FE Civil · Chapter 6 · 4–6 exam questions

FE Civil Dynamics

This chapter tests kinematics, kinetics, energy methods, and momentum principles for particles and rigid bodies.

What the FE tests in Dynamics

Kinematics

Kinematics describes motion without worrying about forces — how fast, how far, what path. Civil engineers use it to derive stopping sight distance on highways, predict projectile trajectories for debris analysis, and calculate the motion of rotating machinery components.

Kinetics & Energy

Kinetics connects forces to the motion they cause. Civil engineers use Newton's second law to compute dynamic forces from earthquake ground acceleration on a building, impact forces on bridge barriers, and work-energy methods to size pump motors, design guardrails for crash energy absorption, and calculate power requirements for construction equipment.

Momentum & Vibrations

Impulse-momentum methods handle short-duration forces like vehicle impacts on bridge piers and water jets on turbine blades. Vibration analysis ensures pedestrian bridges don't resonate with footfalls, buildings don't amplify earthquake motion, and machine foundations don't vibrate excessively.

Key Dynamics formulas

$v = v_0 + at$
Constant AccelerationFE Handbook p. 104
$s = s_0 + v_0 t + \frac{1}{2}at^2$
DisplacementFE Handbook p. 104
$v^2 = v_0^2 + 2a(s - s_0)$
Velocity-PositionFE Handbook p. 104
$\sum F = ma$
Newton's Second LawFE Handbook p. 105
$T_1 + V_1 = T_2 + V_2$
Conservation of EnergyFE Handbook p. 106
$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$
Conservation of MomentumFE Handbook p. 108
$\omega_n = \sqrt{k/m}$
Natural FrequencyFE Handbook p. 112

Sample Dynamics problems

Q1. A truck accelerates uniformly from $10\,\text{m/s}$ to $30\,\text{m/s}$ over a distance of $200\,\text{m}$ . What is the acceleration?

Answer: $2.0\,\text{m/s}^2$

Explain it simply

You have initial velocity, final velocity, and distance — but no time. That combination screams v-squared equation. Plug in and solve for a. Choice A comes from dividing the velocity difference by the distance (20/200). Choice C comes from squaring velocity difference divided by distance incorrectly.

Q2. A projectile is launched from ground level at $40\,\text{m/s}$ at $45\degree$ above the horizontal. What is the total horizontal range? Use $g = 9.81\,\text{m/s}^2$ .

Answer: $163.1\,\text{m}$

Explain it simply

For range, you can either use the range formula $R = v_0^2 \sin(2\theta) / g$ , or work it from first principles: find time of flight from the vertical equation, then multiply by horizontal velocity. At 45 degrees, $\sin(90°) = 1$ , so $R = v_0^2 / g$ . Choice B comes from using $v_0 \sin 45°$ instead of $v_0^2$ . Choice D comes from doubling the correct answer — maybe computing total distance incorrectly.

These are 2 of 1,126 problems across all 15 chapters. The full bank, lessons, mastery tracking, and timed exam simulation live inside the app.

Common Dynamics mistakes on the FE

Using kinematic equations when acceleration is NOT constant — they only work for uniform acceleration.
Forgetting to include all forces (gravity, friction, normals) in ΣF = ma — draw the FBD first.
Mixing up elastic and inelastic collisions — kinetic energy is only conserved in elastic collisions.
Not converting angular units — ω must be in rad/s, not rpm, for most formulas.
Confusing velocity and speed in projectile motion — velocity has direction.

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