Dynamics

4% of exam

Kinematics of particles and rigid bodies, mass moments of inertia, force-acceleration, and work, energy, and power.

2 concepts

Exam areas

A. Kinematics

Kinematics, Newton's Second Law, and Work-Energy

Describe motion with the constant-acceleration and projectile equations, relate force to acceleration through F=ma, then short-circuit the algebra with work-energy and power.

Dynamics splits cleanly into two questions: how a body moves (kinematics) and why it moves that way (kinetics). On the FE Civil exam this topic is worth 4-6 questions, and the fastest way to lose those points is to reach for a force balance when a single energy equation would have finished the problem in one line. The FE Reference Handbook hands you every equation in its Dynamics chapter, so the exam is not testing recall — it is testing whether you can pick the right tool and keep your signs, units, and reference frames straight. This concept builds that judgment: rectilinear and curvilinear kinematics, projectile and relative motion,

\Sigma F = ma

, and the work-energy-power chain that ties forces to speeds without ever solving for time.

Position, velocity, acceleration

Kinematics starts with a position vector

\mathbf{r}(t)

and two derivatives: velocity

\mathbf{v} = d\mathbf{r}/dt

and acceleration

\mathbf{a} = d\mathbf{v}/dt

. In one dimension these are just

v = \dot{s}

and

a = \ddot{s}

. The single most useful relation drops the time variable entirely — multiply

a = dv/dt

ds/ds

to get

a\,ds = v\,dv

, which lets you connect speed and position directly. Whenever a problem gives you a distance and asks for a speed (or vice versa) without mentioning time, this is the equation to reach for.

\mathbf{v} = \frac{d\mathbf{r}}{dt}, \qquad \mathbf{a} = \frac{d\mathbf{v}}{dt}, \qquad a\,ds = v\,dv

Constant-acceleration equations

When acceleration is constant — gravity, steady braking, a uniform thrust — the calculus collapses into three algebraic workhorses. Pick the one that avoids the variable you do not have: use

v = v_0 + a t

when you know time, and the time-free form

v^2 = v_0^2 + 2a(s - s_0)

when you do not. A free-falling body is simply this set with

a = g

pointed down. Choose a positive direction first and hold it for the whole problem; a sign flip mid-solution is the most common kinematics error.

v = v_0 + a\,t, \qquad s = s_0 + v_0 t + \tfrac{1}{2}a t^2, \qquad v^2 = v_0^2 + 2a\,(s - s_0)

Projectile motion

Projectile motion is two independent constant-acceleration problems sharing one clock. Horizontally there is no acceleration, so

x = (v_0\cos\theta)\,t

; vertically gravity acts, so

v_y = v_0\sin\theta - g t

and

y = (v_0\sin\theta)\,t - \tfrac{1}{2}g t^2

. The link between the two axes is time. Apex occurs when

v_y = 0

, and for launch and landing at the same elevation the total flight time is

t_f = 2v_0\sin\theta / g

. Never apply a

\sqrt{}

-style range formula blindly when the launch and landing heights differ — go back to the component equations.

x = (v_0\cos\theta)\,t, \qquad y = (v_0\sin\theta)\,t - \tfrac{1}{2}g t^2

Curvilinear paths and relative motion

On a curved path it is cleaner to resolve acceleration into normal and tangential components: the tangential part

a_t = \dot{v}

changes speed, while the normal part

a_n = v^2/\rho

(always pointing toward the center of curvature) changes direction. For circular motion of radius

r

these become

v = r\omega

a_t = r\alpha

, and

a_n = r\omega^2

. When two bodies move at once, work in a translating frame:

\mathbf{v}_A = \mathbf{v}_B + \mathbf{v}_{A/B}

. Read 'the velocity of A relative to B' as exactly that subscript order, and add the vectors head-to-tail rather than guessing signs.

a_t = \frac{dv}{dt}, \qquad a_n = \frac{v^2}{\rho} = r\omega^2, \qquad \mathbf{v}_A = \mathbf{v}_B + \mathbf{v}_{A/B}

Newton's second law

Kinetics enters through

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

— the resultant of all applied forces equals mass times acceleration, written as one scalar equation per direction. Draw a free-body diagram first, every time: on an incline, resolve weight into components along and perpendicular to the surface so that

\Sigma F = mg\sin\theta - \mu_k N

along the slope and

N = mg\cos\theta

across it. In USCS remember that weight in pounds-force is not mass; convert with

m = W/g

so the units close (

\text{lbf} = \text{slug}\cdot\text{ft/s}^2

\Sigma F_x = m a_x, \qquad \Sigma F_y = m a_y

Work, energy, and the work-energy theorem

The work-energy theorem is the great shortcut: the net work done on a particle equals its change in kinetic energy,

U_{1\to2} = \Delta T = \tfrac{1}{2}m v_2^2 - \tfrac{1}{2}m v_1^2

. Because work depends on distance, not time, you skip the time variable entirely — ideal for 'speed after sliding

d

' problems. Bookkeep work by source: a constant force does

U = F s\cos\theta

, gravity does

U = -W\Delta y

(negative going up), and a spring does

U = -\tfrac{1}{2}k(s_2^2 - s_1^2)

. Friction always removes energy, so its work is negative.

U_{1\to2} = \Delta T = \tfrac{1}{2}m v_2^2 - \tfrac{1}{2}m v_1^2

Conservation of energy and power

When only conservative forces act, mechanical energy is conserved:

T_1 + V_1 = T_2 + V_2

, with gravitational potential

V_g = mgh

and elastic potential

V_e = \tfrac{1}{2}k s^2

. If nonconservative forces (friction, an applied push) are present, carry them as a work term:

T_1 + V_1 + U_{1\to2} = T_2 + V_2

. Power is the time rate of doing work,

P = dU/dt = \mathbf{F}\cdot\mathbf{v}

, and mechanical efficiency is

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

. Watch the units:

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

T_1 + V_1 + U_{1\to2} = T_2 + V_2, \qquad P = \mathbf{F}\cdot\mathbf{v}

Exam strategy

Decide on the method before touching algebra. If the question asks for a speed given a distance (or a distance given a speed) and never mentions time, use work-energy — it is almost always two or three lines. If it asks for time, or for the force/acceleration itself, use

\Sigma F = ma

with a clean free-body diagram. Reserve the constant-acceleration kinematics set for when acceleration is genuinely constant and you need the explicit time history. For projectiles, split into

x

and

y

immediately and let time be the bridge. Lock in a positive direction at the start, label every

g

9.81\ \text{m/s}^2

32.2\ \text{ft/s}^2

, and in USCS convert weight to mass before writing

ma

Key equations

Velocity and acceleration

v = \dfrac{ds}{dt}, \qquad a = \dfrac{dv}{dt} = v\dfrac{dv}{ds}

Definitions for rectilinear motion; the last form (from $a\,ds = v\,dv$ ) eliminates time. $s$ in m or ft, $v$ in m/s or ft/s, $a$ in m/s² or ft/s².

Constant-acceleration set

v = v_0 + a t, \quad s = s_0 + v_0 t + \tfrac{1}{2}a t^2, \quad v^2 = v_0^2 + 2a(s - s_0)

Valid only when $a$ is constant. Use the time-free third form when $t$ is neither given nor asked.

Projectile components

v_x = v_0\cos\theta, \quad v_y = v_0\sin\theta - g t, \quad y = (v_0\sin\theta)t - \tfrac{1}{2}g t^2

Horizontal velocity constant ( $a_x = 0$ ); vertical motion under $a_y = -g$ . Time links the two axes.

Normal and tangential acceleration

a_t = \dfrac{dv}{dt}, \qquad a_n = \dfrac{v^2}{\rho}, \qquad a = \sqrt{a_t^2 + a_n^2}

$\rho$ = radius of curvature (m or ft); $a_n$ points toward the center. For a circle $\rho = r$ .

Circular motion

v = r\omega, \qquad a_t = r\alpha, \qquad a_n = r\omega^2

$\omega$ in rad/s, $\alpha$ in rad/s², $r$ in m or ft. Angles must be in radians.

Relative motion

\mathbf{v}_A = \mathbf{v}_B + \mathbf{v}_{A/B}, \qquad \mathbf{a}_A = \mathbf{a}_B + \mathbf{a}_{A/B}

Translating-frame vector addition; $\mathbf{v}_{A/B}$ is the velocity of $A$ as seen from $B$ .

Newton's second law

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

One scalar equation per coordinate direction. $F$ in N or lbf, $m$ in kg or slug, $a$ in m/s² or ft/s².

Weight

W = mg

Convert between weight and mass; $g = 9.81\ \text{m/s}^2 = 32.2\ \text{ft/s}^2$ . In USCS, $m = W/g$ in slugs.

Kinetic energy (particle)

T = \tfrac{1}{2}m v^2

Joules (N·m) in SI, ft-lbf in USCS. Always uses speed magnitude.

Work-energy theorem

U_{1\to2} = T_2 - T_1 = \tfrac{1}{2}m v_2^2 - \tfrac{1}{2}m v_1^2

Net work of all forces equals change in kinetic energy. Distance-based, time-free.

Work of common forces

U_{F} = F s\cos\theta, \quad U_{grav} = -W\Delta y, \quad U_{spring} = -\tfrac{1}{2}k(s_2^2 - s_1^2)

$\theta$ = angle between force and displacement; gravity work negative when rising; spring work negative when stretched or compressed from its free length.

Potential energy

V_g = mgh, \qquad V_e = \tfrac{1}{2}k s^2

Gravitational ( $h$ above a datum) and elastic ( $s$ = deflection from free length, $k$ in N/m or lbf/ft).

Energy conservation with nonconservative work

T_1 + V_1 + U_{1\to2} = T_2 + V_2

$U_{1\to2}$ is the work of nonconservative forces (negative for friction). Reduces to $T_1+V_1=T_2+V_2$ when only gravity/springs act.

Power and efficiency

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

$P$ in watts (J/s) or ft-lbf/s; $1\ \text{hp} = 746\ \text{W} = 550\ \text{ft-lbf/s}$ .

Worked examples

Projectile range and maximum height

Problem. A projectile leaves the ground at $v_0 = 25\ \text{m/s}$ at $\theta = 40^\circ$ above horizontal, landing back at ground level. Find the horizontal range and the maximum height. Use $g = 9.81\ \text{m/s}^2$ .

Solution. Resolve the launch velocity:

v_x = 25\cos40^\circ = 19.15\ \text{m/s}

and

v_{y0} = 25\sin40^\circ = 16.07\ \text{m/s}

. Time of flight (same launch and landing elevation):

t_f = 2 v_{y0}/g = 2(16.07)/9.81 = 3.276\ \text{s}

. Range:

R = v_x\, t_f = 19.15 \times 3.276 = 62.7\ \text{m}

. Maximum height (apex, where

v_y = 0

H = v_{y0}^2/(2g) = 16.07^2/(2\cdot9.81) = 13.2\ \text{m}

. Sanity check: with the textbook form

R = v_0^2\sin(2\theta)/g = 625\sin80^\circ/9.81 = 62.7\ \text{m}

, matching.

\boxed{R = 62.7\ \text{m},\ H = 13.2\ \text{m}}

R = v_x t_f = 19.15 \times 3.276 = 62.7\ \text{m}

Crate sliding down a rough incline (F = ma plus kinematics)

Problem. A 50 kg crate is released from rest on a $20^\circ$ incline with kinetic friction coefficient $\mu_k = 0.25$ . Find its acceleration and its speed after sliding 8.0 m down the slope.

Solution. Free-body diagram along the incline (positive down-slope):

\Sigma F = mg\sin\theta - \mu_k N

, with

N = mg\cos\theta

. Mass cancels:

a = g(\sin\theta - \mu_k\cos\theta) = 9.81(\sin20^\circ - 0.25\cos20^\circ) = 9.81(0.342 - 0.235) = 1.05\ \text{m/s}^2

. Speed after

d = 8.0\ \text{m}

from rest, time-free kinematics:

v = \sqrt{2 a d} = \sqrt{2(1.05)(8.0)} = 4.10\ \text{m/s}

. Sanity check:

a

is positive, so gravity beats friction and the crate accelerates — consistent with

\tan20^\circ = 0.364 > \mu_k = 0.25

\boxed{a = 1.05\ \text{m/s}^2,\ v = 4.10\ \text{m/s}}

a = g(\sin\theta - \mu_k\cos\theta) = 9.81(\sin20^\circ - 0.25\cos20^\circ) = 1.05\ \text{m/s}^2

Speed at the bottom of a chute (work-energy with friction)

Problem. A 10 kg package starts from rest and slides 5.0 m down a straight chute, dropping 3.0 m in elevation, with $\mu_k = 0.30$ . Find its speed at the bottom.

Solution. Geometry: the chute rises 3.0 m over 5.0 m, so

\sin\theta = 3/5 = 0.6

and

\cos\theta = 4/5 = 0.8

. Gravity does positive work

U_{grav} = mgh = 10(9.81)(3.0) = 294.3\ \text{J}

. Friction does negative work

U_{fric} = -\mu_k(mg\cos\theta)L = -0.30(10)(9.81)(0.8)(5.0) = -117.7\ \text{J}

. Work-energy theorem from rest:

\tfrac{1}{2}m v^2 = U_{grav} + U_{fric} = 294.3 - 117.7 = 176.6\ \text{J}

, so

v = \sqrt{2(176.6)/10} = 5.94\ \text{m/s}

. Sanity check: a frictionless drop would give

v = \sqrt{2gh} = 7.67\ \text{m/s}

; friction reduces it, as expected.

\boxed{v = 5.94\ \text{m/s}}

\tfrac{1}{2}m v^2 = mgh - \mu_k mg\cos\theta\,L \;\Rightarrow\; v = 5.94\ \text{m/s}

Common pitfalls

•Confusing weight and mass in USCS: a 'W = 100 lbf' block has mass $m = W/g = 100/32.2 = 3.11$ slug. Plugging 100 directly into $\Sigma F = ma$ overstates the force by a factor of $g$ .
•Reaching for $\Sigma F = ma$ when no force or time is asked. If the question relates a speed to a distance, work-energy is faster and avoids solving a differential equation.
•Sign errors on gravity's work: $U_{grav} = -W\Delta y$ is negative when the body rises and positive when it falls. Always pick a datum and a positive direction first.
•Using a horizontal-launch range formula when launch and landing elevations differ. Split into $x$ and $y$ components and let time be the link instead.
•Treating friction work as conservative or assigning it the wrong sign. Friction always opposes motion, so its work is negative and energy is lost — it cannot be stored as potential energy.
•Mixing degrees and radians in circular motion. $v = r\omega$ , $a_t = r\alpha$ , and $a_n = r\omega^2$ require $\omega$ in rad/s and $\alpha$ in rad/s².
•Forgetting the normal acceleration on a curve: even at constant speed a body on a curved path accelerates, $a_n = v^2/\rho$ toward the center. Only $a_t = 0$ at constant speed.

References

NCEES FE Reference Handbook — Dynamics (Kinematics, Particle Kinetics, Work and Energy)
Hibbeler, Engineering Mechanics: Dynamics — particle kinematics and energy methods
Beer, Johnston, Vector Mechanics for Engineers: Dynamics

B. Mass moments of inertia

Mass Moments of Inertia and Rotation

Compute mass moment of inertia for standard shapes, shift axes with the parallel-axis theorem, and apply ΣM=Iα and rolling-without-slipping to rotating rigid bodies.

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