Mechanical Comprehension Practice Test 6 for the ASVAB

Answer: B

Correct answer: Second class. Ask what is in the middle. Fulcrum in the middle means first class, load in the middle means second class, and effort in the middle means third class. Here the load is in the middle, so the lever is second class.

• A: A first-class lever has the fulcrum in the middle, not the load.
• B (correct): With the load in the middle (fulcrum, load, effort), the lever is second class.
• C: A third-class lever has the effort in the middle, not the load.
• D: A compound lever links two or more levers in stages; a single bar with the load in the middle is simply second class.

Answer: A

Correct answer: 5. For a ramp, IMA = L / h.

• A (correct): IMA for a ramp is slope length divided by rise, and 10 divided by 2 is 5.
• B: Dividing the rise by the length flips the ratio and gives a value below 1.
• C: Subtracting the rise from the length is not how mechanical advantage is found.
• D: Multiplying the length by the rise does not give a ratio of distances.

Answer: C

Correct answer: 4. IMA equals effort distance divided by load distance: 3.0 m / 0.75 m = 4. Because both distances are in meters, the units cancel and the IMA is a pure number.

• A: Subtracting 0.75 from 3.0 does not give a ratio; IMA is found by dividing the distances.
• B: This divides the load distance by the rope distance, which inverts the ratio.
• C (correct): Dividing effort distance by load distance, 3.0 m / 0.75 m, gives a unitless IMA of 4.
• D: The meters cancel when one distance is divided by the other, so IMA has no unit.

Answer: C

Correct answer: It makes the output turn the same direction as the driver. An idler adds one more external mesh, so the output now turns the same way as the input. Its tooth count cancels in the ratio, so it changes direction but not the overall speed in the ideal model.

• A: The idler's tooth count cancels out when the two mesh ratios are multiplied, so it does not set the speed ratio.
• B: Torque follows the overall speed ratio, which depends only on the first and last gears, so the idler adds no torque gain.
• C (correct): The idler adds a second external mesh, and two reversals bring the output back to the driver's direction while the speed ratio stays the same.
• D: The output speed is set by the driver and output tooth counts alone; adding an idler does not speed it up.

Answer: A

Correct answer: 90 N. The coefficient is

With the new weight, the push needed is

three times as much because the weight tripled.

• A (correct): Tripling the weight triples the normal force and friction, so 30 N becomes 90 N.
• B: This doubles the push, but the weight tripled.
• C: This ignores that more weight raises the normal force and therefore friction.
• D: This divides by 3 instead of multiplying, as if more weight lowered friction.

Answer: B

Equal tooth counts give a one-to-one speed magnitude; external contact reverses direction. This response follows the specified gear geometry and the assumptions in the question.

• A: A doubled speed would require unequal tooth counts or another transmission stage.
• B (correct): Equal tooth counts give a one-to-one speed magnitude; external contact reverses direction.
• C: Neither halving of speed nor same-direction rotation follows the equal external pair.
• D: Equal counts do not make the motion vanish; they simply produce equal speed magnitudes.

Answer: B

Correct answer: 10 m/s. Potential energy lost equals kinetic energy gained, and mass cancels.

• A: This multiplies g by h without the factor 2 or the square root.
• B (correct): Using v = √(2gh) = √(2 × 10 × 5) = √100 = 10 m/s.
• C: This finds v² = 100 but forgets to take the square root.
• D: This leaves out the factor of 2, taking the square root of 50.

Answer: D

Correct answer: 1.5 in. Pounds are already a force, so no conversion is needed. Two identical springs in parallel give

/in, so the sag is

• A: This treats the springs as if they were in series, which halves the stiffness instead of doubling it.
• B: This uses one spring alone; with two springs sharing the load, the support is twice as stiff.
• C: This divides the stiffness by the load instead of dividing the load by the stiffness.
• D (correct): The parallel stiffness is 30 lb/in, and 45 lb divided by 30 lb/in gives 1.5 in.

Answer: B

Correct answer: 450 N. Set the moments equal and solve for the load.

• A: This flips the arms; a load close to the fulcrum can be much heavier than the effort, not lighter.
• B (correct): The effort moment is 90 times 1.5, or 135 N·m, and 135 divided by 0.30 is 450 N.
• C: This is the effort moment in N·m; it still has to be divided by the 0.30 m load arm.
• D: This multiplies the effort by the load arm instead of the effort arm.

Answer: A

Correct answer: 225,600 J. Add the two steps.

Total:

• A (correct): Melting takes 0.6 times 334,000 = 200,400 J, and warming takes 0.6 times 4,200 times 10 = 25,200 J, for 225,600 J in all.
• B: This covers only the melting and leaves out the heat to warm the water by 10 degrees.
• C: This uses the latent heat of vaporization, but melting ice needs the latent heat of fusion.
• D: This covers only the warming and leaves out the large latent heat needed to melt the ice.

Answer: C

Correct answer: 34°C. With the same specific heat, the final temperature is a mass-weighted average.

• A: This weights the masses the wrong way round, giving the hot water the larger weight.
• B: This is the simple average, which applies only when the masses are equal.
• C (correct): The mass-weighted average is (1 × 90 + 4 × 20)/5 = 170/5 = 34°C.
• D: This is just the difference between the two starting temperatures, not a final temperature.

Answer: D

Correct answer: Net force 0 N; net torque 10 N·m. Force test: 25 N up and 25 N down cancel, so the net force is zero. Torque test: both hands turn the wheel the same way.

• A: The forces form a couple; their torques add rather than cancel.
• B: This counts the torque from only one hand.
• C: Opposite forces subtract, so the net force is zero, not their sum.
• D (correct): Up and down forces cancel, but both turn the wheel the same way: 2 × 25 × 0.2 = 10 N·m.

Answer: B

Correct answer: 8 cm/s. On the rod side the effective area is

square centimeters. Speed = flow ÷ area =

/s, faster than the 6 cm/s extending speed.

• A: This divides by the rod's area, but the oil acts on the ring of piston around the rod.
• B (correct): The rod-side area is 40 − 10 = 30 square centimeters, and 240 ÷ 30 = 8 cm/s.
• C: This adds the rod area to the bore area instead of subtracting it.
• D: This divides by the full bore area, which gives the extending speed, not the retracting speed.

Answer: A

Correct answer: 320 N. Upward:

Downward: weight 80 N plus tension T.

Equilibrium:

so T = 320 N.

• A (correct): Buoyancy is 400 N and weight is 80 N, so the rope must pull down 400 − 80 = 320 N.
• B: This adds the weight to the buoyant force, but both the weight and the rope act downward against the buoyancy.
• C: This is the buoyant force alone, ignoring the buoy's own weight that also pulls down.
• D: This is only the buoy's weight, not the force needed to hold it under.

Answer: A

Correct answer: 20 m. Stop time:

Distance up:

From 4 s to 6 s it moves back down from rest:

Total distance is

even though the displacement is only +12 m.

• A (correct): The puck stops at 4 s after 16 m up, then moves 4 m back in the last 2 s, for 16 + 4 = 20 m of path.
• B: This is the displacement, 48 - 36 = 12 m, which subtracts the return trip instead of adding it.
• C: This counts only the trip up the ramp and leaves out the return trip down.
• D: This adds the acceleration term instead of subtracting it, using +36 m where -36 m belongs.