3.3 Potential Energy

1. AIt belongs to the ball alone, since only the ball was raised.
   A student who treats potential energy as something a single object has picks this. Everyday speech says 'the ball has potential energy', but the energy is associated with the gravitational interaction of the ball and Earth; a ball with nothing to interact with has no potential energy.
2. BIt belongs to the ball–Earth system, as the ball and Earth interact. Correct
   Potential energy belongs to a system of two or more objects that interact through a conservative force. Here the interacting objects are the ball and Earth: the gravitational potential energy depends on their separation and belongs to the ball–Earth system, not to the ball alone.
3. CIt is zero until the ball is released and begins to fall.
   A student who reads 'potential' as 'not yet present' picks this. The ball–Earth system has gravitational potential energy while the ball is held at rest, because the energy depends on the positions of the objects, not on whether they are moving.
4. DIt points downward, in the direction of the ball's weight.
   A student who gives potential energy a direction picks this. Potential energy is a scalar: it has a size (and, for some choices of zero, a sign) but no direction. The ball's weight is a force and has a direction; the energy does not.

CED 3.3.A.1 · Read this in Fix

1. AΔU_g is the same for all three paths from P to Q. Correct
   Gravitational potential energy depends only on the positions of the box and Earth. Every path starts at P and ends at Q, so every path gives the same ΔU_g = mgΔy, where Δy is the height of Q above P. The extra rise on path 3 adds to U_g on the way up and is removed again on the way down.
2. BΔU_g is greatest for path 3, which is the longest path.
   A student who thinks ΔU_g grows with the distance traveled picks this. Only the start and end positions matter: path 3 is longer and rises higher on the way, but it ends at Q, at the same height as the other paths.
3. CΔU_g is least for path 2, since the ramp supports the box.
   A student who thinks a ramp reduces the energy needed to raise an object picks this. The ramp lets a smaller force raise the box, but over a longer distance; the box still ends at the same height, so ΔU_g is the same.
4. DΔU_g is zero for every path, as the box ends at rest.
   A student who thinks an object at rest has no potential energy picks this. The box is at rest at both ends, but at Q it is higher than at P, so the box–Earth system has more gravitational potential energy there: ΔU_g is positive.

CED 3.3.A.2 · Read this in Fix

1. AIt is zero, since the book is at rest on the table.
   A student who thinks an object at rest has no potential energy picks this. U_g depends on where the book is relative to the chosen zero, not on whether it moves. The table is below the zero level, so U_g there is negative, not zero.
2. BIt is positive, since a negative potential energy is a mistake.
   A student who thinks potential energy can never be negative picks this. A negative U_g means only that the system is in a configuration with less potential energy than the one chosen as zero; the table is below the shelf, so U_g on the table is negative.
3. CIt is negative, which is valid for this choice of zero. Correct
   The zero of potential energy is the observer's choice. With U_g = 0 for the book on the shelf, the book on the table is 1.2 m lower, so U_g = −mg(1.2 m), a negative value. The choice is valid: only differences in U_g have physical meaning, and they are the same for any choice of zero.
4. DIt is undefined, as the zero of U_g is fixed at the floor.
   A student who thinks the zero of U_g must be at the ground or floor picks this. Any level may be chosen as the zero, whichever makes the analysis simplest. Measuring from the floor is common, but it is a choice, not a rule.

CED 3.3.A.3 · Read this in Fix

1. AU_g is a vector, and the minus sign shows that it points toward the other object.
   A student who gives potential energy a direction picks this. U_g is a scalar; the minus sign is part of its value, not a direction. It is the gravitational force that points from each object toward the other.
2. BThe sign can be dropped, since a potential energy must be a positive amount.
   A student who thinks potential energy cannot be negative picks this. Dropping the sign gives the wrong changes: with the sign, U_g rises toward zero as r increases; without it, U_g would appear to fall. The negative values follow from choosing the zero at infinite separation.
3. CThe minus sign shows that U_g decreases as the separation r increases.
   A student who compares the sizes of the values and ignores their sign picks this. As r increases, the magnitude G(m₁m₂)/r gets smaller, but U_g is negative, so it rises toward zero: U_g increases as the separation increases.
4. DU_g is zero at infinite separation, and the attraction makes it lower at any finite r. Correct
   The equation takes the zero of U_g at infinite separation, where the objects no longer interact. The objects attract, so the system's potential energy increases as they are moved apart and approaches zero from below; at every finite r, U_g is less than zero.

CED 3.3.A.4.ii · Read this in Fix

1. A−5.0 Gm²/d
   A student who adds, for each sphere, its potential energy with each of the other two counts every pair twice and gets 2 × (−2.5 Gm²/d) = −5.0 Gm²/d. Each pair contributes once: three spheres form three pairs, not six.
2. B−2.5 Gm²/d Correct
   The total potential energy is the sum over each pair of spheres: A–B at distance d, B–C at distance d, and A–C at distance 2d. U = −Gm²/d − Gm²/d − Gm²/(2d) = −2.5 Gm²/d.
3. C−0.5 Gm²/d
   A student who treats potential energies like forces lets the A–B and B–C terms cancel, since A and C pull B in opposite directions, leaving only the A–C term, −0.5 Gm²/d. Potential energy is a scalar: all three pair terms are negative, and they add.
4. D−3.0 Gm²/d
   A student who uses the spacing d for every pair gets 3 × (−Gm²/d) = −3.0 Gm²/d. A and C are 2d apart, so their pair contributes −Gm²/(2d), not −Gm²/d.

Working Sum over the three pairs, each with the distance between its own two spheres: U_AB = −Gm²/d, U_BC = −Gm²/d, U_AC = −Gm²/(2d). Total U = −(1 + 1 + 0.5)Gm²/d = −2.5 Gm²/d.

CED 3.3.A.5 · Read this in Fix

1. ATwo blocks, one sliding across the rough top of the other
   A student who thinks every force that does work stores potential energy picks this. The blocks interact through friction, a nonconservative force: the work it does depends on the path, and the energy it transfers is dissipated, not stored as potential energy.
2. BA lump of clay and the wall that squashed it flat on impact
   A student who thinks any deformed object stores elastic energy picks this. The clay stays flat and never pushes back: the forces that flattened it are not conservative, so the energy transferred in the impact is dissipated rather than stored.
3. CA stone thrown upward, the only object in its system
   A student who treats potential energy as a property of one object picks this. A system of a single object has no potential energy, because potential energy is associated with an interaction between objects. The stone and Earth together would form such a system.
4. DA cart and an ideal spring that the cart has compressed Correct
   The force exerted by an ideal spring is conservative: the work it does depends only on how far the spring is compressed, not on how it got there. So the cart–spring system has elastic potential energy, which the spring can return by pushing the cart back.

CED 3.3.A.1 · Read this in Fix

1. AStudent 1's is greater, as student 1's values of U_g are greater.
   A student who thinks ΔU_g depends on the choice of zero picks this. Student 1's two values are each larger than student 2's by the same amount, so the difference between them is the same as student 2's.
2. BStudent 2's is invalid, as U_g on the table comes out as negative.
   A student who thinks potential energy cannot be negative picks this. A negative U_g is valid: it shows that the table is below student 2's zero level. Student 2's ΔU_g = 0 − (−mgΔy) = +mgΔy agrees with student 1's.
3. CStudent 2's is zero, as the book ends at the level where U_g is zero.
   A student who confuses the value of U_g with its change picks this. U_g on the shelf is zero for student 2, but ΔU_g is the final value minus the initial value: 0 − (−mgΔy) = +mgΔy, the same as student 1's result.
4. DStudent 1's equals student 2's, though their U_g values differ. Correct
   Changing the zero adds the same constant to every value of U_g, so the constant cancels in the difference. Both students find ΔU_g = mgΔy, where Δy is the height of the shelf above the table: student 1 from two positive values, student 2 from a negative value and zero.

Working Let the table be at height y_t and the shelf at y_s above the floor, with Δy = y_s − y_t. Student 1: U_i = mgy_t, U_f = mgy_s, so ΔU_g = mg(y_s − y_t) = mgΔy. Student 2: U_i = −mgΔy, U_f = 0, so ΔU_g = 0 − (−mgΔy) = mgΔy. The values of U_g differ by the constant mgy_s; ΔU_g is the same.

CED 3.3.A.3 · Read this in Fix

1. A2.00 J
   A student who multiplies the spring force at full stretch by the stretch, (20.0 N)(0.100 m) = 2.00 J, picks this. The spring force grows from zero as the spring is stretched, so it is not 20.0 N throughout; U_s = (1/2)k(Δx)² includes the factor 1/2.
2. B1.00 J Correct
   Δx is measured from the equilibrium length: Δx = 0.400 m − 0.300 m = 0.100 m. Then U_s = (1/2)k(Δx)² = (1/2)(200 N/m)(0.100 m)² = 1.00 J.
3. C10.0 J
   A student who takes U_s to be proportional to Δx, like the spring force, calculates (1/2)(200 N/m)(0.100 m) = 10.0 J. The stretch must be squared: U_s = (1/2)k(Δx)², which also gives the right unit, (N/m)(m²) = J.
4. D16.0 J
   A student who uses the spring's length for Δx calculates (1/2)(200 N/m)(0.400 m)² = 16.0 J. Δx is how far the spring is stretched from its equilibrium length, 0.100 m, not its total length.

Working Δx is the stretch from the equilibrium length: Δx = 0.400 m − 0.300 m = 0.100 m. U_s = (1/2)k(Δx)² = (1/2)(200 N/m)(0.100 m)² = 1.00 J.

CED 3.3.A.4.i · Read this in Fix

1. A25 N/m
   A student who leaves out the 1/2, taking U_s = k(Δx)² (the spring force at full stretch times the stretch), gets k = 4.0 J ÷ (0.40 m)² = 25 N/m. The spring force grows from zero, so U_s = (1/2)k(Δx)² and k = 2U_s/(Δx)².
2. B20 N/m
   A student who takes U_s to be proportional to Δx, U_s = (1/2)kΔx, gets k = 2(4.0 J) ÷ 0.40 m = 20 N/m. The graph rules this out: it curves upward, so U_s is not proportional to Δx; it grows with (Δx)².
3. C50 N/m Correct
   At the marked point U_s = 4.0 J and Δx = 0.40 m. Rearranging U_s = (1/2)k(Δx)² gives k = 2U_s/(Δx)² = 2(4.0 J)/(0.40 m)² = 50 N/m. The curve agrees: at Δx = 0.20 m, (1/2)(50 N/m)(0.20 m)² = 1.0 J.
4. D10 N/m
   A student who takes k to be the slope of the graph, as it is for a graph of spring force against stretch, calculates 4.0 J ÷ 0.40 m = 10 N/m. This graph shows energy, not force, and its slope is not constant; use U_s = (1/2)k(Δx)² with a point read from the curve.

Working Read the marked point: U_s = 4.0 J at Δx = 0.40 m. From U_s = (1/2)k(Δx)², k = 2U_s/(Δx)² = 2(4.0 J)/(0.40 m)² = 50 N/m. Check at another point: Δx = 0.20 m gives (1/2)(50 N/m)(0.20 m)² = 1.0 J, as the graph shows.

CED 3.3.A.4.i · Read this in Fix

1. A0.50 Correct
   U_s = (1/2)k(Δx)², so U_s is proportional to k and to (Δx)². Doubling k doubles U_s, and halving Δx divides it by 4: U_B/U_A = 2 × (1/2)² = 0.50.
2. B1.00
   A student who takes U_s to be proportional to Δx, like the spring force, finds 2 × 1/2 = 1.00. That is the ratio of the spring forces, kΔx; the energy depends on (Δx)², so halving the stretch divides U_s by 4, not by 2.
3. C2.00
   A student who thinks the stiffer spring stores more energy whatever the stretch uses only the factor 2 from k. Spring B is stretched only half as far, and that divides its energy by 4, which outweighs the doubled k.
4. D0.25
   A student who thinks U_s depends only on the stretch uses only the factor (1/2)² = 1/4 from Δx. U_s also depends on k: spring B is twice as stiff, which doubles the energy for a given stretch.

Working U_A = (1/2)kd². U_B = (1/2)(2k)(d/2)² = (1/2)(2k)(d²/4) = kd²/4. U_B/U_A = (kd²/4)/(kd²/2) = 0.50. Doubling k doubles U_s; halving Δx divides U_s by 4; together × 2/4 = × 0.50.

CED 3.3.A.4.i · Read this in Fix

1. A−0.67 GMm/R
   A student who thinks U_g falls as the objects separate, because GMm/r gets smaller, picks this negative value. U_g is negative, and −GMm/(3R) is greater than −GMm/R, so ΔU_g is positive.
2. B−0.33 GMm/R
   A student who gives the value of U_g at the end of the move, −GMm/(3R), as the change picks this. A change is the final value minus the initial value, and the initial value at the surface, −GMm/R, is not zero: ΔU_g = −GMm/(3R) − (−GMm/R) = +0.67 GMm/R.
3. C+0.67 GMm/R Correct
   The separation of the centers triples, from R to 3R. ΔU_g = U_f − U_i = −GMm/(3R) − (−GMm/R) = +0.67 GMm/R. It is positive because U_g rises toward zero as the separation increases.
4. D+0.89 GMm/R
   A student who thinks U_g falls off with the square of the separation, like the gravitational force, takes U_g at 3R to be one-ninth of its surface value and gets GMm/R − GMm/(9R) = 0.89 GMm/R. U_g is proportional to 1/r, so at 3R it is one-third of the surface value.

Working Center-to-center separation: initially r = R, finally r = R + 2R = 3R. U_g = −GMm/r. ΔU_g = U_f − U_i = −GMm/(3R) − (−GMm/R) = GMm/R − 0.33 GMm/R = +0.67 GMm/R. Positive: the objects are farther apart. (Reporting U_f alone gives −0.33 GMm/R; ignoring the sign of U_g gives −0.67 GMm/R; an inverse-square U_g gives (1 − 1/9) GMm/R = +0.89 GMm/R.)

CED 3.3.A.4.ii · Read this in Fix

1. AU_g increases as the satellite moves farther from the planet. Correct
   The curve rises toward zero as r increases: U_g goes from −8 × 10⁹ J at R to −4 × 10⁹ J at 2R and −2 × 10⁹ J at 4R. Each value is greater than the one before, so moving the satellite away increases U_g.
2. BU_g decreases as the satellite moves away from the planet.
   A student who looks at the size of U_g and ignores its sign picks this. The magnitude does shrink as r grows, but the values are negative: −2 × 10⁹ J is greater than −8 × 10⁹ J, so U_g increases.
3. CU_g is zero when the satellite is at the surface of the planet.
   A student who expects U_g to be zero at the surface, as when heights are measured from the ground, picks this. The graph shows U_g = −8 × 10⁹ J at r = R. For U_g = −G(m₁m₂)/r the zero is at infinite separation, which the curve approaches but does not reach.
4. DDoubling r reduces the magnitude of U_g to one-quarter.
   A student who expects U_g to follow the inverse-square form of the gravitational force picks this. The graph shows −8 × 10⁹ J at R and −4 × 10⁹ J at 2R: doubling r halves the magnitude, as U_g ∝ 1/r predicts.

CED 3.3.A.4.ii · Read this in Fix

1. A+120 J
   A student who takes ΔU_g to equal the work done by the gravitational force picks this. That force does +120 J of work on the crate as it moves down, and the gravitational potential energy of the system decreases by the same amount: ΔU_g = −120 J.
2. B−120 J Correct
   ΔU_g = mgΔy uses the vertical displacement. The crate ends 3.0 m lower, so Δy = −3.0 m and ΔU_g = (4.0 kg)(10 m/s²)(−3.0 m) = −120 J: the system's gravitational potential energy decreases.
3. C+200 J
   A student who uses the distance traveled along the slope, 5.0 m, in place of the change in height picks this: (4.0 kg)(10 m/s²)(5.0 m) = 200 J. ΔU_g depends only on the vertical displacement, Δy = −3.0 m, not on the length of the path, and a distance carries no sign to show that the crate went down.
4. D−160 J
   A student who uses the horizontal side of the ramp as Δy picks this: (4.0 kg)(10 m/s²)(−4.0 m) = −160 J. Horizontal motion does not change U_g; only the 3.0 m vertical side of the triangle is the change in height.

Working Only the vertical displacement matters: the crate ends 3.0 m lower, so Δy = −3.0 m. ΔU_g = mgΔy = (4.0 kg)(10 m/s²)(−3.0 m) = −120 J. The 5.0 m length of the slope and the 4.0 m base do not enter.

CED 3.3.A.4.iii · Read this in Fix

1. AIt is about 1/6 as large, since g is smaller on the Moon. Correct
   ΔU_g = mgΔy. The rock's mass and the height are the same in both places; only g differs. The Moon's field is g = GM/R² ≈ (1/81)/(0.27)² ≈ 1/6 of Earth's, so ΔU_g for the rock–Moon system is about 1/6 of that for the rock–Earth system.
2. BIt is the same, as the rock's mass and Δy are unchanged.
   A student who treats g as the same everywhere picks this. The g in mgΔy is the gravitational field of the body the rock is near; the Moon's field, GM/R², is about (1/81)/(0.27)² ≈ 1/6 of Earth's, so the same lift changes U_g by about one-sixth as much.
3. CIt is zero, as without air the Moon exerts no gravity.
   A student who thinks gravity needs air picks this. The Moon has almost no atmosphere, but it still attracts the rock: a dropped rock falls on the Moon, only with a smaller acceleration. The Moon has mass, so it has a gravitational field, g = GM/R², whatever its atmosphere. So ΔU_g is smaller than on Earth, not zero.
4. DIt is about 1/36 as large, as the rock's mass is also less there.
   A student who thinks the rock's mass is smaller on the Moon multiplies by 1/6 twice. The rock's mass is the same everywhere; only its weight, mg, is smaller on the Moon. ΔU_g is about 1/6 as large, not 1/36.

Working Near a surface g = GM/R², so g_Moon/g_Earth = (1/81)/(0.27)² = 0.17 ≈ 1/6. ΔU_g = mgΔy with the same m and Δy in both places, so ΔU_g,Moon/ΔU_g,Earth ≈ 1/6.

CED 3.3.A.4.iii · Read this in Fix

1. A+0.43 MgL
   A student who uses the horizontal distance, (L/2) cos 30°, as the rise of the center of mass picks this. Only the vertical rise changes U_g, and for a rod at 30° to the floor the center of mass rises (L/2) sin 30° = L/4.
2. B+0.50 MgL
   A student who uses the rise of the lifted end, L sin 30° = L/2, for the whole rod picks this. The rest of the rod rises less than that end, and the end on the floor does not rise at all; for a uniform rod the rise to use is that of the center of mass at its midpoint, L/4.
3. C+1.00 MgL
   A student who divides by sin 30° instead of multiplying picks this: (L/2) ÷ sin 30° = L. The half-length L/2 along the rod is the hypotenuse of the triangle, so the center of mass rises (L/2) sin 30° = L/4. A rise of L would be twice the rise of the lifted end itself, L sin 30° = L/2.
4. D+0.25 MgL Correct
   The rod's center of mass is at its midpoint, so it rises (L/2) sin 30° = L/4. ΔU_g = Mg(L/4) = +0.25 MgL: the gravitational potential energy of the system increases.

Working The rod is uniform, so its center of mass is at its midpoint, a distance L/2 along the rod from the end that stays on the floor, and the rod can be modeled as an object at its center of mass. With the rod at 30° to the floor, the center of mass is (L/2) sin 30° = L/4 higher than at the start. ΔU_g = MgΔy_cm = Mg(L/4) = +0.25 MgL, positive because the center of mass rises.

CED 3.3.A.4.iii · Read this in Fix