Free FE Mechanical Practice Test

59 original FE Mechanical practice questions covering all 14 NCEES knowledge areas, each with a worked solution and the reason every wrong answer is wrong. This is an unofficial short practice set, with no signup.

Practice questions

Mathematics

FEM-01 · Mathematics

You apply one step of Newton's method to f(x) = x³ − 2x − 5, starting from x₀ = 2. What is x₁?

A. 1.900
B. 2.083
C. 2.100
D. 2.500

Answer and explanation

Answer: C. Newton's step is x₁ = x₀ − f(x₀)/f′(x₀). Here f(2) = 8 − 4 − 5 = −1 and f′(x) = 3x² − 2, so f′(2) = 10. Then x₁ = 2 − (−1/10) = 2.100.

Common misses: A adds the correction instead of subtracting it. B differentiates x³ − 2x as 3x² and drops the −2, so f′(2) = 12. D plugs in the constant −5 instead of f(2).

Topic: Mathematics, 1E Numerical methods. Principle: OpenStax Calculus Vol. 1, 4.9 Newton's Method.

Answer and explanation

Answer: C. Newton's step is x₁ = x₀ − f(x₀)/f′(x₀). Here f(2) = 8 − 4 − 5 = −1 and f′(x) = 3x² − 2, so f′(2) = 10. Then x₁ = 2 − (−1/10) = 2.100.

Common misses: A adds the correction instead of subtracting it. B differentiates x³ − 2x as 3x² and drops the −2, so f′(2) = 12. D plugs in the constant −5 instead of f(2).

Topic: Mathematics, 1E Numerical methods. Principle: OpenStax Calculus Vol. 1, 4.9 Newton's Method.

How this set works: 59 questions in NCEES subject order, with more questions in the bigger knowledge areas. Most have one correct answer. Six are "select all that apply" and six ask you to enter a number, two of the formats NCEES says it uses besides standard multiple choice. As on the real exam, there's no partial credit: a select-all question counts only if you pick every correct choice and nothing else (NCEES computer-based testing).

Work them with an approved calculator and the free FE Reference Handbook open, the way you will on exam day. Use g = 9.81 m/s² and a water density of 1,000 kg/m³ unless a question says otherwise.

FEM-02 · Mathematics

A part leaves an oven at 90°C into a room at 20°C. Its temperature follows dT/dt = −k(T − 20), with k = 0.05 per minute. What is its temperature after 20 minutes?

A. 20.0°C
B. 25.8°C
C. 33.1°C
D. 45.8°C

Answer and explanation

Answer: D. This separable equation solves to T(t) = T_room + (T₀ − T_room)e^(−kt). With kt = 0.05 × 20 = 1: T = 20 + 70e^(−1) = 20 + 25.75 = 45.8°C.

Common misses: B is just the decaying difference, 70e^(−1); it forgets to add back the room temperature. C decays the whole 90°C toward zero instead of toward 20°C. A treats the decay as linear: 20 + 70 × (1 − 0.05 × 20) = 20°C.

Topic: Mathematics, 1C Ordinary differential equations. Principle: LibreTexts, Trench, 4.2 Cooling and Mixing (Newton's law of cooling).

Answer and explanation

Answer: D. This separable equation solves to T(t) = T_room + (T₀ − T_room)e^(−kt). With kt = 0.05 × 20 = 1: T = 20 + 70e^(−1) = 20 + 25.75 = 45.8°C.

Topic: Mathematics, 1C Ordinary differential equations. Principle: LibreTexts, Trench, 4.2 Cooling and Mixing (Newton's law of cooling).

FEM-03 · Mathematics

What is the determinant of the matrix with rows (3, 1, 0), (2, 4, 1), and (0, 1, 2)?

A. 17
B. 21
C. 24
D. 25

Answer and explanation

Answer: A. Expand along the first row: 3(4·2 − 1·1) − 1(2·2 − 1·0) + 0 = 3(7) − 1(4) = 21 − 4 = 17.

Common misses: B stops after the first cofactor. D gives the second cofactor a plus sign; the signs alternate +, −, +. C multiplies the main diagonal (3 × 4 × 2), which does not give the determinant for this matrix.

Topic: Mathematics, 1D Linear algebra. Principle: LibreTexts, Interactive Linear Algebra, 4.2 Cofactor Expansions.

Answer and explanation

Answer: A. Expand along the first row: 3(4·2 − 1·1) − 1(2·2 − 1·0) + 0 = 3(7) − 1(4) = 21 − 4 = 17.

Topic: Mathematics, 1D Linear algebra. Principle: LibreTexts, Interactive Linear Algebra, 4.2 Cofactor Expansions.

FEM-04 · Mathematics

What value does this pseudocode output?

S = 0
FOR i = 1 TO 5
    IF i MOD 2 = 1 THEN S = S + i^2
NEXT i
OUTPUT S

A. 9
B. 20
C. 35
D. 55

Answer and explanation

Answer: C. i MOD 2 = 1 is true only for odd i (1, 3, 5), so S = 1² + 3² + 5² = 1 + 9 + 25 = 35.

Common misses: D squares and adds every i from 1 to 5. A adds the odd numbers without squaring them. B squares the even numbers (4 + 16) by reading the test backward.

Topic: Mathematics, 1F Algorithm and logic development. Principle: trace the loop one pass at a time; the MOD (remainder) test selects odd values. Cambridge International, 2026 Pseudocode Guide, §§5.2, 6.1, and 7.1.

Answer and explanation

Answer: C. i MOD 2 = 1 is true only for odd i (1, 3, 5), so S = 1² + 3² + 5² = 1 + 9 + 25 = 35.

Common misses: D squares and adds every i from 1 to 5. A adds the odd numbers without squaring them. B squares the even numbers (4 + 16) by reading the test backward.

Probability and statistics

FEM-05 · Probability and statistics

A worn gearbox can be rebuilt now for $4,000. If you wait a year, there is a 30% chance it fails, costing $12,000, and a 70% chance it only needs a $1,000 repair. What is the expected cost of waiting?

A. $3,600
B. $4,300
C. $6,500
D. $13,000

Answer and explanation

Answer: B. Weight each outcome by its probability: E = 0.30(12,000) + 0.70(1,000) = 3,600 + 700 = $4,300. Ignoring the time value of money, rebuilding now ($4,000) is the cheaper choice on expected cost alone.

Common misses: A keeps only the failure term. C averages the two costs as if they were equally likely. D adds them.

Topic: Probability and Statistics, 2C Expected value in decision making. Principle: OpenStax Introductory Statistics 2e, 4.2 Mean or Expected Value.

Answer and explanation

Common misses: A keeps only the failure term. C averages the two costs as if they were equally likely. D adds them.

Topic: Probability and Statistics, 2C Expected value in decision making. Principle: OpenStax Introductory Statistics 2e, 4.2 Mean or Expected Value.

FEM-06 · Probability and statistics

Fit a least-squares line y = a + bx to the points (1, 2), (2, 4), (3, 5), and (4, 7). Enter the slope b, to one decimal place.

Enter a number.

Answer and explanation

Answer: 1.6. x̄ = 2.5 and ȳ = 4.5. S_xy = Σ(x − x̄)(y − ȳ) = 3.75 + 0.25 + 0.25 + 3.75 = 8.0. S_xx = Σ(x − x̄)² = 2.25 + 0.25 + 0.25 + 2.25 = 5.0. So b = 8.0 ÷ 5.0 = 1.6, and the intercept is a = 4.5 − 1.6(2.5) = 0.5.

Common misses: 1.67 joins the first and last points, (7 − 2) ÷ 3, which ignores the middle data. 0.5 is the intercept, not the slope.

Topic: Probability and Statistics, 2D Regression and curve fitting. Principle: least-squares slope b = S_xy ÷ S_xx; check the regression formulas in your handbook's statistics section. NIST/SEMATECH e-Handbook, 4.4.3.1 Least Squares.

Answer and explanation

Common misses: 1.67 joins the first and last points, (7 − 2) ÷ 3, which ignores the middle data. 0.5 is the intercept, not the slope.

Ethics and professional practice

FEM-07 · Ethics and professional practice

An engineer is asked by her employer to score bids from three pump suppliers. Her brother-in-law owns one of the bidders. Under the NCEES Model Rules, what should she do?

A. Disclose the relationship to her employer
B. Quietly give that bidder a slightly lower score to offset any bias
C. Decline the task without giving a reason
D. Score the bids, since she will judge them on merit

Answer and explanation

Answer: A. §240.15 B.6 requires licensees to disclose to employers or clients any known or potential conflict of interest that could influence, or appear to influence, their judgment. A family tie to a bidder is exactly that. Once it's disclosed, the employer can decide how to handle it.

Common misses: D and B keep the conflict hidden; the rule is about appearance, not only actual bias. C avoids the work but still leaves the employer uninformed.

Topic: Ethics and Professional Practice, 3A Codes of ethics. Principle: NCEES Model Rules, August 2026, §240.15 B.6.

Answer and explanation

Common misses: D and B keep the conflict hidden; the rule is about appearance, not only actual bias. C avoids the work but still leaves the employer uninformed.

Topic: Ethics and Professional Practice, 3A Codes of ethics. Principle: NCEES Model Rules, August 2026, §240.15 B.6.

FEM-08 · Ethics and professional practice

A plant's lead engineer finds that a pressure-relief valve on a new steam system is undersized and could let the vessel over-pressurize. The operations manager overrules her and orders startup on schedule. Under the NCEES Model Rules, what should she do?

A. Start up as ordered, since the manager is responsible for the decision
B. Write a memo to file and start up as ordered
C. Notify her employer or client and any other appropriate authority
D. Resign and say nothing further

Answer and explanation

Answer: C. §240.15 A.3: when a licensee's judgment is overruled under circumstances that endanger public health, safety, or welfare, the licensee notifies the employer or client and other authority as appropriate. A.1 makes public safety the first and foremost responsibility.

Common misses: A and B let the hazard proceed. D removes her from the problem but does not fulfill the notification duty after her judgment is overruled.

Topic: Ethics and Professional Practice, 3B Public health, safety, and welfare. Principle: NCEES Model Rules, August 2026, §240.15 A.

Answer and explanation

Common misses: A and B let the hazard proceed. D removes her from the problem but does not fulfill the notification duty after her judgment is overruled.

Topic: Ethics and Professional Practice, 3B Public health, safety, and welfare. Principle: NCEES Model Rules, August 2026, §240.15 A.

FEM-09 · Ethics and professional practice

Select all that apply. Which statements about U.S. intellectual property are accurate?

A. Copyright can protect the written text of a maintenance manual, but not the maintenance method it describes.
B. A utility patent gives its owner the right to exclude others from making, using, selling, or importing the invention.
C. Information can lose trade-secret protection if its owner doesn't make reasonable efforts to keep it secret.
D. Writing down an idea gives it copyright protection.
E. A U.S. utility patent generally lasts 20 years from the date it is issued.

Answer and explanation

Answers: A, B, C. Copyright covers the expression, not ideas, systems, or methods of operation (A). A patent is a right to exclude (B). Trade-secret status depends on the information staying secret through reasonable efforts (C). You need all three, and no others, to get this right.

Common misses: D: writing it down doesn't make an idea copyrightable; only the particular expression is. E: the term generally runs up to 20 years from the filing date of the application, not the issue date.

Topic: Ethics and Professional Practice, 3C Intellectual property. Principle: U.S. Copyright Office, What Does Copyright Protect?; USPTO, Patent essentials; USPTO, Trade secret policy.

Answer and explanation

Topic: Ethics and Professional Practice, 3C Intellectual property. Principle: U.S. Copyright Office, What Does Copyright Protect?; USPTO, Patent essentials; USPTO, Trade secret policy.

Engineering economics

FEM-10 · Engineering economics

A team has already spent $30,000 on a prototype. Finishing it would cost another $20,000 and produce savings with a present worth of $45,000. Abandoning it now costs nothing more. On economic grounds, what should the team do, and why?

A. Finish it: the net present worth is −$5,000, but the money is already spent
B. Abandon it: the project has already lost $30,000
C. Abandon it: total cost $50,000 exceeds $45,000 in savings
D. Finish it: the relevant net present worth is +$25,000

Answer and explanation

Answer: D. The $30,000 is a sunk cost. It's gone whichever option you pick, so it doesn't affect the decision. Compare only what changes: finishing costs $20,000 more and returns $45,000, a net +$25,000. Abandoning returns $0.

Common misses: C and A count the sunk $30,000. B treats a past loss as a reason to stop.

Topic: Engineering Economics, 4B Cost types (sunk). Principle: OpenStax Principles of Managerial Accounting, 10.1 Identify Relevant Information for Decision-Making.

Answer and explanation

Common misses: C and A count the sunk $30,000. B treats a past loss as a reason to stop.

Topic: Engineering Economics, 4B Cost types (sunk). Principle: OpenStax Principles of Managerial Accounting, 10.1 Identify Relevant Information for Decision-Making.

FEM-11 · Engineering economics

A compressor upgrade costs $12,000 today. It saves $2,500 at the end of each year for 6 years and has a $1,000 salvage value at year 6. At an effective annual interest rate of 8%, what is the net present worth?

A. −$443
B. +$187
C. +$557
D. +$4,000

Answer and explanation

Answer: B. (P/A, 8%, 6) = [(1.08)⁶ − 1] ÷ [0.08(1.08)⁶] ≈ 4.6229, and (P/F, 8%, 6) = (1.08)^(−6) ≈ 0.6302. NPW = −12,000 + 2,500(4.6229) + 1,000(0.6302) = −12,000 + 11,557 + 630 ≈ +$187. The upgrade barely clears 8%.

Common misses: A leaves out the salvage value. C adds the $1,000 salvage without discounting it. D ignores the time value of money: −12,000 + 6 × 2,500 + 1,000.

Topic: Engineering Economics, 4A Time value of money. Principle: Penn State EME 460, Compound Interest Formulas III (P/A); Penn State EME 460, Nominal, Period, and Effective Interest Rates.

Answer and explanation

Common misses: A leaves out the salvage value. C adds the $1,000 salvage without discounting it. D ignores the time value of money: −12,000 + 6 × 2,500 + 1,000.

Topic: Engineering Economics, 4A Time value of money. Principle: Penn State EME 460, Compound Interest Formulas III (P/A); Penn State EME 460, Nominal, Period, and Effective Interest Rates.

Electricity and magnetism

FEM-12 · Electricity and magnetism

A single-phase load draws 15 A rms from a 240 V rms supply at a power factor of 0.80 lagging. What real power does it consume?

A. 2,880 W
B. 3,600 W
C. 4,500 W
D. 2,160 W

Answer and explanation

Answer: A. Real (average) power P = V_rms × I_rms × cos φ = 240 × 15 × 0.80 = 2,880 W.

Common misses: B is the apparent power, 3,600 VA, with no power factor. D is the reactive power, 3,600 × sin φ = 3,600 × 0.60 = 2,160 var. C divides by the power factor instead of multiplying.

Topic: Electricity and Magnetism, 5C AC circuit analysis. Principle: OpenStax University Physics Vol. 2, 15.4 Power in an AC Circuit.

Answer and explanation

Answer: A. Real (average) power P = V_rms × I_rms × cos φ = 240 × 15 × 0.80 = 2,880 W.

Common misses: B is the apparent power, 3,600 VA, with no power factor. D is the reactive power, 3,600 × sin φ = 3,600 × 0.60 = 2,160 var. C divides by the power factor instead of multiplying.

Topic: Electricity and Magnetism, 5C AC circuit analysis. Principle: OpenStax University Physics Vol. 2, 15.4 Power in an AC Circuit.

FEM-13 · Electricity and magnetism

A 4-pole, 60 Hz induction motor runs at 1,746 rpm. What is its slip, in percent, to one decimal place?

Enter a number in percent.

Answer and explanation

Answer: 3.0%. Synchronous speed n_s = 120f/p = 120 × 60 ÷ 4 = 1,800 rpm. Slip = (1,800 − 1,746) ÷ 1,800 = 0.030, or 3.0%.

Common misses: 51.5% uses 3,600 rpm, the synchronous speed of a 2-pole motor. 54 is the speed difference in rpm, not a percentage.

Topic: Electricity and Magnetism, 5D Motors and generators. Principle: LibreTexts, Kuphaldt, 13.7 Tesla Polyphase Induction Motors.

Answer and explanation

Answer: 3.0%. Synchronous speed n_s = 120f/p = 120 × 60 ÷ 4 = 1,800 rpm. Slip = (1,800 − 1,746) ÷ 1,800 = 0.030, or 3.0%.

Common misses: 51.5% uses 3,600 rpm, the synchronous speed of a 2-pole motor. 54 is the speed difference in rpm, not a percentage.

Topic: Electricity and Magnetism, 5D Motors and generators. Principle: LibreTexts, Kuphaldt, 13.7 Tesla Polyphase Induction Motors.

FEM-14 · Electricity and magnetism

A 24 V DC source feeds a 4 Ω resistor in series with a parallel pair of 6 Ω and 12 Ω resistors. How much power does the 12 Ω resistor dissipate?

A. 12 W
B. 24 W
C. 36 W
D. 48 W

Answer and explanation

Answer: A. The parallel pair is 6 × 12 ÷ (6 + 12) = 4 Ω, so the total is 8 Ω and I = 24 ÷ 8 = 3 A. The parallel pair sees 3 A × 4 Ω = 12 V, so the 12 Ω resistor dissipates V²/R = 12² ÷ 12 = 12 W.

Common misses: B is the 6 Ω branch (12² ÷ 6). C is the whole parallel pair (3² × 4). D puts the full 24 V across the 12 Ω resistor.

Topic: Electricity and Magnetism, 5B DC circuit analysis. Principle: OpenStax University Physics Vol. 2, 10.2 Resistors in Series and Parallel.

Answer and explanation

Common misses: B is the 6 Ω branch (12² ÷ 6). C is the whole parallel pair (3² × 4). D puts the full 24 V across the 12 Ω resistor.

Topic: Electricity and Magnetism, 5B DC circuit analysis. Principle: OpenStax University Physics Vol. 2, 10.2 Resistors in Series and Parallel.

Statics

FEM-15 · Statics

A horizontal beam has a pin support at A (x = 0) and a roller at B (x = 6 ft). A downward uniform load of 3 kip/ft acts only from x = 0 to x = 4 ft, and a downward 8 kip point load acts at x = 5 ft. Ignore the beam's weight. What is the upward reaction at B?

A. 10.67 kip
B. 14.67 kip
C. 6.67 kip
D. 9.33 kip

Answer and explanation

Answer: A. Replace the distributed load with its resultant: 3 kip/ft × 4 ft = 12 kip, acting at the middle of the loaded length, x = 2 ft. Take moments about the pin so its reaction drops out: 6R_B = 12(2) + 8(5) = 64, so R_B = 10.67 kip. Check with vertical balance: the pin carries 20 − 10.67 = 9.33 kip.

Common misses: C leaves out the distributed load. D is the reaction at the pin, not the roller. B puts the 12 kip resultant at x = 4 ft, the end of the loaded length, instead of its middle.

Topic: Statics, 6A Resultants of force systems and 6C Equilibrium of rigid bodies. Principle: Penn State Mechanics Map, Finding the Equivalent Point Load (integration); OpenStax University Physics Vol. 1, 12.1 Conditions for Static Equilibrium.

Answer and explanation

Common misses: C leaves out the distributed load. D is the reaction at the pin, not the roller. B puts the 12 kip resultant at x = 4 ft, the end of the loaded length, instead of its middle.

FEM-16 · Statics

A rigid rectangular crate weighs 800 N. Its base is 0.60 m wide in the direction of a push, and its center of gravity is centered over the base. A horizontal push is applied 0.90 m above the level floor and increased slowly from zero. The coefficient of static friction is 0.40, and the crate is tall enough for that push point. What happens first, and at what push?

A. It tips, at 533 N
B. It slides, at 267 N
C. It slides, at 320 N
D. It tips, at 267 N

Answer and explanation

Answer: D. Check both ways it can move. Sliding starts when the push reaches the friction limit: μ_s W = 0.40 × 800 = 320 N. Tipping starts when the push's moment about the front bottom edge matches the weight's: P(0.90) = 800(0.30), so P = 267 N. The tipping threshold comes first. At 267 N the floor only needs to supply 267 N of friction, below the 320 N available, so the crate doesn't slide first.

Common misses: B has the right number but the wrong mode. C checks friction only. A uses the full 0.60 m base width as the weight's lever arm; the weight acts at the center, 0.30 m from the edge.

Topic: Statics, 6C Equilibrium of rigid bodies and 6F Static friction. Principle: OpenStax University Physics Vol. 1, 12.1 Conditions for Static Equilibrium (torque balance, Eq. 12.5); 6.2 Friction.

Answer and explanation

Common misses: B has the right number but the wrong mode. C checks friction only. A uses the full 0.60 m base width as the weight's lever arm; the weight acts at the center, 0.30 m from the edge.

FEM-17 · Statics

An L-shaped area is made of two rectangles. Rectangle 1 is 100 mm wide and 20 mm tall, sitting on the base line. Rectangle 2 is 20 mm wide and 80 mm tall, sitting on top of rectangle 1 at its left end. How far above the base is the centroid?

A. 23.3 mm
B. 32.2 mm
C. 35.0 mm
D. 50.0 mm

Answer and explanation

Answer: B. A₁ = 2,000 mm² with its centroid at y = 10 mm. A₂ = 1,600 mm²; it starts at y = 20 mm, so its centroid is at 20 + 40 = 60 mm. ȳ = (2,000 × 10 + 1,600 × 60) ÷ 3,600 = 116,000 ÷ 3,600 = 32.2 mm.

Common misses: A measures rectangle 2's centroid from its own base (40 mm) instead of from the common base line. C averages the two centroids without weighting by area. D is half the overall height.

Topic: Statics, 6E Centroids. Principle: Engineering Statics, 7.5 Centroids using Composite Parts.

Answer and explanation

Topic: Statics, 6E Centroids. Principle: Engineering Statics, 7.5 Centroids using Composite Parts.

FEM-18 · Statics

A rectangle is 40 mm wide (b) and 60 mm tall (h). What is its area moment of inertia about a horizontal axis 50 mm from its centroid?

A. 0.720 × 10⁶ mm⁴
B. 6.00 × 10⁶ mm⁴
C. 6.32 × 10⁶ mm⁴
D. 6.72 × 10⁶ mm⁴

Answer and explanation

Answer: D. About the centroidal axis, Ī = bh³/12 = 40 × 60³ ÷ 12 = 720,000 mm⁴. The parallel axis theorem adds Ad² = (40 × 60)(50²) = 6,000,000 mm⁴. Total: 6.72 × 10⁶ mm⁴.

Common misses: A is the centroidal value alone. B is the Ad² term alone. C swaps b and h in Ī (60 × 40³ ÷ 12).

Topic: Statics, 6E Moments of inertia. Principle: Engineering Statics, 10.3 Parallel Axis Theorem; 10.2 Moments of Inertia of Common Shapes.

Answer and explanation

Answer: D. About the centroidal axis, Ī = bh³/12 = 40 × 60³ ÷ 12 = 720,000 mm⁴. The parallel axis theorem adds Ad² = (40 × 60)(50²) = 6,000,000 mm⁴. Total: 6.72 × 10⁶ mm⁴.

Common misses: A is the centroidal value alone. B is the Ad² term alone. C swaps b and h in Ī (60 × 40³ ÷ 12).

Topic: Statics, 6E Moments of inertia. Principle: Engineering Statics, 10.3 Parallel Axis Theorem; 10.2 Moments of Inertia of Common Shapes.

FEM-19 · Statics

A simple triangular truss has a pin support at A (0, 0), a roller at B (6 m, 0), and its apex at C (3 m, 4 m). A single 12 kN load acts downward at C. What is the force in the bottom chord AB?

A. 4.5 kN compression
B. 4.5 kN tension
C. 6.0 kN tension
D. 7.5 kN compression

Answer and explanation

Answer: B. By symmetry each support carries 6 kN up. Member AC is 5 m long (a 3-4-5 triangle). At joint A, vertical balance: 6 + F_AC(4/5) = 0, so F_AC = −7.5 kN (compression). Horizontal balance: F_AB + F_AC(3/5) = 0, so F_AB = +4.5 kN. Positive means it pulls on the joint: tension.

Common misses: D is the force in AC, not AB. C is the support reaction. A gets the size right but the sense wrong: the bottom chord ties the two supports together, so it's stretched.

Topic: Statics, 6D Frames and trusses. Principle: Engineering Statics, 6.4 Method of Joints.

Answer and explanation

Common misses: D is the force in AC, not AB. C is the support reaction. A gets the size right but the sense wrong: the bottom chord ties the two supports together, so it's stretched.

Topic: Statics, 6D Frames and trusses. Principle: Engineering Statics, 6.4 Method of Joints.

Dynamics, kinematics, and vibrations

FEM-20 · Dynamics, kinematics, and vibrations

A 20.0 kg crate is already sliding down a 30.0° incline at 2.00 m/s. The coefficient of kinetic friction is 0.200. Ignoring air resistance, what is its speed after it slides another 3.00 m down the incline?

A. 4.65 m/s
B. 4.82 m/s
C. 5.78 m/s
D. 4.39 m/s

Answer and explanation

Answer: B. The normal force on an incline is mg cos 30°, so kinetic friction is μ_k mg cos 30°, acting up the slope. Gravity does positive work and friction does negative work: ½mv² − ½mv₀² = mgs sin 30° − μ_k mgs cos 30°. The mass cancels: v = √[2.00² + 2(9.81)(3.00)(sin 30° − 0.200 cos 30°)] = 4.82 m/s. It speeds up because the downhill pull of gravity beats friction.

Common misses: D drops the starting kinetic energy. A uses the full weight, μ_k mg, as the friction force; on a slope the normal force is smaller than the weight. C ignores friction.

Topic: Dynamics, 7B Kinetic friction and 7D Work-energy of particles. Principle: OpenStax University Physics Vol. 1, 7.3 Work-Energy Theorem; 6.2 Friction.

Answer and explanation

Common misses: D drops the starting kinetic energy. A uses the full weight, μ_k mg, as the friction force; on a slope the normal force is smaller than the weight. C ignores friction.

Topic: Dynamics, 7B Kinetic friction and 7D Work-energy of particles. Principle: OpenStax University Physics Vol. 1, 7.3 Work-Energy Theorem; 6.2 Friction.

FEM-21 · Dynamics, kinematics, and vibrations

A 0.15 kg ball hits a wall at 30 m/s and rebounds straight back at 20 m/s. Contact lasts 5 ms. What is the magnitude of the average net force on the ball?

A. 300 N
B. 900 N
C. 1,500 N
D. 7,500 N

Answer and explanation

Answer: C. Impulse equals change in momentum. Taking motion toward the wall as positive, velocity changes from +30 to −20 m/s, so Δv = −50 m/s. F_avg = mΔv/Δt = 0.15 × (−50) ÷ 0.005 = −1,500 N. Its magnitude is 1,500 N, directed away from the wall.

Common misses: A subtracts the speeds (30 − 20) and ignores the reversal. B uses only the incoming speed, as if the ball stopped dead. D uses 1 ms instead of 5 ms; convert time units carefully.

Topic: Dynamics, 7E Impulse-momentum of particles. Principle: OpenStax University Physics Vol. 1, 9.2 Impulse and Collisions.

Answer and explanation

Common misses: A subtracts the speeds (30 − 20) and ignores the reversal. B uses only the incoming speed, as if the ball stopped dead. D uses 1 ms instead of 5 ms; convert time units carefully.

Topic: Dynamics, 7E Impulse-momentum of particles. Principle: OpenStax University Physics Vol. 1, 9.2 Impulse and Collisions.

FEM-22 · Dynamics, kinematics, and vibrations

A uniform solid cylinder starts from rest and rolls without slipping down a ramp, dropping 2.0 m in height. Neglect energy losses. What is the speed of its center at the bottom?

A. 6.26 m/s
B. 8.86 m/s
C. 4.43 m/s
D. 5.11 m/s

Answer and explanation

Answer: D. Energy is conserved; static friction at the contact point does no work in rolling without slipping. mgh = ½mv² + ½Iω², with I = ½mr² and ω = v/r. That gives mgh = ¾mv², so v = √(4gh/3) = √(4 × 9.81 × 2.0 ÷ 3) = 5.11 m/s.

Common misses: A is a frictionless slide, √(2gh), with no energy going into spin. C is a thin hoop (I = mr²), √(gh). B is √(4gh), which drops the 3.

Topic: Dynamics, 7I Work-energy of rigid bodies. Principle: OpenStax University Physics Vol. 1, 11.1 Rolling Motion.

Answer and explanation

Common misses: A is a frictionless slide, √(2gh), with no energy going into spin. C is a thin hoop (I = mr²), √(gh). B is √(4gh), which drops the 3.

Topic: Dynamics, 7I Work-energy of rigid bodies. Principle: OpenStax University Physics Vol. 1, 11.1 Rolling Motion.

FEM-23 · Dynamics, kinematics, and vibrations

A 12.0 kg mass slides on a frictionless horizontal guide. Two springs act on it in parallel, so both always deflect the same amount: k₁ = 1,800 N/m and k₂ = 3,000 N/m. Ignoring damping, what is its natural frequency?

A. 3.18 Hz
B. 20.0 Hz
C. 1.54 Hz
D. 2.52 Hz

Answer and explanation

Answer: A. Springs in parallel share the same deflection, so their stiffnesses add: k = 1,800 + 3,000 = 4,800 N/m. Then ω_n = √(k/m) = √(4,800 ÷ 12.0) = 20.0 rad/s, and f_n = ω_n ÷ 2π = 3.18 Hz.

Common misses: C combines the springs as if they were in series (k = 1,125 N/m). D uses only the 3,000 N/m spring. B is ω_n in rad/s, not hertz.

Topic: Dynamics, 7K Free vibrations. Principle: OpenStax University Physics Vol. 1, 15.1 Simple Harmonic Motion; MIT OCW 3.11, Introduction to Elasticity (stiffness; Problem 6, springs in series and parallel).

Answer and explanation

Common misses: C combines the springs as if they were in series (k = 1,125 N/m). D uses only the 3,000 N/m spring. B is ω_n in rad/s, not hertz.

FEM-24 · Dynamics, kinematics, and vibrations

Select all that apply. A wheel of radius 0.30 m rolls without slipping, and its center moves at 6.0 m/s. Which statements are true at this instant?

A. The wheel's angular speed is 20 rad/s.
B. The point at the top of the wheel moves at 12 m/s.
C. The point touching the ground moves at 6.0 m/s.
D. Every point on the rim moves at 6.0 m/s.

Answer and explanation

Answers: A, B. Rolling without slipping means v_center = rω, so ω = 6.0 ÷ 0.30 = 20 rad/s. The contact point is momentarily at rest, so the wheel turns about it: the top, twice as far away, moves at 2 × 6.0 = 12 m/s.

Common misses: C: the contact point's velocity is zero; that's what "no slip" means. D: rim points do not all have the same speed relative to the ground. Rim points combine translation with rotation, so their speeds range from 0 at the contact point to 12 m/s at the top.

Topic: Dynamics, 7F Kinematics of rigid bodies. Principle: OpenStax University Physics Vol. 1, 11.1 Rolling Motion.

Answer and explanation

Topic: Dynamics, 7F Kinematics of rigid bodies. Principle: OpenStax University Physics Vol. 1, 11.1 Rolling Motion.

Mechanics of materials

FEM-25 · Mechanics of materials

A straight rod has two segments in series, both with E = 200 GPa. Segment 1 is 300 mm long with a 200 mm² cross section; segment 2 is 200 mm long with a 100 mm² cross section. Equal and opposite 10.0 kN tensile forces act at its ends. Assuming linear elastic behavior and ignoring stress concentration at the step, what is the total elongation?

A. 0.175 mm
B. 0.250 mm
C. 0.075 mm
D. 0.100 mm

Answer and explanation

Answer: A. Both segments carry the same 10.0 kN, so add their stretches. Working in N and mm (200 GPa = 200,000 N/mm²): δ₁ = (10,000 × 300) ÷ (200 × 200,000) = 0.075 mm and δ₂ = (10,000 × 200) ÷ (100 × 200,000) = 0.100 mm. Total: 0.175 mm.

Common misses: C counts only segment 1. D counts only segment 2. B uses the smaller area over the whole 500 mm length.

Topic: Mechanics of Materials, 8C Axial loads and 8I Deformations. Principle: MIT OCW 3.11, Introduction to Elasticity, Eq. 8 (δ = PL/AE); OpenStax University Physics Vol. 1, 12.3 Stress, Strain, and Elastic Modulus.

Answer and explanation

Common misses: C counts only segment 1. D counts only segment 2. B uses the smaller area over the whole 500 mm length.

FEM-26 · Mechanics of materials

A solid circular shaft 30.0 mm in diameter transmits 12.0 kW at a steady 600 rpm. Assuming linear elastic torsion, what is the maximum torsional shear stress?

A. 18.0 MPa
B. 36.0 MPa
C. 72.1 MPa
D. 3.77 MPa

Answer and explanation

Answer: B. First get the torque from power. ω = 600 × 2π ÷ 60 = 62.83 rad/s, so T = P/ω = 12,000 ÷ 62.83 = 191.0 N·m = 191,000 N·mm. For a solid shaft, τ = Tc/J with c = d/2 and J = πd⁴/32, which simplifies to τ = 16T/(πd³) = 16 × 191,000 ÷ (π × 30³) = 36.0 MPa.

Common misses: D treats 600 rpm as 600 rad/s. A uses J = πd⁴/16, twice the correct polar moment. C uses the area moment of inertia, πd⁴/64, where the polar moment belongs.

Topic: Mechanics of Materials, 8E Torsional loads. Principle: MIT OCW 3.11, Torsion, Eqs. 8, 12, 14 (power, J, τ = Tr/J); OpenStax University Physics Vol. 1, 10.8 Work and Power for Rotational Motion.

Answer and explanation

Common misses: D treats 600 rpm as 600 rad/s. A uses J = πd⁴/16, twice the correct polar moment. C uses the area moment of inertia, πd⁴/64, where the polar moment belongs.

FEM-27 · Mechanics of materials

A simply supported beam spans 6.00 ft and carries an 800 lbf point load at midspan. Its rectangular cross section is 2.00 in wide and 4.00 in deep, with the 4.00 in side vertical. Assuming linear elastic bending and ignoring self-weight, what is the maximum bending stress?

A. 2.70 ksi
B. 5.40 ksi
C. 0.225 ksi
D. 1.35 ksi

Answer and explanation

Answer: A. Convert the span first: 6.00 ft = 72.0 in. M_max = PL/4 = 800 × 72.0 ÷ 4 = 14,400 lbf·in. I = bh³/12 = 2.00 × 4.00³ ÷ 12 = 10.67 in⁴, and c = h/2 = 2.00 in. σ = Mc/I = 14,400 × 2.00 ÷ 10.67 = 2,700 psi = 2.70 ksi.

Common misses: C leaves the span in feet, so the moment comes out as 1,200 and gets treated as lbf·in. D uses c = 1.00 in. B uses the full 4.00 in depth as c.

Topic: Mechanics of Materials, 8D Bending loads. Principle: MIT OCW 3.11, Stresses in Beams, Eqs. 5–7; AWC Design Aid 6, Beam Formulas, Fig. 7 (M_max = PL/4).

Answer and explanation

Common misses: C leaves the span in feet, so the moment comes out as 1,200 and gets treated as lbf·in. D uses c = 1.00 in. B uses the full 4.00 in depth as c.

Topic: Mechanics of Materials, 8D Bending loads. Principle: MIT OCW 3.11, Stresses in Beams, Eqs. 5–7; AWC Design Aid 6, Beam Formulas, Fig. 7 (M_max = PL/4).

FEM-28 · Mechanics of materials

At a point in plane stress, σ_x = 80 MPa, σ_y = −20 MPa, and τ_xy = 40 MPa. What is the maximum principal stress?

A. 120 MPa
B. 64.0 MPa
C. 70.0 MPa
D. 94.0 MPa

Answer and explanation

Answer: D. Mohr's circle: center = (80 + (−20)) ÷ 2 = 30 MPa. Radius = √[((80 − (−20)) ÷ 2)² + 40²] = √(50² + 40²) = 64.0 MPa. σ₁ = 30 + 64.0 = 94.0 MPa (and σ₂ = 30 − 64.0 = −34.0 MPa).

Common misses: B is the radius, which equals the maximum in-plane shear stress. C adds τ to the center. A adds σ_x and τ_xy.

Topic: Mechanics of Materials, 8B Stress transformations and Mohr's circle. Principle: LibreTexts, Roylance, 3.3 Tensor Transformations (Mohr's circle).

Answer and explanation

Common misses: B is the radius, which equals the maximum in-plane shear stress. C adds τ to the center. A adds σ_x and τ_xy.

Topic: Mechanics of Materials, 8B Stress transformations and Mohr's circle. Principle: LibreTexts, Roylance, 3.3 Tensor Transformations (Mohr's circle).

FEM-29 · Mechanics of materials

A steel bar fits snugly between two rigid walls with no initial stress. Its temperature increases by 50°C. Assume elastic axial response and no buckling. With E = 200 GPa and α = 12 × 10⁻⁶ /°C, what stress develops?

A. 0 MPa
B. 60 MPa compression
C. 120 MPa compression
D. 120 MPa tension

Answer and explanation

Answer: C. The walls stop the free expansion, so the stress is the one that would push the bar back: σ = EαΔT = 200,000 MPa × 12 × 10⁻⁶ × 50 = 120 MPa. The walls push in, so it's compression.

Common misses: A is the answer if the bar were free to expand; here it isn't. D has the right size but the wrong sense; heating a restrained bar compresses it. B halves the result for no physical reason.

Topic: Mechanics of Materials, 8G Temperature changes and 8K Statically indeterminate systems. Principle: OpenStax University Physics Vol. 2, 1.3 Thermal Expansion (thermal stress).

Answer and explanation

Topic: Mechanics of Materials, 8G Temperature changes and 8K Statically indeterminate systems. Principle: OpenStax University Physics Vol. 2, 1.3 Thermal Expansion (thermal stress).

Material properties and processing

FEM-30 · Material properties and processing

Two materials are linearly elastic up to yield. Material 1 has E = 200 GPa and a yield strength of 250 MPa. Material 2 has E = 100 GPa and a yield strength of 300 MPa. What is the ratio of material 2's modulus of resilience to material 1's?

A. 1.20
B. 1.44
C. 2.00
D. 2.88

Answer and explanation

Answer: D. The modulus of resilience is the area under the stress-strain curve up to yield. For a straight line, that triangle is U_r = σ_y²/(2E). Material 1: 250² ÷ (2 × 200,000) = 0.156 MPa. Material 2: 300² ÷ (2 × 100,000) = 0.450 MPa. Ratio: 0.450 ÷ 0.156 = 2.88. A lower modulus lets material 2 strain more before it yields, so both strength and stiffness matter.

Common misses: A uses only the yield-strength ratio. B squares that ratio but ignores the moduli. C uses only the modulus ratio.

Topic: Material Properties, 9A Properties and 9B Stress-strain diagrams. Principle: LibreTexts, Roylance, 1.4 Stress-Strain Curves (strain energy; area to yield); MIT OCW 3.11, Stress-Strain Curves (modulus of resilience).

Answer and explanation

Common misses: A uses only the yield-strength ratio. B squares that ratio but ignores the moduli. C uses only the modulus ratio.

FEM-31 · Material properties and processing

A medium-carbon steel gear is fully austenitized and then water-quenched to room temperature rapidly enough to suppress ferrite, pearlite, and bainite formation. What is the main resulting microstructure, and what's the usual next step?

A. Coarse pearlite; no further treatment needed
B. Martensite; temper it to recover toughness
C. Ferrite; normalize it to harden it
D. Austenite; slow-cool it to keep it

Answer and explanation

Answer: B. Quenching too fast for carbon to diffuse out traps it, forming martensite: very hard but brittle. Tempering, reheating below the transformation range, trades some hardness for toughness.

Common misses: A describes slow cooling (annealing), which produces ferrite and pearlite in medium-carbon steel. C: ferrite is soft, and normalizing doesn't harden by quenching. D: slow cooling does not preserve the high-temperature austenite; some austenite can remain after quenching.

Topic: Material Properties, 9G Phase transformation and heat treating. Principle: University of Cambridge, Bhadeshia, Martensite in Steels; University of Cambridge, Bhadeshia, Tempered Martensite.

Answer and explanation

Topic: Material Properties, 9G Phase transformation and heat treating. Principle: University of Cambridge, Bhadeshia, Martensite in Steels; University of Cambridge, Bhadeshia, Tempered Martensite.

FEM-32 · Material properties and processing

Select all that apply. An annealed ductile metal is plastically cold worked (deformed below its recrystallization temperature) and isn't heat treated afterward. Compared with its annealed state, which changes are generally expected?

A. Higher dislocation density
B. Higher yield strength
C. Lower ductility
D. Young's modulus roughly doubles
E. The plastic deformation disappears once the load is removed

Answer and explanation

Answers: A, B, C. Cold work multiplies dislocations. They tangle and block each other, so more stress is needed to keep deforming the metal (higher yield strength, the basis of work hardening), and less ductility is left before fracture. You need all three, and only those three.

Common misses: D: work hardening changes how much stress it takes to deform the metal permanently, not its elastic stiffness. E: plastic strain is permanent; only the elastic part springs back.

Topic: Material Properties, 9A Properties and 9F Manufacturing processes. Principle: IIT Kanpur Virtual Labs, Strain and Work Hardening, Theory.

Answer and explanation

Common misses: D: work hardening changes how much stress it takes to deform the metal permanently, not its elastic stiffness. E: plastic strain is permanent; only the elastic part springs back.

Topic: Material Properties, 9A Properties and 9F Manufacturing processes. Principle: IIT Kanpur Virtual Labs, Strain and Work Hardening, Theory.

FEM-33 · Material properties and processing

A unidirectional composite has 60% carbon fiber (E_f = 230 GPa) and 40% epoxy (E_m = 3.5 GPa) by volume. Assume continuous, perfectly bonded fibers and equal longitudinal strain in both phases. What is its elastic modulus along the fibers?

A. 8.6 GPa
B. 94 GPa
C. 117 GPa
D. 139 GPa

Answer and explanation

Answer: D. Along the fibers, both phases stretch together, so the rule of mixtures applies: E = V_fE_f + V_mE_m = 0.60(230) + 0.40(3.5) = 138 + 1.4 = 139.4 GPa ≈ 139 GPa.

Common misses: A uses the inverse rule of mixtures, a series-model estimate, instead of the longitudinal rule. B swaps the volume fractions. C averages the two moduli without weighting.

Topic: Material Properties, 9E Engineered materials. Principle: MIT OCW 3.11, Introduction to Composites, Stiffness, Eqs. 1–2.

Answer and explanation

Answer: D. Along the fibers, both phases stretch together, so the rule of mixtures applies: E = V_fE_f + V_mE_m = 0.60(230) + 0.40(3.5) = 138 + 1.4 = 139.4 GPa ≈ 139 GPa.

Common misses: A uses the inverse rule of mixtures, a series-model estimate, instead of the longitudinal rule. B swaps the volume fractions. C averages the two moduli without weighting.

Topic: Material Properties, 9E Engineered materials. Principle: MIT OCW 3.11, Introduction to Composites, Stiffness, Eqs. 1–2.

FEM-34 · Material properties and processing

Select all that apply. A steel fence post is hot-dip galvanized (coated with zinc). Which statements are true?

A. Zinc is more active than steel, so it corrodes first and protects the steel.
B. A small scratch through the coating still leaves the nearby exposed steel protected while zinc remains.
C. In a galvanic couple, the more noble metal is the one that corrodes faster.
D. Galvanizing protects only as a paint-like barrier; any scratch removes all protection.

Answer and explanation

Answers: A, B. Zinc is oxidized more readily than iron, so it acts as a sacrificial anode. That's why galvanizing keeps working at a small scratch: the surrounding zinc still corrodes in place of the exposed steel.

Common misses: C has it backward: the more active metal is the anode and corrodes. D ignores the sacrificial action, which is the point of using zinc.

Topic: Material Properties, 9I Corrosion mechanisms and control. Principle: OpenStax Chemistry 2e, 17.6 Corrosion.

Answer and explanation

Common misses: C has it backward: the more active metal is the anode and corrodes. D ignores the sacrificial action, which is the point of using zinc.

Topic: Material Properties, 9I Corrosion mechanisms and control. Principle: OpenStax Chemistry 2e, 17.6 Corrosion.

Fluid mechanics

FEM-35 · Fluid mechanics

An open tank holds water 3.0 m deep. Atmospheric pressure is 101.3 kPa. What is the absolute pressure at the bottom?

A. 29.4 kPa
B. 71.9 kPa
C. 101.3 kPa
D. 130.7 kPa

Answer and explanation

Answer: D. Gauge pressure at depth is ρgh = 1,000 × 9.81 × 3.0 = 29,430 Pa = 29.4 kPa. Absolute = gauge + atmospheric = 29.4 + 101.3 = 130.7 kPa.

Common misses: A is gauge pressure. B subtracts instead of adds. C ignores the water.

Topic: Fluid Mechanics, 10B Fluid statics. Principle: OpenStax University Physics Vol. 1, 14.1 Fluids, Density, and Pressure; 14.2 Measuring Pressure.

Answer and explanation

Answer: D. Gauge pressure at depth is ρgh = 1,000 × 9.81 × 3.0 = 29,430 Pa = 29.4 kPa. Absolute = gauge + atmospheric = 29.4 + 101.3 = 130.7 kPa.

Common misses: A is gauge pressure. B subtracts instead of adds. C ignores the water.

Topic: Fluid Mechanics, 10B Fluid statics. Principle: OpenStax University Physics Vol. 1, 14.1 Fluids, Density, and Pressure; 14.2 Measuring Pressure.

FEM-36 · Fluid mechanics

Water flows through a horizontal pipe that narrows from 100 mm to 50 mm in diameter. The velocity in the large section is 2.0 m/s. Ignoring losses, what is the pressure drop from the large to the small section?

A. 32 kPa
B. 60 kPa
C. 6.0 kPa
D. 30 kPa

Answer and explanation

Answer: D. Continuity: velocity scales with the inverse of area, so halving the diameter quadruples it: v₂ = 2.0 × (100/50)² = 8.0 m/s. Bernoulli with no height change: p₁ − p₂ = ½ρ(v₂² − v₁²) = ½ × 1,000 × (64 − 4) = 30,000 Pa = 30 kPa.

Common misses: C scales velocity with diameter (v₂ = 4 m/s) instead of area. B forgets the ½. A uses only v₂² and ignores the upstream velocity.

Topic: Fluid Mechanics, 10C Energy. Principle: OpenStax University Physics Vol. 1, 14.5 Fluid Dynamics (continuity); 14.6 Bernoulli's Equation.

Answer and explanation

Common misses: C scales velocity with diameter (v₂ = 4 m/s) instead of area. B forgets the ½. A uses only v₂² and ignores the upstream velocity.

Topic: Fluid Mechanics, 10C Energy. Principle: OpenStax University Physics Vol. 1, 14.5 Fluid Dynamics (continuity); 14.6 Bernoulli's Equation.

FEM-37 · Fluid mechanics

A liquid with density 1,000 kg/m³ and viscosity 0.00100 Pa·s flows at an average 0.100 m/s through a straight horizontal pipe 10.0 mm in diameter and 20.0 m long. The flow is fully developed; ignore minor losses. What is the flow regime, and what is the frictional pressure drop?

A. Laminar; 160 Pa
B. Laminar; 640 Pa
C. Turbulent; 160 Pa
D. Turbulent; 640 Pa

Answer and explanation

Answer: B. Check the regime first: Re = ρvd/μ = 1,000 × 0.100 × 0.0100 ÷ 0.00100 = 1,000. That's well below about 2,000, so the flow is laminar and Poiseuille's law applies. Written with average velocity, it gives Δp = 32μLv/d² = 32 × 0.00100 × 20.0 × 0.100 ÷ 0.0100² = 640 Pa.

Common misses: A gets the regime right but loses a factor of 4, usually by mixing radius and diameter in the Poiseuille relation. C and D call a Reynolds number of 1,000 turbulent.

Topic: Fluid Mechanics, 10D Internal flow. Principle: OpenStax University Physics Vol. 1, 14.7 Viscosity and Turbulence (Poiseuille's law, Eq. 14.19; Reynolds number, Eq. 14.20).

Answer and explanation

Common misses: A gets the regime right but loses a factor of 4, usually by mixing radius and diameter in the Poiseuille relation. C and D call a Reynolds number of 1,000 turbulent.

Topic: Fluid Mechanics, 10D Internal flow. Principle: OpenStax University Physics Vol. 1, 14.7 Viscosity and Turbulence (Poiseuille's law, Eq. 14.19; Reynolds number, Eq. 14.20).

FEM-38 · Fluid mechanics

A pump moves 0.0150 m³/s of water while adding 18.0 m of total head. Water density is 998 kg/m³, and pump efficiency is 0.720. What shaft input power does the pump need, in kW? Round to two decimal places.

Enter a number in kW.

Answer and explanation

Answer: 3.67 kW. Power delivered to the water is ρgQH = 998 × 9.81 × 0.0150 × 18.0 = 2,643 W. Pump efficiency is water power ÷ shaft power, so the shaft must supply more: 2,643 ÷ 0.720 = 3,671 W = 3.67 kW.

Common misses: 2.64 kW is the water power, the pump's output. 1.90 kW multiplies by the efficiency, which would make the input smaller than the output. 3,671 is the right number in watts, not kilowatts.

Topic: Fluid Mechanics, 10G Power and efficiency. Principle: U.S. DOE and Hydraulic Institute, Improving Pumping System Performance: A Sourcebook for Industry, Section 1 (pump efficiency = fluid power ÷ shaft power).

Answer and explanation

FEM-39 · Fluid mechanics

A centrifugal pump at 1,750 rpm delivers 30 L/s against 20 m of head and requires 8.0 kW of shaft input power. Its speed is cut to 1,450 rpm with the same impeller. Assume negligible static head and unchanged pump efficiency. Using the affinity laws, what shaft input power is required?

A. 4.55 kW
B. 5.49 kW
C. 6.63 kW
D. 8.00 kW

Answer and explanation

Answer: A. Power scales with the cube of speed: P₂ = 8.0 × (1,450/1,750)³ = 8.0 × 0.569 = 4.55 kW. (Flow drops to 24.9 L/s and head to 13.7 m.)

Common misses: C scales power linearly with speed (that's the flow law). B uses the square (the head law). D assumes power doesn't change.

Topic: Fluid Mechanics, 10I Scaling laws for pumps. Principle: U.S. DOE, Pumping Systems Tip Sheet #12, Control Strategies for Centrifugal Pumps with Variable Flow Rate Requirements, pp. 1–2.

Answer and explanation

Answer: A. Power scales with the cube of speed: P₂ = 8.0 × (1,450/1,750)³ = 8.0 × 0.569 = 4.55 kW. (Flow drops to 24.9 L/s and head to 13.7 m.)

Common misses: C scales power linearly with speed (that's the flow law). B uses the square (the head law). D assumes power doesn't change.

Topic: Fluid Mechanics, 10I Scaling laws for pumps. Principle: U.S. DOE, Pumping Systems Tip Sheet #12, Control Strategies for Centrifugal Pumps with Variable Flow Rate Requirements, pp. 1–2.

FEM-40 · Fluid mechanics

Air (γ = 1.4, R = 287 J/kg·K) at 250 K moves at 380 m/s. What is the Mach number?

A. 0.83
B. 1.20
C. 1.42
D. 1.44

Answer and explanation

Answer: B. Speed of sound a = √(γRT) = √(1.4 × 287 × 250) = 316.9 m/s. M = V/a = 380 ÷ 316.9 = 1.20, so the flow is supersonic.

Common misses: C leaves γ out of the speed of sound. A inverts the ratio (a/V). D squares the ratio. And always use kelvins in √(γRT).

Topic: Fluid Mechanics, 10F Compressible flow. Principle: NASA Glenn, Speed of Sound; NASA Glenn, Role of Mach Number in Compressible Flows.

Answer and explanation

Answer: B. Speed of sound a = √(γRT) = √(1.4 × 287 × 250) = 316.9 m/s. M = V/a = 380 ÷ 316.9 = 1.20, so the flow is supersonic.

Common misses: C leaves γ out of the speed of sound. A inverts the ratio (a/V). D squares the ratio. And always use kelvins in √(γRT).

Topic: Fluid Mechanics, 10F Compressible flow. Principle: NASA Glenn, Speed of Sound; NASA Glenn, Role of Mach Number in Compressible Flows.

Thermodynamics

FEM-41 · Thermodynamics

A heat engine operates between a 327°C source and a 27°C sink. What is the highest thermal efficiency it could possibly have?

A. 91.7%
B. 100%
C. 8.3%
D. 50%

Answer and explanation

Answer: D. The Carnot efficiency is the upper limit: η = 1 − T_C/T_H, with absolute temperatures. T_H = 600 K and T_C = 300 K, so η = 1 − 300/600 = 50%. Any claim above 50% for these temperatures breaks the second law.

Common misses: A plugs in Celsius: 1 − 27/327. C is 27/327, also in Celsius. B divides the temperature difference by T_C instead of T_H: 300/300.

Topic: Thermodynamics, 11C Laws of thermodynamics. Principle: OpenStax University Physics Vol. 2, 4.5 The Carnot Cycle.

Answer and explanation

Common misses: A plugs in Celsius: 1 − 27/327. C is 27/327, also in Celsius. B divides the temperature difference by T_C instead of T_H: 300/300.

Topic: Thermodynamics, 11C Laws of thermodynamics. Principle: OpenStax University Physics Vol. 2, 4.5 The Carnot Cycle.

FEM-42 · Thermodynamics

Steam flows steadily through a turbine at 5.0 kg/s. The enthalpy drops from 3,230 kJ/kg at the inlet to 2,560 kJ/kg at the exit, and the turbine loses 50 kW of heat to its surroundings. Changes in kinetic and potential energy are negligible. What is the power output?

A. 3,350 kW
B. 3,400 kW
C. 670 kW
D. 3,300 kW

Answer and explanation

Answer: D. Steady-flow energy balance: Q̇ − Ẇ = ṁ(h₂ − h₁). With Q̇ = −50 kW (heat leaving): Ẇ = ṁ(h₁ − h₂) + Q̇ = 5.0 × 670 − 50 = 3,300 kW. Heat lost is energy that doesn't become work.

Common misses: A ignores the heat loss. B adds it with the wrong sign. C reports the 670 kJ/kg enthalpy drop as power, omitting both the mass flow and the heat loss.

Topic: Thermodynamics, 11E Performance of components. Principle: Yan, Introduction to Engineering Thermodynamics, 5.2.3 Energy conservation equation.

Answer and explanation

Common misses: A ignores the heat loss. B adds it with the wrong sign. C reports the 670 kJ/kg enthalpy drop as power, omitting both the mass flow and the heat loss.

Topic: Thermodynamics, 11E Performance of components. Principle: Yan, Introduction to Engineering Thermodynamics, 5.2.3 Energy conservation equation.

FEM-43 · Thermodynamics

Select all that apply. In one complete cycle, a refrigerator removes 600 kJ from a cold space at 250 K and takes in 200 kJ of work. It rejects heat to surroundings at 300 K, with no other energy transfers. Which statements are true?

A. It rejects 800 kJ to the surroundings.
B. Its coefficient of performance as a refrigerator is 3.
C. A reversible refrigerator working between these temperatures would have a coefficient of performance of 5.
D. Credited as a heat pump, the same cycle has a coefficient of performance of 3.
E. Rejecting more heat than the work it takes in breaks the second law.

Answer and explanation

Answers: A, B, C. Over a full cycle the energy in equals the energy out: Q_H = 600 + 200 = 800 kJ. Refrigerator COP = Q_C/W = 600 ÷ 200 = 3. The reversible (Carnot) limit uses absolute temperatures: T_C/(T_H − T_C) = 250 ÷ 50 = 5. The real machine's 3 sits below that limit, as it must.

Common misses: D: as a heat pump, the useful output is the rejected heat, so COP = 800 ÷ 200 = 4. E: the extra rejected heat is the heat pulled from the cold space; it doesn't have to come from the work.

Topic: Thermodynamics, 11C Laws of thermodynamics and 11G Refrigeration and heat pump cycles. Principle: OpenStax University Physics Vol. 2, 4.3 Refrigerators and Heat Pumps; 4.5 The Carnot Cycle (Eq. 4.6).

Answer and explanation

FEM-44 · Thermodynamics

Air (an ideal gas with γ = 1.4) at 300 K is compressed isentropically through a pressure ratio of 8. What is the exit temperature, to the nearest kelvin?

Enter a number in K.

Answer and explanation

Answer: 543 K. T₂/T₁ = (p₂/p₁)^((γ−1)/γ) = 8^(0.4/1.4) = 8^0.2857 = 1.811. T₂ = 300 × 1.811 = 543 K.

Common misses: 2,400 K multiplies by the pressure ratio as if T were proportional to p. 1,325 K uses the exponent 1/γ instead of (γ − 1)/γ.

Topic: Thermodynamics, 11D Processes. Principle: NASA Glenn, Isentropic Compression or Expansion.

FEM-45 · Thermodynamics

Methane (CH₄) burns completely with the stoichiometric amount of air. Treat air as O₂ + 3.76 N₂. What is the air-fuel ratio by mass, to the nearest whole number?

A. 2
B. 4
C. 10
D. 17

Answer and explanation

Answer: D. Balance: CH₄ + 2(O₂ + 3.76 N₂) → CO₂ + 2H₂O + 7.52 N₂. The air is 2 × (32.0 + 3.76 × 28.0) ≈ 274.6 kg per 16.04 kg of fuel, so A/F ≈ 17.1 by mass.

Common misses: C is the molar ratio, 2 × 4.76 = 9.52 moles of air per mole of fuel. A is moles of O₂ per mole of fuel. B is the mass ratio of O₂ alone (64 ÷ 16).

Topic: Thermodynamics, 11K Combustion. Principle: MIT OCW 2.60J, Lecture 5, Chemical Thermodynamics (methane-air stoichiometry).

Answer and explanation

Answer: D. Balance: CH₄ + 2(O₂ + 3.76 N₂) → CO₂ + 2H₂O + 7.52 N₂. The air is 2 × (32.0 + 3.76 × 28.0) ≈ 274.6 kg per 16.04 kg of fuel, so A/F ≈ 17.1 by mass.

Common misses: C is the molar ratio, 2 × 4.76 = 9.52 moles of air per mole of fuel. A is moles of O₂ per mole of fuel. B is the mass ratio of O₂ alone (64 ÷ 16).

Topic: Thermodynamics, 11K Combustion. Principle: MIT OCW 2.60J, Lecture 5, Chemical Thermodynamics (methane-air stoichiometry).

FEM-46 · Thermodynamics

Select all that apply. Moist air is cooled in a duct coil at constant total pressure, with no moisture added and without reaching its dew point, so no water condenses. Which statements are true?

A. The humidity ratio stays the same.
B. The relative humidity rises.
C. The dew-point temperature stays the same.
D. The relative humidity falls.

Answer and explanation

Answers: A, B, C. No water is added or removed, so the humidity ratio (mass of water per mass of dry air) is constant; on a psychrometric chart the process is a horizontal line. At constant total pressure, that unchanged humidity ratio also means unchanged water-vapor partial pressure, so the dew point doesn't move either. Saturation vapor pressure falls as the air cools, so the same water-vapor partial pressure is a larger fraction of saturation: relative humidity rises.

Common misses: D gets the direction backward. Relative humidity falls when you heat air at constant humidity ratio and total pressure.

Topic: Thermodynamics, 11I Psychrometrics and 11J HVAC processes. Principle: ASHRAE Handbook—Fundamentals (2025), Chapter 1 Psychrometrics, §§5, 6 and 10.

Answer and explanation

Common misses: D gets the direction backward. Relative humidity falls when you heat air at constant humidity ratio and total pressure.

Topic: Thermodynamics, 11I Psychrometrics and 11J HVAC processes. Principle: ASHRAE Handbook—Fundamentals (2025), Chapter 1 Psychrometrics, §§5, 6 and 10.

Heat transfer

FEM-47 · Heat transfer

A plane wall 2.00 m² in area and 0.0400 m thick has k = 0.200 W/m·K. Hot fluid at 80.0°C flows on one side (h = 50.0 W/m²·K) and cold fluid at 20.0°C on the other (h = 10.0 W/m²·K). Assume steady one-dimensional heat flow and ignore radiation. What is the heat transfer rate?

A. 188 W
B. 375 W
C. 400 W
D. 600 W

Answer and explanation

Answer: B. Heat passes through three resistances in series: hot-side film, wall, cold-side film. R = 1/(h_hot·A) + L/(kA) + 1/(h_cold·A) = 0.010 + 0.100 + 0.050 = 0.160 K/W. Then q = ΔT/R = 60.0 ÷ 0.160 = 375 W.

Common misses: A leaves out the 2.00 m² area. C drops the hot-side film resistance. D counts the wall's conduction resistance alone.

Topic: Heat Transfer, 12A Conduction and 12B Convection. Principle: MIT Unified Engineering, 17.2 Combined Conduction and Convection, Eq. 17.21.

Answer and explanation

Common misses: A leaves out the 2.00 m² area. C drops the hot-side film resistance. D counts the wall's conduction resistance alone.

Topic: Heat Transfer, 12A Conduction and 12B Convection. Principle: MIT Unified Engineering, 17.2 Combined Conduction and Convection, Eq. 17.21.

FEM-48 · Heat transfer

A plate with a total exposed radiating area of 0.50 m² (emissivity 0.80) at 500 K is completely surrounded by a large isothermal enclosure at 300 K. What is the net radiation heat loss? Use σ = 5.67 × 10⁻⁸ W/m²·K⁴.

A. 36 W
B. 60 W
C. 1,234 W
D. 1,542 W

Answer and explanation

Answer: C. q = εσA(T_s⁴ − T_sur⁴) = 0.80 × 5.67 × 10⁻⁸ × 0.50 × (500⁴ − 300⁴) = 2.268 × 10⁻⁸ × (6.25 × 10¹⁰ − 0.81 × 10¹⁰) = 1,234 W.

Common misses: A raises the temperature difference to the fourth power, (500 − 300)⁴, which isn't the same as the difference of fourth powers. B uses Celsius temperatures. D leaves out the emissivity (a blackbody).

Topic: Heat Transfer, 12C Radiation. Principle: OpenStax University Physics Vol. 2, 1.6 Mechanisms of Heat Transfer (radiation).

Answer and explanation

Answer: C. q = εσA(T_s⁴ − T_sur⁴) = 0.80 × 5.67 × 10⁻⁸ × 0.50 × (500⁴ − 300⁴) = 2.268 × 10⁻⁸ × (6.25 × 10¹⁰ − 0.81 × 10¹⁰) = 1,234 W.

Topic: Heat Transfer, 12C Radiation. Principle: OpenStax University Physics Vol. 2, 1.6 Mechanisms of Heat Transfer (radiation).

FEM-49 · Heat transfer

In a counterflow heat exchanger, oil cools from 150°C to 90°C while water heats from 30°C to 70°C. What is the log mean temperature difference?

A. 55.8°C
B. 60.0°C
C. 69.5°C
D. 70.0°C

Answer and explanation

Answer: C. In counterflow, pair the hot inlet with the cold outlet: ΔT₁ = 150 − 70 = 80°C and ΔT₂ = 90 − 30 = 60°C. LMTD = (80 − 60) ÷ ln(80/60) = 20 ÷ 0.2877 = 69.5°C.

Common misses: A pairs the ends as in parallel flow (120°C and 20°C). D is the arithmetic mean, which overestimates when the ends differ. B uses the smaller end difference alone.

Topic: Heat Transfer, 12E Heat exchangers. Principle: MIT Unified Engineering, 18.5 Heat Exchangers.

Answer and explanation

Answer: C. In counterflow, pair the hot inlet with the cold outlet: ΔT₁ = 150 − 70 = 80°C and ΔT₂ = 90 − 30 = 60°C. LMTD = (80 − 60) ÷ ln(80/60) = 20 ÷ 0.2877 = 69.5°C.

Common misses: A pairs the ends as in parallel flow (120°C and 20°C). D is the arithmetic mean, which overestimates when the ends differ. B uses the smaller end difference alone.

Topic: Heat Transfer, 12E Heat exchangers. Principle: MIT Unified Engineering, 18.5 Heat Exchangers.

FEM-50 · Heat transfer

A copper sphere 10 mm in diameter (ρ = 8,900 kg/m³, c = 385 J/kg·K, k = 400 W/m·K) at 200°C is dropped into 25°C air with h = 50 W/m²·K. Treat convection to the air as the only heat-transfer path. Using a lumped-capacitance model, about how long until it reaches 50°C?

A. 114 s
B. 158 s
C. 222 s
D. 667 s

Answer and explanation

Answer: C. For a sphere, V/A = D/6 = 0.001667 m. The Biot number h(V/A)/k = 50 × 0.001667 ÷ 400 ≈ 0.0002, tiny, so the lumped model is reasonable. Time constant τ = ρc(V/A)/h = 8,900 × 385 × 0.001667 ÷ 50 = 114 s. Then t = τ ln[(T₀ − T∞)/(T − T∞)] = 114 × ln(175/25) = 114 × ln 7 = 222 s.

Common misses: A is the time constant itself, which only brings the temperature difference down to 37% of its start. B measures temperatures from 0°C instead of from the 25°C air: ln(200/50). D uses V/A = D/2 instead of D/6 for a sphere, tripling τ.

Topic: Heat Transfer, 12D Transient processes. Principle: MIT Unified Engineering, 18.3 Transient Heat Transfer.

Answer and explanation

Topic: Heat Transfer, 12D Transient processes. Principle: MIT Unified Engineering, 18.3 Transient Heat Transfer.

Measurements, instrumentation, and controls

FEM-51 · Measurements, instrumentation, and controls

A plant with G(s) = 10/(s + 2) is placed in a unity negative-feedback loop. What is the steady-state output for a unit step input?

A. 0.833
B. 1.00
C. 5.00
D. 0.167

Answer and explanation

Answer: A. Closed loop: T(s) = G/(1 + G) = 10/(s + 12). By the final value theorem, the step response settles at T(0) = 10/12 = 0.833.

Common misses: C is the open-loop gain G(0) = 10/2. B assumes the output always matches the input; a type-0 loop leaves a steady-state error. D is that error, 1 − 0.833.

Topic: Measurements and Controls, 13B Control systems and 13C Dynamic system response. Principle: MIT OCW 2.004, Lecture 24 (unity feedback and final value theorem).

Answer and explanation

Answer: A. Closed loop: T(s) = G/(1 + G) = 10/(s + 12). By the final value theorem, the step response settles at T(0) = 10/12 = 0.833.

Common misses: C is the open-loop gain G(0) = 10/2. B assumes the output always matches the input; a type-0 loop leaves a steady-state error. D is that error, 1 − 0.833.

Topic: Measurements and Controls, 13B Control systems and 13C Dynamic system response. Principle: MIT OCW 2.004, Lecture 24 (unity feedback and final value theorem).

FEM-52 · Measurements, instrumentation, and controls

A stable first-order system has G(s) = 4/(2s + 1), with time in seconds, and starts at rest. A unit step input is applied at t = 0. When does the output first reach 3.60? Enter seconds, rounded to two decimal places.

Enter a number in seconds.

Answer and explanation

Answer: 4.61 s. Read off the gain and time constant: K = 4 and τ = 2 s, so the step response is y(t) = 4(1 − e^(−t/2)). Set 3.60 = 4(1 − e^(−t/2)), so e^(−t/2) = 0.10 and t = 2 ln 10 = 4.61 s. That's 90% of the final value, reached at about 2.3 time constants.

Common misses: 2.00 s is the time constant, when the output reaches only 63% of its final value (also what you get using log base 10). 4.39 s is the 10%-to-90% rise time, τ ln 9 ≈ 2.2τ, a different measure.

Topic: Measurements and Controls, 13C Dynamic system response. Principle: University of Illinois ECE 486 Handbook, Lecture 6, Rise Time (first-order step response).

Answer and explanation

Topic: Measurements and Controls, 13C Dynamic system response. Principle: University of Illinois ECE 486 Handbook, Lecture 6, Rise Time (first-order step response).

FEM-53 · Measurements, instrumentation, and controls

Power is computed as P = VI from V = 120 V ± 1% and I = 5.0 A ± 2%, with independent standard uncertainties. What is the uncertainty in P using first-order root-sum-square combination?

A. ± 6.0 W
B. ± 12 W
C. ± 13.4 W
D. ± 18 W

Answer and explanation

Answer: C. For a product, combine the relative uncertainties in quadrature: √(1%² + 2%²) = 2.24%. P = 600 W, so the uncertainty is 0.0224 × 600 = ±13.4 W.

Common misses: D adds the percentages (3%) instead of combining them in quadrature. B uses only the current's 2%. A uses only the voltage's 1%.

Topic: Measurements, 13D Measurement uncertainty. Principle: NIST, Essentials of Expressing Measurement Uncertainty, Combining uncertainty components (Eq. 6 and simplified product rule).

Answer and explanation

Answer: C. For a product, combine the relative uncertainties in quadrature: √(1%² + 2%²) = 2.24%. P = 600 W, so the uncertainty is 0.0224 × 600 = ±13.4 W.

Common misses: D adds the percentages (3%) instead of combining them in quadrature. B uses only the current's 2%. A uses only the voltage's 1%.

Topic: Measurements, 13D Measurement uncertainty. Principle: NIST, Essentials of Expressing Measurement Uncertainty, Combining uncertainty components (Eq. 6 and simplified product rule).

FEM-54 · Measurements, instrumentation, and controls

A 120 Ω strain gauge with a gauge factor of 2.1 is bonded to a part that strains 500 microstrain. What is the change in the gauge's resistance, in ohms, to three decimal places?

Enter a number in Ω.

Answer and explanation

Answer: 0.126 Ω. GF = (ΔR/R)/ε, so ΔR = GF × ε × R = 2.1 × (500 × 10⁻⁶) × 120 = 0.126 Ω. Changes this small are why strain gauges are read with a bridge circuit.

Common misses: 126,000 leaves microstrain as 500 instead of 500 × 10⁻⁶. 0.00105 is ΔR/R, the fractional change, not ohms.

Topic: Measurements, 13A Sensors and transducers. Principle: LibreTexts, Roylance, 5.2 Experimental Solutions (strain gages).

Answer and explanation

Answer: 0.126 Ω. GF = (ΔR/R)/ε, so ΔR = GF × ε × R = 2.1 × (500 × 10⁻⁶) × 120 = 0.126 Ω. Changes this small are why strain gauges are read with a bridge circuit.

Common misses: 126,000 leaves microstrain as 500 instead of 500 × 10⁻⁶. 0.00105 is ΔR/R, the fractional change, not ohms.

Topic: Measurements, 13A Sensors and transducers. Principle: LibreTexts, Roylance, 5.2 Experimental Solutions (strain gages).

Mechanical design and analysis

FEM-55 · Mechanical design and analysis

At the critical point of a ductile steel shaft (S_y = 300 MPa), bending gives σ = 80 MPa and torsion gives τ = 40 MPa. What is the factor of safety against yielding by the distortion-energy (von Mises) theory?

A. 2.50
B. 2.65
C. 2.83
D. 3.75

Answer and explanation

Answer: C. For a normal stress plus a shear stress, σ′ = √(σ² + 3τ²) = √(80² + 3 × 40²) = √11,200 = 105.8 MPa. n = S_y/σ′ = 300 ÷ 105.8 = 2.83.

Common misses: B is the maximum-shear-stress (Tresca) result: τ_max = √(40² + 40²) = 56.6 MPa, and n = (S_y/2)/τ_max = 2.65. It's more conservative, but the question asked for distortion energy. D ignores the torsion. A adds σ and τ directly.

Topic: Mechanical Design, 14B Failure theories. Principle: IIT Kharagpur, Static Failure Theories, pp. 23–27.

Answer and explanation

Answer: C. For a normal stress plus a shear stress, σ′ = √(σ² + 3τ²) = √(80² + 3 × 40²) = √11,200 = 105.8 MPa. n = S_y/σ′ = 300 ÷ 105.8 = 2.83.

Topic: Mechanical Design, 14B Failure theories. Principle: IIT Kharagpur, Static Failure Theories, pp. 23–27.

FEM-56 · Mechanical design and analysis

A closed-end thin-walled cylinder has a 200 mm inside diameter, a 4.00 mm wall, and 2.00 MPa internal gauge pressure. The wall material is ductile with a 130 MPa tensile yield strength. Using thin-wall stresses based on the inside diameter, and neglecting radial stress and end effects, what is the factor of safety against yielding by the von Mises (distortion-energy) theory?

A. 1.73
B. 2.60
C. 3.00
D. 5.20

Answer and explanation

Answer: C. Hoop stress: pD/(2t) = 2.00 × 200 ÷ 8.00 = 50.0 MPa. Axial stress: pD/(4t) = 25.0 MPa. With the radial stress neglected, σ′ = √(σ₁² + σ₂² − σ₁σ₂) = √(50² + 25² − 50 × 25) = 43.3 MPa, so n = 130 ÷ 43.3 = 3.00.

Common misses: A adds the two stresses. B uses the hoop stress alone; for this stress state it's also the maximum-shear-stress (Tresca) result, but the question asks for von Mises. D uses the axial stress alone.

Topic: Mechanical Design, 14B Failure theories and 14E Pressure vessels. Principle: MIT OCW 3.11, Pressure Vessels, Eqs. 2–3; MIT OCW 3.11, Yield and Plastic Flow (von Mises criterion).

Answer and explanation

Topic: Mechanical Design, 14B Failure theories and 14E Pressure vessels. Principle: MIT OCW 3.11, Pressure Vessels, Eqs. 2–3; MIT OCW 3.11, Yield and Plastic Flow (von Mises criterion).

FEM-57 · Mechanical design and analysis

A hole is specified 25.000 to 25.021 mm and its mating shaft 24.980 to 24.993 mm. What is the maximum clearance?

A. 0.007 mm
B. 0.020 mm
C. 0.028 mm
D. 0.041 mm

Answer and explanation

Answer: D. Maximum clearance = largest hole − smallest shaft = 25.021 − 24.980 = 0.041 mm. (Minimum clearance = smallest hole − largest shaft = 25.000 − 24.993 = 0.007 mm. Both are positive, so this is a clearance fit.)

Common misses: A is the minimum clearance. C pairs the largest hole with the largest shaft. B pairs the smallest hole with the smallest shaft.

Topic: Mechanical Design, 14J Manufacturability (limits and fits). Principle: University of Illinois ME 170, Lecture Class Assignment 12, p. 1, Tolerance Analysis.

Answer and explanation

Common misses: A is the minimum clearance. C pairs the largest hole with the largest shaft. B pairs the smallest hole with the smallest shaft.

Topic: Mechanical Design, 14J Manufacturability (limits and fits). Principle: University of Illinois ME 170, Lecture Class Assignment 12, p. 1, Tolerance Analysis.

FEM-58 · Mechanical design and analysis

A ball bearing has a basic dynamic load rating C = 30 kN and carries an equivalent load P = 5.0 kN at 1,500 rpm. What is its basic rating life L₁₀ in hours?

A. 400 h
B. 2,400 h
C. 4,360 h
D. 144,000 h

Answer and explanation

Answer: B. For ball bearings, L₁₀ = (C/P)³ million revolutions = (30/5.0)³ = 216 million revolutions. At 1,500 rpm, that's 216 × 10⁶ ÷ (1,500 × 60) = 2,400 hours. L₁₀ is the life 90% of a large group of identical bearings are expected to reach.

Common misses: C uses the roller-bearing exponent, 10/3. A uses an exponent of 2. D divides revolutions by rpm (giving minutes) and reports the result as hours.

Topic: Mechanical Design, 14F Bearings. Principle: NSK, Dynamic Load Ratings and Fatigue Life.

Answer and explanation

Common misses: C uses the roller-bearing exponent, 10/3. A uses an exponent of 2. D divides revolutions by rpm (giving minutes) and reports the result as hours.

Topic: Mechanical Design, 14F Bearings. Principle: NSK, Dynamic Load Ratings and Fatigue Life.

FEM-59 · Mechanical design and analysis

A motor delivers 12.0 N·m of torque at 900 rpm into a gearbox with a speed ratio of 3.00 (input speed ÷ output speed) and an efficiency of 0.900 (output power ÷ input power). What is the steady output torque?

A. 10.8 N·m
B. 32.4 N·m
C. 40.0 N·m
D. 3.60 N·m

Answer and explanation

Answer: B. The output turns at 900 ÷ 3.00 = 300 rpm. Power is torque × speed, and 90% of it gets through: T_out ω_out = 0.900 T_in ω_in. So T_out = 0.900 × 12.0 × 3.00 = 32.4 N·m. Speed goes down by 3, torque goes up by almost 3.

Common misses: D divides by the speed ratio instead of multiplying. A ignores the speed ratio. C divides by the efficiency, which would put out more power than goes in.

Topic: Mechanical Design, 14H Power transmission. Principle: OpenStax University Physics Vol. 1, 10.8 Work and Power for Rotational Motion (P = τω); UC San Diego MAE 3, Gear Ratios.

Answer and explanation

Common misses: D divides by the speed ratio instead of multiplying. A ignores the speed ratio. C divides by the efficiency, which would put out more power than goes in.

Topic: Mechanical Design, 14H Power transmission. Principle: OpenStax University Physics Vol. 1, 10.8 Work and Power for Rotational Motion (P = τω); UC San Diego MAE 3, Gear Ratios.

Review your answers and choose what to study next

Your score describes how you did on these 59 questions. It doesn't predict your NCEES result. NCEES converts your correct answers to a scaled score, compares it with a standard set by subject-matter experts, and reports pass or fail; it doesn't publish a passing score (NCEES exam scoring).

Score on these 59 practice questions: ____ / 59 correct; ____ incorrect; ____ unanswered. Give each correct answer one point. For numerical responses, use the requested unit and rounding; a select-all answer must match the complete correct set. Keep questions marked for review separate from this raw score.

Mark each miss in the table below, including any question you got right only by guessing.

Review your answers and choose what to study next
NCEES knowledge area	Questions on the exam (NCEES range)	Questions here
1 Mathematics	6–9	FEM-01 to FEM-04 (4)
2 Probability and Statistics	4–6	FEM-05 to FEM-06 (2)
3 Ethics and Professional Practice	4–6	FEM-07 to FEM-09 (3)
4 Engineering Economics	4–6	FEM-10 to FEM-11 (2)
5 Electricity and Magnetism	5–8	FEM-12 to FEM-14 (3)
6 Statics	9–14	FEM-15 to FEM-19 (5)
7 Dynamics, Kinematics, and Vibrations	10–15	FEM-20 to FEM-24 (5)
8 Mechanics of Materials	9–14	FEM-25 to FEM-29 (5)
9 Material Properties and Processing	7–11	FEM-30 to FEM-34 (5)
10 Fluid Mechanics	10–15	FEM-35 to FEM-40 (6)
11 Thermodynamics	10–15	FEM-41 to FEM-46 (6)
12 Heat Transfer	7–11	FEM-47 to FEM-50 (4)
13 Measurements, Instrumentation, and Controls	5–8	FEM-51 to FEM-54 (4)
14 Mechanical Design and Analysis	10–15	FEM-55 to FEM-59 (5)
Total	110	59

The ranges come from the NCEES FE Mechanical specification, effective with the July 2020 exams. Our counts loosely follow them; this isn't an NCEES exam form. With two to six questions per area, an area tally is a prompt for what to review next, not a measure of how well you know the whole subject.

Then turn each miss into a specific fix. Write down what actually went wrong, not just the topic name:

Review your answers and choose what to study next
Question	What went wrong	What I'll change	Reworked correctly without the solution?
Example: FEM-38	Used the water power as the shaft power	Check which side of the machine the question asks about before applying efficiency	Yes / Not yet
Your miss:

Three steps that work for most misses:

Rework the question from a blank page. Name the unknown, its units, and the governing idea before you touch a number.
Review the principle, not just the answer. Use the source under the explanation, then find the same formula in your handbook and note the search term that found it.
Try a fresh problem on the same idea from your coursework or another lawful source. Coming back to this question later tests your memory of the answer as much as the method, so don't treat a repeat score as proof you're ready.

Got one right but slowly? Count it as a miss for planning. Speed matters on a 110-question exam.

Use the handbook and pace your practice

Practice with the handbook you'll get on exam day. NCEES supplies the FE Reference Handbook on screen as a searchable PDF, and it's the only reference allowed. You search it with a box on the left side of the screen; Ctrl+F doesn't work (NCEES Examinee Guide, May 2026, p. 10). Download the current version free through a MyNCEES account. You may print it for personal study but not share or post it (NCEES help: exam reference handbooks).
Use your exam calculator. You can bring one NCEES-approved calculator, and an on-screen TI-30XS is also available (Examinee Guide, p. 8). Check your exact model against the calculator policy linked at the bottom of the NCEES exams page.
Go untimed first, then time yourself. The exam gives you 320 minutes for 110 questions, about 2 minutes 55 seconds each on average. At that pace, this set takes about 2 hours 52 minutes (our arithmetic: 320 ÷ 110 × 59). That's an average, not a limit per question, and our questions aren't calibrated to NCEES difficulty.
Expect more formats on the real exam. NCEES also uses point-and-click and drag-and-drop questions, all scored right or wrong (NCEES computer-based testing). This page covers single-answer, select-all, and enter-a-number questions only.
Never leave a question blank. Your result is based on the number you get right, with nothing taken off for wrong answers (NCEES exam scoring).
Review before you submit the first half. After about half the questions, you review and submit them, and you can't go back to them (Examinee Guide, pp. 11–12).

FE Mechanical exam facts

FE Mechanical exam facts
Detail	FE Mechanical
Questions	110
Exam time	5 hours 20 minutes, plus a 25-minute scheduled break
Full appointment	6 hours on the NCEES FE page; the May 2026 Examinee Guide lists 5 hours 55 minutes (2-minute nondisclosure agreement + 8-minute tutorial + exam + break)
Fee	$225, paid to NCEES; your licensing board may charge its own application fee
Where and when	Computer-based, at NCEES-approved Pearson test centers, year-round
Reference	NCEES FE Reference Handbook, on screen; no personal references
Units	SI and U.S. Customary
Results	Pass or fail, typically in 7–10 days; a diagnostic report by knowledge area if you don't pass

Sources: NCEES FE exam page; FE Mechanical specification; Examinee Guide, May 2026, pp. 3, 14, 16. Your licensing board decides whether you're eligible to take the exam; check it through the state selector on the NCEES FE exam page.

Common questions

Are these questions as hard as the real exam? We can't promise that. They're written to the current specification's topics and use realistic setups, but they haven't been calibrated against NCEES questions, so treat your score as feedback on these questions only.

How many questions do I need right to pass? NCEES doesn't publish a number. The passing score varies slightly with exam-form difficulty, and first-time and repeat takers are held to the same standard (Examinee Guide, p. 14). Be wary of any site that quotes an exact passing percentage.

Is this the current specification? The FE Mechanical specification used for this page took effect with the July 2020 exams. Before you test, check the specification linked for Mechanical on the NCEES FE exam page.

Next steps

Plan the rest of your prep: FE exam prep: subject maps for all seven FE exams and a 12-week plan
Ready to book? How NCEES exam registration works
Need testing accommodations? How to request NCEES accommodations (you request them during registration)
After the exam: NCEES exam results · NCEES retake policy
Taking a different FE exam? Free FE Electrical and Computer practice test · Free FE Chemical practice test

Sources and verification

Last verified: October 7, 2026. We checked the exam facts on this page against the NCEES FE exam page, the FE Mechanical specification, the May 2026 NCEES Examinee Guide, and the NCEES scoring, computer-based testing, and handbook-access pages. We checked the ethics questions against the August 2026 NCEES Model Rules. Teaching-source passages were checked against the principles cited, all 59 questions were re-solved against their stated assumptions, and the numerical answers and wrong-answer calculations were recalculated.

AI-assisted tools were used to develop and check this content. This is editorial source checking, not professional engineering review.

Official sources:

The teaching source for each question is linked directly beneath it. How we check exam facts: our methodology. Found an error? Our corrections page explains how to report it.

Written by the Castleport Test Prep Editorial Team.

Castleport Test Prep is an independent exam prep publisher, not affiliated with, endorsed by, or approved by the National Council of Examiners for Engineering and Surveying (NCEES), Pearson VUE, or any state engineering licensing board. The practice questions on this page are original and unofficial; they are not NCEES exam questions. Exam and credential names identify their subjects; trademarks belong to their respective owners. This resource doesn't guarantee an exam result or determine eligibility or licensure.