Free practice problems and solutions for the FE exams

Question 1.

If a train is travelling at 55 m/s at its maximum acceleration and takes 2 minutes to reach 70 m/s, how long would it take for the train to stop when traveling at 40 m/s, if it can decelerate four times faster than it can accelerate?

60 s
120 s
80 s
280 s

Answer 1: c: 80s

Solution:

Apply Equations of motion to calculate the maximum acceleration:

V = U + A x T

70 = 55 + 120 x a

Maximum acceleration, a = 0.125 m/s²

Maximum deceleration = 4 x 0.125 = 0.5 m/s²

Now apply equations of motion to train starting at 70 m/s and coming to rest at this rate of deceleration:

V = U + A x T

0 = 40 + 0.5 x T

Which gives the time taken, T = 80 s

Question 2.

An engineer is designing the runway for an airport. The lowest rate at which the planes using this runway can accelerate is 4 m/s. Based upon a takeoff speed of 65 m/s, what is the minimum length of the runway?

480 m
528 m
220 m
762 m

Answer 2: b. 528 m

Solution:

Apply equations of motion to calculate the distance the plane travels, while accelerating from rest to 65 m/s

V² = U² + 2 x A x

65² = 0 + 2 x 4 x S

Which gives the distance, S = 528 m

Question 3.

The external work done by a pump that delivers 2,500 cubic meters of water per hour against a total head of 5.9 meters is most nearly:

30 kW
40 kW
50 kW
60 kW

Answer 3: b. 40 kW

Solution:

Apply the pump power equation to solve this problem

P = Q x Y x H / (E / 100)

The flow rate, Q must be converted to standard units:

Q = 2,500 m³/hour = 2,500 / (60 x 60) m³/s = 0.694 m³/s

The specific weight of fluid (water), Y = density x g = 1000 kg/m³ x 9.81 m/s² = 9,800 N/m³

Therefore

P = 0.694 x 5.9 x 9,800 / (100/100)

P = 40,000 W = 40 kW

Question 4.

When determining the strength of commercial materials using the testing techniques of the American Society for Testing Materials, the American Iron and Steel Institute, or the Society of Automotive Engineers, the results of the common tensile test are used to determine the material characteristics listed below EXCEPT:

The fatigue strength
The tensile strength
The modulus of elasticity
The yield strength

Answer 4: a. The fatigue strength.

Question 5.

To measure the muzzle velocity of a bullet fired from a rifle, two cardboard discs were mounted on a long axle which was rotated at constant speed. The discs were one meter apart along the axle which was rotating at a speed of 3,000 rpm. The rifle was mounted parallel to the axle and fired toward the discs. The resulting bullet hole in the second disc was displaced 20 degrees with respect to the hole in the first disc. The muzzle velocity was most nearly:

600 m/s
700 m/s
800 m/s
900 m/s

Answer 5: d. 900 m/s

Solution:

If we calculate the time it takes the axle to rotate 20, then we can calculate the time it takes the bullet to travel the distance between the two discs.

The discs rotate at 3,00 rpm which is equal to 3,000 / 60 = 50 revolutions per second

20º is 20º / 360º = 0.0556 of a complete revolution

Therfore the disc rotates 20º in 0.0556 / 50 = 0.0011 s

Which means the bullet travels 1 m in the same amount of time

The linear velocity of the bullet is therefore 1 / 0.0011 m/s = 900 m/s

Question 6.

If the sum of $12,000 is borrowed and the debtor is obligated to pay the creditor $900 for each year the loan is in existence, then, the simple interest is most nearly:

2.0%
5.0%
7.5%
12.5%

Answer 6: c. 7.5%

Solution:

The $900 per year represents the interest alone and doesn’t include payment of the principal balance

Therefore the question is simply a matter of calculating the percentage of $12,000 that is equal to $900

900 / 12,000 = 0.075 = 7.5%

Question 7.

The statement of those given below which is NOT CORRECT is:

A moment is defined as the product of the magnitude of the force and the perpendicular distance from a point to the line of action of a force.
A scalar quantity is not completely described by its magnitude.
For a vector quantity, both the magnitude and direction should be specified.
Whenever a number of vectors add to give a resultant that is zero, the components of the resultant vector must separately be zero.

Answer 7: b. A scalar quantity is not completely described by its magnitude.

Question 8.

A car travels around a curve of radius 200 m at a speed of 14 m/s and the force produced by the tires is 965 N. Calculate the car’s weight.

985 kg
456 kg
885 kg
725 kg

Answer 8: a. 985 kg

Solution:

Use the formula for centripetal force:

F = M*R*(W^2)

965 = M x 200 x W²

Calculate the angular velocity, W from the angular velocity equation W = V/R

W = 14 / 200 = 0.07

Therefore 965 = M x 200 x 0.07²

So the mass, M = 985 kg

Question 9.

A curve of a road has a radius of 50 m. What is the correct banking angle required to eliminate the frictional force for vehicles traveling around this curve at a speed of 10 m/s?

16.8º
6.5º
8.2º
11.5º

Answer 9: d. 11.5º

Solution:

Use the formula for banking of curves:

A= ATAN((V^2)/(G*R))

A = ATAN ((10²⁾ / (9.8 x 50))

A = 11.5

So the desired banking angle is 11.5º

Question 10.

A Cannon is located on the roof of a building 60 m high. The cannon fires a projectile at 25 m/s at an angle of 53 degrees above the horizontal. The projectile lands on the ground at what horizontal distance from the point it was fired? Effects of air resistance can be ignored.

110 m
75 m
90 m
60 m

Answer 10: c. 90 m

Solution:

First, calculate the X and Y components of the initial velocity

V_x = V x cos53º V_y = V x sin 53º

V_x = 25 x 0.6 V_y = 25 x 0.8

V_x = 15 m/s V_y = 20 m/s

Vertical motion:

Calculate the time taken for the projectile to reach its maximum height:

V_final = V_initial – g x T

0 = 20 – 9.8 x T

T = 2 s

Calculate the maximum height:

Y = V_y x T x SIN(A) – g x T² / 2

H = 20– 9.8 x 2²/ 2

H = 20 m

H_max – 20 + 60 = 80 m

Free fall from the maximum height:

H = ½ x g x T²

80 = ½ x 9.8 x T²

T = 4 s

Therefore the total time = 4 + 2 = 6 s

Horizontal motion:

X = V x T = 15 x 6

X = 90 m

Question 11.

Using the Gibbs Phase Rule, determine how many degrees of freedom exist at the triple point of water.

Answer 11: a. 0

Solution:

The Gibbs Phase Rule states that:

F = C – P + 2

Where:

F = Degrees of Freedom

C = Number of Components

P = Number of phases (solid, liquid or gas)

At the triple point of water, C = 1 (water) and P = 3 (solid, liquid, gas)

Therefore, F = 1 – 3 + 2 = 0