ASVAB Mechanical Comprehension Practice Test 2

Answer: A

Correct answer: Joule. Work is force times the distance moved in the direction of the force, so its unit is a newton multiplied by a meter, named the joule. Because work is energy transferred, energy uses the same unit. The tempting mix-up is the watt: power tells how fast work is done, so 1 watt means 1 joule each second. Horsepower is also a power unit, and the newton measures force alone.

• A (correct): Work and energy share the joule, defined as one newton acting through one meter.
• B: The newton measures force; work needs that force multiplied by a distance.
• C: Horsepower is a power rating of about 746 W, describing a rate of doing work.
• D: The watt is a unit of power, equal to one joule per second, not an amount of work.

Answer: A

Correct answer: 3 m/s². Average acceleration is change in velocity over time:

• A (correct): The velocity changes by 12 m/s in 4 s, and 12 divided by 4 is 3 m/s².
• B: This divides the time by the change in velocity, which is upside down.
• C: This divides the final speed of 14 m/s by the time and ignores the starting speed of 2 m/s.
• D: This multiplies the change in velocity by the time instead of dividing.

Answer: C

Correct answer: It raises the coolant's boiling point so it stays liquid above 100°C. A liquid boils when its vapor pressure equals the pressure above it. Raising the pressure therefore raises the boiling point, so pressurized coolant stays liquid well above 100°C.

• A: Flow is driven by the water pump, and the radiator is still needed to release the heat to the air.
• B: Pressure in this range does not meaningfully change specific heat; its effect here is on the boiling point.
• C (correct): Higher pressure above a liquid raises its boiling point, letting the coolant run hotter without boiling.
• D: Pressure raises, not lowers, the boiling point, and boiling coolant would cool the engine poorly.

Answer: B

Correct answer: 4,000 Pa. Use

2,400 N ÷ 0.6 m² = 4,000 Pa.

• A: This drops a zero; 2,400 divided by 0.6 is 4,000.
• B (correct): Pressure is force over area: 2,400 ÷ 0.6 = 4,000 Pa.
• C: This multiplies the force by the area instead of dividing.
• D: This divides the area by the force, which inverts the pressure formula.

Answer: C

Correct answer: 16 N·m. The torque ratio is the diameter ratio, driven over driver.

The output turns at one quarter of the motor speed, so power is unchanged.

• A: This divides the torque by the ratio, which happens only when the driven pulley is smaller and speeds up.
• B: This multiplies by the square of the diameter ratio, but torque scales with the simple ratio of 4.
• C (correct): The larger pulley turns 300 / 75 = 4 times slower, so in an ideal drive the torque rises by the same factor: 4 times 4 = 16 N·m.
• D: A belt shares rim speed between pulleys, but torque changes inversely with rpm, so it does not stay the same.

Answer: B

Correct answer: 0.2 m/s. Rope and load move for the same time, so their speeds have the same ratio as their distances. Load speed = 0.8

• A: Dividing by 2 uses the wrong number of supporting segments.
• B (correct): Load speed equals rope speed divided by the segment count: 0.8 / 4 = 0.2 m/s.
• C: Multiplying by 4 makes the load move faster than the rope, but a force-multiplying tackle slows the load.
• D: Equal speeds would mean an IMA of 1, not a four-segment tackle.

Answer: D

Correct answer: 6 m/s. The same volume per second passes every section:

• A: This squares the area ratio, but speed changes in direct proportion to the area ratio.
• B: This treats the area ratio of 4 as if it were a diameter ratio and uses only its square root.
• C: This flips the area ratio; a smaller passage makes the oil speed up, not slow down.
• D (correct): By continuity, 8 × 1.5 = 2 × v, so v = 12 ÷ 2 = 6 m/s.

Answer: A

Correct answer: 24 N. Measure deflection from natural length and convert to meters: 31 cm minus 25 cm is 6 cm, or 0.06 m.

• A (correct): The extension is 31 cm minus 25 cm, or 0.06 m, and 400 times 0.06 gives 24 N.
• B: This multiplies by 6 cm without converting to meters, giving an answer 100 times too large.
• C: This uses the natural length of 0.25 m as if it were the deflection.
• D: This uses the full loaded length of 0.31 m instead of the change from natural length.

Answer: B

Correct answer: 4 N·m clockwise. Left moment:

N·m counterclockwise. Right moment:

N·m clockwise. They oppose, so subtract:

N·m in the clockwise direction of the larger moment.

• A: The larger moment comes from the right-hand weight, which turns a level beam clockwise, not counterclockwise.
• B (correct): The right weight gives 24 N·m clockwise and the left gives 20 N·m counterclockwise; the difference is 4 N·m clockwise.
• C: This adds the two moments, but they turn the beam in opposite directions and partly cancel.
• D: This adds opposing moments and also picks the direction of the smaller one.

Answer: D

Conical bevel gear geometry is suitable for intersecting shafts and changing shaft direction. This response follows the specified gear geometry and the assumptions in the question.

• A: A coupling joins shaft ends and does not itself provide a meshing angle change.
• B: A rack translates and does not constitute a pair of intersecting rotating shafts.
• C: Standard spur pairs generally use parallel rather than intersecting axes.
• D (correct): Conical bevel gear geometry is suitable for intersecting shafts and changing shaft direction.

Answer: C

Correct answer: 1 m/s east. Displacement is

m and total time is 40 s + 20 s = 60 s.

The positive sign means 1 m/s east.

• A: This uses only the first leg, 80 m in 40 s, and ignores the walk back.
• B: This divides the 60 m displacement by only the 40 s of the first leg.
• C (correct): Displacement is 80 m minus 20 m, or +60 m, and 60 m divided by 60 s is 1 m/s east.
• D: This uses total distance (100 m) over 60 s, which is average speed, not average velocity.

Answer: D

Correct answer: 150 N·m. A weight acts straight down, so its moment arm is the horizontal distance from the pivot: 1.5 m × cos 60° = 0.75 m. The moment is

N·m.

• A: This uses sin 60 degrees, which gives the vertical height of the point rather than its horizontal distance from the pivot.
• B: This uses 150 cm without converting to 1.5 m, giving a value 100 times too large.
• C: This uses the full distance along the bar, but for a hanging weight the moment arm is the horizontal distance to the pivot.
• D (correct): The horizontal moment arm is 1.5 m × cos 60 degrees = 0.75 m, and 200 N × 0.75 m = 150 N·m.

Answer: B

Correct answer: About 2.5 m/s². Forward part:

Upward part: 100 N, so

N and friction is

• A: This treats the whole 200 N rope force as horizontal.
• B (correct): The rope's lift cuts the normal force to 300 N, so friction is 75 N against a 173 N forward pull.
• C: This ignores the rope's upward part, using 400 N as the normal force and 100 N of friction.
• D: This uses the correct forward pull but ignores friction altogether.

Answer: D

Correct answer: No; it needs about 85 N of friction but at most about 82 N is available. Compare tan θ with the coefficient: tan 25° ≈ 0.47 is greater than 0.45, so the block slides. In forces: needed friction = 200 × sin 25° ≈ 85 N, while the maximum is 0.45 × 200 × cos 25° ≈ 82 N.

• A: Only the downhill component of weight, about 85 N, must be balanced by friction.
• B: This uses the full weight as the normal force; on a slope the normal force is only W cos θ.
• C: The normal force is perpendicular to the slope and cannot hold the block against the downhill pull.
• D (correct): Needed friction is 200 sin 25° ≈ 85 N; the limit is 0.45 × 200 cos 25° ≈ 82 N, so friction falls short and the block slides.

Answer: A

Correct answer: 1200 N, with the sides moved 1.5 cm apart. IMA =

Spreading force = 150

and the spread = 12 cm / 8 = 1.5 cm. Check: 150 N × 0.12 m = 18 J in and 1200 N × 0.015 m = 18 J out.

• A (correct): IMA is 8, so force is 150 times 8, or 1200 N, and the spread is 12 cm divided by 8, or 1.5 cm.
• B: Dividing the force by the IMA reverses its effect; a wedge multiplies the push.
• C: The output cannot move as far as the input when the force is multiplied; that would create work.
• D: Multiplying the input distance by the IMA would make output work far exceed input work.