Mechanical Comprehension Practice Test 3 for the ASVAB

Answer: B

Correct answer: 35 m. Distance is the total path length, so the two legs simply add: 25 m + 10 m = 35 m. Direction does not matter for distance.

• A: This counts only the first leg and leaves out the 10 m walked back.
• B (correct): Distance adds the path lengths without signs: 25 m plus 10 m is 35 m.
• C: 15 m east is the displacement, the net change in position, not the distance walked.
• D: Distance is never negative, and 15 m is the size of the displacement, not the path length.

Answer: C

Correct answer: It keeps moving in a straight line at a constant speed. Newton's first law: an object with zero net force keeps a constant velocity. A puck that is already moving keeps going straight at the same speed.

• A: Speeding up is an acceleration, which would require a net forward force.
• B: Slowing down is a change in velocity, which would require a net backward force.
• C (correct): With no net force, Newton's first law says the velocity does not change, so speed and direction stay the same.
• D: Objects do not need a force to keep moving; they need a force to change their motion.

Answer: D

Correct answer: An extension spring, loaded by pulling. Tension, or extension, springs are wound with the coils tight together and have hooks or loops so they can be attached and pulled. Compression springs instead have gaps between coils and flat ends so they can be pushed shorter.

• A: A torsion spring resists twisting about its axis; hooks on each end and closed coils point to a spring that is pulled.
• B: A compression spring has open gaps between coils and flat ends so it has room to be squeezed.
• C: A leaf spring is a stack of flat strips that bend, not a coil with hooks.
• D (correct): Closed coils with hooks or loops are the signature of a tension (extension) spring, which is pulled and pulls back.

Answer: D

Correct answer: They run more quietly and smoothly. Helical teeth are cut at an angle, so each tooth comes into contact gradually instead of all at once. More teeth share the load, which makes the drive quieter and smoother than a spur pair, at the cost of axial thrust on the bearings.

• A: The angled teeth actually push each gear along its shaft, creating axial thrust the bearings must resist.
• B: Self-locking is a feature of many worm sets, not of helical gears.
• C: Joining intersecting shafts is the job of bevel gears; helical gears connect parallel shafts.
• D (correct): Angled teeth engage gradually and share the load among more teeth, so helical gears are quieter and smoother.

Answer: B

Correct answer: 6 m/s. From rest, all the net work becomes kinetic energy.

So v² = 36 and

• A: This divides the work by the mass, skipping the one-half and the square root.
• B (correct): Setting 360 = ½ × 20 × v² gives v² = 36, so v = 6 m/s.
• C: This leaves out the one-half, giving v² = 18 and a speed that is too low.
• D: This finds v² correctly but forgets to take the square root.

Answer: D

Correct answer: 3.98 N·m. Set input work per turn equal to output work:

• A: Multiplying by 2 pi instead of dividing makes the torque far too large.
• B: Leaving the lead in millimeters instead of meters inflates the answer a thousandfold.
• C: This is the work per turn in joules; it still needs to be divided by 2 pi.
• D (correct): Work per turn is 5000 N times 0.005 m, or 25 J, and dividing by 2 pi gives about 3.98 N·m.

Answer: C

Correct answer: Skidding tires rely on kinetic friction, which is lower than the static limit. A rolling tire's contact patch does not slip on the road, so braking uses static friction up to ${\mu }_{s}N$. When the wheel locks, the patch slides and only the smaller kinetic friction acts, so the deceleration is lower and the stopping distance longer.

• A: In the simple model friction does not depend on contact area.
• B: Locking the wheels does not change the car's weight on the road.
• C (correct): A rolling tire's contact patch is momentarily at rest, so it can use the larger static friction.
• D: Skidding does not change the car's mass.

Answer: D

Correct answer: A toothed timing belt. Friction belts lose a little speed to creep and can slip under load. A timing belt has teeth that engage toothed pulleys, so the shafts keep an exact speed ratio, which is why engines use it to time the camshaft.

• A: Crossing the belt reverses rotation and adds wear, but it is still a friction drive that can slip.
• B: A flat belt grips by friction only, so creep and slip prevent an exact speed ratio.
• C: A V-belt grips well by wedging into its groove, but it still relies on friction and can creep or slip.
• D (correct): Its molded teeth mesh with grooved pulleys like gears, so it cannot slip in normal use and keeps an exact ratio.

Answer: D

Correct answer: 200 N. The lower block and hook move with the compressor, so total lifted weight = 570 N + 30 N = 600 N. Ideal effort = 600

• A: This adds the block weight after dividing, but the block is shared by the three segments like the load.
• B: Dividing 600 N by 2 uses the wrong segment count for a three-segment tackle.
• C: This leaves out the 30 N block and hook, which the rope must also lift.
• D (correct): The rope lifts 570 N + 30 N = 600 N, and 600 N / 3 = 200 N.

Answer: B

Correct answer: 6,800 Pa. Convert 5 cm to 0.05 m, then

• A: This adds the atmosphere, but the height difference gives gauge pressure.
• B (correct): Gauge pressure is 13,600 × 10 × 0.05 = 6,800 Pa.
• C: This uses water's density instead of mercury's.
• D: This uses 0.5 m for 5 cm, a factor of ten too large.

Answer: A

Correct answer: 100 W.

• A (correct): $P=kA\Delta T/d=(0.04)(5)(25)/0.05=100$ W.
• B: This multiplies by the thickness instead of dividing by it.
• C: This leaves out the 25 K temperature difference.
• D: This uses 0.005 m for the thickness, a power-of-ten slip.

Answer: B

Correct answer: 180 N. Weight:

Buoyancy:

Upward forces must balance the weight:

so the bottom pushes up with 180 N.

• A: This is the buoyant force, not the support from the pond bottom.
• B (correct): The rock weighs 300 N and gets 120 N of buoyancy, so the bottom supplies the remaining 180 N.
• C: This ignores the buoyant force, which still acts on a rock resting on the bottom.
• D: This adds the buoyant force to the weight, but buoyancy helps hold the rock up.

Answer: C

Correct answer: 12 strokes. Volume per stroke =

cubic centimeters. Volume to raise the ram 3 cm =

cubic centimeters. Strokes =

• A: This divides the 120 cubic centimeters needed by the area ratio instead of by the 10 cubic centimeters delivered per stroke.
• B: This is just the area ratio; it does not account for the lift height or the stroke volume.
• C (correct): Each stroke moves 2 × 5 = 10 cubic centimeters, the ram needs 40 × 3 = 120 cubic centimeters, and 120 ÷ 10 = 12.
• D: This multiplies the 3 cm lift by the area ratio of 20 but ignores the 5 cm the pump piston moves on each stroke.

Answer: C

Correct answer: 324 N. Input work =

Output work =

Load =

• A: This is the load for a perfect lever; it ignores the 10 percent of input work lost to friction.
• B: This flips the distance ratio, as though the load moved farther than the handle.
• C (correct): Input work is 54 J, 90 percent of that is 48.6 J, and 48.6 J ÷ 0.15 m = 324 N.
• D: This divides by the efficiency instead of multiplying, which would make the output work larger than the input work.

Answer: C

Correct answer: 100 N. In series, advantages multiply: IMA =

Efficiencies multiply:

AMA =

so effort =

• A: Multiplying the ideal effort by the efficiency lowers it, but losses raise the real effort.
• B: Adding the two IMAs instead of multiplying them understates the combined advantage.
• C (correct): Total IMA is 60 times 5, or 300; total efficiency is 0.5 times 0.8, or 0.40; AMA is 120, so effort is 12,000 divided by 120, or 100 N.
• D: This is the ideal effort and ignores the losses in both stages.