Friction Lesson Test for ASVAB Mechanical

Answer: D

Correct answer: 70 N. The cabinet stays at rest, so the horizontal forces balance. Static friction adjusts to equal the 70 N push, pointing opposite it, as long as that is under the 150 N limit.

• A: This subtracts the push from the limit, but friction does not equal the leftover capacity.
• B: A push with no motion still needs an equal friction force to keep the forces balanced.
• C: This is the maximum static friction, which is reached only when the cabinet is about to start moving.
• D (correct): Static friction only grows as large as needed to balance the push, so it matches the 70 N.

Answer: C

Correct answer: It has no unit. The coefficient is

Both forces are in newtons, so the units cancel and the coefficient is dimensionless.

• A: That is a unit of work or torque, which requires multiplying force by distance.
• B: That is a unit of gravitational field strength, not of a force ratio.
• C (correct): Newtons divided by newtons cancel, so the coefficient is a pure ratio.
• D: Newtons appear in both the top and bottom of the ratio, so they cancel out.

Answer: A

Correct answer: Perpendicular to the ramp surface, away from it. The normal force always acts at right angles to the contact surface, pushing out of it. On a ramp it therefore tilts back from vertical by the same angle as the ramp.

• A (correct): Normal means perpendicular, so the push comes straight out of the ramp face.
• B: The surface pushes outward on the box; it does not pull the box down.
• C: A force along the surface would be friction, not the normal force.
• D: That is true only on a level surface; on a ramp the normal force tilts with the surface.

Answer: B

Correct answer: 588 N. Weight is

• A: This divides the mass by g instead of multiplying.
• B (correct): Weight is mass times g, and 60 times 9.8 gives 588 N.
• C: This treats the mass in kilograms as if it were already a force in newtons.
• D: This adds g to the mass instead of multiplying them.

Answer: C

Correct answer: 180 N. Use

• A: This misplaces the decimal; 0.3 times 600 is 180, not 18.
• B: This subtracts the friction from the normal force, which is not a friction value.
• C (correct): Kinetic friction is the coefficient times the normal force: 0.3 times 600 N.
• D: This divides the normal force by the coefficient instead of multiplying.

Answer: C

Correct answer: The static coefficient is larger than the kinetic one. Getting an object moving requires exceeding ${\mu }_{s}N$, while keeping it moving takes only ${\mu }_{k}N$. The drop in required force shows that ${\mu }_s$ is larger than ${\mu }_k$ for this pair of surfaces.

• A: Equal coefficients would make the starting push and the moving push the same.
• B: The coefficients describe the pair of surfaces; weight changes the normal force, not the coefficients.
• C (correct): Starting requires beating the larger static limit; sliding then meets only the smaller kinetic friction.
• D: If kinetic friction were larger, keeping the desk moving would take more effort, not less.

Answer: D

Correct answer: It replaces sliding contact with rolling contact. In a ball bearing the balls roll between the races instead of the shaft rubbing against a sleeve. Rolling resistance is usually far smaller than sliding friction, so less power is lost to heat.

• A: Bearings still need clean, lubricated races; running them dry raises resistance and wear.
• B: More normal force would raise friction, not lower it.
• C: A bearing does not change the shaft's mass or weight.
• D (correct): Rolling surfaces do not scrub past each other, so resistance is far lower than sliding friction.

Answer: D

Correct answer: It is the same in both positions. In the simple model

with no area term. The brick's weight and the coefficient stay the same, so the friction force is the same whichever face touches the board.

• A: Higher pressure on the narrow edge is offset by the smaller area, so friction is unchanged.
• B: The brick still presses on the board with its full weight, so friction is still present.
• C: The larger face spreads the same weight over more area, so the pressure drops and the effects cancel.
• D (correct): Friction depends on the normal force and coefficient, which do not change when the brick is turned.

Answer: A

Correct answer: It forms a film that lowers friction and wear. A lubricant places a thin fluid film between the surfaces, so they no longer grind directly against each other. This sharply lowers friction, cuts wear, and helps remove heat.

• A (correct): The parts ride on the lubricant layer instead of grinding directly together.
• B: Reducing friction reduces heating, and the lubricant also helps carry heat away.
• C: Grease does not add any push between the parts; it separates the surfaces.
• D: Lubricants are used to reduce friction, not increase grip.

Answer: A

Correct answer: Static friction from the ground. Friction acts at the contact surface to oppose slipping between the surfaces. Your foot pushes backward on the ground, and by Newton's third law the ground pushes forward on the foot; because the sole does not slide, that grip is static friction. Many people assume friction always opposes motion, but it opposes relative sliding at the contact, so here it points in the direction you walk. On ice, too little static friction makes walking hard.

• A (correct): Correct: the shoe pushes backward against the ground and the ground provides forward static traction.
• B: Kinetic friction would require relative sliding between shoe and ground at the contact patch.
• C: Walking without slipping still requires traction; zero friction would make propulsion difficult.
• D: Friction opposes relative slip at the contact, not necessarily the person's forward velocity.

Answer: B

Correct answer: 320 N. First find the normal force:

• A: This divides the normal force by the coefficient instead of multiplying.
• B (correct): The normal force is 800 N, and 0.4 times 800 N gives the 320 N limit.
• C: This multiplies the coefficient by the mass in kilograms instead of by the normal force in newtons.
• D: This subtracts 320 N from the weight, which is not the friction limit.

Answer: B

Correct answer: 0.30. Steady speed means zero net force, so kinetic friction equals the 72 N pull. On a level floor

• A: This divides the weight by the friction force, inverting the ratio.
• B (correct): At steady speed friction equals the 72 N pull, and 72 divided by 240 is 0.30.
• C: This subtracts 0.30 from 1, which has no physical meaning here.
• D: This simply moves the decimal in the pull force instead of dividing by the normal force.

Answer: C

Correct answer: 4 m/s². Subtract friction from the push to get the net force, then divide by mass.

• A: This divides the friction force alone by the mass.
• B: This adds friction to the push instead of subtracting it.
• C (correct): The net force is 140 minus 60, or 80 N, and 80 N divided by 20 kg is 4 m/s².
• D: This ignores friction and divides the full push by the mass.

Answer: D

Correct answer: 4 m/s². Friction is ${\mu }_{k}mg$, and dividing by the mass gives the deceleration.

• A: This divides the correct deceleration by the mass a second time.
• B: This is the friction force in newtons, not the acceleration; it still needs to be divided by the mass.
• C: This uses the coefficient alone and forgets to multiply by g.
• D (correct): With friction the only horizontal force, the deceleration is the coefficient times g, and the mass cancels.

Answer: B

Correct answer: 60 N. A downward push adds to the normal force:

• A: This multiplies the coefficient by the press alone and leaves out the box's weight.
• B (correct): The downward press raises the normal force to 300 N, and 0.2 times 300 N is 60 N.
• C: This uses only the box's weight and ignores the extra downward press.
• D: This subtracts the press from the weight, but a downward push adds to the normal force.

Answer: D

Correct answer: 114 N. An upward pull subtracts from the normal force:

• A: This ignores the hoist and uses the full weight as the normal force.
• B: This multiplies the coefficient by the hoist force instead of by the normal force.
• C: This adds the upward pull to the weight, but an upward force reduces the normal force.
• D (correct): The upward pull reduces the normal force to 380 N, and 0.3 times 380 N is 114 N.

Answer: A

Correct answer: 300 N. The component of weight along a ramp is $W\sin \theta$.

• A (correct): The downhill component is the weight times sin 30 degrees, which is 0.5 times 600 N.
• B: This uses cos 30 degrees, which gives the component pressing into the ramp.
• C: This divides by sin 30 degrees instead of multiplying by it.
• D: The full weight acts along the surface only on a vertical wall.

Answer: D

Correct answer: 250 N. On a ramp the normal force is

• A: The normal force equals the full weight only on level ground.
• B: This divides by cos 60 degrees instead of multiplying by it.
• C: This uses sin 60 degrees, which gives the component pulling the box down the ramp.
• D (correct): The normal force balances the perpendicular weight component, 500 times cos 60 degrees, which is 250 N.

Answer: C

Correct answer: 0.58. At the angle of repose,

• A: This uses sin 30 degrees instead of the tangent.
• B: This is 1 divided by tan 30 degrees, the inverted ratio.
• C (correct): At the angle of repose the coefficient equals tan 30 degrees, which is about 0.58.
• D: This uses cos 30 degrees instead of the tangent.

Answer: D

Correct answer: Up the hill, along the road surface. With nothing else pushing along the slope, the car tends to roll or slip downhill. Static friction opposes that tendency, so it points up the hill with a size equal to the downhill component of weight.

• A: Friction always acts along the contact surface, not vertically.
• B: That would add to gravity's downhill pull and the car could not stay parked.
• C: A parked car on a slope needs static friction to stay at rest.
• D (correct): Gravity tends to pull the car downhill, so static friction acts uphill to hold it.

Answer: C

Correct answer: Forward, in the direction the belt is moving. Friction opposes relative slipping, not motion over the ground. As the belt speeds up, the box would slip backward relative to it, so static friction points forward and accelerates the box along with the belt.

• A: Friction opposes slipping between surfaces, and here the box would slip backward relative to the belt.
• B: Static friction acts between surfaces that are not slipping; it is what speeds the box up.
• C (correct): Without friction the box would lag behind the belt, so static friction drags it forward.
• D: Friction acts along the contact surface; the upward push is the normal force.

Answer: C

Correct answer: 7 m/s². The road can supply at most

• A: This divides the coefficient by g instead of multiplying.
• B: This misplaces the decimal; 0.7 times 10 is 7.
• C (correct): The maximum deceleration is the coefficient times g, 0.7 times 10.
• D: This divides g by the coefficient instead of multiplying.

Answer: A

Correct answer: 32 m. Use the sliding-distance relation.

• A (correct): The distance is 8 squared divided by 2 times 0.1 times 10, which is 64 divided by 2.
• B: This uses the speed without squaring it.
• C: This forgets the factor of 2 in the denominator.
• D: This leaves the coefficient out of the denominator, dividing 64 by 20.

Answer: A

Correct answer: 600 J. First find the friction force:

The heat equals friction force times sliding distance.

• A (correct): Friction is 0.2 times 250 N, or 50 N, and 50 N times 12 m is 600 J.
• B: This adds the friction force and the distance instead of multiplying them.
• C: This multiplies the weight by the distance instead of the friction force.
• D: This divides the friction force by the distance instead of multiplying.

Answer: C

Correct answer: 72 m. Ideal braking distance is

so it is proportional to speed squared. Doubling speed from 15 to 30 m/s multiplies the distance by

giving

• A: This triples the distance, which matches no rule for braking.
• B: This assumes distance grows in direct proportion to speed, but speed is squared.
• C (correct): Braking distance grows with the square of speed, so doubling speed multiplies 18 m by 4.
• D: This multiplies by 8, as if distance grew with the cube of speed.

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

Correct answer: 600 N. The vertical part of the push is

N downward. The floor must support both this and the weight.

• A: This adds 200 times cos 30 degrees, which is the forward part of the push, not the downward part.
• B (correct): The downward part of the push is 200 times sin 30 degrees, or 100 N, which adds to the 500 N weight.
• C: This ignores the downward part of the angled push.
• D: This subtracts the downward part, but a downward push increases the normal force.

Answer: B

Correct answer: 30 N. The horizontal component of an angled pull is $F\cos \theta$.

• A: This divides by cos 60 degrees instead of multiplying.
• B (correct): The horizontal part is 60 times cos 60 degrees, which is 0.5 times 60 N.
• C: This uses sin 60 degrees, which gives the upward part of the pull.
• D: Only part of an angled pull acts horizontally.

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: B

Correct answer: Increasing the belt tension. A belt drives a pulley only through friction. Tension creates the normal force between belt and pulley, so raising the tension raises the maximum friction the contact can supply before slipping.

• A: A smoother face usually lowers the friction coefficient.
• B (correct): More tension presses the belt harder against the pulley, raising the normal force and the available friction.
• C: Less tension lowers the normal force and reduces grip.
• D: Oil lowers the friction coefficient and would make the slipping worse.

Answer: A

Correct answer: 250 N. The normal force is 600 N, so the static limit is

Since 250 N is below 300 N, the crate does not move and static friction equals the push, 250 N.

• A (correct): The static limit is 300 N; the 250 N push is below it, so the crate stays put and static friction matches the push.
• B: This is the static limit, which applies only when the crate is just about to slip.
• C: This is the gap between the push and the static limit, not a friction force.
• D: This is kinetic friction, but the crate never starts sliding.

Answer: B

Correct answer: 2 m/s². The normal force is 500 N. Starting takes

while sliding friction is

• A: This ignores friction and divides the whole push by the mass.
• B (correct): Sliding friction is only 150 N, so the 250 N starting push leaves 100 N of net force on 50 kg.
• C: This divides the kinetic friction alone by the mass.
• D: This assumes the crate moves at constant speed, but the push now exceeds kinetic friction.

Answer: D

Correct answer: About 6.7 m/s². For sliding downhill,

• A: This subtracts the coefficient times g without multiplying by cos 60 degrees.
• B: This includes only gravity's component along the chute and leaves out friction.
• C: This adds friction, but for downhill sliding friction acts uphill and opposes motion.
• D (correct): Gravity's part along the chute is 8.66 m/s², and friction removes 0.4 times 5, or 2 m/s².

Answer: C

Correct answer: About 8.8 m/s². Sliding uphill, friction acts down the slope along with gravity's component.

• A: This adds the coefficient times g without multiplying by cos 45 degrees.
• B: This subtracts friction, which is right for sliding down, but here friction points downhill too.
• C (correct): Moving uphill, both gravity's component and friction point downhill, so their effects add.
• D: This includes gravity's component along the ramp but ignores friction.

Answer: A

Correct answer: 100 N, pointing down the ramp. Gravity's pull along the ramp is

N downhill. The 300 N push up the ramp exceeds it by 100 N, so static friction supplies 100 N down the ramp. The limit is

so the crate stays at rest.

• A (correct): The push exceeds gravity's 200 N downhill pull by 100 N, so the crate tends to move uphill and friction holds it back.
• B: This ignores gravity's downhill component, which already balances part of the push.
• C: This is the static limit, which is reached only when the crate is about to slip.
• D: Friction opposes the tendency to slip, and here the net tendency is uphill, not downhill.

Answer: A

Correct answer: It slips, because tan 35 degrees is greater than 0.6. The box stays only if

Here

which is greater than 0.6, so the downhill pull exceeds the static limit and the box slips, whatever its weight.

• A (correct): Resting requires the tangent of the angle to be no more than the static coefficient, and 0.70 exceeds 0.6.
• B: The test compares the tangent, not the sine, with the coefficient; both sides contain the weight times cos as well.
• C: The cosine sets the normal force; it is not compared directly with the coefficient.
• D: Weight cancels from the slipping condition, so a heavier box slips at the same angle.

Answer: D

Correct answer: 62.5 m. Use the wet coefficient in the braking-distance relation.

• A: This forgets the factor of 2 in the denominator.
• B: This uses the speed without squaring it.
• C: This uses the dry-road coefficient of 0.8 instead of the wet value.
• D (correct): On the wet road, 25 squared divided by 2 times 0.5 times 10 is 625 divided by 10.

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: 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: B

Correct answer: 3.2 m. The sliding distance is

with no mass term. Doubling the launch speed multiplies the distance by 4, so

• A: This doubles the distance as if it grew in direct proportion to speed.
• B (correct): Mass cancels, so only speed matters; doubling the speed multiplies the distance by 4.
• C: This also multiplies by 4 for the heavier mass, but mass cancels in sliding problems.
• D: This assumes nothing changes, but the higher speed lengthens the stop.

Answer: D

Correct answer: 360 J. Steady speed means zero net force, so kinetic friction equals the 45 N pull. The heat produced equals friction force times sliding distance: 45 N times 8 m = 360 J.

• A: This assumes balanced forces mean no friction; steady speed only means friction exactly matches the 45 N pull.
• B: This halves the work, as if the force averaged out; kinetic friction stays a constant 45 N the whole way.
• C: This divides force by distance; heat from friction is force multiplied by sliding distance.
• D (correct): At steady speed the pull balances kinetic friction, so friction is 45 N, and 45 N times 8 m gives 360 J of heat.

Answer: A

Correct answer: 60 N. The downhill pull is

The static limit is

so the crate holds, and static friction is exactly 60 N pointing up the ramp.

• A (correct): Static friction only needs to balance the downhill weight component, 120 sin 30° = 60 N, which is below the limit.
• B: This is the maximum static friction, 0.7 times 120 cos 30°; the actual friction is only what is needed to hold the crate.
• C: This is the weight component pressing into the ramp, which sets the normal force, not the friction.
• D: This uses the full weight; on a 30 degree ramp only half the weight acts down the slope.

Answer: C

Correct answer: To increase friction so the hand grips without slipping. Roughening a surface, as with knurling, sandpaper or textured flooring, usually raises the friction coefficient. A higher coefficient lets static friction hold the tool in the hand with less squeezing.

• A: Smooth, polished or lubricated surfaces lower friction; a cut texture does the opposite.
• B: The normal force comes from how hard you squeeze, not from the surface pattern.
• C (correct): Roughening a surface raises the friction coefficient, giving the hand more grip.
• D: The texture is there for grip; the handle is not meant to move against the hand at all.

Answer: A

Correct answer: It is the same. In the dry-friction model,

The coefficient and the normal force are unchanged, so the friction force is the same at any sliding speed.

• A (correct): Kinetic friction equals the kinetic coefficient times the normal force, and neither depends on sliding speed.
• B: This treats dry friction like slow fluid drag, which grows in proportion to speed.
• C: This treats dry friction like fast air drag, which grows roughly with the square of speed.
• D: Sliding faster does not reduce dry friction in the simple model; the force stays fixed.

Answer: B

Correct answer: About 37 degrees. An object on an incline starts to slip when

Here

The crate weight cancels out, so a heavier crate would slip at the same angle.

• A: This uses 1 divided by 0.75; the condition is tan θ equal to the coefficient itself, not its reciprocal.
• B (correct): Slipping starts when the tangent of the angle reaches the static coefficient, and tan 37° ≈ 0.75.
• C: Objects do not all slip at 45 degrees; that is the slip angle only when the coefficient is 1.0.
• D: At 30 degrees the tangent is only 0.58, which is less than 0.75, so static friction still holds the crate.

Answer: B

Correct answer: 0.20. Split the pull: horizontal part

upward part

Steady speed means friction equals 80 N, and the normal force is

• A: This divides 80 N by the full 460 N weight, ignoring that the rope lifts 60 N and lowers the normal force.
• B (correct): The forward pull 80 N equals friction, and the normal force is 460 − 60 = 400 N, so 80 ÷ 400 = 0.20.
• C: This uses the whole 100 N rope tension as friction; only the 80 N horizontal component balances friction.
• D: This adds the 60 N rope component to the weight; an upward pull reduces the normal force instead.

Answer: D

Correct answer: 50 N. Treat each contact separately. At the sled-snow contact, the normal force carries everything above it:

Kinetic friction there is

• A: This uses only the sled weight; the box presses down through the sled, so its weight adds to the normal force on the snow.
• B: This uses only the box weight; the snow must support the sled too.
• C: The box not slipping on the sled is a separate contact; the sled itself is sliding on the snow, so kinetic friction acts there.
• D (correct): The snow supports both the sled and the box, so the normal force is 500 N and friction is 0.10 × 500 = 50 N.

Answer: C

Correct answer: 3 m/s². Weight and normal force:

Friction:

• A: This divides the full 120 N push by the mass, forgetting to subtract friction.
• B: This divides the 75 N friction force by the mass instead of the net force.
• C (correct): Friction on the bin is 0.5 × 150 = 75 N, net force is 120 − 75 = 45 N, and 45 ÷ 15 = 3 m/s².
• D: This adds friction to the push; kinetic friction opposes the sliding, so it must be subtracted.

Answer: A

Correct answer: 7.5 N. With

drag grows in direct proportion to speed. Find

N·s/m, then

• A (correct): Drag is proportional to speed here, and 0.5 is 2.5 times 0.2, so the drag is 2.5 × 3 N = 7.5 N.
• B: This treats the oil drag like dry sliding friction, which does not depend on speed.
• C: This scales drag with the square of speed, a pattern for fast air drag, not slow motion in thick fluid.
• D: This scales the drag down instead of up; a faster probe meets more fluid drag.

Answer: B

Correct answer: 0 m/s², because it stays at rest. First decide whether the toolbox slips. It stays put if

Here 0.75 is less than 0.80, so static friction balances the downhill pull and the acceleration is zero.

The kinetic formula $g(\sin \theta -{\mu }_k\cos \theta )$ is used only once an object is already sliding.

• A: This applies the sliding formula with the kinetic coefficient, but the toolbox never starts sliding, so kinetic friction does not apply.
• B (correct): Since tan 37° ≈ 0.75 is less than the static coefficient 0.80, static friction can hold the toolbox in place.
• C: This uses g sin 37° and ignores friction entirely.
• D: This uses g cos 37°, the component that presses into the ramp, as if it pulled the toolbox along the slope.