2.7 Kinetic and Static Friction

1. AIt is the same at both instants, as the suitcase is moving at both.
   A student who thinks kinetic friction acts whenever an object moves relative to the ground picks this. What matters is motion relative to the belt: at instant 2 the suitcase moves with the belt, the surfaces do not slide, and there is no kinetic friction.
2. BIt acts forward at instants 1 and 2, as a force must keep it moving.
   A student who believes a moving object needs a force in its direction of motion to keep it moving picks this. At instant 1 kinetic friction does point forward, but at instant 2 the suitcase moves at constant velocity with the belt: no force is needed to keep it moving, and because the surfaces do not slide there is no kinetic friction (nor any static friction).
3. CIt is exerted at instant 1 only, while the suitcase slides on the belt. Correct
   Kinetic friction occurs only while two surfaces in contact move relative to each other. At instant 1 the belt moves faster than the suitcase, so the surfaces slide and kinetic friction acts, pointing forward and speeding the suitcase up. At instant 2 the suitcase and belt have the same velocity, so there is no kinetic friction.
4. DIt is zero at both instants, since friction cannot speed the suitcase up.
   A student who believes friction can only oppose an object's motion picks this, and decides the belt simply 'carries' the suitcase. While the suitcase slides, it moves backward relative to the belt, so kinetic friction on it points forward and is what speeds it up.

Working Kinetic friction needs the two surfaces in contact to move relative to each other. Instant 1: the belt moves faster than the suitcase, so the suitcase slides backward relative to the belt and kinetic friction, magnitude μ_k F_N, acts on it in the belt's direction of motion, speeding it up. Instant 2: suitcase and belt have the same velocity, so there is no relative motion and no kinetic friction (and, with the belt at constant speed, no static friction either). So kinetic friction acts at instant 1 only.

CED 2.7.A.1 · Read this in Fix

1. APlacing a heavy load on top of the block as it slides
   A student who thinks the coefficient grows with the object's weight picks this. A load increases the normal force and so the friction force, but the ratio F_f/F_N, μ_k, stays the same because the surfaces are unchanged.
2. BTurning the block onto a face with a smaller area
   A student who thinks the area of contact affects friction picks this. Every face is the same sanded wood, and neither the friction force nor μ_k depends on the area of contact.
3. CSliding the block on a sheet of glass laid on the table Correct
   The coefficient of kinetic friction depends on the materials of the two surfaces in contact. Sliding on glass replaces a wood–wood pair with a wood–glass pair, which has a different μ_k.
4. DPushing the block so that it slides at a much greater speed
   A student who thinks kinetic friction grows with speed picks this. In the friction model used in AP Physics 1, μ_k depends only on the materials of the two surfaces, not on how fast they slide.

CED 2.7.A.2.i · Read this in Fix

1. AFriction pushes the book backward, opposite to its motion.
   A student who thinks friction always opposes an object's velocity picks this. If the only horizontal force on the book pointed backward, the book could not speed up with the truck. Without friction the book would slide backward relative to the bed, so static friction points forward.
2. BNo friction acts, since the book does not slide on the bed.
   A student who thinks friction acts only on sliding objects picks this. Static friction acts between surfaces that are not moving relative to each other. Without a forward friction force the book could not speed up with the truck.
3. CKinetic friction acts on the book, since it moves along the road.
   A student who thinks kinetic friction acts whenever an object moves relative to the ground picks this. Kinetic friction needs sliding between the two surfaces in contact, and the book does not slide on the bed.
4. DStatic friction from the bed pushes the book forward. Correct
   The book and the bed are not moving relative to each other, so any friction is static. The book speeds up with the truck, so a forward horizontal force must act on it, and the only horizontal force is friction from the bed: static friction, directed forward.

CED 2.7.B.1 · Read this in Fix

1. AIt is moving relative to the ground beside the track.
   A student who thinks sliding means moving relative to the ground picks this. A suitcase carried along at the train's velocity moves past the ground but does not slip on the floor; slipping is motion relative to the surface it touches.
2. BIts velocity differs from the velocity of the floor beneath it. Correct
   Slipping and sliding mean that the two surfaces in contact move relative to each other. The suitcase slips on the floor exactly when its velocity is not the same as the floor's.
3. CThe train speeds up, and the suitcase tends to stay behind.
   A student who takes a tendency to slip for slipping picks this. While the train speeds up, static friction can speed the suitcase up with it; the suitcase slips only if the friction needed exceeds μ_s F_N.
4. DThe floor exerts a friction force on the suitcase.
   A student who thinks friction acts only on sliding objects picks this. Static friction can act on the suitcase while it does not slip, for example while the train speeds up gently.

CED 2.7.B.2.i · Read this in Fix

1. AOne coefficient describes friction before and after slipping.
   A student who thinks a pair of surfaces has one coefficient of friction picks this. The graph shows two different values: a maximum of 12 N before slipping and a steady 8 N afterward, with the same normal force, so μ_s and μ_k differ.
2. BNo friction acts on the block until it begins to slide.
   A student who thinks friction acts only on sliding objects picks this. The graph shows friction rising with the pull from zero while the block is still at rest: that is static friction matching the pull.
3. COnce sliding, the friction force rises as the pull is increased.
   A student who thinks kinetic friction adjusts to the applied force picks this. After the block starts to slide, the graph is level at 8 N while the pull increases from 12 N to 16 N: kinetic friction stays at μ_k F_N.
4. DThe coefficient μ_s is larger than μ_k for these surfaces. Correct
   The largest friction before sliding, the peak of 12 N, is μ_s F_N; the steady friction once the block slides, 8 N, is μ_k F_N. The normal force is the same throughout, so μ_s is larger than μ_k, by a factor of 12/8 = 1.5.

CED 2.7.B.3 · Read this in Fix

1. AIt is correct, as friction opposes the block's motion to the right.
   A student who thinks friction always opposes an object's velocity relative to the ground picks this. The friction exerted by the board depends on the block's motion relative to the board, which is to the left, so friction on the block points to the right.
2. BIt should also include a force of motion pointing to the right.
   A student who believes a moving object needs a force in its direction of motion picks this. No object exerts such a 'force of motion', so no fourth arrow belongs on the diagram. The only horizontal force on the block is friction from the board, and it should point to the right.
3. CThe normal force arrow should tilt right, as the board drags the block.
   A student who treats the normal force as the whole push of the board picks this. The board's drag on the block is friction, parallel to the surface; the normal force stays perpendicular to the board, straight up.
4. DThe friction arrow should point to the right, not to the left. Correct
   Kinetic friction on the block points opposite to the block's motion relative to the board. The board moves right faster than the block, so relative to the board the block slides to the left, and the friction on it points to the right. That friction is what drags the block forward.

CED 2.7.A.1.i · Read this in Fix

1. AF₃ > F₁ > F₂
   A student who thinks a larger area of contact gives more friction picks this, ranking brick 1 above brick 2 because it lies on its largest face. The friction force does not depend on the area of contact, so F₁ = F₂.
2. BF₃ > F₁ = F₂ Correct
   Friction depends on μ_k, which is the same for every arrangement, and on the normal force. Arrangements 1 and 2 each press on the floor with the weight of one brick, so F₁ = F₂ even though their areas of contact differ; arrangement 3 presses with the weight of two bricks, so F₃ is twice as large.
3. CF₂ > F₃ > F₁
   A student who thinks friction depends on pressure picks this: brick 2 presses its weight onto the smallest area and arrangement 3 has twice the weight on the largest area. Friction depends on the normal force, not on the force per area, so F₁ = F₂ and F₃ is the largest.
4. DF₁ = F₂ = F₃
   A student who thinks the friction force is fixed by the materials alone picks this. The materials fix μ_k, but the friction force is μ_k F_N, and the two stacked bricks press on the floor with twice the normal force, so F₃ is twice F₁.

Working Each arrangement slides on the same floor with the same brick material, so μ_k is the same. On a level floor with no vertical acceleration, F_N equals the weight of the arrangement: mg for 1 and 2, 2mg for 3. |F_f,k| = μ_k F_N, so F₁ = F₂ = μ_k mg and F₃ = 2μ_k mg. The area of contact does not affect the friction force, so F₃ > F₁ = F₂.

CED 2.7.A.1.ii · Read this in Fix

1. A6.0 N Correct
   The block slides, so the friction is kinetic: |F_f,k| = μ_k F_N. The push is horizontal, so F_N = mg = 20 N, and the friction force is 0.30 × 20 N = 6.0 N. It is less than the 8.0 N push, so the block speeds up.
2. B9.0 N
   A student who uses the coefficient of static friction for a block sliding over a stationary floor calculates 0.45 × 20 N = 9.0 N. The surfaces are sliding relative to each other, so μ_k applies.
3. C8.0 N
   A student who thinks kinetic friction adjusts to equal the push picks 8.0 N. Kinetic friction is μ_k F_N = 6.0 N whatever the push; the push is larger, so the net force is 2.0 N to the right and the block speeds up.
4. D2.4 N
   A student who multiplies the coefficient by the push instead of the normal force calculates 0.30 × 8.0 N = 2.4 N. Friction is μ_k times the normal force, F_N = mg = 20 N here.

Working The block slides on the floor, so kinetic friction acts. The push is horizontal, so vertically F_N = mg = (2.0 kg)(10 m/s²) = 20 N. |F_f,k| = μ_k F_N = 0.30 × 20 N = 6.0 N, directed to the left. (The push, 8.0 N, is larger, so the block speeds up; μ_s is not used because the surfaces are sliding.)

CED 2.7.A.2 · Read this in Fix

1. AThe friction force is fixed by the two materials alone.
   A student who thinks friction is set only by the materials picks this. The materials were the same in every trial, yet the friction force rose from about 2.4 N to about 10 N as the normal force increased.
2. BThe coefficient μ_k increases as mass is added to the block.
   A student who thinks the coefficient grows with the object's mass picks this. The ratio F_f/F_N is about 0.25 in every trial, the constant slope of the line, so μ_k does not change as mass is added; only the friction force does.
3. CThe friction force increases as the block slides faster.
   A student who thinks kinetic friction grows with speed picks this. The block moved at the same slow speed in every trial, so the data cannot show any effect of speed; what changed between trials was the normal force.
4. DThe friction force is directly proportional to the normal force. Correct
   The data points lie on a straight line through the origin: doubling the normal force from 20 N to 40 N doubles the friction force from about 5 N to about 10 N. This is |F_f,k| = μ_k|F_N| with a constant μ_k = 0.25, the slope of the line.

Working The best-fit line is straight and passes through the origin, so F_f is directly proportional to F_N. Its slope, (12 N)/(48 N) = 0.25, is μ_k, the same for every trial.

CED 2.7.A.2 · Read this in Fix

1. Aμ_k(mg − T sinθ) Correct
   The upward component of the rope's pull, T sinθ, supports part of the box's weight, so the floor pushes up with F_N = mg − T sinθ. The box slides, so the friction is kinetic: μ_k F_N = μ_k(mg − T sinθ).
2. Bμ_s(mg − T sinθ)
   A student who uses the static coefficient for an object sliding over a stationary floor picks this. The box slides relative to the floor, so μ_k applies; μ_s would only give the maximum static friction if the box were not sliding.
3. Cμ_k(mg + T sinθ)
   A student who adds the vertical component of the pull to the weight picks this. The rope pulls up, so it reduces the normal force: F_N + T sinθ − mg = 0 gives F_N = mg − T sinθ.
4. Dμ_k(mg − T cosθ)
   A student who swaps sine and cosine picks this. T cosθ is the horizontal component of the pull, along the floor; the vertical component, which changes the normal force, is T sinθ.

Working Perpendicular to the floor the box has no acceleration: F_N + T sinθ − mg = 0, so F_N = mg − T sinθ. The box slides, so kinetic friction acts: |F_f,k| = μ_k F_N = μ_k(mg − T sinθ). Units: μ_k has no unit, so the expression is in newtons.

CED 2.7.A.2.ii · Read this in Fix

1. AArrow P
   A student who thinks the normal force always points straight up picks this. That is true only on a level surface. On a ramp the normal force is perpendicular to the ramp's surface, tilted from the vertical by the ramp's angle.
2. BArrow R Correct
   The normal force is the component of the ramp's contact force perpendicular to the ramp's surface, and it is directed away from the surface. Arrow R is at right angles to the ramp and points away from it.
3. CArrow Q
   A student who thinks the normal force points into the surface picks this. Arrow Q is the direction of the force the block exerts on the ramp. The force exerted on the block by the ramp points the opposite way, away from the ramp.
4. DArrow S
   A student who thinks the normal force is the block's weight pressing on the ramp picks this. Arrow S is the direction of the gravitational force exerted by Earth; the normal force is a different force, exerted by the ramp, perpendicular to its surface.

CED 2.7.A.2.ii · Read this in Fix

1. A5.0 N Correct
   Static friction takes the value needed to prevent sliding. The book is at rest, so vertically the upward static friction balances the downward gravitational force: F_f,s = mg = 0.50 kg × 10 m/s² = 5.0 N. This is below the maximum, 0.80 × 9.0 N = 7.2 N, so the book stays put.
2. B7.2 N
   A student who thinks static friction always has its maximum value calculates μ_s F_N = 0.80 × 9.0 N = 7.2 N. That is only the largest friction the wall can exert; the book needs just 5.0 N to stay at rest.
3. C9.0 N
   A student who thinks friction always opposes and balances the push picks 9.0 N. The push is horizontal and is balanced by the wall's normal force. Friction is parallel to the wall, vertical, and balances the book's weight.
4. D0.0 N
   A student who thinks friction acts only on sliding objects picks zero. With no friction, nothing would balance the 5.0 N gravitational force and the book would slide down; static friction acts precisely because it is not sliding.

Working Horizontally, the wall's normal force balances the push: F_N = 9.0 N. The maximum static friction is μ_s F_N = 0.80 × 9.0 N = 7.2 N. Vertically the book is at rest, so static friction balances the gravitational force: F_f,s = mg = (0.50 kg)(10 m/s²) = 5.0 N, directed upward. 5.0 N ≤ 7.2 N, so the book does not slip.

CED 2.7.B.2 · Read this in Fix

1. Aμ_s mg cosθ
   A student who thinks static friction always takes its maximum value picks this. μ_s mg cosθ is the largest static friction the ramp can exert. The actual value is whatever keeps the block at rest, P − mg sinθ.
2. BP + mg sinθ
   A student who adds the magnitudes of the push and of the component of the gravitational force along the ramp, as if friction had to oppose both, picks this. The push acts up the ramp and mg sinθ down it, so they partly cancel: friction supplies only the difference, P − mg sinθ, down the ramp.
3. CP − mg sinθ Correct
   Without friction the push would move the block up the ramp, since P > mg sinθ, so static friction points down the ramp. The block is at rest, so the forces along the ramp balance: P = mg sinθ + F_f,s, which gives F_f,s = P − mg sinθ.
4. DP − mg cosθ
   A student who swaps sine and cosine takes mg cosθ as the component of the gravitational force along the ramp. mg cosθ is the component perpendicular to the ramp; the component along it is mg sinθ.

Working Along the ramp, without friction the push P up the ramp would exceed the component of the gravitational force down the ramp, mg sinθ, so the block would tend to slide up; static friction therefore points down the ramp. The block is at rest, so along the ramp (up positive): P − mg sinθ − F_f,s = 0, giving F_f,s = P − mg sinθ. This must be no more than μ_s mg cosθ, which is only the maximum.

CED 2.7.B.2 · Read this in Fix

1. A18 N
   A student who takes the normal force to equal the weight calculates 0.90 × 20 N = 18 N. On a ramp the normal force balances only the component of the gravitational force perpendicular to the surface, 16 N.
2. B11 N
   A student who swaps the components uses the 12 N component along the ramp as the normal force: 0.90 × 12 N ≈ 11 N. The normal force balances the component perpendicular to the surface, 16 N.
3. C14 N Correct
   The maximum static friction is μ_s F_N. Perpendicular to the ramp the block does not accelerate, so the normal force balances the 16 N perpendicular component: F_N = 16 N. Then F_f,s,max = 0.90 × 16 N ≈ 14 N, more than the 12 N actually needed, so the block stays at rest.
4. D20 N
   A student who thinks the greatest static friction a surface can exert equals the object's weight picks 20 N. The maximum is μ_s F_N = 0.90 × 16 N ≈ 14 N; the weight enters only through the normal force.

Working The block has no acceleration perpendicular to the ramp, so the normal force balances the perpendicular component of the gravitational force: F_N = 16 N (= mg cos 37° = 20 N × 0.80). F_f,s,max = μ_s F_N = 0.90 × 16 N = 14.4 N ≈ 14 N. (The static friction actually exerted is 12 N, which is less than this, so the block stays at rest.)

CED 2.7.B.2.ii · Read this in Fix

1. A2.8 N
   A student who thinks one coefficient governs both starting and keeping an object sliding uses μ_k: 0.35 × 8.0 N = 2.8 N. That push keeps the box sliding at constant velocity once it moves; starting it needs μ_s F_N = 4.8 N.
2. B4.8 N Correct
   Static friction matches the push until the push reaches the maximum static friction, μ_s F_N. On a level floor F_N = mg = 8.0 N, so the box starts to slide when the push reaches 0.60 × 8.0 N = 4.8 N.
3. C8.0 N
   A student who thinks the push must overcome the box's weight picks 8.0 N. On a level floor the weight is perpendicular to the push; it affects friction only through the normal force, and the box slides once the push exceeds μ_s F_N = 4.8 N.
4. D0.0 N
   A student who thinks no friction acts on a box that is not sliding concludes that any push starts it moving. Static friction balances the push while the box is at rest, up to its maximum of 4.8 N.

Working While the box is at rest, static friction equals the push. The box starts to slide when the push reaches the maximum static friction: F_N = mg = (0.80 kg)(10 m/s²) = 8.0 N, F_f,s,max = μ_s F_N = 0.60 × 8.0 N = 4.8 N. (Once sliding, the friction drops to μ_k F_N = 2.8 N.)

CED 2.7.B.2.ii · Read this in Fix

1. Ag(cosθ − μ_k sinθ)
   A student who swaps sine and cosine, taking the gravitational component along the ramp as mg cosθ and the normal force as mg sinθ, picks this. Check a limit: for a level surface, θ = 0, it would give an acceleration of g. Along the ramp the component is mg sinθ, and F_N = mg cosθ.
2. Bg(sinθ − μ_k sinθ)
   A student who takes the friction force as μ_k times the component of the gravitational force along the ramp, μ_k mg sinθ, picks this. Kinetic friction is μ_k times the normal force, mg cosθ, which is set by how hard the surfaces press together, not by the force tending to move the block.
3. Cg(sinθ − μ_k cosθ) Correct
   Perpendicular to the ramp, F_N = mg cosθ. The block is sliding, so kinetic friction acts: μ_k mg cosθ, up the ramp. Along the ramp, ma = mg sinθ − μ_k mg cosθ, so a = g(sinθ − μ_k cosθ).
4. Dg(sinθ − μ_s cosθ)
   A student who uses the coefficient of static friction because the block started from rest picks this. Once the block slides over the ramp, the friction is kinetic, so μ_k applies, however the motion began.

Working Take axes along and perpendicular to the ramp. Perpendicular to the ramp the block has no acceleration, so F_N = mg cosθ. The block slides over the ramp, so the friction is kinetic: |F_f,k| = μ_k F_N = μ_k mg cosθ, directed up the ramp, opposite to the sliding. Along the ramp, taking down the ramp as positive: ma = mg sinθ − μ_k mg cosθ, so a = g(sinθ − μ_k cosθ). The mass cancels. (Sine and cosine swapped: g(cosθ − μ_k sinθ). Friction taken as μ_k times the component of gravity along the ramp: g(sinθ − μ_k sinθ). Static coefficient used because the block started from rest: g(sinθ − μ_s cosθ).)

CED 2.7.A.2 · Read this in Fix

1. A2.0
   A student who thinks one coefficient describes both starting and keeping an object sliding picks this: doubling the normal force doubles the push, 2 × 1 = 2.0. Once the stack slides, the friction is kinetic, μ_k F_n, and μ_k = 0.35 is less than μ_s = 0.50, so the push needed is only 1.4P.
2. B0.7
   A student who thinks friction is fixed by the two materials alone, so stacking a second crate does not change it, picks this, keeping only the change from μ_s to μ_k: 0.35/0.50 = 0.7. Friction is proportional to the normal force; the stack presses on the floor twice as hard, so the friction and the push are doubled: 1.4P.
3. C2.8
   A student who thinks the coefficient of friction grows with the load, doubling μ_k along with the weight, picks this: 2 × 2 × 0.70 = 2.8. The coefficient depends on the materials in contact, not on the load; only the normal force doubles, giving 2 × 0.35/0.50 = 1.4.
4. D1.4 Correct
   Starting the single crate takes a push equal to the maximum static friction, P = μ_s mg. Keeping the stack sliding at constant velocity takes a push equal to the kinetic friction, μ_k(2m)g, since the net force is zero. The ratio is 2 × (0.35/0.50) = 1.4: the doubled normal force doubles the friction, and the switch from static to kinetic friction multiplies it by 0.70.

Working One crate, mass m: the push that just starts it sliding equals the maximum static friction force, P = μ_s F_n = μ_s mg. Stack, mass 2m, sliding at constant velocity: the net force is zero, so the push equals the kinetic friction force, μ_k F_n = μ_k(2m)g. Ratio: μ_k(2m)g / (μ_s mg) = 2(μ_k/μ_s) = 2(0.35/0.50) = 1.4. Two factors change: the normal force doubles (×2) and the coefficient changes from μ_s to μ_k (×0.70).

CED 2.7.B.3 · Read this in Fix