ASVAB Mechanical Comprehension: Practice Test 5

Answer: D

Correct answer: It stays permanently longer than before. Past the elastic limit, deformation becomes plastic, meaning permanent. The spring takes a set and stays longer than its original length, and it will give less force at a given position from then on.

• A: Overstretching yields the metal in the stretched direction, so it does not end up shorter.
• B: Full recovery happens only for elastic deformation, which ends at the elastic limit.
• C: An overloaded spring is weakened and gives less force at the same position, not more.
• D (correct): Beyond the elastic limit the metal yields, leaving a permanent set so the spring does not fully return.

Answer: D

Correct answer: 60 kPa.

Pa = 60 kPa. Gauge pressure leaves out the atmosphere.

• A: This adds atmospheric pressure, giving absolute rather than gauge pressure.
• B: This adds an extra factor of ten to the depth calculation.
• C: This drops a factor of ten; each meter of fresh water adds about 10 kPa, not 1 kPa.
• D (correct): Water pressure is 1,000 × 10 × 6 = 60,000 Pa, which is 60 kPa.

Answer: C

Correct answer: 75%. Efficiency = output ÷ input ×

The other 200 J warms the jack's moving parts.

• A: This divides input by output; an efficiency above 100% signals an inverted ratio.
• B: This misplaces the decimal when changing 0.75 into a percentage.
• C (correct): Useful output over input is 600 ÷ 800 = 0.75, or 75%.
• D: This is the fraction lost to friction and heat, not the useful fraction.

Answer: D

Correct answer: The fixed point the bar turns around. Every lever has a fulcrum, an effort and a load. The fulcrum is the fixed pivot point, usually drawn as a triangle or hinge, and the bar rotates around it.

• A: That is the effort, the force you supply, not the pivot.
• B: That is the load, the resistance the lever works against.
• C: A distance is not a part of the lever; the fulcrum is a physical point on which the bar turns.
• D (correct): The fulcrum is the pivot, the one point of the bar that stays put while the rest rotates about it.

Answer: D

Correct answer: It is cut in half. Screw IMA is $2\pi r/\ell$. Halving $\ell$ doubles the IMA, so the ideal effort is cut in half. The tradeoff is that twice as many turns are needed for the same lift.

• A: A smaller lead gives more advantage, so the effort goes down, not up.
• B: The IMA changes in direct proportion to the lead, so the effort halves rather than quarters.
• C: The lead directly affects how far the load rises per turn, so it changes the IMA.
• D (correct): Lead is in the denominator of the screw IMA, so halving the lead doubles the IMA and halves the ideal effort.

Answer: D

Correct answer: It extends slower, but its maximum force is the same. Cylinder speed is flow rate divided by piston area,

A flow control cuts Q, so the piston moves more slowly. It does not change the piston area, and in the ideal model the force available at full pressure stays the same.

• A: Speed depends on flow, so cutting the flow does change the speed.
• B: Less oil reaching the cylinder each second cannot make it move faster.
• C: In the ideal model, restricting flow changes only speed; full pressure, and so full force, is still available when the piston stalls against a load.
• D (correct): Speed equals flow divided by piston area, so less flow means a slower piston, while the force available at full pressure is unchanged.

Answer: A

Correct answer: 36 N·m. Multiply input torque by the ratio, then by the efficiency.

N·m.

• A (correct): Ideal torque is 4 times 12 = 48 N·m, and 75 percent of that is 36 N·m.
• B: This divides the input torque by the ratio, which is what a speed-increasing drive would do.
• C: This is the ideal torque; it ignores the 75 percent efficiency.
• D: This divides the ideal torque by the efficiency, making the output larger than the ideal value.

Answer: A

Correct answer: 13 m/s. The area under an acceleration-time graph is the change in velocity:

Add it to the starting velocity:

• A (correct): The area 2.5 times 4 gives a 10 m/s gain, and 3 + 10 = 13 m/s.
• B: This is only the change in velocity from the graph area; the starting 3 m/s must be added.
• C: This uses one-half a t squared, which gives a distance from rest, not a velocity.
• D: This adds the acceleration value directly to the velocity without multiplying by time.

Answer: B

Correct answer: 30 cm long and 3 cm thick. Wedge IMA is length divided by thickness:

The largest IMA, 10, needs the smallest effort.

• A: Its IMA is only 5, so it needs twice the effort of the best wedge.
• B (correct): Its IMA is 30 divided by 3, or 10, the largest of the four, so it needs the least effort.
• C: Its IMA is just 4, the smallest of the group, so it needs the most effort.
• D: Its IMA is 8, which is good but still less than 10.

Answer: C

Correct answer: An internal leak past a worn piston seal. Slow drift under load with no visible oil and a steady reservoir points to oil bypassing the piston inside the cylinder, such as past a worn seal, rather than leaking outside.

• A: Thick oil makes motion sluggish when commanded; it does not let a held load sink.
• B: An external leak would leave visible oil and lower the reservoir level, neither of which is seen.
• C (correct): Oil slipping past a worn seal inside the cylinder lets it drift under load without any visible oil loss.
• D: A high relief setting raises the maximum pressure but would not make a held load creep down.

Answer: C

Correct answer: 300 N. Only the vertical parts of the two tensions support the weight:

• A: Half the weight would be correct only if the ropes hung straight up.
• B: This uses cos 30° instead of sin 30°, but only the vertical parts of the tensions hold up the sign.
• C (correct): T = W ÷ (2 sin θ) = 300 ÷ (2 × 0.5) = 300 N.
• D: This divides the weight by sin 30° but forgets that two ropes share the load.

Answer: C

Correct answer: 17:1. Solve

for the ratio.

The ratio must be at least this, so round up to 17:1, which gives

N·m.

• A: This ignores the 80 percent efficiency; 14:1 would give only 0.80 times 14 times 3, or 33.6 N·m.
• B: This multiplies the target by the efficiency instead of dividing by it, giving far too small a ratio.
• C (correct): The ratio must be at least 40 divided by (0.80 times 3), or about 16.7, so the next whole number, 17, is needed.
• D: This rounds 16.7 down; 16:1 delivers 0.80 times 16 times 3, or 38.4 N·m, which falls short.

Answer: D

Correct answer: 160 J. Friction is

giving a deceleration of 2.5 m/s². The crate slides

• A: This uses mass times speed squared and forgets the one-half in kinetic energy.
• B: This multiplies the mass by the speed, which is momentum, not energy.
• C: This multiplies the 50 N friction force by the speed, treating 4 m/s as a distance.
• D (correct): Friction is 50 N and the crate slides 3.2 m, so the heat is 160 J, equal to its starting kinetic energy.

Answer: D

Correct answer: 3,200 J. Useful work:

Since efficiency is output over input, input work is

As a check, the worker pulls 8 m of rope with

• A: This is the useful output work done on the crate; the losses add to the input.
• B: This multiplies the output work by the efficiency, giving less input than output.
• C: This multiplies the full load by the 8 m of rope pulled, as if each segment needed the whole load.
• D (correct): Output work is 1,200 times 2 = 2,400 J; input work = 2,400 / 0.75 = 3,200 J.

Answer: C

Correct answer: 2.3 mm. First find the temperature rise:

Then

m = 2.3 mm.

• A: This is a power-of-ten slip when converting 0.0023 m to millimeters.
• B: This is a power-of-ten slip in the other direction.
• C (correct): The heat raises the rod by 54,000/(1.2 × 900) = 50°C, and $(23\times 10^{-6})(2)(50)=0.0023$ m.
• D: This finds the temperature rise as 54,000/900 = 60°C, leaving out the rod's mass.