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10 FE practice problems: dynamics: kinematics and work-energy, with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten original problems practice dynamics: kinematics and work-energy, a topic from the FE Civil exam specification, using the Dynamics, Projectile Motion part of the FE Reference Handbook 10.6. Each problem gives the situation and the values with units. Work it with the handbook PDF open and commit to an answer before you open the solution. Every solution shows the handbook page, the equation, the substitution with units, a size check, and the mistake behind each wrong option, so a wrong pick tells you exactly what to fix.

How to use these problems

Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.

The problems

Problem 1 · FE Civil, Dynamics: kinematics and work-energy

A ball leaves level ground at 26 m/s at 20° above the horizontal. Neglecting air resistance, the horizontal distance to where it lands on the same level is most nearly:

  • A 13.5 m
  • B 22.2 m
  • C 44.3 m
  • D 4.03 m

Handbook: Dynamics, Projectile Motion, FE Reference Handbook 10.6

Show the worked solution
Answer
C (44.3 m)
Given
v0 = 26 m/s, theta = 20°, g = 9.807 m/s² (handbook)
Find
range (m)
Handbook
Dynamics, Projectile Motion, page 106
Equation
x = v0 cos θ · t, y = v0 sin θ · t - gt²/2; range R = v0² sin 2θ/g
Substitute
  1. Time of flight t = 2v0 sin θ/g = 2(26)(sin 20°)/9.807 = 1.814 s
  2. R = v0 cos θ · t = v0² sin 2θ/g = (26)²(sin 40°)/9.807 = 44.3 m
Result
44.3 m, 3 significant figures
Check
the range is below the 45° maximum v0²/g = 68.9 m.
Why the others are wrong
  • A: used g = 32.174 in a SI problem
  • B: divided by 2g instead of g
  • D: is the maximum height, not the range

Problem 2 · FE Civil, Dynamics: kinematics and work-energy

A crate starts from rest and slides 7 m down a ramp inclined at 24°. The coefficient of kinetic friction is 0.20. The crate's speed at the bottom is most nearly:

  • A 3.92 m/s
  • B 30.8 m/s
  • C 5.55 m/s
  • D 7.47 m/s

Handbook: Dynamics, Principle of Work and Energy, FE Reference Handbook 10.6

Show the worked solution
Answer
C (5.55 m/s)
Given
L = 7 m, theta = 24°, mu = 0.20, g = 9.807 m/s² (handbook)
Find
speed at the bottom (m/s)
Handbook
Dynamics, Principle of Work and Energy, page 107
Equation
T2 + V2 = T1 + V1 + U(1→2), with U(1→2) = -μ(W cos θ)L
Substitute
  1. Drop in height: h = L sin θ = 7 × sin 24° = 2.847 m
  2. Friction work per unit weight: μL cos θ = 0.20 × 7 × cos 24° = 1.279 m
  3. ½v² = g(h - μL cos θ): v = √[2 × 9.807 × (2.847 - 1.279)] = 5.55 m/s
Result
5.55 m/s, 3 significant figures
Check
v is below the frictionless value √(2gh) = 7.47 m/s; the mass cancels.
Why the others are wrong
  • A: left out the 2 in v² = 2g(h - μL cos θ)
  • B: stopped at v² (no square root)
  • D: left out the work done by friction

Problem 3 · FE Civil, Dynamics: kinematics and work-energy

A ball leaves level ground at 130 ft/sec at 25° above the horizontal. Neglecting air resistance, the horizontal distance to where it lands on the same level is most nearly:

  • A 402 ft
  • B 1,320 ft
  • C 222 ft
  • D 525 ft

Handbook: Dynamics, Projectile Motion, FE Reference Handbook 10.6

Show the worked solution
Answer
A (402 ft)
Given
v0 = 130 ft/sec, theta = 25°, g = 32.174 ft/sec² (handbook)
Find
range (ft)
Handbook
Dynamics, Projectile Motion, page 106
Equation
x = v0 cos θ · t, y = v0 sin θ · t - gt²/2; range R = v0² sin 2θ/g
Substitute
  1. Time of flight t = 2v0 sin θ/g = 2(130)(sin 25°)/32.174 = 3.415 s
  2. R = v0 cos θ · t = v0² sin 2θ/g = (130)²(sin 50°)/32.174 = 402 ft
Result
402 ft, 3 significant figures
Check
the range is below the 45° maximum v0²/g = 525 ft.
Why the others are wrong
  • B: used g = 9.807 in a USCS problem
  • C: used sin θ instead of sin 2θ
  • D: is the maximum range at 45°, ignoring the actual angle

Problem 4 · FE Civil, Dynamics: kinematics and work-energy

A ball leaves level ground at 74 ft/sec at 55° above the horizontal. Neglecting air resistance, the horizontal distance to where it lands on the same level is most nearly:

  • A 170 ft
  • B 57.1 ft
  • C 160 ft
  • D 80.0 ft

Handbook: Dynamics, Projectile Motion, FE Reference Handbook 10.6

Show the worked solution
Answer
C (160 ft)
Given
v0 = 74 ft/sec, theta = 55°, g = 32.174 ft/sec² (handbook)
Find
range (ft)
Handbook
Dynamics, Projectile Motion, page 106
Equation
x = v0 cos θ · t, y = v0 sin θ · t - gt²/2; range R = v0² sin 2θ/g
Substitute
  1. Time of flight t = 2v0 sin θ/g = 2(74)(sin 55°)/32.174 = 3.768 s
  2. R = v0 cos θ · t = v0² sin 2θ/g = (74)²(sin 110°)/32.174 = 160 ft
Result
160 ft, 3 significant figures
Check
the range is below the 45° maximum v0²/g = 170 ft.
Why the others are wrong
  • A: is the maximum range at 45°, ignoring the actual angle
  • B: is the maximum height, not the range
  • D: divided by 2g instead of g

Problem 5 · FE Civil, Dynamics: kinematics and work-energy

A ball leaves level ground at 122 ft/sec at 75° above the horizontal. Neglecting air resistance, the horizontal distance to where it lands on the same level is most nearly:

  • A 759 ft
  • B 231 ft
  • C 463 ft
  • D 216 ft

Handbook: Dynamics, Projectile Motion, FE Reference Handbook 10.6

Show the worked solution
Answer
B (231 ft)
Given
v0 = 122 ft/sec, theta = 75°, g = 32.174 ft/sec² (handbook)
Find
range (ft)
Handbook
Dynamics, Projectile Motion, page 106
Equation
x = v0 cos θ · t, y = v0 sin θ · t - gt²/2; range R = v0² sin 2θ/g
Substitute
  1. Time of flight t = 2v0 sin θ/g = 2(122)(sin 75°)/32.174 = 7.325 s
  2. R = v0 cos θ · t = v0² sin 2θ/g = (122)²(sin 150°)/32.174 = 231 ft
Result
231 ft, 3 significant figures
Check
the range is below the 45° maximum v0²/g = 463 ft.
Why the others are wrong
  • A: used g = 9.807 in a USCS problem
  • C: is the maximum range at 45°, ignoring the actual angle
  • D: is the maximum height, not the range

Problem 6 · FE Civil, Dynamics: kinematics and work-energy

A vehicle traveling at 50 km/h brakes to a stop with a constant deceleration of 4.2 m/s². The braking distance is most nearly:

  • A 45.9 m
  • B 23.0 m
  • C 1.65 m
  • D 298 m

Handbook: Dynamics, Constant Acceleration, FE Reference Handbook 10.6

Show the worked solution
Answer
B (23.0 m)
Given
v_kmh = 50 km/h, a = 4.2 m/s²
Find
braking distance (m)
Handbook
Dynamics, Constant Acceleration, page 105
Equation
v² = v0² + 2a(s - s0), so s = v0²/(2a) to stop
Substitute
  1. v0 = 50 km/h ÷ 3.6 = 13.89 m/s
  2. v² = v0² - 2as with v = 0: s = v0²/(2a) = (13.89)²/(2 × 4.2) = 23.0 m
Result
23.0 m, 3 significant figures
Check
stopping time t = v0/a = 3.31 s, and s = v0·t/2 = 23.0 m, the same answer.
Why the others are wrong
  • A: left out the 2 in v0²/(2a)
  • C: did not square the speed
  • D: used the speed in km/h as if it were m/s

Problem 7 · FE Civil, Dynamics: kinematics and work-energy

A ball leaves level ground at 33 m/s at 65° above the horizontal. Neglecting air resistance, the horizontal distance to where it lands on the same level is most nearly:

  • A 85.1 m
  • B 111 m
  • C 45.6 m
  • D 101 m

Handbook: Dynamics, Projectile Motion, FE Reference Handbook 10.6

Show the worked solution
Answer
A (85.1 m)
Given
v0 = 33 m/s, theta = 65°, g = 9.807 m/s² (handbook)
Find
range (m)
Handbook
Dynamics, Projectile Motion, page 106
Equation
x = v0 cos θ · t, y = v0 sin θ · t - gt²/2; range R = v0² sin 2θ/g
Substitute
  1. Time of flight t = 2v0 sin θ/g = 2(33)(sin 65°)/9.807 = 6.099 s
  2. R = v0 cos θ · t = v0² sin 2θ/g = (33)²(sin 130°)/9.807 = 85.1 m
Result
85.1 m, 3 significant figures
Check
the range is below the 45° maximum v0²/g = 111 m.
Why the others are wrong
  • B: is the maximum range at 45°, ignoring the actual angle
  • C: is the maximum height, not the range
  • D: used sin θ instead of sin 2θ

Problem 8 · FE Civil, Dynamics: kinematics and work-energy

A vehicle traveling at 60 km/h brakes to a stop with a constant deceleration of 7.8 m/s². The braking distance is most nearly:

  • A 1.07 m
  • B 35.6 m
  • C 17.8 m
  • D 231 m

Handbook: Dynamics, Constant Acceleration, FE Reference Handbook 10.6

Show the worked solution
Answer
C (17.8 m)
Given
v_kmh = 60 km/h, a = 7.8 m/s²
Find
braking distance (m)
Handbook
Dynamics, Constant Acceleration, page 105
Equation
v² = v0² + 2a(s - s0), so s = v0²/(2a) to stop
Substitute
  1. v0 = 60 km/h ÷ 3.6 = 16.67 m/s
  2. v² = v0² - 2as with v = 0: s = v0²/(2a) = (16.67)²/(2 × 7.8) = 17.8 m
Result
17.8 m, 3 significant figures
Check
stopping time t = v0/a = 2.14 s, and s = v0·t/2 = 17.8 m, the same answer.
Why the others are wrong
  • A: did not square the speed
  • B: left out the 2 in v0²/(2a)
  • D: used the speed in km/h as if it were m/s

Problem 9 · FE Civil, Dynamics: kinematics and work-energy

A crate starts from rest and slides 30 ft down a ramp inclined at 42°. The coefficient of kinetic friction is 0.17. The crate's speed at the bottom is most nearly:

  • A 22.9 ft/sec
  • B 32.4 ft/sec
  • C 35.9 ft/sec
  • D 1,050 ft/sec

Handbook: Dynamics, Principle of Work and Energy, FE Reference Handbook 10.6

Show the worked solution
Answer
B (32.4 ft/sec)
Given
L = 30 ft, theta = 42°, mu = 0.17, g = 32.174 ft/sec² (handbook)
Find
speed at the bottom (ft/sec)
Handbook
Dynamics, Principle of Work and Energy, page 107
Equation
T2 + V2 = T1 + V1 + U(1→2), with U(1→2) = -μ(W cos θ)L
Substitute
  1. Drop in height: h = L sin θ = 30 × sin 42° = 20.07 ft
  2. Friction work per unit weight: μL cos θ = 0.17 × 30 × cos 42° = 3.790 ft
  3. ½v² = g(h - μL cos θ): v = √[2 × 32.174 × (20.07 - 3.790)] = 32.4 ft/sec
Result
32.4 ft/sec, 3 significant figures
Check
v is below the frictionless value √(2gh) = 35.9 ft/sec; the mass cancels.
Why the others are wrong
  • A: left out the 2 in v² = 2g(h - μL cos θ)
  • C: left out the work done by friction
  • D: stopped at v² (no square root)

Problem 10 · FE Civil, Dynamics: kinematics and work-energy

A crate starts from rest and slides 30 m down a ramp inclined at 34°. The coefficient of kinetic friction is 0.08. The crate's speed at the bottom is most nearly:

  • A 17.4 m/s
  • B 12.0 m/s
  • C 290 m/s
  • D 17.0 m/s

Handbook: Dynamics, Principle of Work and Energy, FE Reference Handbook 10.6

Show the worked solution
Answer
D (17.0 m/s)
Given
L = 30 m, theta = 34°, mu = 0.08, g = 9.807 m/s² (handbook)
Find
speed at the bottom (m/s)
Handbook
Dynamics, Principle of Work and Energy, page 107
Equation
T2 + V2 = T1 + V1 + U(1→2), with U(1→2) = -μ(W cos θ)L
Substitute
  1. Drop in height: h = L sin θ = 30 × sin 34° = 16.78 m
  2. Friction work per unit weight: μL cos θ = 0.08 × 30 × cos 34° = 1.990 m
  3. ½v² = g(h - μL cos θ): v = √[2 × 9.807 × (16.78 - 1.990)] = 17.0 m/s
Result
17.0 m/s, 3 significant figures
Check
v is below the frictionless value √(2gh) = 18.1 m/s; the mass cancels.
Why the others are wrong
  • A: used sin θ for the normal force
  • B: left out the 2 in v² = 2g(h - μL cos θ)
  • C: stopped at v² (no square root)

A new problem posts every day inside the lab, with the full worked solution the same evening.

Frequently asked questions

Where is dynamics: kinematics and work-energy in the FE Reference Handbook?

Look in the Dynamics, Projectile Motion part of FE Reference Handbook 10.6. Each solution gives the exact page, so you can practice finding it in the PDF the way you will on exam day.

Are these real FE exam questions?

No. They are original problems generated from the handbook formulas and checked by code. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee accepts.

How do I check my answer before opening the solution?

Check the units of your result and whether its size makes sense for the situation. Each solution ends with the same kind of check, so you can compare your habit with ours.

Sources

  1. NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
  2. NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.