All FE Civil practice problems

FE Civil Dynamics practice problems

6 free FE Civil Dynamics practice problems with answers and full worked solutions. Dynamics carries 4–6 of the 110 questions on the FE Civil exam (NCEES specification effective July 2020). Pick an answer to check it, then read the worked solution and why each wrong choice is wrong.

6 worked Dynamics problems

  1. Question 1 of 6

    A body starts at 15 ft/s and accelerates uniformly at 16 ft/s^2 over 110 ft. Its final velocity is most nearly:

    Answer choices for question 1

    Show the answer and the reasoning

    Why D

    Given: v0 = 15 ft/s, a = 16 ft/s^2, s = 110 ft; no time is given. Relation: v^2 = v0^2 + 2as, the constant-acceleration form without time. Substitute: v^2 = 15^2 + 2(16)(110) = 3745. Result: v = 61.20 ft/s.

    Why A is tempting

    Halved the result, which would violate the energy the acceleration supplies over that distance.

    Why B is tempting

    Dropped the factor of two.

    Why C is tempting

    Ignored the initial velocity.

    Source NCEES FE Reference Handbook 10.6 — Dynamics, Kinematics and Kinetics

  2. Question 2 of 6

    A solid circular disk weighs 100 lb and has a radius of 1.25 ft. Its mass moment of inertia about its central axis is most nearly:

    Solid circular disk weighing 100 lb with a radius of 1.25 ft, spinning about the axis through its centre.
    Answer choices for question 2

    Show the answer and the reasoning

    Why B

    Given: W = 100 lb, so m = 3.1056 slugs; R = 1.25 ft. Relation: solid disk about its axis, I = (1/2) m R^2 (a thin ring would be m R^2). Substitute: I = 0.5(3.1056)(1.25)^2. Result: I = 2.426 slug-ft^2.

    Why A is tempting

    Did not square the radius.

    Why C is tempting

    Omitted the factor of one half. That expression describes a thin ring, whose mass all sits at the rim.

    Why D is tempting

    Used the weight rather than the mass in slugs.

    Source NCEES FE Reference Handbook 10.6 — Dynamics, Kinematics and Kinetics

  3. Question 3 of 6

    A horizontal force of 220 lb acts on a body weighing 1500 lb on a frictionless surface. Its acceleration is most nearly:

    Answer choices for question 3

    Show the answer and the reasoning

    Why D

    Given: F = 220 lb on a frictionless surface; m = 1500/32.2 = 46.5839 slugs. Relation: a = F/m. Substitute: a = 220/46.5839. Result: a = 4.723 ft/s^2.

    Why A is tempting

    Inverted the relation.

    Why B is tempting

    Used g = 9.81, the SI value, with a weight in pounds.

    Why C is tempting

    Rounded g to 30 ft/s^2.

    Source NCEES FE Reference Handbook 10.6 — Dynamics, Kinematics and Kinetics

  4. Question 4 of 6

    A body weighing 2,400 lb moves at 15 ft/s. Its kinetic energy is most nearly:

    Answer choices for question 4

    Show the answer and the reasoning

    Why A

    Given: W = 2,400 lb, so m = 74.534 slugs; v = 15 ft/s. Relation: KE = (1/2) m v^2, with the mass in slugs. Substitute: KE = 0.5(74.534)(15)^2. Result: KE = 8385 ft-lb.

    Why B is tempting

    Omitted the factor of one half.

    Why C is tempting

    Multiplied weight by velocity, which is neither energy nor momentum in consistent units.

    Why D is tempting

    Multiplied by gravity again after already converting to mass.

    Source NCEES FE Reference Handbook 10.6 — Dynamics, Kinematics and Kinetics

  5. Question 5 of 6

    A body moving at 30 ft/s undergoes a constant acceleration of 4 ft/s^2 for 1.0 s. The distance it covers is most nearly:

    Answer choices for question 5

    Show the answer and the reasoning

    Why C

    Given: v0 = 30 ft/s, a = 4 ft/s^2, t = 1.0 s. Relation: s = v0 t + (1/2) a t^2. Substitute: s = 30(1.0) + 0.5(4)(1.0)^2. Result: s = 32.00 ft.

    Why A is tempting

    Ignored the initial velocity.

    Why B is tempting

    Ignored the acceleration entirely.

    Why D is tempting

    Omitted the factor of one half on the acceleration term.

    Source NCEES FE Reference Handbook 10.6 — Dynamics, Kinematics and Kinetics

  6. Question 6 of 6

    A body weighing 375 lb has a mass moment of inertia of 80 slug-ft^2 about a given axis. Its radius of gyration about that axis is most nearly:

    Answer choices for question 6

    Show the answer and the reasoning

    Why B

    Given: I = 80 slug-ft^2, m = 375/32.2 = 11.6460 slugs. Relation: k = sqrt(I/m), the radius at which the whole mass would give the same I. Substitute: k = sqrt(80/11.6460). Result: k = 2.621 ft.

    Why A is tempting

    Inverted the ratio inside the radical.

    Why C is tempting

    Never took the square root.

    Why D is tempting

    Multiplied rather than divided.

    Source NCEES FE Reference Handbook 10.6 — Dynamics, Kinematics and Kinetics

Going further

Frequently asked questions

How many Dynamics questions are on the FE Civil exam?

Between 4 and 6 of the 110 questions, according to the NCEES FE Civil specification effective July 2020.

Are these real NCEES exam questions?

No. NCEES does not release its exam questions. These are original FE-style problems written to the NCEES specification and the FE Reference Handbook; every calculated key is checked by two independent solution methods.

Are these Dynamics practice problems free?

Yes. All 6 problems on this page, with answers and full worked solutions, are free and need no account.

Check where you stand, free

The quick diagnostic is eight FE Civil questions in about ten minutes, with no account. The free mock is a full 110-question, 5 hour 20 minute FE Civil paper with a worked solution for every question.