \text{N}

. Kinetic friction opposes motion:

F_f = \mu_k N = 0.30 \times 245.25 = 73.6\ \text{N}

a = \frac{F_{ap} - \mu_k mg}{m} = \frac{120 - (0.30)(25)(9.81)}{25} = 1.86\ \text{m/s}^2

Speed at the bottom of an incline (work-energy theorem)

Problem. A $12\ \text{kg}$ block is released from rest and slides $5.0\ \text{m}$ down a $30^\circ$ incline with kinetic friction coefficient $\mu_k = 0.25$

Perfectly plastic collision (conservation of momentum)

Problem. A $2000\ \text{kg}$ truck moving at $15\ \text{m/s}$ strikes a stationary $1000\ \text{kg}$ car; the two lock together. Find the common velocity after impact and the kinetic energy lost.

Solution. No significant external impulse acts during the brief impact, so momentum is conserved. Bodies stick (perfectly plastic,

e=0

USCS force with the gc factor

Problem. A $50\ \text{lbm}$ casting on a frictionless conveyor must be accelerated at $8.0\ \text{ft/s}^2$ . What horizontal force in $\text{lbf}$ is required?

Solution. In USCS, Newton's second law carries the conversion constant

g_c

Common pitfalls

•Writing friction as $\mu m g$ on an incline. The normal force there is $N = mg\cos\theta$ , so $F_f = \mu m g\cos\theta$ ; using $\mu m g$ overestimates friction.
•Dropping the $g_c$ factor in USCS. With lbm and lbf you must use $F = ma/g_c$ ; it only 'disappears' when the acceleration happens to equal $g$
•Treating kinetic energy as linear in $v$ . $T = \tfrac{1}{2}mv^2$ — doubling speed quadruples kinetic energy; a missing square is the most common work-energy mistake.
•Using work-energy when force and time are given (or impulse-momentum when force and distance are given). Match the theorem to what the problem supplies before computing.
•Assuming kinetic energy is conserved in every collision. Momentum is always conserved (no external impulse), but KE is conserved only for a perfectly elastic impact ( $e=1$ ); plastic impacts dissipate energy.
•Forgetting that a normal force and a perpendicular tension do no work. Including them in $\int \mathbf{F}\cdot d\mathbf{r}$ corrupts the energy balance.
•Mixing power units: $1\,\text{hp} = 550\ \text{ft·lbf/s} = 746\ \text{W}$ . Multiplying a force in lbf by a velocity in ft/s gives ft·lbf/s, not hp.

References

NCEES FE Reference Handbook — Dynamics
NCEES FE Reference Handbook — Dynamics (Laws of Friction) — Laws of Friction: $F \\le \\mu N$; below impending motion $F < \\mu_s N$ (no slip), at impending motion $F = \\mu_s N$, while sliding $F = \\mu_k N$. Printed in the Dynamics section, consistent with section 1.
Hibbeler, Engineering Mechanics: Dynamics — Work-energy and impulse-momentum methods; source of the handbook's notation.

C. Electricity, current, and voltage laws

Electricity: Charge, Ohm’s and Kirchhoff’s Laws

Charge, current, voltage and power, Ohm's law, series and parallel resistance, and Kirchhoff's two laws — the toolkit for any resistive DC circuit on the FE.

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Required whenever mass is in lbm and force in lbf; $g_c$ reconciles the two. Weight $W = mg/g_c$ .

Friction opposes relative motion. $N$ = actual normal force ( $mg\cos\theta$ on an incline). $\mu_s,\mu_k$ dimensionless.

Links velocity, acceleration, time, and displacement for constant $a$ . Use with $\sum F = ma$ to bridge force and motion.

is squared — never linear.

Net work equals change in kinetic energy. Time-free — ideal for force-and-distance to speed problems.

$V = mgh + \\tfrac{1}{2}ks^2$ . $U_{1\\to 2}^{nc}$ = work of nonconservative forces only (negative for friction; zero for conservative systems). Distinct from the work-energy theorem $U_{1\\to 2}$ , which is the net work of all forces — do not equate the two.

Impulse of net force equals change in momentum. For constant $F$ : $F\,\Delta t = m\,\Delta v$ .

Holds when external impulse is negligible during impact. KE conserved only if $e=1$ .

$e=1$ perfectly elastic, $e=0$ perfectly plastic (bodies stick). Applied along the line of impact.

Power in W (or ft·lbf/s); $1\,\text{hp}=746\,\text{W}$ . Force and velocity must be collinear.

. Apply

\sum F_x = ma

along the floor:

120 - 73.6 = 25\,a

, so

a = \frac{120 - 73.575}{25} = \frac{46.43}{25} = 1.86\ \text{m/s}^2.

Sanity check: with no friction the acceleration would be

120/25 = 4.80\ \text{m/s}^2

, so friction must reduce it —

1.86\ \text{m/s}^2

is appropriately smaller. Final answer:

a = 1.86\ \text{m/s}^2

in the direction of the applied force.

. Find its speed at the bottom. Use

g = 9.81\ \text{m/s}^2

.

Solution. Forces along the

5\ \text{m}

path: gravity component

mg\sin\theta

aids motion; friction

\mu_k N = \mu_k mg\cos\theta

opposes it. The normal force is

N = mg\cos\theta

. Net work over distance

d

:

U_{1\to 2} = mgd(\sin\theta - \mu_k\cos\theta) = (12)(9.81)(5)(\sin 30^\circ - 0.25\cos 30^\circ).

U_{1\to 2} = 588.6\,(0.5 - 0.2165) = 588.6 \times 0.2835 = 167\ \text{J}.

Work-energy theorem from rest:

U_{1\to 2} = \tfrac{1}{2}mv^2

, so

v = \sqrt{\frac{2U_{1\to 2}}{m}} = \sqrt{\frac{2(166.86)}{12}} = \sqrt{27.81} = 5.27\ \text{m/s}.

Sanity check: a frictionless drop of the same vertical height

h = 5\sin 30^\circ = 2.5\ \text{m}

gives

v=\sqrt{2gh}=7.00\ \text{m/s}

; friction must lower this, and

5.27\ \text{m/s}

is correctly less. Final answer:

v = 5.27\ \text{m/s}

.

v = \sqrt{2gd(\sin\theta - \mu_k\cos\theta)} = 5.27\ \text{m/s}

):

v' = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{(2000)(15) + (1000)(0)}{3000} = 10.0\ \text{m/s}.

Kinetic energy before:

T_1 = \tfrac{1}{2}(2000)(15)^2 = 225{,}000\ \text{J}

. Kinetic energy after:

T_2 = \tfrac{1}{2}(3000)(10)^2 = 150{,}000\ \text{J}

. Energy lost:

\Delta T = 225{,}000 - 150{,}000 = 75{,}000\ \text{J} = 75.0\ \text{kJ}.

Sanity check: momentum after

= 3000 \times 10 = 30{,}000\ \text{kg·m/s}

equals momentum before

= 2000 \times 15 = 30{,}000\ \text{kg·m/s}

. Energy is not conserved in a plastic impact, so a positive loss is expected. Final answers:

v' = 10.0\ \text{m/s}

,

\Delta T = 75.0\ \text{kJ}

dissipated.

v' = \frac{m_1 v_1}{m_1+m_2}, \qquad \Delta T = T_1 - T_2 = 75.0\ \text{kJ}

:

F = \frac{m a}{g_c} = \frac{(50\ \text{lbm})(8.0\ \text{ft/s}^2)}{32.174\ \tfrac{\text{lbm}\cdot\text{ft}}{\text{lbf}\cdot\text{s}^2}} = 12.4\ \text{lbf}.

The

\text{lbm}\cdot\text{ft/s}^2

in the numerator cancels against

g_c

to leave

\text{lbf}

. Sanity check: the casting weighs

W = mg/g_c = 50\ \text{lbf}

at standard gravity; since the required acceleration (

8\ \text{ft/s}^2

) is about one-quarter of

g

(

32.2\ \text{ft/s}^2

), the force should be roughly one-quarter of the weight —

12.4\ \text{lbf}

matches. Final answer:

F = 12.4\ \text{lbf}

.

F = \frac{m a}{g_c} = \frac{(50)(8.0)}{32.174} = 12.4\ \text{lbf}

, masking the error.