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

A particle starts from rest and accelerates uniformly at 3 m/s^2. What is its velocity after 4 seconds?

• a.0.75 m/s
• b.12 m/s✓
• c.7 m/s
• d.24 m/s

With constant acceleration, v = v0 + a·t = 0 + (3)(4) = 12 m/s. This is the basic kinematic equation for linear motion starting from rest.

What is the kinetic energy of a 10 kg mass moving at 4 m/s?

• a.40 J
• b.80 J✓
• c.160 J
• d.20 J

Kinetic energy KE = (1/2) m v^2 = 0.5 x 10 x 4^2 = 0.5 x 10 x 16 = 80 J. Note the velocity is squared, so KE grows rapidly with speed.

What is the linear momentum of a 5 kg object moving at 10 m/s?

• a.100 kg·m/s
• b.2 kg·m/s
• c.50 kg·m/s✓
• d.15 kg·m/s

Linear momentum p = m·v = 5 kg x 10 m/s = 50 kg·m/s. Momentum is conserved in the absence of external impulse, the basis of impact and collision analysis.

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.