3.2 Work

1. AEnergy transferred into or out of the system by the force as the system moves. Correct
   Work is a transfer of energy between a system and its surroundings by a force, and it happens only while the point where the force is exerted moves with a component along the force. Positive work transfers energy in; negative work transfers it out.
2. BThe force exerted on the system multiplied by the time for which it is exerted.
   A student who thinks work depends on how long a force is exerted picks this. Work is the force component along the displacement times the displacement; time does not appear, and a force on an object that does not move does no work however long it is exerted.
3. CEnergy that the force puts inside the system, which the system then contains as work.
   A student who thinks of work as something a system holds picks this. Work is energy in transit; the energy transferred shows up as a form of the system’s energy, such as kinetic energy, not as stored ‘work’.
4. DThe effort needed to exert the force, whether or not the system ever moves at all.
   A student who uses the everyday meaning of work as effort picks this. A force on a system that does not move transfers no energy to it, however tiring the force is to exert.

CED 3.2.A.1 · Read this in Fix

1. AW_N > 0, W_g < 0, W_net > 0
   A student who takes the sign of each work from the direction of the force (up as positive) picks this. The sign depends on the angle between each force and the displacement, which is downward: the upward forces do negative work.
2. BW_N = 0, W_g > 0, W_net > 0
   A student who thinks the normal force never does work picks this. The floor moves down with the box, so the normal force is exerted through a displacement opposite to it and does negative work, large enough to make the net work negative.
3. CW_N < 0, W_g > 0, W_net < 0 Correct
   The box moves downward. The normal force points up, opposite to the displacement, so W_N < 0; gravity points down, along the displacement, so W_g > 0. The diagram shows F_N longer than F_g, so the net force is upward, opposite to the motion, and W_net < 0: the box is slowing down.
4. DW_N < 0, W_g < 0, W_net < 0
   A student who gives every force’s work the sign of the net work picks this. The box is slowing, so the net work is negative, but gravity points along the downward displacement and does positive work.

Working Displacement downward. F_N up: θ = 180°, W_N < 0. F_g down: θ = 0°, W_g > 0. Diagram: |F_N| > |F_g|, so F_net is up, opposite to the displacement: W_net = W_N + W_g < 0.

CED 3.2.A.2 · Read this in Fix

1. AIt is zero, as the tension is perpendicular to the ball’s velocity at each instant. Correct
   The tension always points toward the center of the circle, and the velocity is always tangent to it, so the angle between the tension and each small displacement is 90° and T⃗ · dr⃗ = 0 throughout. The tension changes the direction of the ball’s motion but not its speed or kinetic energy.
2. BIt is positive, because the string pulls on the ball all the time that it moves.
   A student who thinks every force on a moving object does work picks this. A force does work only through its component along the motion; the tension has no such component at any point.
3. CIt is T times the straight-line distance between the start and end of the quarter turn.
   A student who multiplies the full force by the size of the displacement, ignoring their directions, picks this. At every instant the tension is at 90° to the ball’s motion, so it does no work.
4. DIt is a vector pointing to the circle’s center, the way the tension points.
   A student who treats work as a vector picks this. Work is a scalar, the dot product of force and displacement, and here it is zero.

CED 3.2.A.3.iv · Read this in Fix

1. A7.2 m/s
   A student who uses only the work done by the applied force picks this: K_f = 10 J + 120 J = 130 J, v = 7.2 m/s. Friction does −40 J of work, which must be included in the net work.
2. B7.7 m/s
   A student who writes the change in kinetic energy as (1/2)m(Δv)² picks this: 80 J = (1/2)(5.0 kg)(Δv)², so Δv = 5.7 m/s and v = 7.7 m/s. ΔK is the difference of the squares of the speeds.
3. C5.7 m/s
   A student who sets the net work equal to the final kinetic energy picks this: (1/2)(5.0 kg)v² = 80 J. The box already had 10 J of kinetic energy at the start, so K_f = 90 J.
4. D6.0 m/s Correct
   Net work: W = (30 N)(4.0 m) − (0.20)(5.0 kg)(10 m/s²)(4.0 m) = 120 J − 40 J = 80 J. K_i = (1/2)(5.0)(2.0)² = 10 J, so K_f = 90 J and v = √(2 × 90 J/5.0 kg) = 6.0 m/s.

Working F_f = μ_k mg = 0.20 × 5.0 × 10 = 10 N. W_net = (30 − 10)(4.0) = 80 J. K_i = ½(5.0)(2.0²) = 10 J; K_f = 90 J; v = √(2·90/5.0) = 6.0 m/s.

CED 3.2.A.4 · Read this in Fix

1. A40 J Correct
   The work is the area under a graph of F_∥, the component along the track: F_∥ = F cos 60° = 0.5F. The area under the F graph is (1/2)(2 m)(20 N) + (3 m)(20 N) = 20 J + 60 J = 80 N·m, so W = 0.5 × 80 J = 40 J.
2. B80 J
   A student who takes the area under the graph of the full force F picks this. Only the component along the displacement, F cos 60°, does work, so the area must be multiplied by 0.5.
3. C69 J
   A student who uses sin 60° for the component along the track picks this: 80 J × 0.866. The rope is at 60° to the track, so the parallel component is F cos 60°.
4. D50 J
   A student who treats the force as constant at its largest value, 20 N, over the whole 5 m picks this: (20 N)(5 m)(0.5). For the first 2 m the force is smaller, so the area under the graph is less.

Working Area under F–x: triangle ½(2)(20) = 20 N·m + rectangle (3)(20) = 60 N·m → 80 N·m. F_∥ = F cos 60° ⇒ W = 0.5 × 80 = 40 J.

CED 3.2.A.5 · Read this in Fix

1. AThe weight and the normal force act on the sled as it moves, so they transfer energy out of it.
   A student who thinks every force on a moving object does work picks this. The weight and the normal force are vertical and the displacement is horizontal, so each does zero work; neither transfers energy.
2. BFriction on the sled is opposite to its displacement, so friction does negative work on it. Correct
   Kinetic friction from the ground points opposite to the sled’s motion, so the angle between the friction force and the displacement is 180° and W_f = −F_f d. Negative work done by an external force transfers energy out of the system, which is why the sled’s kinetic energy decreases.
3. CFriction does positive work on the sled, and positive work removes energy from a system.
   A student who treats every amount of work as positive picks this. Friction is opposite to the displacement, so its work is negative, and it is negative work that transfers energy out of a system; positive work transfers energy in.
4. DThe sled’s motion uses up the energy it was given, so energy leaves even without a force.
   A student who thinks motion itself uses up energy (the impetus view) picks this. Without a force doing work, the sled’s kinetic energy would stay constant; the energy leaves because friction does negative work.

CED 3.2.A.1 · Read this in Fix

1. AIt is greatest on Ramp 3, the longest.
   A student who thinks gravity’s work grows with the distance traveled picks this. On a longer ramp the component of gravity along the ramp is smaller, mg sin θ, and over the length h/sin θ it does the same work, mgh.
2. BIt is greatest on Ramp 1, the steepest.
   A student who compares forces instead of work picks this. The component of gravity along Ramp 1 is the largest, but it acts over the shortest distance; the product is mgh on every ramp.
3. CIt is the same on all three ramps. Correct
   Gravity is a conservative force, so the work it does depends only on the initial and final positions: W_g = mgh for each block, because each drops the same height h. The ramp’s length and steepness change how the work is spread along the path, not its total.
4. DIt is zero: gravity is conservative.
   A student who thinks a conservative force does no work picks this. Gravity does positive work mgh on each block as it descends; ‘conservative’ means only that this work does not depend on the path.

Working W_g = (mg sin θ)(L) along a ramp of angle θ and length L = h/sin θ, so W_g = mgh on every ramp. Distractors: longest ramp (Ramp 3) from W = mgL; steepest ramp (Ramp 1) from the largest force component; zero from ‘conservative means no work’.

CED 3.2.A.1.i · Read this in Fix

1. A(π/2)mgR
   A student who multiplies the weight by the distance traveled along the wire picks this: mg × (πR/2). Gravity is not along the wire at most points; only its tangential component does work, and the total is mg times the drop in height, R.
2. B√2 mgR
   A student who multiplies the full weight by the straight-line distance from A to B, √2R, ignoring the 45° angle between them, picks this. The dot product gives mg(√2R)cos 45° = mgR.
3. C0
   A student who thinks a conservative force does no work picks this. Gravity does positive work on the descending bead; its work is zero only for a path that returns to the starting height.
4. DmgR Correct
   Gravity is conservative, so its work depends only on the change in height. Along the arc, at angle φ below the horizontal radius the tangential component of gravity is mg cos φ and dr = R dφ, so W = ∫₀^π/2 mgR cos φ dφ = mgR: the bead drops a height R, from the level of O to B.

Working Let φ be the angle of the radius to the bead below the horizontal line OA. The bead’s path element has length R dφ and is tangent to the circle; the component of mg along it is mg cos φ. W = ∫₀^π/2 mg cos φ · R dφ = mgR [sin φ]₀^π/2 = mgR (sympy-checked), which equals mg times the drop in height from A (level with O) to B (R below O). Distractors: mg × arc length πR/2 = (π/2)mgR; mg × chord √2R ignoring the angle = √2 mgR; ‘conservative means no work’ = 0.

CED 3.2.A.1.i · Read this in Fix

1. AGravity does negative work while the ball rises and equal positive work as it falls back. Correct
   On the way up gravity is opposite to the displacement, W = −mgh; on the way down it is along it, W = +mgh. The two cancel because the ball returns to its starting height: for a conservative force the work around a closed path is zero.
2. BGravity is a conservative force, so it does no work on the ball at any point in its flight.
   A student who thinks conservative forces do no work picks this. Gravity does work on each part of the flight, −mgh going up and +mgh coming down; only the total for the closed path is zero.
3. CGravity’s work is force times time, and the rise and the fall take equal times, so they cancel.
   A student who thinks work is force multiplied by time picks this. The rise and the fall do take equal times, but work is the force component along the displacement times the displacement; the cancellation comes from the equal and opposite displacements, not from the times.
4. DThe work gravity does as the ball rises is stored in the ball and returned as it falls.
   A student who thinks of work as something a system stores picks this. Work is a transfer, not a store; and a ball on its own has no potential energy. Gravity simply does negative work on the way up and positive work on the way down.

CED 3.2.A.1.ii · Read this in Fix

1. AYes: U_f = μ_k mg x works, as friction’s work depends only on the displacement x.
   A student who thinks friction’s work depends only on the start and end points picks this. Moving 2 m away and 2 m back gives zero displacement but work −μ_k mg(4 m), so U_f = μ_k mg x fails.
2. BNo: friction’s work depends on the path, so U_f would have no single value at each point. Correct
   A potential energy must have one value at each position, so that the work between two positions is fixed. Friction’s work between two points is −F_f times the path length, which differs from path to path, so no U_f can match it. Only conservative forces have a potential energy associated with them.
3. CYes: every force that does work on a block has a potential energy associated with it.
   A student who thinks any force that does work has a potential energy picks this. Only conservative forces, whose work is path independent, have potential energies.
4. DNo: friction destroys the block’s energy rather than transferring it out of the block.
   A student who thinks nonconservative forces destroy energy picks this. Friction transfers energy out of the block (mainly into thermal energy); the reason no U_f exists is that friction’s work depends on the path.

CED 3.2.A.1.iii · Read this in Fix

1. A0 J
   A student who thinks every force does zero work when the ball returns to its starting point picks this. That holds for gravity, but air resistance opposes the motion on the way up and on the way down, so its work is negative on both legs.
2. B−0.60 J
   A student who treats kinetic energy as proportional to speed picks this: (1/2)(0.30 kg)(12 m/s − 16 m/s) = −0.60. Kinetic energy depends on the square of the speed.
3. C−17 J Correct
   Gravity is conservative, so its work over the round trip is zero. By the work–energy theorem, W_air = ΔK = (1/2)(0.30 kg)(12 m/s)² − (1/2)(0.30 kg)(16 m/s)² = 21.6 J − 38.4 J = −16.8 J ≈ −17 J. Air resistance is nonconservative: it opposes the motion on both legs, so its work around the closed path is negative, not zero.
4. D+2.4 J
   A student who finds the change in kinetic energy from the change in speed picks this: (1/2)(0.30 kg)(12 m/s − 16 m/s)² = +2.4 J. ΔK is the difference of the squares, (1/2)m(v_f² − v_i²), which is negative here.

Working Round trip: W_g = 0 (conservative force, same start and end height). W_net = W_air = ΔK = (1/2)(0.30)(12² − 16²) J = 0.15 × (144 − 256) J = −16.8 J ≈ −17 J (2 s.f.).

CED 3.2.A.1.iv · Read this in Fix

1. AKinetic friction: it destroys some of the block’s energy, so energy is not conserved.
   A student who thinks nonconservative forces destroy energy picks this. Friction is nonconservative, but energy is conserved: friction transfers mechanical energy into thermal energy. The defining property is the path dependence of its work.
2. BGravity: it changes the block’s kinetic energy, so it does not conserve its energy.
   A student who thinks a conservative force does no work on an object picks this. Gravity does change the block’s kinetic energy, but its work depends only on the change in height, so it is conservative.
3. CThe spring force: it does negative work on the block while the block compresses the spring.
   A student who thinks a force that does negative work is nonconservative picks this. The spring force does negative work during compression and positive work during extension; its work depends only on the initial and final stretch, so it is conservative.
4. DKinetic friction: its work on the block depends on the length of the path the block takes. Correct
   Kinetic friction always opposes the block’s motion relative to the incline, so its work is −F_f times the path length and depends on the path; it is nonconservative. Gravity and the spring force do work that depends only on the block’s initial and final positions.

CED 3.2.A.1.v · Read this in Fix

1. ANegative: friction opposes motion, so it takes energy away from the box as the box moves.
   A student who thinks friction always does negative work picks this. Friction opposes the relative motion (or tendency to slip) of the surfaces; here it pushes the box forward, along its displacement.
2. BPositive: friction on the box points forward, the direction in which the box moves. Correct
   The box speeds up with the truck, so the net horizontal force on it, the static friction from the bed, points forward. The box moves forward relative to the road, so the friction force and the displacement are in the same direction and the work is positive; it is this work that increases the box’s kinetic energy.
3. CZero: the box does not slide on the truck bed, so static friction does no work on the box.
   A student who thinks static friction cannot do work picks this. In the road’s frame the box, and the point where friction is exerted on it, move forward, so the forward friction force does positive work.
4. DEqual to μ_s mg times the distance the box moves, whatever the truck’s acceleration.
   A student who thinks static friction is always at its maximum, μ_s F_N, picks this. Static friction is only as large as needed to accelerate the box with the truck, ma, so the work depends on the acceleration.

CED 3.2.A.2 · Read this in Fix

1. A2.33 Correct
   W = ∫_L^2L Cx² dx = (C/3)[(2L)³ − L³] = (C/3)(8L³ − L³) = (7/3)CL³, so k = 2.33.
2. B4.00
   A student who multiplies the force at the final position, C(2L)² = 4CL², by the displacement L picks this. The force grows over the interval, so the work must be found by integration.
3. C2.50
   A student who multiplies the average of the end forces, (CL² + 4CL²)/2, by L picks this. That shortcut is valid only for a force that is linear in x; for Cx² it overestimates the work.
4. D2.67
   A student who evaluates the antiderivative at the final position only, C(2L)³/3 = 8CL³/3, picks this. That is the work from x = 0 to 2L; the work from 0 to L, CL³/3, must be subtracted.

Working W = ∫_L^2L Cx² dx = C[x³/3]_L^2L = C(8L³ − L³)/3 = 7CL³/3 (sympy-checked); k = 7/3 = 2.33. Distractors: F(2L)·L = 4CL³ → 4.00; average force [(CL² + 4CL²)/2]·L = 5CL³/2 → 2.50; upper limit only, C(2L)³/3 = 8CL³/3 → 2.67.

CED 3.2.A.3 · Read this in Fix

1. A1.00
   A student who treats the force as constant at its starting value, GMm/R², and multiplies by the displacement R picks this. The force falls as 1/r², so the work must be found by integration.
2. B0.63
   A student who multiplies the average of the end forces, (GMm/R² + GMm/4R²)/2 = 5GMm/(8R²), by R picks this. Averaging the end values works only for a force that changes linearly with distance.
3. C0.00
   A student who thinks no force does work when the speed does not change picks this. The external force does positive work and gravity does equal negative work; only the net work is zero.
4. D0.50 Correct
   The external force balances gravity, so it points away from the planet with magnitude GMm/r². W = ∫_R^2R (GMm/r²) dr = GMm[1/R − 1/(2R)] = GMm/(2R), so k = 0.50.

Working F_ext = +GMm/r² (outward, balancing gravity). W = ∫_R^2R GMm r⁻² dr = GMm[−1/r]_R^2R = GMm(1/R − 1/(2R)) = GMm/(2R) (sympy-checked); k = 0.50. Distractors: F(R)·R = GMm/R → 1.00; average of end forces × R = (1 + 1/4)/2 = 5/8 → 0.63; ‘constant speed means no work’ → 0.00.

CED 3.2.A.3 · Read this in Fix

1. A1
   A student who thinks the work is proportional to the stretch picks this, since both intervals have the same length x. The force is larger during the second interval, so more work is done.
2. B3 Correct
   The external force is kx′ at extension x′, so W(0→x) = ∫₀^x kx′ dx′ = (1/2)kx². W(x→2x) = (1/2)k(2x)² − (1/2)kx² = (3/2)kx² = 3W, so n = 3.
3. C4
   A student who uses the work from 0 to 2x, (1/2)k(2x)² = 4W, without subtracting the work already done from 0 to x picks this.
4. D2
   A student who multiplies the force at the end of each interval by its length picks this: kx·x for the first and k(2x)·x for the second, a ratio of 2. The force varies over each interval, so the work must be integrated.

Working W(0→x) = ∫₀^x kx′ dx′ = kx²/2 = W. W(x→2x) = ∫_x^2x kx′ dx′ = k(4x² − x²)/2 = 3kx²/2 = 3W (sympy-checked). n = 3. Distractors: proportional to stretch → 1; upper limit only, k(2x)²/2 ÷ kx²/2 → 4; end force × interval, 2kx²/kx² → 2.

CED 3.2.A.3 · Read this in Fix

1. A23.0 J
   A student who adds the component products as positive amounts picks this: 15.0 J + 8.00 J. The y-component of the force is opposite to the y-component of the displacement, so its contribution is −8.00 J.
2. B26.9 J
   A student who multiplies the magnitudes, |F⃗||Δr⃗| = (5.39 N)(5.00 m), picks this. That ignores the angle between the vectors; the dot product includes cos θ.
3. C7.00 J Correct
   W = F⃗ · Δr⃗ = F_xΔx + F_yΔy = (5.00 N)(3.00 m) + (−2.00 N)(4.00 m) = 15.0 J − 8.00 J = 7.00 J. The y-component of the force is opposite to the y-component of the displacement, so that part of the work is negative.
4. D26.0 J
   A student who uses sin θ instead of cos θ for the angle θ between the vectors picks this: (5.39 N)(5.00 m)(0.965) = 26.0 J. The dot product is AB cos θ, which here gives 7.00 J.

Working F⃗ · Δr⃗ = (5.00)(3.00) + (−2.00)(4.00) = 15.0 − 8.00 = 7.00 J. |F⃗| = √29 N = 5.385 N, |Δr⃗| = 5.00 m, cos θ = 7.00/(5.385 × 5.00) = 0.260, sin θ = 0.966. Distractors: 15.0 + 8.00 = 23.0 J; |F||Δr| = 26.9 J; |F||Δr| sin θ = |5.00 × 4.00 − (−2.00)(3.00)| = 26.0 J.

CED 3.2.A.3.i · Read this in Fix

1. AW₁ > W₃ > W₂ Correct
   Only the component of each force along the displacement does work. F⃗₁ is along it: W₁ = Fd. F⃗₃ is perpendicular to it: W₃ = 0. F⃗₂ is at 120° to the displacement (60° above the leftward direction): W₂ = Fd cos 120° = −Fd/2. So W₁ > W₃ > W₂.
2. BW₁ > W₂ > W₃
   A student who ranks the sizes of the work without signs picks this: Fd, Fd/2 and 0. F⃗₂ has a component opposite to the displacement, so its work is negative and less than W₃ = 0.
3. CW₁ = W₂ = W₃
   A student who multiplies each full force by the displacement, ignoring direction, picks this. Only the parallel component does work, and the three forces have different parallel components.
4. DW₃ > W₂ > W₁
   A student who uses sin θ in place of cos θ picks this: Fd sin 90° = Fd for F⃗₃, Fd sin 120° ≈ 0.87Fd for F⃗₂ and Fd sin 0° = 0 for F⃗₁. The parallel component is F cos θ.

Working θ₁ = 0°: W₁ = Fd. θ₂ = 180° − 60° = 120°: W₂ = −0.5Fd. θ₃ = 90°: W₃ = 0. Ranking W₁ > W₃ > W₂. Distractors: magnitudes Fd, 0.5Fd, 0 → W₁ > W₂ > W₃; full force → all Fd; sin θ: 0, 0.87Fd, Fd → W₃ > W₂ > W₁.

CED 3.2.A.3.ii · Read this in Fix

1. A1.41
   A student who takes the work as the full force times the distance, Fd, picks this: v = √(2Fd/m). The vertical component of the pull is perpendicular to the displacement and does no work.
2. B1.10
   A student who uses sin 37° for the parallel component picks this: W = 0.60Fd and v = √(1.2Fd/m). The angle is measured from the horizontal, the direction of the displacement, so the parallel component is F cos 37°.
3. C0.89
   A student who writes the kinetic energy as mv² picks this: 0.80Fd = mv², so v = √(0.80Fd/m). With K = (1/2)mv² the speed is √2 times larger.
4. D1.26 Correct
   Only the horizontal component of the pull, F cos 37° = 0.80F, is along the displacement; gravity and the normal force are perpendicular to it. W_net = 0.80Fd = (1/2)mv², so v = √(1.6Fd/m) = 1.26√(Fd/m).

Working W_net = W_rope = (F cos 37°)d = 0.80Fd (F_g, F_N ⟂ displacement). ΔK = (1/2)mv² = 0.80Fd ⇒ v = √(1.6Fd/m) = 1.265√(Fd/m); k = 1.26 (sympy-checked). Distractors: √2 = 1.41 (full F); √1.2 = 1.10 (sin); √0.8 = 0.89 (no ½).

CED 3.2.A.3.ii · Read this in Fix

1. A48.0 J
   A student who uses cos 37° picks this: (2.00)(10)(3.00)(4/5) = 48.0 J. The 37° is the angle between the slope and the horizontal, not between gravity and the displacement; the parallel component of gravity is mg sin 37°.
2. B36.0 J Correct
   The angle between gravity (straight down) and the displacement (down the slope) is 90° − 37° = 53°, so W = mgd cos 53° = mgd sin 37° = (2.00 kg)(10 m/s²)(3.00 m)(3/5) = 36.0 J. This equals mg times the drop in height, 1.80 m.
3. C60.0 J
   A student who multiplies the full weight by the distance along the slope picks this: (20.0 N)(3.00 m). Only the component of gravity along the slope does work; equivalently, the block drops only 1.80 m.
4. D3.60 J
   A student who uses the mass in place of the weight picks this: (2.00)(3.00)(3/5) = 3.60, leaving out g. The gravitational force is mg = 20.0 N.

Working F_g = mg = 20.0 N. Angle between F_g and the displacement down the slope = 53°. W = (20.0 N)(3.00 m)cos 53° = (20.0)(3.00)(0.60) = 36.0 J (= mg × drop of (3.00 m)(3/5) = 1.80 m).

CED 3.2.A.3.iii · Read this in Fix

1. A3.00W
   A student who uses the full force times the distance, F(3d), picks this. Only the component of the force along the displacement, F cos 60°, does work.
2. B2.60W
   A student who uses sin 60° instead of cos 60° picks this: F(3d)(0.866) = 2.60Fd. The parallel component of a force at θ to the displacement is F cos θ.
3. C1.50W Correct
   W = Fd cos θ. First trial: W = Fd. Second trial: F(3d)cos 60° = F(3d)(0.500) = 1.50Fd = 1.50W. The parallel component is halved and the distance is tripled.
4. D0.50W
   A student who judges the work from the size of the parallel force component alone picks this: F cos 60° is half of F, so the work is taken to halve. The distance triples as well, so the work increases by a factor of 1.50.

Working W₂/W₁ = [F(3d)cos 60°]/[Fd cos 0°] = 3 × 0.500 = 1.50. Distractors: 3 (no angle); 3 sin 60° = 2.60; cos 60° alone = 0.50.

CED 3.2.A.3.iii · Read this in Fix

1. A1.10
   A student who leaves friction out of the net work picks this: 0.60mgL = (1/2)mv², v = √(1.2gL). Friction does −0.20mgL of work and must be included.
2. B0.89 Correct
   F_N = mg cos 37° = 0.80mg, so friction is 0.25 × 0.80mg = 0.20mg. Net work: (mg sin 37°)L − (0.20mg)L = 0.60mgL − 0.20mgL = 0.40mgL = (1/2)mv², so v = √(0.80gL) = 0.89√(gL).
3. C0.84
   A student who takes the normal force as mg picks this: friction 0.25mg, net work 0.35mgL, v = √(0.70gL). On the incline F_N = mg cos 37° = 0.80mg.
4. D0.63
   A student who writes the kinetic energy as mv² picks this: 0.40mgL = mv², v = √(0.40gL). With K = (1/2)mv² the speed is √2 times larger.

Working Along the slope: W_g = mgL sin 37° = 0.60mgL; F_N = mg cos 37° = 0.80mg; W_f = −μ_k F_N L = −0.20mgL; W_N = 0. W_net = 0.40mgL = ½mv² ⇒ v = √(0.80gL) = 0.894√(gL) (sympy-checked). Distractors: no friction √1.2 = 1.10; F_N = mg, √(2(0.60 − 0.25)) = √0.70 = 0.84; no ½, √0.40 = 0.63.

CED 3.2.A.4 · Read this in Fix

1. A−mv₀²
   A student who writes the kinetic energy as mv² picks this. With K = (1/2)mv², the object loses (1/2)mv₀² of kinetic energy, and that is the work done on it by the resistive force.
2. B−mv₀/2
   A student who treats kinetic energy as proportional to speed picks this: ΔK = (1/2)m(0 − v₀). Kinetic energy depends on the square of the speed, and −mv₀/2 does not even have the units of energy.
3. C−bv₀
   A student who takes the value of the force at the start as the work picks this. −bv₀ is the initial force, in newtons; the work is the force integrated over the whole displacement.
4. D−mv₀²/2 Correct
   The resistive force is the only force with a component along the motion, so the net work equals the work done by it. By the work–energy theorem, W = ΔK = 0 − (1/2)mv₀² = −mv₀²/2. (Directly: v = v₀e^−bt/m, and W = ∫F_r v dt = −b∫₀^∞ v₀²e^−2bt/m dt = −mv₀²/2.)

Working Work–energy: W_r = ΔK = −½mv₀². Check by integration (sympy): m dv/dt = −bv ⇒ v = v₀e^−bt/m; W = ∫₀^∞ (−bv)(v) dt = −bv₀² · m/(2b) = −mv₀²/2. Distractors: K = mv² → −mv₀²; K ∝ v → −mv₀/2 (units kg·m/s); initial force −bv₀ (units N).

CED 3.2.A.4 · Read this in Fix

1. A1.00v
   A student who thinks the mass cancels, as it does in free fall, picks this. Here the same work gives the same kinetic energy, and the larger mass then needs a smaller speed.
2. B0.25v
   A student who treats kinetic energy as proportional to speed picks this: (1/2)(4m)v′ = (1/2)mv gives v′ = v/4. Because K = (1/2)mv², the speed changes by the square root of the mass factor.
3. C0.50v Correct
   By the work–energy theorem, W = (1/2)mv² for both objects, so v = √(2W/m) ∝ 1/√m. Multiplying the mass by 4 divides the speed by √4 = 2: the final speed is 0.50v.
4. D2.00v
   A student who applies the dependence the wrong way round picks this, multiplying by √4 instead of dividing. Since v = √(2W/m), a larger mass gives a smaller speed.

Working (1/2)mv² = W = (1/2)(4m)v′² ⇒ v′ = v/√4 = 0.50v. Distractors: mass cancels → 1.00v; K ∝ v → 0.25v; √4 applied as a multiplier → 2.00v.

CED 3.2.A.4 · Read this in Fix

1. AIt is F times 0.30 m, the displacement of block 1, the point where F is exerted. Correct
   The work done by an external force is its parallel component times the displacement of its point of application. F is exerted on block 1, which moves 0.30 m, so W = F(0.30 m). The system changes shape as the spring stretches, so this differs from F times the displacement of the center of mass.
2. BIt is F times 0.18 m, the displacement of the center of mass of the two-block system.
   A student who uses the center of mass’s displacement picks this: (0.30 m + 0.06 m)/2 = 0.18 m for identical blocks. The point of application, block 1, moves 0.30 m, and that is the displacement that sets the work done by F.
3. CIt is F times 0.24 m, the distance by which the spring has been stretched.
   A student who uses the change in the spring’s length picks this: 0.30 m − 0.06 m. The stretch would equal the displacement of the pulled end only if block 2 were fixed.
4. DIt is F times the time between the two instants, the time for which F has been exerted.
   A student who thinks work is a force multiplied by the time for which it is exerted picks this. Work is the force component along the displacement times the displacement of the point of application, here F(0.30 m); time does not appear.

Working W_F = F × (displacement of point of application) = F(0.30 m). x_cm shift = (0.30 + 0.06)/2 = 0.18 m (equal masses). Spring stretch = 0.30 − 0.06 = 0.24 m.

CED 3.2.A.4.i · Read this in Fix

1. APositive: the wall’s force is what speeds her up, so it must transfer energy to her.
   A student who thinks any force that speeds a system up must do work on it picks this. The wall’s force does accelerate her, but its point of application does not move, so it transfers no energy.
2. BPositive: it equals the wall’s force multiplied by the time her hands push on the wall.
   A student who thinks work depends on how long a force is exerted picks this. Work involves the displacement of the point of application, which is zero here; force times time is a different quantity.
3. CZero: her hands, where the wall’s force is exerted, do not move while she pushes the wall. Correct
   Work is done by a force only if its point of application moves. Her hands stay against the wall during the push, so the wall’s force does no work, even though it is the external force that accelerates her. She cannot be modeled as an object, because her center of mass moves while the point of application does not; her kinetic energy comes from her own internal energy.
4. DZero: the wall’s force on her and her force on the wall cancel, so no net force acts on her.
   A student who thinks a third-law pair cancels picks this. The two forces are exerted on different objects; the wall’s force is unbalanced on her and accelerates her. The work is zero for a different reason: her hands do not move.

CED 3.2.A.4.ii · Read this in Fix

1. A48.0 J
   A student who multiplies the friction force by the straight-line distance from A to B, 4.00 m, picks this. Friction opposes the motion all along the curved path, so the path length, 6.28 m, must be used.
2. B75.4 J Correct
   The box slides, so kinetic friction acts: F_f = μ_k mg = (0.400)(3.00 kg)(10 m/s²) = 12.0 N, opposite to the motion at every point. The energy dissipated is F_f times the length of the path, a semicircle of radius 2.00 m: (12.0 N)(π × 2.00 m) = 75.4 J.
3. C7.54 J
   A student who uses the mass in place of the weight picks this: (0.400)(3.00)(6.28) = 7.54, leaving out g. The friction force is μ_k mg = 12.0 N.
4. D94.2 J
   A student who uses the coefficient of static friction for the sliding box picks this: (0.500)(30.0 N)(6.28 m). A sliding box experiences kinetic friction, μ_k F_N.

Working F_N = mg = 30.0 N (horizontal rope). F_f = μ_k F_N = 12.0 N. Path length = πR = π(2.00 m) = 6.283 m. |ΔE_mech| = F_f × path = 75.4 J.

CED 3.2.A.4.iii · Read this in Fix

1. A0.60
   A student who uses sin 37° for the normal force picks this: F_N = 0.60mg. The normal force balances the component of gravity perpendicular to the incline, mg cos 37°.
2. B1.00
   A student who takes the normal force as mg picks this. On an incline the normal force balances only the perpendicular component of gravity, 0.80mg.
3. C1.25
   A student who sets the vertical component of the normal force equal to mg picks this: F_N = mg/cos 37° = 1.25mg. The vertical forces on the block need not balance (the block can accelerate along the incline, and friction has a vertical component); perpendicular to the incline there is no acceleration, so F_N = mg cos 37° = 0.80mg.
4. D0.80 Correct
   Perpendicular to the incline the block has no acceleration, so F_N = mg cos 37° = 0.80mg and F_f = μ_k F_N = 0.80μ_k mg. The energy dissipated is the friction force times the path length L: 0.80μ_k mgL, so k = 0.80.

Working Perpendicular: F_N − mg cos 37° = 0 ⇒ F_N = 0.80mg. F_f = 0.80μ_k mg. |ΔE_mech| = F_f L = 0.80μ_k mgL; k = 0.80. Distractors: sin → 0.60; F_N = mg → 1.00; F_N = mg/cos 37° → 1.25.

CED 3.2.A.4.iii · Read this in Fix

1. AW₁ > W₂ = W₃
   A student who adds areas without signs picks this: 6 J, 1.5 J + 1.5 J = 3 J and 3 J. Area below the axis is negative work, so W₂ = 0 and W₃ = −3 J.
2. BW₃ > W₁ > W₂
   A student who ranks the intervals by the slope of the graph picks this: +1.5 N/m, 0 and −3 N/m. The slope tells how quickly the force changes, not the work it does.
3. CW₁ > W₂ > W₃ Correct
   The work in each interval is the signed area between the graph and the x-axis. Interval 1: (3 N)(2 m) = +6 J. Interval 2: the line falls from 3 N to −3 N, so equal triangles above and below the axis cancel: 0 J. Interval 3: the triangle from −3 N to 0 lies below the axis: −3 J. So W₁ > W₂ > W₃.
4. DW₁ > W₃ > W₂
   A student who ranks by the value of the force at the end of each interval picks this: 3 N, 0 and −3 N for intervals 1, 3 and 2. The work is the area under the graph over the interval, not its height at one point.

Working Signed areas: interval 1 (0–2 m): 3 × 2 = +6 J. Interval 2 (2–4 m): F from +3 to −3, zero at 3 m: +1.5 − 1.5 = 0 J. Interval 3 (4–6 m): F from −3 to 0: −½(2)(3) = −3 J. Distractors: magnitudes 6, 3, 3; slopes 0, −3, +1.5 N/m; end values 3, −3, 0 N.

CED 3.2.A.5 · Read this in Fix

1. A2.24
   A student who adds the area below the axis as positive work picks this: total 2.5F₀x₀, v = √5 √(F₀x₀/m). The force is opposite to the motion between 3x₀ and 4x₀, so that area is negative work.
2. B2.83
   A student who uses the largest force, F₀, as if it were exerted over the whole 4x₀ picks this: W = 4F₀x₀, v = √8 √(F₀x₀/m). The work is the area under the graph, not the peak force times the displacement.
3. C1.22
   A student who writes the kinetic energy as mv² picks this: mv² = 1.5F₀x₀, v = √1.5 √(F₀x₀/m). With K = (1/2)mv² the speed is √2 times larger.
4. D1.73 Correct
   The work is the signed area under the graph: (1/2)F₀x₀ from 0 to x₀, F₀x₀ from x₀ to 2x₀, +(1/2)F₀x₀ from 2x₀ to 3x₀ and −(1/2)F₀x₀ from 3x₀ to 4x₀, a total of (3/2)F₀x₀. (1/2)mv² = (3/2)F₀x₀ gives v = √3 √(F₀x₀/m) = 1.73√(F₀x₀/m).

Working Areas: ½F₀x₀ + F₀x₀ + ½F₀x₀ − ½F₀x₀ = 1.5F₀x₀ = W_net = ½mv² ⇒ v = √(3F₀x₀/m) = 1.732√(F₀x₀/m) (sympy-checked). Distractors: magnitudes 2.5 → √5 = 2.24; F₀ × 4x₀ → √8 = 2.83; no ½ → √1.5 = 1.22.

CED 3.2.A.5 · Read this in Fix

1. Av₀/√(bg) Correct
   The friction force is μ_k mg = bmgx, so its work over the slide is −∫₀ᴰ bmgx dx = −bmgD²/2. By the work–energy theorem this equals the change in kinetic energy, −(1/2)mv₀², so D² = v₀²/(bg) and D = v₀/√(bg).
2. Bv₀/√(2bg)
   A student who multiplies the friction force at the end of the slide, bmgD, by the whole distance D picks this: bmgD² = (1/2)mv₀². The force grows from zero, so its work is the integral of the force over the path, bmgD²/2, half as much.
3. Cv₀√(2/(bg))
   A student who writes the block’s kinetic energy as mv₀², leaving out the ½, picks this: bmgD²/2 = mv₀² gives D = v₀√(2/(bg)). The block’s initial kinetic energy is (1/2)mv₀².
4. D√(v₀/(bg))
   A student who takes kinetic energy to be proportional to speed rather than to its square picks this: bmgD²/2 = (1/2)mv₀ gives D = √(v₀/(bg)), which does not even have the units of length.

Working F_N = mg on the level patch, so the friction force has magnitude μ_k mg = bmgx, opposite the motion. Work done by friction over the slide: W_f = ∫₀ᴰ (−bmgx) dx = −bmgD²/2. Work–energy theorem (friction is the only force with a component along the motion): W_f = ΔK = 0 − (1/2)mv₀², so bmgD²/2 = (1/2)mv₀² and D = v₀/√(bg) (sympy-checked; b in 1/m, so D is in m). Distractors: friction at the end of the slide times D, bmgD² = (1/2)mv₀², D = v₀/√(2bg); K written as mv₀², bmgD²/2 = mv₀², D = v₀√(2/(bg)); K taken as proportional to v₀, bmgD²/2 = (1/2)mv₀, D = √(v₀/(bg)).

CED 3.2.A.3 · Read this in Fix