3.3 Potential Energy

1. AIt belongs to the ball alone, because the ball has mass and is held high above the floor.
   A student who thinks a single object can have potential energy of its own picks this. Potential energy needs two or more interacting objects: the ball’s gravitational potential energy is shared with Earth, which exerts the gravitational force on it.
2. BIt belongs to the ball–floor system, because the height is measured up from the floor.
   A student who mistakes the reference level for the interacting partner picks this. The floor only marks where U_g = 0 is chosen; the gravitational interaction, and so the potential energy, is between the ball and Earth.
3. CIt belongs to the ball–Earth system and depends on how far apart the ball and Earth are. Correct
   The gravitational force is a conservative interaction between the ball and Earth, so the potential energy belongs to the ball–Earth system, a system of two interacting objects. Its value depends on their relative positions, here the height of the ball above Earth’s surface. The ball on its own has no potential energy.
4. DIt belongs to the ball–Earth system and is directed downward, the way Earth pulls.
   A student who treats potential energy as a vector picks this. The system is right, but potential energy is a scalar: it has a value, which may be positive or negative, and no direction.

CED 3.3.A.1 · Read this in Fix

1. AIt has the same value at every point of the orbit, as the separation never changes. Correct
   Potential energy is a scalar associated with the positions of the objects in the system. U_g = −Gm₁m₂/r depends only on the separation r of Earth’s center and the satellite, which is constant in a circular orbit, so U_g has the same value all the way around.
2. BIt changes direction around the orbit, as it points from the satellite toward Earth.
   A student who treats potential energy as a vector picks this. The gravitational force on the satellite changes direction around the orbit, but potential energy is a scalar and has no direction.
3. CIt is zero all the way around, since a weightless satellite feels no gravitational pull.
   A student who thinks gravity does not act on objects in orbit picks this. Earth’s gravitational force is what keeps the satellite in orbit; the system has U_g = −Gm₁m₂/r, which is negative, not zero, with the usual zero at infinite separation.
4. DIt is undefined, since the satellite is not at a height above any ground to measure from.
   A student who thinks potential energy means height above the ground picks this. Gravitational potential energy depends on the separation of the interacting objects; U_g = −Gm₁m₂/r applies at any separation, with no ground needed.

CED 3.3.A.2 · Read this in Fix

1. ATheir values of U_g differ, but their values of ΔU_g agree, as only the zero level differs. Correct
   Changing the zero adds the same constant to every value of U_g. Student 1’s values are positive and Student 2’s are negative, but both find the same decrease in U_g as the rock falls, because only differences in U_g carry physical meaning.
2. BTheir values of U_g and of ΔU_g both differ, because ΔU_g is measured from the chosen zero.
   A student who thinks the choice of zero changes the physics picks this. ΔU_g = U_f − U_i; shifting both values by the same constant leaves the difference unchanged.
3. COnly Student 1’s values can be right, because a rock–Earth system cannot have negative U_g.
   A student who thinks potential energy cannot be negative picks this. Student 2’s negative values mean only that the rock below the cliff top has less potential energy than the configuration chosen as zero.
4. DTheir values of U_g agree, because U_g is set by the rock’s actual height, not by a choice.
   A student who thinks potential energy has one true value picks this. The zero is the observer’s choice, so the two students’ values of U_g differ by a constant (only their values of ΔU_g agree).

CED 3.3.A.3 · Read this in Fix

1. Abd⁴/4
   A student who sets ΔU equal to the work done by the conservative force, dropping the minus sign, picks this. The force does positive work, bd⁴/4, so U decreases by that amount.
2. B−bd⁴/4 Correct
   ΔU = −∫F_x dx from x = d to x = 0: −∫_d⁰ (−bx³) dx = [bx⁴/4] from d to 0 = −bd⁴/4. The force points toward x = 0 and does positive work as the object moves there, so U decreases.
3. C−bd⁴
   A student who treats the force as constant at its starting value, F = −bd³, and uses ΔU = −FΔx = −(−bd³)(−d) picks this. The force shrinks to zero as the object approaches x = 0, so it must be integrated.
4. D0
   A student who thinks a conservative force does no work, so that U cannot change, picks this. Its work is zero only when the system returns to its initial configuration; here the object moves from x = d to x = 0.

Working ΔU = −∫_d⁰ F_x dx = −∫_d⁰ (−bx³) dx = [bx⁴/4]_d⁰ = 0 − bd⁴/4 = −bd⁴/4. Distractors: ΔU = +∫F dx → +bd⁴/4; force at x = d treated as constant, −F(d)Δx = −(−bd³)(−d) → −bd⁴; ‘conservative means no change’ → 0.

CED 3.3.A.4 · Read this in Fix

1. A3bd²−2ad Correct
   F_x = −dU/dx = −(2ax − 3bx²) = 3bx² − 2ax, so at x = d the force is 3bd² − 2ad.
2. Bbd²−ad
   A student who uses the average slope of U(x) from the origin, −[U(d) − U(0)]/d = −(ad² − bd³)/d, picks this. The force at x = d depends on the slope of the graph at that point, not on the slope of a line from the origin.
3. Cbd⁴/4−ad³/3
   A student who integrates U instead of differentiating it, −∫U dx from 0 to d, picks this. Its units are J·m, not N: the force is the negative derivative of U.
4. Dbd³−ad²
   A student who applies the minus sign to the value of U rather than to its slope, F = −U(d), picks this. This has units of energy; the force needs the derivative, −dU/dx.

Working F_x = −dU/dx = −(2ax − 3bx²) = 3bx² − 2ax; at x = d: 3bd² − 2ad. Distractors: −U(d)/d → bd² − ad; −∫₀^d U dx → bd⁴/4 − ad³/3; −U(d) → bd³ − ad². (check.key reproduces the leading coefficient 3 of the key, from −d(−bx³)/dx = 3bx².)

CED 3.3.A.5 · Read this in Fix

1. ANo force acts on the object at that instant, because its velocity is zero then.
   A student who thinks zero velocity means zero force picks this. The force depends on the object’s position: away from equilibrium it is not zero, even at the instant the object is at rest.
2. BA force in the +x direction accelerates the object farther from the equilibrium position.
   A student who has the stable and unstable cases the wrong way round picks this. A force in the direction of the displacement is the mark of an unstable equilibrium; at a stable one the force points back toward equilibrium.
3. CA force in the −x direction accelerates the object back to the equilibrium position. Correct
   At a stable equilibrium position, a small displacement results in a force opposite to the displacement. The object was displaced in the +x direction, so the force is in the −x direction and accelerates it back toward the equilibrium position.
4. DThe object stays at rest where it is released, because its equilibrium position is stable.
   A student who takes ‘stable’ in its everyday sense of ‘not moving’ picks this. Stable means that the force after a small displacement points back toward equilibrium, so the released object accelerates back.

CED 3.3.A.6.i · Read this in Fix

1. AThe block’s mass and the spring’s stretch or compression
   A student who expects every energy of the system to contain the block’s mass, as mgΔy and (1/2)mv² do, picks this. U_s = (1/2)k(Δx)² does not involve the block’s mass.
2. BThe block’s speed and the spring’s stretch or compression
   A student who thinks potential energy depends on motion picks this. Potential energy depends only on positions; the block’s speed determines its kinetic energy, a separate quantity.
3. CThe spring constant and the spring’s stretch or compression Correct
   U_s = (1/2)k(Δx)²: the elastic potential energy is set by the spring constant k and the deformation Δx from the relaxed length, physical properties of the spring and its configuration.
4. DThe spring constant and the block’s height above the floor
   A student who thinks potential energy is always about height picks this. The block moves horizontally; the elastic potential energy depends on the spring’s stretch or compression, not on any height.

CED 3.3.A.7 · Read this in Fix

1. A−3.3 × 10⁻¹⁰ J
   A student who thinks sphere B blocks the interaction of A and C picks this: −G(2.0 + 3.0) J. Every pair interacts whatever lies between them, so the A–C pair adds −1.0G J.
2. B−8.0 × 10⁻¹⁰ J
   A student who adds each sphere’s potential energy with each of the other two picks this, which counts every pair twice: −2(6.0G) J. Each pair contributes once.
3. C−2.6 × 10⁻¹⁰ J
   A student who uses 1/r², as for the force, picks this: −G(2.0/1.0² + 3.0/3.0² + 6.0/2.0²) J. Gravitational potential energy depends on 1/r.
4. D−4.0 × 10⁻¹⁰ J Correct
   Add the potential energies of the three pairs, each counted once: U = −G[(1.0)(2.0)/1.0 + (1.0)(3.0)/3.0 + (2.0)(3.0)/2.0] J = −G(2.0 + 1.0 + 3.0) J = −6.0G J = −4.0 × 10⁻¹⁰ J.

Working Pairs: AB (2.0 kg², 1.0 m), AC (3.0 kg², 3.0 m), BC (6.0 kg², 2.0 m). U = −G(2.0 + 1.0 + 3.0) = −(6.67 × 10⁻¹¹)(6.0) J = −4.0 × 10⁻¹⁰ J. Distractors: AC omitted → −3.3 × 10⁻¹⁰ J; pairs doubled → −8.0 × 10⁻¹⁰ J; 1/r² → −2.6 × 10⁻¹⁰ J.

CED 3.3.A.8 · Read this in Fix

1. AIt increases as the crate moves, since friction stores energy in the crate–floor system.
   A student who thinks friction stores energy as potential energy, as springs and gravity do, picks this. Kinetic friction is not conservative: its work depends on the path, so no potential energy is associated with it.
2. BIt does not exist, since the work done by friction depends on the path the crate takes. Correct
   A potential energy can be associated only with a conservative force, whose work depends only on the initial and final configurations. Kinetic friction does negative work on the crate on every part of the loop, and a longer loop means more negative work, so no potential energy can describe it. The mechanical energy it removes is dissipated, not stored.
3. CIt returns to its starting value, since friction does zero net work around a closed loop.
   A student who thinks every force’s work depends only on the start and end points picks this. Kinetic friction opposes the crate’s motion on every part of the loop, so its work is negative throughout and does not cancel to zero.
4. DIt is zero at every point, since the crate stays at the same height above the ground.
   A student who thinks potential energy is always about height picks this. If a potential energy for friction were zero everywhere, friction would do no work at all; it does negative work all the way around the loop, so no potential energy can be associated with it.

CED 3.3.A.1 · Read this in Fix

1. AU_P equals U_∞, because the two pairs’ contributions cancel at P, as the two forces do.
   A student who treats potential energy as a vector picks this. The forces from the two large spheres point in opposite directions and cancel, but potential energies are scalars: both pair energies are lower at P than far away, and they add.
2. BU_P is greater than U_∞, because U_g of a pair is larger when the spheres are closer.
   A student who uses U_g = +Gm₁m₂/r, without the minus sign, picks this. Gravitational potential energy decreases as attracting objects move closer together, so U_P is lower than U_∞.
3. CU_P and U_∞ can be compared only after a zero of potential energy has been chosen.
   A student who thinks the choice of zero affects comparisons picks this. Moving the zero shifts U_P and U_∞ by the same constant, so U_P − U_∞ = −2GMm/d whatever zero is chosen.
4. DU_P is less than U_∞, because each large sphere’s pair with the small one has lower U when closer. Correct
   Potential energy is a scalar, so the two large-sphere–small-sphere pairs add rather than cancel. Bringing the small sphere from very far away to P lowers the potential energy of each of those pairs by GMm/d, while the pair of large spheres is unchanged, so U_P − U_∞ = −2GMm/d < 0. A difference in U does not depend on the choice of zero.

Working Pairs: the M–M pair is the same in both configurations. Each M–m pair has separation d at P (U = −GMm/d with zero at infinity) and a very large separation far away (U ≈ 0). U_P − U_∞ = −2GMm/d < 0, the same for any choice of zero.

CED 3.3.A.2 · Read this in Fix

1. AStudent 1’s force is larger, because the solid curve is higher at x = 2 m.
   A student who reads the force from the value of U instead of its slope picks this. The solid curve is higher, but the two curves are equally steep at x = 2 m, so the forces are equal.
2. BStudent 2’s force is zero, because the dashed curve has U = 0 at x = 2 m.
   A student who thinks the force is zero wherever U = 0 picks this. The force is zero only where the slope of U(x) is zero; at x = 2 m the dashed curve is sloping down, so the force there is not zero.
3. CThe forces cannot be compared, because each depends on that student’s zero.
   A student who thinks the choice of zero changes the forces picks this. A change of zero shifts the whole curve by a constant and leaves its slope, and so the force, unchanged.
4. DThe forces are equal, as the two curves have the same slope at x = 2 m. Correct
   The dashed curve is the solid curve moved down by a constant, which is all a change of zero does. F_x = −dU/dx depends on the slope, and the slopes are equal at every x, so both students find the same force at x = 2 m (in the +x direction, since U decreases with x there).

Working U₂(x) = U₁(x) − C for a constant C, so dU₂/dx = dU₁/dx at every x. At x = 2 m both curves slope downward (slope −1 J/m for the curves drawn), so F_x = −dU/dx = +1 N for both students.

CED 3.3.A.3 · Read this in Fix

1. Ak(x − d)²/2
   A student who measures the stretch from the point chosen as U = 0, rather than from the relaxed length, picks this. In U_s = (1/2)k(Δx)², Δx is always measured from the relaxed length; the choice of zero only shifts U by a constant.
2. Bkx²/2
   A student who thinks potential energy has one fixed value for each configuration picks this. kx²/2 is the energy with the zero at the relaxed length; with the student’s zero, every value is lower by (1/2)kd².
3. Ck(x² − d²)/2 Correct
   With the zero at the relaxed length, U_s = (1/2)kx². Choosing U = 0 at stretch d subtracts the same constant, (1/2)kd², from every value: U = (1/2)kx² − (1/2)kd² = k(x² − d²)/2. As a check, it gives U = 0 at x = d.
4. Dkd(x − d)
   A student who treats the spring force as constant, at its value kd where U = 0, and multiplies by the further stretch x − d picks this. The spring force grows as the stretch grows, so the energy must be found from ∫kx dx.

Working U(x) − U(d) = −∫_d^x (−kx′) dx′ = (1/2)k(x² − d²); with U(d) = 0, U(x) = k(x² − d²)/2. Distractors: stretch measured from x = d → k(x − d)²/2; zero choice ignored → kx²/2; force at d treated as constant → kd(x − d).

CED 3.3.A.3 · Read this in Fix

1. AC/(2r²)
   A student who sets ΔU equal to the work done by the conservative force, ΔU = +∫F_r dr, picks this. The attractive force does positive work as the particles come together from infinity, so U decreases from zero to a negative value.
2. BC/r²
   A student who treats the force as constant and multiplies it by the separation, U = −F_r r = −(−C/r³)(r), picks this. The force changes with r, so U must be found by integrating it.
3. C−C/(2r²) Correct
   The radial component of the attractive force is F_r = −C/r³. U(r) − U(∞) = −∫ from ∞ to r of F_r dr′ = ∫ from ∞ to r of C/r′³ dr′ = −C/(2r²). It is negative because bringing the particles together from infinity lowers the energy of an attracting pair.
4. D−3C/r⁴
   A student who differentiates the force instead of integrating it, U = −dF_r/dr, picks this. Its units are N/m, not J: energy comes from integrating force over distance.

Working F_r = −C/r³ (attractive). U(r) = −∫_∞^r F_r dr′ = ∫_∞^r C r′⁻³ dr′ = [−C/(2r′²)]_∞^r = −C/(2r²). Distractors: sign dropped → C/(2r²); U = −F_r r → C/r²; U = −dF_r/dr → −3C/r⁴ (units N/m).

CED 3.3.A.4 · Read this in Fix

1. A+12 J
   A student who sets ΔU equal to the work done by the conservative force picks this. The force does +12 J of work, so U decreases by 12 J.
2. B−16 J
   A student who counts the area below the x-axis as positive picks this: 12 J + 2 J + 2 J = 16 J. Between 4 m and 5 m the force is negative, so its work there is −2 J and must be subtracted.
3. C+20 J
   A student who multiplies the final force by the whole displacement picks this: ΔU = −F(5 m)Δx = −(−4 N)(5 m) = +20 J. The force varies, so the work is the area under the graph, not one value of F times Δx.
4. D−12 J Correct
   The work done by the force is the signed area under the F–x graph: (4 N)(3 m) = 12 J from 0 to 3 m, +2 J from 3 m to 4 m and −2 J from 4 m to 5 m, a total of 12 J. ΔU = −W = −12 J.

Working W = ∫F dx = (4 N)(3 m) + (1/2)(1 m)(4 N) − (1/2)(1 m)(4 N) = 12 J. ΔU = −W = −12 J. Distractors: ΔU = +W → +12 J; area below the axis counted positive → −16 J; −F(5 m)Δx = −(−4 N)(5 m) → +20 J.

CED 3.3.A.4 · Read this in Fix

1. A−4.0 N
   A student who takes the force equal to the slope, F_x = +dU/dx, picks this. The minus sign in F_x = −dU/dx makes the force point toward decreasing U, which here is the +x direction.
2. B+4.0 N Correct
   Between x = 2 m and x = 6 m the graph is a straight line with slope (−6 J − 10 J)/(6 m − 2 m) = −4.0 J/m. F_x = −dU/dx = +4.0 N: the force points in the +x direction, toward lower potential energy.
3. C−2.0 N
   A student who uses the value of U instead of its slope picks this: U = 2 J at x = 4 m, and F = −U is taken as −2.0 N. The force comes from the slope of the graph at x = 4 m, which is −4.0 J/m.
4. D+2.0 N
   A student who uses the average slope from x = 0 to x = 4 m picks this: (2 J − 10 J)/(4 m) = −2.0 J/m, giving +2.0 N. The force at x = 4 m depends on the slope at that point, which is −4.0 J/m.

Working Slope at x = 4 m: (−6 J − 10 J)/(6 m − 2 m) = −4.0 J/m. F_x = −dU/dx = +4.0 N. Distractors: F = +slope → −4.0 N; F = −U(4 m) = −2 J → −2.0 N; average slope from x = 0 to 4 m, (2 J − 10 J)/(4 m) = −2.0 J/m → +2.0 N.

CED 3.3.A.5 · Read this in Fix

1. APoint B
   A student who thinks the force is greatest at the bottom of the well picks this. B is a local minimum, where the slope of U(x), and so the force, is zero.
2. BPoint C
   A student who reads the force from the value of U picks this. U is highest at C, but the graph is nearly flat there, so the force is small.
3. CPoint D
   A student who thinks every force grows with distance from the equilibrium position, like a spring force, picks this. D is the farthest point from the minimum at B, but the graph is only gently sloping there.
4. DPoint A Correct
   The magnitude of the force is |dU/dx|, the steepness of the graph. The graph is steepest at A. It is flat at B (a minimum, where the force is zero), nearly flat at C, close to the top of a broad maximum, and only gently sloping at D.

Working |F_x| = |dU/dx|. Slopes of the curve drawn: A about −4 J per unit x (steepest); B 0 (minimum); C about +0.3; D about +0.7. Greatest force at A.

CED 3.3.A.5 · Read this in Fix

1. A×256
   A student who takes the force to be proportional to U itself picks this: U ∝ x⁴ grows by 4⁴ = 256. The force depends on the slope of U(x), which is proportional to x³.
2. B×64 Correct
   F_x = −dU/dx = −4βx³, so the magnitude of the force is proportional to x³. Multiplying x by 4 multiplies the force by 4³ = 64.
3. C×4
   A student who assumes the force is proportional to the distance from the equilibrium position at x = 0, as for a spring, picks this. Here U ∝ x⁴, so the force is proportional to x³, not x.
4. D×1024
   A student who integrates U instead of differentiating it picks this: ∫βx⁴ dx ∝ x⁵, which grows by 4⁵ = 1024. The force is the negative derivative of U, which is proportional to x³.

Working F_x = −4βx³ → |F(4d)|/|F(d)| = 4³ = 64. Distractors: ∝ U ∝ x⁴ → 256; ∝ x → 4; ∝ ∫U dx ∝ x⁵ → 1024.

CED 3.3.A.5 · Read this in Fix

1. A80 N/m Correct
   At the marked point, U = 3.6 J when Δx = 0.30 m. U_s = (1/2)k(Δx)², so k = 2U/(Δx)² = 2(3.6 J)/(0.30 m)² = 80 N/m.
2. B40 N/m
   A student who takes the stored energy as the force times the stretch, U = (kΔx)(Δx), picks this: k = U/(Δx)² = 40 N/m. The spring force grows from zero as the spring is stretched, so the energy is (1/2)k(Δx)².
3. C24 N/m
   A student who takes the slope of this graph as the spring constant picks this: at 0.30 m the slope is 2U/Δx = 24 J/m. The slope of a U–Δx graph is the spring force kΔx; k is the slope of a force–stretch graph.
4. D12 N/m
   A student who treats the stored energy as equal to the spring force, U = kΔx, picks this: k = U/Δx = 12 N/m. Force and energy are different quantities; U_s = (1/2)k(Δx)².

Working Read the marked point: U = 3.6 J at Δx = 0.30 m. k = 2U/(Δx)² = 2(3.6 J)/(0.30 m)² = 80 N/m. Distractors: U = kΔx² → 40 N/m; tangent slope at 0.30 m, 2U/Δx = 24 J/m, taken as k → 24 N/m; U = kΔx → 12 N/m.

CED 3.3.A.6 · Read this in Fix

1. AA small displacement either way along the track gives a net force that points away from P. Correct
   P is the top of a hump: the net force on the ball there is zero, so it can stay at rest, but U_g of the ball–Earth system is greater at P than at nearby points. After a small displacement either way, the net force along the track points away from P, in the direction of the displacement. P is an unstable equilibrium position.
2. BA small displacement either way along the track gives a net force directed back toward P.
   A student who thinks a balanced position on top is stable picks this. On a hump, a displaced ball is pushed farther downhill, away from P; a restoring force would need a dip in the track.
3. CP is not an equilibrium position, because the ball would roll away if it were disturbed.
   A student who thinks every equilibrium must be stable picks this. The net force at P is zero, so P is an equilibrium position; rolling away after a disturbance makes it unstable, not a non-equilibrium.
4. DThe net force on the ball is zero at all points close to P, since P is an equilibrium position.
   A student who pictures equilibrium as a region rather than a point picks this. The net force is zero only exactly at P; a small displacement gives a small, nonzero force away from P.

CED 3.3.A.6.ii · Read this in Fix

1. A−m/k
   A student who writes the gravitational term with the mass in place of the weight, U_g = my, picks this: ky + m = 0. The block–Earth term is mgy; m/k does not even have units of length.
2. B−mg/k Correct
   The system’s potential energy is the sum for its two interacting pairs: U(y) = (1/2)ky² + mgy. Stable equilibrium is at the local minimum: dU/dy = ky + mg = 0 gives y = −mg/k, and d²U/dy² = k > 0 confirms a minimum.
3. C−(1/2) mg/k
   A student who writes the spring energy as force times stretch, U_s = (ky)(y) = ky², picks this: 2ky + mg = 0. The spring force grows from zero as the spring stretches, so U_s = (1/2)ky².
4. D0
   A student who counts only the spring’s elastic potential energy picks this: (1/2)ky² is least at y = 0. The block–Earth pair adds mgy, which moves the minimum of the total down to y = −mg/k.

Working U(y) = (1/2)ky² + mgy. dU/dy = ky + mg = 0 → y = −mg/k; d²U/dy² = k > 0 (minimum → stable). Distractors: mass used for weight, U_g = my → y = −m/k; U_s = ky² → 2ky + mg = 0 → y = −(1/2)mg/k; spring term only → y = 0.

CED 3.3.A.6.iii · Read this in Fix

1. APoint Q
   A student who has stable and unstable equilibrium the wrong way round picks this. Q is a local minimum of U(x), where a displaced object is pushed back: a stable equilibrium position.
2. BPoint R
   A student who thinks the force is zero wherever U = 0 picks this. The graph crosses U = 0 at R but is steep there, so the force at R is large and R is not an equilibrium position.
3. CPoint S
   A student who looks for the highest value of U on the whole graph picks this. U is greatest at S, but the graph is steep there, so the force is not zero and S is not an equilibrium position.
4. DPoint P Correct
   P is a local maximum of U(x): the slope there is zero, so the force is zero, and U is higher than at nearby points, so a small displacement either way gives a force away from P. That is an unstable equilibrium position.

Working Equilibrium needs dU/dx = 0: only P (local maximum) and Q (local minimum). Unstable: local maximum → P.

CED 3.3.A.6.iv · Read this in Fix

1. A12 J
   A student who squares only the extra stretch picks this: (1/2)(600 N/m)(0.20 m)² = 12 J. Δx in U_s = (1/2)k(Δx)² is measured from the relaxed length, so the energies at 0.30 m and 0.10 m must be subtracted.
2. B36 J
   A student who multiplies the final spring force, (600 N/m)(0.30 m) = 180 N, by the extra stretch of 0.20 m picks this. The force grows from 60 N to 180 N during the stretch, so force times distance overestimates the energy.
3. C27 J
   A student who takes the change to be the final energy picks this: (1/2)(600 N/m)(0.30 m)² = 27 J. The spring already stored (1/2)(600 N/m)(0.10 m)² = 3 J, which must be subtracted.
4. D24 J Correct
   ΔU_s = (1/2)k(x₂² − x₁²) = (1/2)(600 N/m)[(0.30 m)² − (0.10 m)²] = (300 N/m)(0.080 m²) = 24 J. Both stretches are measured from the relaxed length.

Working ΔU_s = (1/2)(600 N/m)[(0.30 m)² − (0.10 m)²] = 24 J. Distractors: (1/2)k(0.20 m)² = 12 J; (180 N)(0.20 m) = 36 J; (1/2)k(0.30 m)² = 27 J.

CED 3.3.A.7.i · Read this in Fix

1. A(3/2)kd²
   A student who adds the spring constants, k_eq = 3k, picks this: (1/2)(3k)d². Springs in series combine as 1/k_eq = 1/k + 1/(2k), so k_eq = 2k/3.
2. B(3/8)kd²
   A student who gives each spring half of the stretch picks this: (1/2)k(d/2)² + (1/2)(2k)(d/2)² = (3/8)kd². The springs carry equal forces, so the softer spring stretches twice as far as the stiffer one.
3. C(1/3)kd² Correct
   The springs carry the same force F, so they stretch by F/k and F/(2k). These add to d: 3F/(2k) = d, so F = 2kd/3 and the stretches are 2d/3 and d/3. U = (1/2)k(2d/3)² + (1/2)(2k)(d/3)² = (2/9)kd² + (1/9)kd² = (1/3)kd², the same as (1/2)k_eq d² with k_eq = 2k/3.
4. D(2/3)kd²
   A student who multiplies the final force, F = (2k/3)d, by the total stretch d picks this. The force grows from zero during the stretch, so the energy is (1/2)Fd = (1/3)kd².

Working F = k x₁ = 2k x₂, x₁ + x₂ = d → x₁ = 2d/3, x₂ = d/3. U = (1/2)k(4d²/9) + (1/2)(2k)(d²/9) = kd²/3. Distractors: k_eq = 3k → (3/2)kd²; x₁ = x₂ = d/2 → (3/8)kd²; Fd with F = 2kd/3 → (2/3)kd².

CED 3.3.A.7.i · Read this in Fix

1. AU_X = 2U_Y Correct
   Each spring is deformed by F/k_spring: X by F/k and Y by F/(2k). U_X = (1/2)k(F/k)² = F²/(2k) and U_Y = (1/2)(2k)(F/(2k))² = F²/(4k), so U_X = 2U_Y. Under equal forces, the softer spring stores more energy.
2. BU_X = 0.5U_Y
   A student who thinks the stiffer spring always stores more energy, in proportion to k, picks this. Under the same force the stiffer spring deforms only half as much, and U_s = F²/(2k) is smaller for larger k.
3. CU_X = U_Y
   A student who thinks equal forces mean equal stored energies picks this. Force and energy are different quantities: the two springs are deformed by different amounts, so they store different energies.
4. DU_X = −2U_Y
   A student who thinks a compressed spring stores negative energy picks this. U_s = (1/2)k(Δx)² contains the square of the deformation, so U_Y is positive: U_Y = F²/(4k).

Working Δx_X = F/k, Δx_Y = F/(2k). U_X = F²/(2k), U_Y = F²/(4k) → U_X = 2U_Y. Distractors: U ∝ k → U_Y = 2U_X (U_X = 0.5U_Y); equal forces → equal energies; compression taken as negative energy → U_X = −2U_Y.

CED 3.3.A.7.i · Read this in Fix

1. AU_B is two times U_A.
   A student who takes the energy to be proportional to the stretch, like the force, picks this: (4)(1/2) = 2. U_s depends on the square of the stretch, so halving the stretch divides the energy by 4.
2. BU_B is equal to U_A. Correct
   U_s = (1/2)k(Δx)². For B, the spring constant is 4 times as large and the stretch is half as large: U_B = (1/2)(4k)(x/2)² = (1/2)kx² = U_A.
3. CU_B is four times U_A.
   A student who scales the energy by the spring constant alone picks this. The stretch is halved as well, which divides the energy by 4 and cancels the factor of 4 from k.
4. DU_B is a quarter of U_A.
   A student who scales the energy by the stretch alone picks this: (1/2)² = 1/4. The spring constant is 4 times as large, which multiplies the energy by 4.

Working U_B/U_A = (4k/k)(1/2)² = 1. Distractors: U ∝ kΔx → 2; U ∝ k only → 4; U ∝ (Δx)² only → 1/4.

CED 3.3.A.7.i · Read this in Fix

1. A1.2 × 10¹⁰ J Correct
   The separations are measured from Earth’s center: r₁ = 8.0 × 10⁶ m and r₂ = 2.0 × 10⁷ m. ΔU_g = GMm(1/r₁ − 1/r₂) = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(4.0 × 10²)(1.25 × 10⁻⁷ − 5.0 × 10⁻⁸) J = 1.2 × 10¹⁰ J. It is positive: the satellite moves farther from Earth.
2. B8.8 × 10¹⁰ J
   A student who uses the altitudes as r picks this. In U_g = −Gm₁m₂/r, r is the distance between centers, so Earth’s radius must be added to each altitude.
3. C4.8 × 10¹⁰ J
   A student who uses mgΔy with Earth’s surface field of about 10 N/kg picks this: (4.0 × 10² kg)(10 N/kg)(1.2 × 10⁷ m). The field is much weaker at these distances, so the exact change is far smaller.
4. D−8.0 × 10⁹ J
   A student who takes the change to be the final potential energy, −GMm/r₂, picks this. ΔU_g = U_f − U_i, and U_i = −GMm/r₁ is more negative than U_f, so the change is positive.

Working r₁ = 6.4 × 10⁶ m + 1.6 × 10⁶ m = 8.0 × 10⁶ m; r₂ = 6.4 × 10⁶ m + 1.36 × 10⁷ m = 2.0 × 10⁷ m. ΔU_g = GMm(1/r₁ − 1/r₂) = (1.60 × 10¹⁷ N·m²)(7.5 × 10⁻⁸ m⁻¹) = 1.2 × 10¹⁰ J. Distractors: altitudes as r → 8.8 × 10¹⁰ J; mgΔy with g = 10 N/kg → 4.8 × 10¹⁰ J; −GMm/r₂ → −8.0 × 10⁹ J.

CED 3.3.A.7.ii · Read this in Fix

1. AIts magnitude becomes one-ninth as large, so U_g also increases.
   A student who uses a 1/r² dependence, as for the gravitational force, picks this. U_g is proportional to 1/r, so its magnitude becomes one-third as large.
2. BIts magnitude becomes one-third as large, so U_g increases. Correct
   U_g = −Gm₁m₂/r: tripling r makes the magnitude one-third as large. U_g is negative, so a smaller magnitude means a larger value: U_g increases toward zero as the spheres move apart.
3. CIts magnitude becomes one-third as large, and so U_g decreases.
   A student who drops the minus sign, U_g = +Gm₁m₂/r, picks this. With the minus sign, a smaller magnitude means a less negative, larger U_g.
4. DIts magnitude becomes three times as large, so U_g decreases.
   A student who takes the magnitude of U_g to grow in proportion to separation, as mgΔy grows with height, picks this; with the negative sign kept, a larger magnitude means a lower U_g. The magnitude is proportional to 1/r, so it becomes one-third as large.

Working |U_g| ∝ 1/r → ×1/3; U_g < 0, so U_g increases (from −X to −X/3). Distractors: ∝ 1/r² → ×1/9; sign dropped → decreases; ∝ r → ×3 (more negative).

CED 3.3.A.7.ii · Read this in Fix

1. A2.4 × 10⁶ J
   A student who uses the length of the trail as Δy picks this: (60 kg)(10 N/kg)(4000 m). The change in gravitational potential energy depends only on the change in vertical position, not on the path.
2. B1.8 × 10⁴ J
   A student who multiplies the mass by the change in height, leaving out g, picks this: (60 kg)(300 m). kg·m is not a unit of energy; the weight mg = 600 N is needed.
3. C1.8 × 10⁵ J Correct
   Near Earth’s surface ΔU_g = mgΔy, where Δy is the change in vertical position: 1500 m − 1200 m = 300 m. ΔU_g = (60 kg)(10 N/kg)(300 m) = 1.8 × 10⁵ J. The length of the trail does not matter.
4. D9.0 × 10⁵ J
   A student who uses the final elevation above sea level picks this: (60 kg)(10 N/kg)(1500 m), the potential energy at the summit with U_g = 0 at sea level. The change is the difference between the summit and trailhead values.

Working Δy = 1500 m − 1200 m = 300 m. ΔU_g = mgΔy = (60 kg)(10 N/kg)(300 m) = 1.8 × 10⁵ J. Distractors: path length → 2.4 × 10⁶ J; mΔy → 1.8 × 10⁴ J; mg(1500 m) → 9.0 × 10⁵ J.

CED 3.3.A.7.iii · Read this in Fix

1. AΔU_exact = ΔU_approx, because mgΔy is exact for every height above Earth.
   A student who thinks mgΔy holds at any height picks this. It treats the field as constant, which is accurate only for heights small compared with Earth’s radius; here the field weakens noticeably along the way.
2. BΔU_exact > ΔU_approx, because the gravitational field grows stronger with height.
   A student who thinks gravity is stronger higher up picks this. The field strength GM/r² decreases as r increases, so the exact change is smaller than mgh.
3. CΔU_exact < 0 < ΔU_approx, because U_g found from −Gm₁m₂/r is negative.
   A student who confuses the sign of U_g with the sign of its change picks this. U_g is negative at every height, but it increases toward zero as the object rises, so ΔU_exact is positive.
4. DΔU_exact < ΔU_approx, because the gravitational field weakens with height. Correct
   mgh assumes the field keeps its surface value all the way up. The field strength GM/r² decreases with distance from Earth’s center, so less energy is stored per meter at greater heights: ΔU_exact = mgh·R/(R + h), which is less than mgh.

Working ΔU_exact = GMm(1/R − 1/(R + h)) = GMmh/(R(R + h)) = mgh·R/(R + h) with g = GM/R². R/(R + h) < 1, so 0 < ΔU_exact < mgh.

CED 3.3.A.7.iii · Read this in Fix

1. A−2.0U₀
   A student who counts only the fourth sphere’s nearest neighbors picks this, leaving out the sphere at the opposite corner. Every pair interacts gravitationally, so the diagonal pair adds −U₀/√2.
2. B−2.7U₀ Correct
   The total potential energy is the sum over pairs, and only the three pairs that include the fourth sphere change. Two of the other spheres are a distance s away and one is a distance √2 s away (the diagonal): ΔU = −Gm²/s − Gm²/s − Gm²/(√2 s) = −(2 + 1/√2)U₀ ≈ −2.7U₀.
3. C−5.4U₀
   A student who gives the total potential energy of the final four-sphere system, −(4 + √2)U₀, picks this. The question asks for the change; the three pairs among the spheres already in place are unchanged.
4. D−2.1U₀
   A student who adds the three pair energies as vectors pointing from the fourth sphere toward the others picks this: the two side pairs combine at right angles to √2 U₀ along the diagonal, plus U₀/√2 along the diagonal, giving −2.1U₀. Potential energies are scalars and add as numbers.

Working New pairs: two at distance s, one at √2 s. ΔU = −Gm²(2/s + 1/(√2 s)) = −(2 + 1/√2)U₀ = −2.71U₀. Distractors: nearest neighbors only → −2.0U₀; total of all six pairs, −(4 + √2)U₀ → −5.4U₀; vector sum, −(√2 + 1/√2)U₀ → −2.1U₀.

CED 3.3.A.8 · Read this in Fix

1. A−GMm/(2R)
   A student who takes ΔU as +∫F⃗ · dr⃗ picks this. That makes ΔU = +GMm/(2R) and U = −GMm/R + GMm/(2R). Gravity does positive work on the object as it moves toward the center, so U must decrease, not increase.
2. B−3GMm/(2R) Correct
   At the surface U_g = −GMm/R. Moving in to the center, ΔU = −∫_R⁰ F_r dr with F_r = −GMmr/R³, which gives ΔU = −GMm/(2R): gravity does positive work as the object moves toward the center, so U decreases. U at the center is −GMm/R − GMm/(2R) = −3GMm/(2R).
3. C−2GMm/R
   A student who uses ΔU_g = mgΔy with the surface value g = GM/R² over the whole distance R picks this: ΔU = −GMm/R, so U = −2GMm/R. Inside the planet the field weakens steadily to zero at the center, so the true decrease is only half as large.
4. D−GMm/R
   A student who thinks a conservative force does no work, so that U stays the same, picks this: the surface value −GMm/R. Gravity does positive work on the object as it moves from the surface to the center, so U decreases.

Working At the surface (r = R), outside-sphere form with U_g = 0 at infinity: U(R) = −GMm/R. Inside, the radial component of the force is F_r = −GMmr/R³ (toward the center). From the surface to the center: ΔU = −∫_R⁰ F_r dr = −∫_R⁰ (−GMmr/R³) dr = (GMm/R³)[r²/2]_R⁰ = −GMm/(2R). U(0) = U(R) + ΔU = −GMm/R − GMm/(2R) = −3GMm/(2R). Distractors: ΔU = +∫F_r dr = +GMm/(2R) gives −GMm/(2R); mgΔy with surface g = GM/R² and Δy = −R gives ΔU = −GMm/R and U(0) = −2GMm/R; ΔU = 0 gives −GMm/R. Checked with sympy (G = 1.3, M = 2.1, m = 0.7, R = 1.9: −1.509, −0.503, −2.012, −1.006).

CED 3.3.A.4 · Read this in Fix

1. A3.0 J
   A student who counts only the spring’s elastic potential energy picks this: the stretch goes from 0.10 m to 0.40 m, so (1/2)(40 N/m)[(0.40 m)² − (0.10 m)²] = 3.0 J. The block–Earth pair also belongs to the system, and its potential energy falls by (0.40 kg)(10 m/s²)(0.30 m) = 1.2 J as the block moves down.
2. B4.2 J
   A student who thinks potential energy cannot be negative, and so adds the size 1.2 J of the gravitational change, picks this: 3.0 J + 1.2 J = 4.2 J. The block moves down, so U_g changes by −1.2 J, which brings the total to 1.8 J.
3. C1.8 J Correct
   At equilibrium the spring is stretched mg/k = 0.10 m. Relative to equilibrium, the elastic energy rises by (1/2)(40 N/m)[(0.40 m)² − (0.10 m)²] = 3.0 J and the gravitational energy falls by (0.40 kg)(10 m/s²)(0.30 m) = 1.2 J. The sum of the two pair energies is 1.8 J, which equals (1/2)(40 N/m)(0.30 m)².
4. D4.8 J
   A student who finds the elastic energy as force times stretch, (kΔx)(Δx), picks this: (40 N/m)[(0.40 m)² − (0.10 m)²] − 1.2 J = 4.8 J. The spring force varies with stretch, so U_s = (1/2)k(Δx)², and the total is 1.8 J.

Working At equilibrium the spring is stretched d = mg/k = (0.40 kg)(10 m/s²)/(40 N/m) = 0.10 m. Relative to equilibrium, ΔU_s = (1/2)(40 N/m)[(0.40 m)² − (0.10 m)²] = 3.0 J and ΔU_g = −(0.40 kg)(10 m/s²)(0.30 m) = −1.2 J, so U = 1.8 J = (1/2)(40 N/m)(0.30 m)².

CED 3.3.A.8 · Read this in Fix