FE Other Disciplines (NCEES Fundamentals of Engineering) — All Questions

A rigid body is in static equilibrium when:

• a.Both the sum of forces and the sum of moments are zero✓
• b.Neither condition is required
• c.The sum of forces is zero only
• d.The sum of moments is zero only

Complete static equilibrium requires both the net force and the net moment (about any point) to be zero. Force balance alone prevents translation; moment balance alone prevents rotation; both are needed.

Two parallel forces of 30 N and 40 N act in the same direction on a rigid body. What is the magnitude of their resultant?

• a.1200 N
• b.50 N
• c.70 N✓
• d.10 N

For parallel forces in the same direction, the resultant is the arithmetic sum: 30 + 40 = 70 N. (Only if they opposed each other would you subtract, giving 10 N.) Its line of action lies between them, closer to the larger force.

The center of gravity and the centroid of a body coincide when:

• a.They always differ
• b.They never coincide
• c.The body is painted
• d.The material has uniform density✓

The centroid is a purely geometric center, while the center of gravity is the weight-averaged center. They coincide when the body has uniform density (and a uniform gravitational field), so weight is proportional to volume.

A car starts from rest and accelerates uniformly at 2 m/s^2 over a distance of 25 m. What is its final speed?

• a.100 m/s
• b.5 m/s
• c.10 m/s✓
• d.50 m/s

Using v^2 = v0^2 + 2·a·s = 0 + 2(2)(25) = 100, so v = 10 m/s. This kinematic relation connects velocity and displacement without needing time.

A ball is thrown straight up at 20 m/s. How long does it take to reach its highest point (g = 9.81 m/s^2)?

• a.2.04 s✓
• b.0.51 s
• c.1.02 s
• d.4.08 s

At the peak the velocity is zero: t = v0/g = 20/9.81 = 2.04 s. The total flight time back to the launch height would be twice this, 4.08 s.

A ball is thrown straight up at 20 m/s. What maximum height does it reach (g = 9.81 m/s^2)?

• a.2.04 m
• b.40.8 m
• c.10.2 m
• d.20.4 m✓

Maximum height h = v0^2/(2g) = 20^2/(2 x 9.81) = 400/19.62 = 20.4 m. At the top all kinetic energy has converted to gravitational potential energy.

A projectile is launched at 20 m/s at 45 degrees over level ground. What is its horizontal range (g = 9.81 m/s^2)?

• a.20.4 m
• b.400 m
• c.40.8 m✓
• d.81.5 m

Range R = v0^2·sin(2·theta)/g = 20^2·sin(90)/9.81 = 400/9.81 = 40.8 m. A 45-degree launch angle maximizes range for a given speed on level ground.

What net force is needed to accelerate a 10 kg mass at 5 m/s^2?

• a.15 N
• b.500 N
• c.2 N
• d.50 N✓

Newton's second law: F = m·a = 10 x 5 = 50 N. Net force is directly proportional to acceleration for a fixed mass.

What is the weight of a 50 kg object on Earth (g = 9.81 m/s^2)?

• a.5 N
• b.50 N
• c.4905 N
• d.490.5 N✓

Weight W = m·g = 50 x 9.81 = 490.5 N. Mass (kg) is a fixed quantity of matter, while weight (N) is the gravitational force on that mass and varies with g.

A block slides on a horizontal surface with kinetic friction coefficient 0.4. What is its deceleration (g = 9.81 m/s^2)?

• a.9.81 m/s^2
• b.39.2 m/s^2
• c.3.92 m/s^2✓
• d.0.4 m/s^2

On a level surface, friction deceleration a = mu·g = 0.4 x 9.81 = 3.92 m/s^2. It is independent of mass because both the friction force and inertia scale with mass.

A vehicle moving at 20 m/s decelerates uniformly at 5 m/s^2. What distance does it travel before stopping?

• a.20 m
• b.4 m
• c.80 m
• d.40 m✓

Using v^2 = v0^2 - 2·a·s with v = 0: s = v0^2/(2a) = 400/(2 x 5) = 40 m. Stopping distance grows with the square of speed, a key road-safety relationship.

An object moves in a circle of radius 5 m at a constant speed of 10 m/s. What is its centripetal acceleration?

• a.20 m/s^2✓
• b.4 m/s^2
• c.2 m/s^2
• d.50 m/s^2

Centripetal acceleration a_c = v^2/r = 10^2/5 = 20 m/s^2, directed toward the center. Even at constant speed, circular motion is accelerated because the velocity direction keeps changing.

A 2 kg object moves in a circle of radius 5 m at 10 m/s. What centripetal force is required?

• a.80 N
• b.20 N
• c.4 N
• d.40 N✓

Centripetal force F = m·v^2/r = 2 x 10^2/5 = 40 N, directed toward the center. This inward net force is what continually changes the object's direction of motion.

A point on a rotating wheel moves at 10 m/s at a radius of 2 m. What is the angular velocity?

• a.0.2 rad/s
• b.20 rad/s
• c.2 rad/s
• d.5 rad/s✓

Angular velocity omega = v/r = 10/2 = 5 rad/s. Linear (tangential) speed equals angular velocity times radius, so points farther from the axis move faster.

A shaft rotates at 300 rpm. What is its angular velocity in radians per second?

• a.3.14 rad/s
• b.31.4 rad/s✓
• c.5 rad/s
• d.300 rad/s

omega = 300 rev/min x (2·pi rad/rev) x (1 min/60 s) = 300 x 2·pi/60 = 31.4 rad/s. Converting rpm to rad/s multiplies by 2·pi/60.

A constant 20 N force pushes an object 10 m in the direction of the force. How much work is done?

• a.2000 J
• b.200 J✓
• c.100 J
• d.20 J

Work W = F·d·cos(theta) = 20 x 10 x cos(0) = 200 J. Work equals force times displacement along the force's direction; a force perpendicular to motion does zero work.

A 100 N force moves an object at a constant velocity of 5 m/s in the direction of the force. What power is delivered?

• a.20 W
• b.500 W✓
• c.2500 W
• d.105 W

Power P = F·v = 100 x 5 = 500 W. Power is the rate of doing work; for a constant force it equals the force times the velocity along its line of action.

A constant force of 10 N acts on an object for 5 seconds. What impulse does it deliver?

• a.15 N·s
• b.2 N·s
• c.50 N·s✓
• d.250 N·s

Impulse J = F·t = 10 x 5 = 50 N·s. By the impulse-momentum theorem, this equals the change in the object's linear momentum.

A 2 kg cart moving at 3 m/s collides and sticks to a stationary 1 kg cart. What is their common velocity after the perfectly inelastic collision?

• a.3 m/s
• b.6 m/s
• c.1.5 m/s
• d.2 m/s✓

Conservation of momentum: (2 x 3) + (1 x 0) = (2 + 1)·v, so v = 6/3 = 2 m/s. In a perfectly inelastic collision the objects move together and momentum (not kinetic energy) is conserved.

A spring with stiffness 200 N/m is compressed 0.1 m. How much elastic potential energy is stored?

• a.2 J
• b.10 J
• c.1 J✓
• d.0.1 J

Spring PE = (1/2)·k·x^2 = 0.5 x 200 x 0.1^2 = 0.5 x 200 x 0.01 = 1 J. The energy grows with the square of the deflection.

A 1 kg mass on a spring of stiffness 100 N/m oscillates. What is the period of the simple harmonic motion?

• a.6.28 s
• b.1.26 s
• c.0.314 s
• d.0.628 s✓

Period T = 2·pi·sqrt(m/k) = 2·pi·sqrt(1/100) = 2·pi x 0.1 = 0.628 s. The period increases with mass and decreases with spring stiffness.

What is the period of a simple pendulum of length 1 m undergoing small oscillations (g = 9.81 m/s^2)?

• a.2.0 s✓
• b.6.28 s
• c.1.0 s
• d.4.0 s

Period T = 2·pi·sqrt(L/g) = 2·pi·sqrt(1/9.81) = 2·pi x 0.319 = 2.0 s. For small angles the period depends only on length and g, not on mass or amplitude.

An object is dropped from rest and falls freely for 3 seconds. How far does it fall (g = 9.81 m/s^2)?

• a.14.7 m
• b.29.4 m
• c.44.1 m✓
• d.88.3 m

Distance s = (1/2)·g·t^2 = 0.5 x 9.81 x 3^2 = 0.5 x 9.81 x 9 = 44.1 m. Distance in free fall grows with the square of the elapsed time.

An object dropped from rest falls freely for 3 seconds. What is its velocity at that moment (g = 9.81 m/s^2)?

• a.19.6 m/s
• b.9.81 m/s
• c.44.1 m/s
• d.29.4 m/s✓

Velocity v = g·t = 9.81 x 3 = 29.4 m/s. In free fall velocity increases linearly with time, while distance increases with time squared.

What is the gravitational potential energy of a 5 kg mass raised 10 m above a reference level (g = 9.81 m/s^2)?

• a.490.5 J✓
• b.50 J
• c.4905 J
• d.98 J

Potential energy PE = m·g·h = 5 x 9.81 x 10 = 490.5 J. It represents the work done against gravity to lift the mass to that height.

A mass is released from rest and falls 20 m. Using energy conservation, what is its speed just before impact (g = 9.81 m/s^2)?

• a.20 m/s
• b.19.8 m/s✓
• c.14 m/s
• d.39.6 m/s

Equating potential and kinetic energy, m·g·h = (1/2)·m·v^2 gives v = sqrt(2·g·h) = sqrt(2 x 9.81 x 20) = sqrt(392.4) = 19.8 m/s. Mass cancels out of the equation.

A rotating body has a mass moment of inertia of 2 kg·m^2 and spins at 3 rad/s. What is its rotational kinetic energy?

• a.9 J✓
• b.6 J
• c.18 J
• d.3 J

Rotational KE = (1/2)·I·omega^2 = 0.5 x 2 x 3^2 = 0.5 x 2 x 9 = 9 J. It is the rotational analog of (1/2)·m·v^2, with moment of inertia replacing mass and angular velocity replacing linear velocity.

A body with mass moment of inertia 4 kg·m^2 is given an angular acceleration of 2 rad/s^2. What torque is required?

• a.6 N·m
• b.8 N·m✓
• c.16 N·m
• d.2 N·m

The rotational form of Newton's second law: torque = I·alpha = 4 x 2 = 8 N·m. Moment of inertia plays the role of rotational inertia, resisting angular acceleration.

A point at radius 0.5 m experiences an angular acceleration of 4 rad/s^2. What is its tangential acceleration?

• a.8 m/s^2
• b.4 m/s^2
• c.2 m/s^2✓
• d.0.5 m/s^2

Tangential acceleration a_t = r·alpha = 0.5 x 4 = 2 m/s^2. It is the rate of change of tangential speed, separate from the centripetal (radial) acceleration.

What is the minimum speed at the top of a vertical circular loop of radius 2 m for an object to maintain contact (g = 9.81 m/s^2)?

• a.2 m/s
• b.19.6 m/s
• c.9.81 m/s
• d.4.43 m/s✓

At minimum speed, gravity alone supplies the centripetal force: m·g = m·v^2/r, so v = sqrt(g·r) = sqrt(9.81 x 2) = 4.43 m/s. Below this speed the object falls away from the track.

A collision in which kinetic energy is fully conserved has a coefficient of restitution equal to:

• a.0.5
• b.1✓
• c.0
• d.2

A perfectly elastic collision conserves kinetic energy and has a coefficient of restitution e = 1. A perfectly inelastic collision has e = 0 (objects stick together); real collisions fall between.

Two cars approach each other head-on, one at 60 km/h and the other at 40 km/h. What is their relative (closing) speed?

• a.100 km/h✓
• b.20 km/h
• c.60 km/h
• d.40 km/h

For objects moving toward each other, the closing speed is the sum of the individual speeds: 60 + 40 = 100 km/h. If they moved in the same direction, the relative speed would be the difference, 20 km/h.

The linear momentum of a system is conserved:

• a.When friction acts on it
• b.When no net external force acts on it✓
• c.Whenever the speed is constant
• d.Only when speed increases

Linear momentum is conserved when the net external force on the system is zero. Internal forces (such as those during a collision) cannot change the total momentum; only external forces can.

A 2 m long bar stretches 2 mm under load. What is the axial strain?

• a.1
• b.0.001✓
• c.0.01
• d.0.0005

Strain = deformation/original length = 0.002 m/2 m = 0.001. Strain is dimensionless, expressing the fractional change in length.

A material with Young's modulus 200 GPa experiences an axial strain of 0.001 within its elastic range. What is the resulting stress?

• a.2 MPa
• b.20 MPa
• c.200 MPa✓
• d.200 GPa

Hooke's law: stress = E·strain = 200x10^9 x 0.001 = 200x10^6 Pa = 200 MPa. Stress is proportional to strain only within the linear elastic region.

A shear force of 5000 N acts over a cross-sectional area of 0.001 m^2. What is the average shear stress?

• a.50 MPa
• b.5 MPa✓
• c.0.5 MPa
• d.2.5 MPa

Average shear stress tau = V/A = 5000/0.001 = 5,000,000 Pa = 5 MPa. Shear stress acts parallel to the cross-section, unlike normal stress which acts perpendicular to it.

A component has a yield strength of 250 MPa and is designed to a working stress of 100 MPa. What is the factor of safety?

• a.0.4
• b.2.5✓
• c.150
• d.2.0

Factor of safety = failure (yield) stress/allowable working stress = 250/100 = 2.5. A factor above 1 provides margin against uncertainties in loads, materials, and analysis.

A solid circular shaft 0.05 m in diameter transmits a torque of 1000 N·m. What is the maximum shear stress at its surface?

• a.40.7 MPa✓
• b.81.5 MPa
• c.20.4 MPa
• d.4.07 MPa

Torsion formula tau = T·r/J, with J = pi·d^4/32 = pi(0.05)^4/32 = 6.14x10^-7 m^4 and r = 0.025 m. tau = 1000 x 0.025/6.14x10^-7 = 40.7 MPa. Shear stress is maximum at the outer radius.

What is the polar moment of inertia of a solid circular shaft of diameter d?

• a.pi·d^4/64
• b.pi·d^4/32✓
• c.pi·d^4/12
• d.pi·d^3/16

For a solid circular cross-section, J = pi·d^4/32. (The area moment of inertia about a diameter is pi·d^4/64, exactly half of J, since J = Ix + Iy for the two perpendicular axes.)

A rectangular beam (width 0.05 m, height 0.1 m) carries a bending moment of 10 kN·m. What is the maximum bending stress?

• a.120 MPa✓
• b.240 MPa
• c.60 MPa
• d.12 MPa

Bending stress sigma = M·c/I, with I = b·h^3/12 = 0.05(0.1)^3/12 = 4.17x10^-6 m^4 and c = 0.05 m. sigma = 10,000 x 0.05/4.17x10^-6 = 120 MPa. Maximum stress occurs at the extreme fiber farthest from the neutral axis.

What is the section modulus of a rectangular cross-section 0.05 m wide and 0.1 m tall about its horizontal centroidal axis?

• a.8.33x10^-4 m^3
• b.4.17x10^-5 m^3
• c.8.33x10^-5 m^3✓
• d.1.67x10^-4 m^3

Section modulus S = I/c = b·h^2/6 = 0.05(0.1)^2/6 = 0.05 x 0.01/6 = 8.33x10^-5 m^3. Bending stress can be computed directly as sigma = M/S.

A bar with Poisson's ratio 0.3 is stretched to an axial strain of 0.001. What is the lateral (transverse) strain?

• a.-0.0001
• b.-0.001
• c.-0.0003✓
• d.-0.003

Lateral strain = -nu x axial strain = -0.3 x 0.001 = -0.0003. The negative sign shows the bar contracts laterally as it stretches axially, which is the Poisson effect.

A material has Young's modulus 200 GPa and Poisson's ratio 0.25. What is its shear modulus?

• a.100 GPa
• b.200 GPa
• c.80 GPa✓
• d.40 GPa

For an isotropic material, G = E/(2(1 + nu)) = 200/(2(1 + 0.25)) = 200/2.5 = 80 GPa. The three elastic constants E, G, and nu are interdependent for isotropic materials.

A fully restrained steel bar (E = 200 GPa, alpha = 12x10^-6 /degC) is heated by 50 degC. What thermal stress develops?

• a.240 MPa
• b.120 MPa✓
• c.60 MPa
• d.12 MPa

For full restraint, thermal stress sigma = E·alpha·(delta T) = 200x10^9 x 12x10^-6 x 50 = 120x10^6 Pa = 120 MPa (compressive). If the bar were free to expand, no stress would arise.

A pin-ended column (E = 200 GPa, I = 1x10^-6 m^4) is 2 m long. What is the Euler critical buckling load?

• a.1974 kN
• b.247 kN
• c.987 kN
• d.493 kN✓

Euler buckling load P_cr = pi^2·E·I/L^2 = pi^2 x 200x10^9 x 1x10^-6/2^2 = 9.87 x 200,000/4 = 493,000 N = 493 kN. Critical load drops with the square of the effective length.

A simply supported beam (E = 200 GPa, I = 1x10^-5 m^4) of span 4 m carries a 10 kN load at midspan. What is the maximum deflection?

• a.6.67 mm✓
• b.13.3 mm
• c.0.67 mm
• d.3.33 mm

Central-load deflection delta = P·L^3/(48·E·I) = 10,000 x 4^3/(48 x 200x10^9 x 1x10^-5) = 640,000/9.6x10^7 = 0.00667 m = 6.67 mm. Deflection grows with the cube of the span.

A cantilever beam (E = 200 GPa, I = 1x10^-5 m^4) of length 3 m carries a 5 kN load at its free end. What is the tip deflection?

• a.45 mm
• b.7.5 mm
• c.22.5 mm✓
• d.11.25 mm

Cantilever end deflection delta = P·L^3/(3·E·I) = 5000 x 3^3/(3 x 200x10^9 x 1x10^-5) = 135,000/6x10^6 = 0.0225 m = 22.5 mm. A cantilever deflects far more than a simply supported beam of the same span.

A cantilever beam 3 m long carries a 5 kN point load at its free end. What is the maximum bending moment?

• a.30 kN·m
• b.7.5 kN·m
• c.5 kN·m
• d.15 kN·m✓

For an end-loaded cantilever, the maximum moment occurs at the fixed support: M = P·L = 5 x 3 = 15 kN·m. The bending-moment diagram is triangular, largest at the wall.

A cantilever beam 3 m long carries a uniform load of 4 kN/m over its full length. What is the maximum bending moment?

• a.18 kN·m✓
• b.36 kN·m
• c.6 kN·m
• d.9 kN·m

For a uniformly loaded cantilever, maximum moment at the support = w·L^2/2 = 4 x 3^2/2 = 18 kN·m. The total load (12 kN) acts at the centroid 1.5 m from the wall, giving the same result.

A simply supported beam of span 4 m carries a uniform load of 6 kN/m. What is the maximum bending moment?

• a.48 kN·m
• b.6 kN·m
• c.12 kN·m✓
• d.24 kN·m

For a simply supported beam with a uniform load, maximum moment at midspan = w·L^2/8 = 6 x 4^2/8 = 12 kN·m. This peaks at the center, where shear is zero.