3.2 Work

1. AIt destroys 40 J of the kinetic energy that the cart had.
   A student who thinks energy is used up when negative work is done picks this. Energy is not destroyed: the 40 J leaves the cart and is transferred to the student's hand and body.
2. BIt transfers 40 J of energy into the cart, since work is an amount.
   A student who thinks work cannot be negative, and so ignores the sign, picks this. The minus sign matters: negative work means energy is transferred out of the system, and the cart slows down.
3. CIt transfers no energy, as the cart pushes back equally on the hand.
   A student who thinks the two forces of an interaction cancel picks this. The cart's push acts on the hand, not on the cart, so it cannot cancel the hand's force on the cart. The hand's force does −40 J of work on the cart, and the cart loses 40 J of kinetic energy.
4. DIt transfers 40 J of energy out of the cart and to the student. Correct
   Work is energy transferred into or out of a system by a force. Negative work means energy leaves the system: the hand's force takes 40 J of energy from the cart, so the cart's kinetic energy decreases by 40 J (it is the only force doing work on the cart here). The energy goes to the student.

Working W_hand = −40 J < 0, so 40 J of energy leaves the cart (the system) and goes to the student. The hand's force is the only force doing work on the cart, so ΔK_cart = −40 J.

CED 3.2.A.1 · Read this in Fix

1. AThe tension does positive work and gravity does negative work on the box.
   A student who thinks gravity always does negative work picks this. Here the box moves down, in the direction of the gravitational force, so gravity does positive work; the upward tension does the negative work.
2. BThe tension does negative work and gravity does positive work. Correct
   The box moves down. The tension points up, opposite to the displacement, so its work is negative; the gravitational force points down, along the displacement, so its work is positive. At constant speed the two works add to zero.
3. CThe tension and gravity both do positive work on the box as it moves.
   A student who thinks work cannot be negative picks this. The tension points opposite to the box's displacement, so cos θ = −1 and its work is negative: it takes energy out of the box.
4. DNeither force does any work, as the box's speed does not change.
   A student who thinks no force does work at constant speed picks this. The box's kinetic energy is constant, so the NET work is zero, but each force does work: +mgd by gravity and −mgd by the tension.

CED 3.2.A.2 · Read this in Fix

1. AW₂ > W₁ > 0
   A student who multiplies the student's force by the distance moved, whatever the direction, picks this: mg × 4.0 m > mg × 1.0 m. While carrying, the force is perpendicular to the displacement and does no work.
2. BW₁ > W₂ = 0 Correct
   While lifting, the student's upward force is parallel to the upward displacement, so W₁ = mg × 1.0 m > 0. While carrying at constant velocity, the student's force is vertical (it balances gravity) and the displacement is horizontal: θ = 90°, so W₂ = 0.
3. CW₁ = W₂ > 0
   A student who thinks exerting a force is the same as doing work picks this: the student holds the backpack up with a force of mg in both cases. Equal forces need not do equal work; during the carry the displacement has no component along the force.
4. DW₁ = W₂ = 0
   A student who thinks no work is done at constant speed picks this. During the lift the student's upward force is along the upward displacement and does positive work, mg × 1.0 m, even though the net work is zero.

Working Lift: F = mg upward, d = 1.0 m upward, θ = 0°: W₁ = mg(1.0 m) > 0. Carry: F = mg upward, d = 4.0 m horizontal, θ = 90°: W₂ = mg(4.0 m) cos 90° = 0. So W₁ > W₂ = 0.

CED 3.2.A.3.i · Read this in Fix

1. A√(4gh) Correct
   Two forces act on the crate. The tension does +3mgh of work and the gravitational force does −mgh. By the work-energy theorem, (1/2)mv² − 0 = 3mgh − mgh = 2mgh, so v² = 4gh and v = √(4gh).
2. B√(6gh)
   A student who uses only the tension's work, (1/2)mv² = 3mgh, picks this. The change in kinetic energy equals the NET work, and gravity does −mgh on the rising crate.
3. C√(8gh)
   A student who adds the magnitudes of both works, 3mgh + mgh = 4mgh, picks this. Gravity points opposite to the upward displacement, so its work is −mgh and it reduces the kinetic energy gained.
4. D√(2gh)
   A student who writes the kinetic energy as mv², dropping the 1/2, picks this: mv² = 2mgh. With K = (1/2)mv², the net work 2mgh gives v² = 4gh.

Working W_T = (3mg)h, W_g = −mgh. ΔK = (1/2)mv² = 3mgh − mgh = 2mgh → v² = 4gh → v = √(4gh). Tension only: (1/2)mv² = 3mgh → √(6gh). Magnitudes added: (1/2)mv² = 4mgh → √(8gh). Dropped 1/2: mv² = 2mgh → √(2gh).

CED 3.2.A.4 · Read this in Fix

1. A3F₀x₀/2 Correct
   The work is the area under the graph. From 0 to x₀ the area is a rectangle, F₀x₀; from x₀ to 2x₀ it is a triangle, (1/2)F₀x₀. The total is 3F₀x₀/2.
2. B2F₀x₀
   A student who multiplies the largest force by the whole displacement, F₀ × 2x₀, picks this. That is the area of a rectangle enclosing the graph; the force falls to zero over the second half, so the true area is smaller.
3. CF₀x₀
   A student who averages the first and last values of the force, (F₀ + 0)/2, and multiplies by 2x₀ picks this. That average holds only for a force that changes steadily over the whole displacement; here it stays at F₀ for the first half.
4. D−F₀/x₀
   A student who takes the slope of the graph picks this. −F₀/x₀ is the slope of the sloping part, in N/m; the work is the area under the graph, in N·m = J.

Working Area = F₀x₀ (rectangle, 0 to x₀) + (1/2)(x₀)(F₀) (triangle, x₀ to 2x₀) = 3F₀x₀/2. In units of F₀x₀/2 the options are 3 (key), 4 (largest force times 2x₀) and 2 (mean of end values times 2x₀).

CED 3.2.A.5 · Read this in Fix

1. A|W₃| > |W₂| > |W₁| > 0
   A student who thinks gravity's work grows with the length of the path picks this. Ramp 3 is the longest, but only the vertical component of the displacement is along the gravitational force, and it is h for every ramp.
2. B|W₁| = |W₂| = |W₃| > 0 Correct
   The gravitational force is conservative, so its work depends only on the start and end configurations, not on the path. Along a ramp of length L inclined at angle φ, the angle between the downward gravitational force and the displacement up the ramp is 90° + φ, so W = mgL cos(90° + φ) = −mgL sin φ = −mgh, because L sin φ = h is the vertical rise. Every ramp gives W = −mgh: equal, nonzero magnitudes.
3. C|W₁| > |W₂| > |W₃| > 0
   A student who thinks a gentle ramp reduces the work against gravity picks this. A gentler ramp needs a smaller push, but over a longer distance; the gravitational force does work of magnitude mgh along every ramp.
4. D|W₁| = |W₂| = |W₃| = 0
   A student who thinks no force does work on an object moving at constant speed picks this. It is the NET work that is zero at constant speed; gravity does −mgh on each box, and the worker's push does +mgh.

Working Along each ramp, the angle between the downward gravitational force mg and the displacement L (up the ramp) gives W_g = mgL cos(90° + φ) = −mgL sin φ = −mgh, where φ is the ramp angle and L sin φ = h. All three ramps rise the same h, so |W₁| = |W₂| = |W₃| = mgh > 0.

CED 3.2.A.1.i · Read this in Fix

1. AIt is zero, because the ball ends at the height from which it started. Correct
   The gravitational force is conservative, so its total work is zero when the ball–Earth system returns to its initial configuration. Gravity does −mgh on the way up and +mgh on the way down; the two cancel.
2. BIt is negative, because gravity opposes the ball's motion throughout.
   A student who thinks gravity always does negative work picks this. Gravity opposes the motion only on the way up. On the way down the ball moves in the direction of the gravitational force, and gravity does positive work.
3. CIt is positive, since gravity acts over the whole distance traveled.
   A student who adds up the work over the total distance, 2h, as if every part were positive picks this. The work on the way up is negative and the work on the way down is positive; they cancel.
4. DIt is negative, since the ball is caught more slowly than it was thrown.
   A student who thinks the ball's energy is used up during its flight picks this. With negligible air resistance nothing takes energy from the ball–Earth system; the ball returns to the hand with its launch speed, and gravity's total work is zero.

CED 3.2.A.1.ii · Read this in Fix

1. APotential energy is stored by height, and friction is a contact force.
   A student who thinks potential energy just means height picks this. The force of an ideal spring is also a contact force, and it has a potential energy because its work is path-independent. What rules out friction is the path dependence of its work.
2. BFriction does negative work, but a potential energy is a positive value.
   A student who thinks every potential energy must be positive picks this. The sign of a potential energy depends on where its zero is chosen; the gravitational force also does negative work, on a rising object, and has a potential energy. The deciding test is whether the work is path-independent.
3. CFriction's work between two points depends on the path taken. Correct
   A potential energy can be associated only with a conservative force, whose work depends only on the start and end configurations. Friction's work is −F_f times the path length, so it differs for different paths between the same two points; no potential energy can be defined for it.
4. DFriction does no work on the box, since it acts against the box's motion.
   A student who thinks a resisting force does no work picks this. Friction does do work on the sliding box: negative work, since it points opposite to the displacement. The reason it has no potential energy is that its work depends on the path.

CED 3.2.A.1.iii · Read this in Fix

1. A−84 J Correct
   Kinetic friction points opposite to the crate's motion on each leg (θ = 180°), so on every leg it does negative work equal to −F_f times the leg's length: −12 N × (3.0 m + 4.0 m) = −84 J. Friction is nonconservative, so the longer path gives more negative work than the straight path would.
2. B−60 J
   A student who uses the straight-line displacement from P to Q, 5.0 m, picks this. That would be friction's work along path A. Friction's work depends on the path: along path B the crate slides 7.0 m.
3. C+84 J
   A student who thinks work cannot be negative picks this. Friction points opposite to the crate's motion on both legs, so cos θ = cos 180° = −1 and its work is negative; it transfers energy out of the crate.
4. D0.0 J
   A student who thinks that a force opposing the motion, such as friction, does no work, and that only forces that drive the motion do work, picks this. Friction points opposite to the crate's motion on each leg (θ = 180°), so it does negative work on each: −12 N × (3.0 m + 4.0 m) = −84 J.

Working On each leg F_f is opposite to the displacement (θ = 180°). W_f = −F_f(d₁ + d₂) = −12 N × (3.0 m + 4.0 m) = −84 J. (Using the straight-line 5.0 m gives −60 J; ignoring the sign gives +84 J; treating friction as doing no work gives 0.0 J.)

CED 3.2.A.1.iv · Read this in Fix

1. ALess, as gravity does negative work on the ball going up and coming down.
   A student who thinks gravity always does negative work picks this. Gravity does negative work on the way up but positive work on the way down, and over the round trip its work is zero. The loss of speed is due to air resistance.
2. BEqual, as air resistance does positive work going up but negative work coming down.
   A student who thinks air resistance always points upward picks this: an upward force on the rising ball would do positive work. Air resistance points opposite to the velocity, so it points down while the ball rises and up while it falls: its work is negative on both parts of the trip.
3. CEqual, as the ball returns to the height from which it was thrown.
   A student who thinks every force does zero work over a round trip picks this. That is true of conservative forces such as gravity, but air resistance is nonconservative and does negative work throughout the flight.
4. DLess, as air resistance does negative work on the ball going up and coming down. Correct
   Air resistance is a nonconservative force that points opposite to the ball's velocity: down while it rises, up while it falls. It does negative work on both parts of the trip, so the net work on the ball is negative, and by the work-energy theorem the ball returns with less kinetic energy and less speed.

CED 3.2.A.1.v · Read this in Fix

1. A300 J
   A student who adds the two ropes' works as perpendicular vectors, √(300² + 400²) = 500 J, and then subtracts 200 J picks this. The ropes' forces are vectors, but their works are scalars and simply add: 700 J from the ropes.
2. B900 J
   A student who treats every work as a positive amount picks this, adding 200 J for friction. Friction does −200 J: it transfers energy out of the sled, so it reduces the net work.
3. C500 J Correct
   Work is a scalar, so the works add as signed numbers whatever the directions of the forces: 300 J + 400 J + (−200 J) = 500 J. The gravitational and normal forces are perpendicular to the level motion and do no work.
4. D700 J
   A student who leaves friction out, thinking a resisting force does no work, picks this. The stem gives friction's work as −200 J, and it must be included in the net work.

Working W_net = W₁ + W₂ + W_f + W_g + W_N = 300 J + 400 J − 200 J + 0 + 0 = 500 J.

CED 3.2.A.2 · Read this in Fix

1. A400 J
   A student who multiplies the whole tension by the distance, 50 N × 8.0 m, picks this. The rope pulls partly upward, and only the horizontal component, 40 N, is along the displacement.
2. B240 J
   A student who uses sin 37° for the parallel component picks this. The angle is measured from the horizontal displacement, so the parallel component is T cos 37° = 40 N, not T sin 37° = 30 N.
3. C560 J
   A student who thinks both components of the tension do work on the moving sled, (40 N + 30 N) × 8.0 m, picks this. The vertical component is perpendicular to the displacement, so it does no work.
4. D320 J Correct
   Only the component of the tension parallel to the displacement does work: F_∥ = T cos 37° = 50 N × 0.80 = 40 N, so W = F_∥d = 40 N × 8.0 m = 320 J. The vertical component, 30 N, is perpendicular to the displacement and does no work.

Working F_∥ = T cos 37° = 50 N × 0.80 = 40 N. W = F_∥d = 40 N × 8.0 m = 320 J. (T d = 400 J; T sin 37° d = 240 J; (T cos 37° + T sin 37°)d = 560 J.)

CED 3.2.A.3.i · Read this in Fix

1. APositive, since the tension is exerted on the puck the whole time 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 displacement, and the tension has no such component at any instant.
2. BEqual to T times πr, the distance the puck travels around the circle.
   A student who multiplies the force by the distance traveled, ignoring the angle, picks this. At every instant the angle between the tension and the motion is 90°, so cos θ = 0 and the work is zero.
3. CZero, since the tension is perpendicular to the puck's velocity at every instant. Correct
   The tension always points along the string toward O, and the velocity is always along the circle, at 90° to the string. A force perpendicular to the displacement does no work, so the tension changes only the direction of the puck's motion and its kinetic energy stays the same.
4. DNonzero, since the tension keeps changing the direction of the puck's velocity.
   A student who thinks changing the direction of motion takes work picks this. Kinetic energy, (1/2)mv², depends only on speed; a force perpendicular to the velocity turns the puck without changing its speed.

CED 3.2.A.3.ii · Read this in Fix

1. AThe push does more work than friction, since it must keep the crate moving.
   A student who thinks a net force is needed to keep an object moving picks this. At constant speed the push only balances friction; if the push did more work than friction, the net work would be positive and the crate would speed up.
2. BNo force does any work on the crate, since the crate's speed does not change.
   A student who thinks constant speed means no work is done picks this. The push does positive work and friction does negative work; only their sum is zero.
3. CThe total work on the crate is the work done by the push alone.
   A student who leaves out every force but the applied one picks this. Friction does negative work on the crate, and it must be added in; the total is zero, not the positive work of the push.
4. DThe push does positive work, and the net work on the crate is zero. Correct
   The push is along the displacement, so it does positive work; friction is opposite to it and does equal negative work; gravity and the normal force are perpendicular and do none. The crate's kinetic energy does not change, so by the work-energy theorem the net work is zero.

CED 3.2.A.4 · Read this in Fix

1. A7.0 × 10² J Correct
   By the work-energy theorem, W_net = ΔK = (1/2)(50 kg)[(6.0 m/s)² − (8.0 m/s)²] = 25 kg × (36 − 64) m²/s² = −700 J. The magnitude is 7.0 × 10² J; the net work is negative because the sled slows down.
2. B1.0 × 10² J
   A student who squares the change in speed, (1/2)(50 kg)(2.0 m/s)² = 100 J, picks this. Kinetic energy depends on the square of the speed, so ΔK = (1/2)m(v_f² − v_i²), not (1/2)m(Δv)².
3. C1.4 × 10³ J
   A student who writes the kinetic energy as mv², dropping the 1/2, picks this: 50 kg × 28 m²/s² = 1400 J. With K = (1/2)mv² the change is half of that.
4. D9.0 × 10² J
   A student who sets the net work equal to the final kinetic energy, (1/2)(50 kg)(6.0 m/s)² = 900 J, picks this. The sled did not start from rest, so the net work equals the CHANGE in kinetic energy, K_f − K_i.

Working K_i = (1/2)(50)(8.0)² = 1600 J; K_f = (1/2)(50)(6.0)² = 900 J. W_net = ΔK = 900 J − 1600 J = −700 J; magnitude 7.0 × 10² J.

CED 3.2.A.4 · Read this in Fix

1. A2
   A student who thinks equal increases in speed need equal work picks this: v to 3v is two steps of v, so 2W. Kinetic energy grows with the square of the speed, so each step of v costs more than the one before.
2. B8 Correct
   The work needed equals the change in kinetic energy. From rest to v: W = (1/2)mv². From v to 3v: ΔK = (1/2)m(3v)² − (1/2)mv² = 9W − W = 8W.
3. C4
   A student who squares the change in speed, (1/2)m(2v)² = 4W, picks this. The change in kinetic energy is (1/2)m(3v)² − (1/2)mv², the difference of the squares, not the square of the difference.
4. D9
   A student who sets the work equal to the final kinetic energy, (1/2)m(3v)² = 9W, picks this. The cart already has kinetic energy W at speed v, so only 8W more is needed.

Working W = (1/2)mv². W(v → 3v) = (1/2)m(9v²) − (1/2)mv² = 8 × (1/2)mv² = 8W. Linear: 2; (1/2)m(2v)²: 4; final K only: 9.

CED 3.2.A.4 · Read this in Fix

1. A+4mv²
   A student who thinks work cannot be negative picks this. The hand pushes opposite to the cart's displacement, so its work is negative: it takes 4mv² of energy out of the cart.
2. B+2mv²
   A student who squares the change in speed, (1/2)m(v − 3v)² = 2mv², picks this. Kinetic energy depends on the square of the speed, so the change is (1/2)mv² − (1/2)m(3v)², the difference of the squares.
3. C−4mv² Correct
   The hand's force is the only force that does work on the cart (the gravitational and normal forces are perpendicular to the level track), and it points opposite to the displacement. So the work it does equals the change in kinetic energy: (1/2)mv² − (1/2)m(3v)² = −4mv². The negative sign shows energy transferred out of the cart.
4. D−8mv²
   A student who writes the kinetic energy as mv², dropping the 1/2, picks this: mv² − m(3v)² = −8mv². With K = (1/2)mv² the change is half of that.

Working Only the hand's force does work (F_g and F_N are perpendicular to the displacement). W_hand = ΔK = (1/2)mv² − (1/2)m(3v)² = (1/2)m(v² − 9v²) = −4mv². Sign dropped: +4mv²; (1/2)m(Δv)² = (1/2)m(2v)² = +2mv²; 1/2 dropped: m(v² − 9v²) = −8mv².

CED 3.2.A.4.i · Read this in Fix

1. AThe gain in kinetic energy equals the pull times the displacement. Correct
   The best-fit line passes through the origin with slope 4.0 J/m: for example K = 3.2 J at x = 0.80 m, and 4.0 N × 0.80 m = 3.2 J. So ΔK = F_∥x, as the work-energy theorem predicts when the pull is the only force doing work.
2. BThe cart's speed increases in direct proportion to its displacement.
   A student who takes kinetic energy to be proportional to speed picks this. The graph shows K proportional to x, and K = (1/2)mv², so v² is proportional to x: the speed grows more slowly than the displacement.
3. CThe cart gains an equal amount of kinetic energy during each second of the pull.
   A student who reads the horizontal axis as time picks this. The axis is displacement: the cart gains equal kinetic energy over equal distances, and because it speeds up it covers each distance in less time.
4. DPart of the pull's work is used up simply in keeping the cart moving along.
   A student who thinks some force is needed just to keep an object moving picks this. The data rule it out: all of the pull's work, 4.0 N times the displacement, appears as kinetic energy.

Working Slope of best-fit line = 3.2 J ÷ 0.80 m ≈ 4.0 J/m = 4.0 N, equal to the pull. Check a point: 4.0 N × 0.60 m = 2.4 J, matching the data point at x = 0.60 m.

CED 3.2.A.4.i · Read this in Fix

1. AIt is positive, as the student's center of mass moves in the force's direction.
   A student who uses the displacement of the center of mass picks this. The student's center of mass rises, but the point where the normal force is exerted, the soles of the feet, does not move, so the force does no work.
2. BIt is negative, as the student's feet push downward on the floor.
   A student who judges the work from the force the student exerts on the floor picks this. That downward force is exerted on the floor, not on the student. The normal force on the student is upward, and its point of application does not move.
3. CIt is positive, as only external work can give the student kinetic energy.
   A student who treats the student as an object picks this. The student's center of mass and the point of application of the normal force do not move the same distance, so the object model fails; the kinetic energy comes from energy stored inside the student's body.
4. DIt is zero, as the point where the force is exerted does not move. Correct
   Work depends on the displacement of the point of application. The normal force is exerted on the soles of the feet, which stay at rest on the floor, so the force does no work. The student cannot be modeled as an object: the kinetic energy comes from energy stored in the student's body, released as the legs straighten.

CED 3.2.A.4.ii · Read this in Fix

1. A50 J
   A student who takes the normal force to be the full weight, 50 N, picks this: 0.25 × 50 N × 4.0 m. On a ramp the surface supports only the component of the weight perpendicular to it, mg cos 37° = 40 N.
2. B30 J
   A student who uses sin 37° for the normal force picks this: 0.25 × 50 N × 0.60 × 4.0 m. The normal force balances the component of the weight perpendicular to the ramp, which is mg cos 37°.
3. C40 J Correct
   The normal force on the box is F_N = mg cos 37° = 50 N × 0.80 = 40 N, so the friction force is F_f = μ_k F_N = 0.25 × 40 N = 10 N. Friction is exerted over the 4.0 m path, so it dissipates F_f d = 10 N × 4.0 m = 40 J of mechanical energy.
4. D24 J
   A student who multiplies the friction force by the vertical drop, 10 N × 2.4 m, picks this. Only gravity's work depends on the height alone; friction acts along the whole 4.0 m path.

Working F_N = mg cos 37° = 5.0 × 10 × 0.80 = 40 N. F_f = 0.25 × 40 = 10 N. |ΔE_mech| = F_f d = 10 N × 4.0 m = 40 J. (F_N = mg: 50 J; F_N = mg sin 37°: 30 J; vertical drop 4.0 × 0.60 = 2.4 m: 24 J.)

CED 3.2.A.4.iii · Read this in Fix

1. AMore on ramp S, as that box moves faster and friction grows with speed.
   A student who thinks kinetic friction increases with speed picks this. In the model used here F_f = μ_k F_N, which does not depend on speed; the steep ramp has the smaller normal force and the shorter path.
2. BMore on ramp G, as that box slides a longer path with a larger normal force. Correct
   Energy dissipated = F_f × path length = μ_k(mg cos θ)(h/sin θ). On the gentle ramp the path h/sin θ is longer and the normal force mg cos θ is larger, so friction dissipates more mechanical energy there.
3. CThe same on both ramps, as the boxes drop through the same height.
   A student who treats friction like gravity, whose work depends only on the height, picks this. Friction's work depends on the path length along the surface, which is longer on ramp G.
4. DNone on either ramp, as friction only resists the boxes' motion.
   A student who thinks a resisting force does no work picks this. Friction points opposite to each box's motion along the whole ramp, so it does negative work and dissipates mechanical energy on both ramps.

Working E_diss = F_f L = μ_k mg cos θ × h/sin θ = μ_k mgh/tan θ. A smaller θ (ramp G) gives a larger 1/tan θ, so E_diss(G) > E_diss(S).

CED 3.2.A.4.iii · Read this in Fix

1. A40 J
   A student who counts the area below the axis as positive work picks this: 32 J + 8 J. Below the axis the force is opposite to the displacement, so that area is negative work.
2. B32 J
   A student who ignores the part where the force opposes the motion, thinking an opposing force does no work, picks this. From 4.0 m to 6.0 m the hand's force does −8 J, which must be included.
3. C24 J Correct
   The work is the signed area between the graph and the x-axis. From 0 to 4.0 m: +8 N × 4.0 m = +32 J. From 4.0 m to 6.0 m the force points in −x while the cart moves in +x: −4 N × 2.0 m = −8 J. Total: 32 J − 8 J = 24 J.
4. D48 J
   A student who multiplies the largest force by the whole displacement, 8 N × 6.0 m, picks this. The force is not 8 N over the whole motion: it reverses at x = 4.0 m, and the work is the signed area under the graph.

Working W = (8 N)(4.0 m) + (−4 N)(2.0 m) = 32 J − 8 J = 24 J. (Area below axis counted positive: 40 J; negative part ignored: 32 J; 8 N × 6.0 m: 48 J.)

CED 3.2.A.5 · Read this in Fix

1. A1
   A student who thinks equal extra stretches need equal work picks this. The force grows with the stretch, so the second x₀ of stretch is done against larger forces and takes more work than the first.
2. B2
   A student who treats the force as constant at its value at x₀, kx₀, over the extra stretch picks this: kx₀ × x₀ = kx₀² = 2W. The force keeps growing, to 2kx₀, during the extra stretch, so the area is larger.
3. C4
   A student who gives the total work from zero to 2x₀ picks this. That total, 4W, includes the W already done to reach x₀; the extra work from x₀ to 2x₀ is 4W − W.
4. D3 Correct
   The work is the area under the graph of force against stretch, a straight line through the origin. From 0 to x₀ the area is a triangle, W = (1/2)kx₀². From 0 to 2x₀ it is (1/2)k(2x₀)² = 4W, so the area from x₀ to 2x₀ is 4W − W = 3W.

Working F = kx. W = area 0 → x₀ = (1/2)kx₀². Area 0 → 2x₀ = (1/2)k(2x₀)² = 4W. Extra = 4W − W = 3W. Equal work per stretch: 1; constant force kx₀ over x₀: kx₀² = 2W; total to 2x₀: 4.

CED 3.2.A.5 · Read this in Fix