9 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 9
A ball is thrown straight upward. Air resistance is negligible. A student chooses the ball alone as the system. Which statement correctly describes the energy of this system while the ball is rising?
Answer and reasoning
AIts kinetic energy is gradually converted into gravitational potential energy of the ball. A student who thinks a single object has its own gravitational potential energy picks this. Ug belongs to the ball–Earth system; with the ball alone as the system there is no Ug to convert into, and the decrease in K is energy transferred out by gravity's work.
BIts kinetic energy decreases as Earth’s gravitational force does negative work on it.Correct A system of the ball alone can have only kinetic energy. Earth is outside the system, so its gravitational force is external; it points opposite to the ball's upward displacement, does negative work, and transfers energy out of the system, so K decreases.
CIts total energy stays constant, since energy is conserved in every interaction. A student who thinks conservation of energy makes every system's energy constant picks this. Energy is conserved overall, but this system's energy decreases: Earth, outside it, does negative work on the ball.
DIts kinetic energy decreases as the upward force from the throw is used up. A student who thinks the ball carries a force from the throw picks this. Once the ball leaves the hand, no upward force acts on it; the only force is gravity, and it is gravity's negative work that reduces K.
Which of the following systems can have potential energy?
Answer and reasoning
AA block and the ideal spring that it is compressingCorrect An ideal spring changes shape reversibly and pushes back on the block with a conservative force, so the block–spring system has elastic potential energy, Us = (1/2)k(Δx)², as well as kinetic energy.
BA single rock on its own, held at rest high above the ground A student who thinks an object on its own has gravitational potential energy picks this. Ug belongs to the rock–Earth system; a system of the rock alone has only kinetic energy, here zero.
CA crate and the rough floor across which the crate slides A student who thinks friction stores energy as potential energy picks this. Friction is nonconservative: the energy it removes becomes thermal energy and sound, and no potential energy is associated with it.
DA lump of clay on its own, squashed flat by a blow A student who thinks any change of shape stores potential energy picks this. Clay does not spring back; its deformation is permanent, so the energy of the blow is dissipated as thermal energy and sound, not stored.
The bar chart shows the kinetic energy K and the gravitational potential energy Ug of a cart–Earth system at three points, P, Q and R, on a track. Which ranking of the system's mechanical energy at the three points is correct?
Answer and reasoning
AEQ > EP > ER A student who treats a system's energy as its kinetic energy alone picks this, ranking the K bars (5, 2, 1 J). Mechanical energy includes Ug: Q's total, 7 J, is less than P's, 8 J.
BEP > ER > EQ A student who equates mechanical energy with potential energy picks this, ranking the Ug bars (6, 4, 2 J). Mechanical energy adds K: R has 5 J in total and Q has 7 J.
CEP = EQ = ER A student who thinks mechanical energy is always conserved picks this without adding the bars. The totals are 8 J, 7 J and 5 J, so this system's mechanical energy changes along the track.
DEP > EQ > ERCorrect Mechanical energy is the sum of the kinetic and potential energies: 2 + 6 = 8 J at P, 5 + 2 = 7 J at Q, and 1 + 4 = 5 J at R.
Working Mechanical energy = K + Ug: P, 2 J + 6 J = 8 J; Q, 5 J + 2 J = 7 J; R, 1 J + 4 J = 5 J. So EP > EQ > ER.
The diagram shows a block held at rest against a compressed spring at the bottom of a frictionless incline. The block is released, is pushed up the incline by the spring, and leaves the spring. Use g = 10 m/s². How far along the incline from its release point does the block travel before it momentarily stops?
Answer and reasoning
A2.0 m A student who uses the distance along the incline as the rise in height picks this: 4.0 J = mgd gives 2.0 m. That is the vertical rise; the distance along a 30° incline is twice as long.
B8.0 m A student who takes the spring's energy as k(Δx)² = 8.0 J, the largest spring force times the compression, picks this. The force falls to zero as the spring extends, so the stored energy is (1/2)k(Δx)² = 4.0 J.
C4.0 mCorrect The spring stores (1/2)kΔx² = (1/2)(800)(0.10)² = 4.0 J, which all becomes gravitational potential energy of the block–Earth part of the system at the highest point. Moving d along the incline raises the block d sin 30°, so 4.0 = 0.20 × 10 × d × 0.50, giving d = 4.0 m.
D2.3 m A student who finds the rise along the incline with cos 30° picks this: 4.0 J = mgd cos 30° gives 2.3 m. The rise is the side opposite the 30° angle, d sin 30°.
Working System: block, spring and Earth; frictionless and no external work, so mechanical energy is constant. Us initially = (1/2)(800 N/m)(0.10 m)² = 4.0 J; K = 0 at release and at the highest point. After moving a distance d along the incline the block has risen d sin 30°: 4.0 J = (0.20 kg)(10 m/s²)(d)(0.50), so d = 4.0 m.
A block slides down a rough incline that is fixed to the ground, speeding up as it goes. Air resistance is negligible. A student wants to choose a system whose total energy stays constant as the block slides. Which choice of system, with its reason, is correct?
Answer and reasoning
ABlock and Earth: gravity is internal, so the mechanical energy is conserved. A student who thinks mechanical energy is always conserved picks this. The incline is outside this system, and the kinetic friction it exerts on the block does negative work, so energy leaves the block–Earth system as the incline warms.
BBlock and incline: friction is internal, and the block has its own Ug. A student who thinks the block has its own gravitational potential energy picks this. Ug belongs to the block–Earth pair; with Earth outside this system, Earth's gravitational force on the block is external and does positive work, so the system's energy increases.
CIncline and Earth: neither of them moves, so their energy stays the same. A student who thinks energy belongs only to moving objects picks this. The block is outside this system, and kinetic friction between block and incline warms the incline, so this system's energy increases even though nothing in it moves.
DBlock, incline and Earth: friction and gravity are then both internal forces.Correct Every object that exerts a force on the block is then inside the system. Gravity's effect appears as ΔUg, and kinetic friction between block and incline turns mechanical energy into thermal energy within the system. No external force does work, so the total energy is constant.
A person lifts a box of mass m straight up from rest with a constant upward acceleration of magnitude g/2, where g is the magnitude of the gravitational field. Air resistance is negligible. By how much has the total energy of the box–Earth system increased when the box has risen a height h?
Answer and reasoning
A1.0 mgh A student who equates the system's mechanical energy with its potential energy picks this, counting only ΔUg = mgh. The box is also speeding up: v² = 2(g/2)h, so it gains 0.5mgh of kinetic energy as well.
B1.5 mghCorrect The increase equals the energy transferred in by the person's work. The person pushes with F = m(g + g/2) = 1.5mg through h, so W = 1.5mgh. It shows up as ΔUg = mgh plus ΔK = (1/2)m(gh) = 0.5mgh.
C2.0 mgh A student who writes the kinetic energy as mv² picks this: ΔK = m(gh) = mgh, added to ΔUg = mgh. With K = (1/2)mv², ΔK = 0.5mgh and the total is 1.5mgh.
D0.5 mgh A student who thinks the work done on a system all becomes kinetic energy picks this, taking the increase to be ΔK = 0.5mgh. Earth is in the system, so the system also gains ΔUg = mgh.
Working The person's force is the only external force doing work. Newton's second law: F − mg = m(g/2), so F = (3/2)mg, and W = Fh = 1.5mgh. Check by energy: ΔUg = mgh; v² = 2(g/2)h = gh, so ΔK = (1/2)mgh; ΔE = 1.5mgh.
A sled slides across level snow and comes to rest. Which statement about the sled's kinetic energy is correct?
Answer and reasoning
AIt was used up by friction, so that it no longer exists anywhere at all. A student who thinks energy is used up picks this. Energy is never destroyed: the kinetic energy is still there as thermal energy and sound, even though it is spread out and hard to notice.
BIt was turned into the friction force that brought the sled to a stop. A student who treats force and energy as the same kind of thing picks this. Friction is a force (in newtons) exerted by the snow while the sled slides; it transferred the energy by doing negative work but is not itself energy.
CIt became thermal energy of the sled, the snow and the air, and some sound.Correct Every interaction conserves energy. Kinetic friction and air resistance are nonconservative, so they dissipate the sled's kinetic energy as thermal energy of the surfaces and the air, and as sound; the total energy is unchanged.
DIt was stored as potential energy of the system made up of the sled and the snow. A student who thinks friction stores energy as potential energy picks this. Friction is nonconservative, so no potential energy is associated with it; the energy cannot come back as kinetic energy.
A 2.0 kg cart moves along a frictionless track. The graph shows the gravitational potential energy U of the cart–Earth system as a function of the cart's horizontal position x. The cart is released from rest at x = 1.0 m. Air resistance is negligible. What is the cart's speed at x = 3.0 m?
Answer and reasoning
A2.4 m/sCorrect The cart–Earth system's mechanical energy is constant. It starts with U = 8 J and K = 0; at x = 3.0 m, U = 2 J, so K = 6 J. Then (1/2)(2.0)v² = 6, so v = √6 ≈ 2.4 m/s.
B2.8 m/s A student who thinks all the potential energy has become kinetic energy picks this: K = 8 J gives v = √8 ≈ 2.8 m/s. The system still has U = 2 J at x = 3.0 m; only the decrease, 6 J, has become kinetic energy.
C1.7 m/s A student who writes the kinetic energy as mv² picks this: (2.0)v² = 6 J gives √3 ≈ 1.7 m/s. With K = (1/2)mv², v = √6 ≈ 2.4 m/s.
D6.0 m/s A student who treats kinetic energy as proportional to speed picks this: (1/2)(2.0)v = 6 J gives 6.0 m/s. K = (1/2)mv², so the speed is √6 ≈ 2.4 m/s.
Working No external work and no nonconservative forces, so K + U is constant. From the graph U(1.0 m) = 8 J and U(3.0 m) = 2 J. K at 3.0 m = 8 J − 2 J = 6 J = (1/2)(2.0 kg)v², so v = √6 ≈ 2.4 m/s.
A person pushes a box in a straight line across a level floor at a constant speed v for a distance d; the floor exerts a kinetic friction force on the box. The energy the person transfers into the box–floor system is E₁. The person then pushes the same box across the same floor at a constant speed 2v for the same distance d, transferring energy E₂. What is E₂/E₁?
Answer and reasoning
AE₂/E₁ = 2 A student who thinks a larger force is needed to keep an object moving faster picks this. At constant velocity the push only has to balance kinetic friction, which does not depend on speed.
BE₂/E₁ = ½ A student who thinks the energy transferred depends on how long the push lasts picks this, since the second push takes half the time. Work depends on force and displacement, both the same here.
CE₂/E₁ = 4 A student who takes the energy transferred to be the box's kinetic energy, (1/2)mv², picks this: doubling v multiplies it by 4. The box's kinetic energy does not change during either push; the person's work, F d, is dissipated as thermal energy, and it is the same in both.
DE₂/E₁ = 1Correct At constant velocity the net force is zero, so the push equals the kinetic friction force, μk FN, which is the same at any speed. The person's work, F d, is therefore the same, and it is all dissipated as thermal energy and sound.
Working At constant velocity the push equals the kinetic friction force, μk FN = μk mg, which does not depend on speed. Energy transferred = work done by the push = μk mg d in both cases. E₂/E₁ = 1.
In preparation: 0 of 9 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
3.4.A.1 Single-object system Fix
Single-object system
A system made up of one object modeled with no internal structure. Its only energy is kinetic energy; every force on it is external, so an interaction such as gravity appears as work done on the system, not as potential energy.
Translational kinetic energy (K)
Energy a system has because its objects move: K = (1/2)mv² for each object. A scalar that is never negative. SI unit: joule (J).
Students often think A single object has its own gravitational potential energy, so an object on its own stores energy by being high up and turns it into kinetic energy as it falls. In fact No. Gravitational potential energy belongs to a system of objects that attract each other, such as a ball and Earth. A system made up of the ball alone has only kinetic energy; Earth's gravitational force on it is an external force that does work on it.
Students often think A thrown object carries a force from the throw that keeps it moving and is gradually used up, so it slows down as that force runs out. In fact No. Once the ball leaves the hand, the hand exerts no force on it. With air resistance negligible, the only force on the ball is Earth's gravitational force, which slows it on the way up.
3.4.A.2 Potential energy (U) Fix
Potential energy (U)
Energy associated with the configuration of a system whose objects interact through conservative forces or that can change its shape reversibly. It belongs to the system, not to one object. SI unit: J.
Gravitational potential energy (Ug)
Potential energy of a system of objects that attract each other gravitationally. For two approximately spherical masses Ug = −Gm₁m₂/r (zero at infinite separation); for an object near a planet's surface ΔUg = mgΔy. SI unit: J.
Elastic potential energy (Us)
Potential energy of a system containing an ideal spring stretched or compressed by Δx from its equilibrium length: Us = (1/2)k(Δx)². SI unit: J.
Students often think Friction between two objects stores energy as potential energy of the pair, so a system with friction inside it has potential energy, as a system with a spring does. In fact No. Potential energy is associated only with conservative forces, such as gravity and ideal spring forces. Friction is nonconservative: the mechanical energy it removes becomes thermal energy and sound and cannot be fully recovered as kinetic energy.
Students often think Any deformation stores potential energy, whether or not the object can spring back, so a squashed lump of clay stores energy as a compressed spring does. In fact No. Only a shape change that is reversible, so that the object springs back and returns the energy, as an ideal spring does, gives the system potential energy. A permanent change of shape, as when clay is squashed, does not store energy that can be returned as kinetic energy.
3.4.B.1 Mechanical energy Fix
Mechanical energy
The sum of a system's kinetic and potential energies, for example K + Ug + Us. SI unit: J.
Students often think Kinetic energy is proportional to speed, so the square in K = (1/2)mv² can be left out and speed is proportional to the energy that produced it. In fact No. K = (1/2)mv², so kinetic energy depends on the square of the speed: doubling the speed multiplies the kinetic energy by 4, and multiplying the kinetic energy by 4 only doubles the speed.
Students often think Mechanical energy is the energy stored in the system's configuration, its potential energy, so comparing potential energies compares mechanical energies, and kinetic energy can be left out. In fact No. Mechanical energy is the sum of kinetic and potential energies. A system with a large potential energy and little kinetic energy can have less mechanical energy than one with less potential energy and more kinetic energy.
3.4.B.2 Energy conversion Fix
Energy conversion
A change of energy from one type to another inside a system, such as Us to K or Ug to K. A conversion alone leaves the system's total energy unchanged.
Students often think Kinetic energy is mv², the product of mass and the square of speed, without the factor 1/2. In fact No. Translational kinetic energy is K = (1/2)mv². Leaving out the factor 1/2 doubles every kinetic energy and makes every speed found from an energy too small by a factor of √2.
Students often think An object moving in a circle feels an outward (centrifugal) force, so the net force on it points away from the center and the tension is less than the weight at the bottom. In fact No. The bob is moving in a circle, so the net force on it points toward the center of the circle, up toward the pivot, with magnitude mv²/L. The tension is therefore greater than the weight there.
3.4.B.3 System Fix
System
An object or group of objects chosen for analysis; everything else is its surroundings. The choice decides which forces are internal and which are external, which types of energy the system can have, and whether its total energy can change.
System with constant total energy
A system chosen so that no net energy crosses its boundary, typically by including the objects that interact so that the forces between them are internal. Energy can still change type inside it.
Students often think Conservation of energy means that the total energy of any system stays constant, whatever objects are chosen as the system. In fact No. Energy is conserved in all interactions, but the energy of a particular system changes whenever energy is transferred into or out of it, for example by work done by an external force. Only some choices of system have constant total energy.
3.4.B.4 Energy transfer by work Fix
Energy transfer by work
Energy moved into or out of a system when an external force does work on it. Positive work transfers energy in and negative work transfers it out; the change in the system's total energy equals the energy transferred. SI unit: J.
Students often think The work done by a variable force is its largest (or final) value times the distance moved, so the energy stored in a spring is k(Δx)² and the work done by any force is F × d with the final F. In fact No. The work done by a variable force is W = ∫F dx, the area under the graph of F against x. A single value of the force times the distance gives the wrong work: for a force that rises from zero, such as an ideal spring’s, the largest (final) value gives k(Δx)² instead of (1/2)k(Δx)²; for a force that falls, the final value understates the work.
Students often think The work done on a system by an outside force all becomes kinetic energy, so the energy transferred equals the change in kinetic energy, whatever else the system contains. In fact No. Work done on a system changes its total energy, which can include potential energy (and thermal energy). Only for a single object, which has no potential energy, does the net work done on it all go into kinetic energy.
3.4.C.1 Conservation of energy Fix
Conservation of energy
In every interaction, energy is neither created nor destroyed: it changes type within a system or is transferred between a system and its surroundings.
Students often think Energy is used up as objects move or rub against each other: the energy an object loses no longer exists anywhere. In fact No. Every interaction conserves energy. The kinetic energy that seems to disappear becomes thermal energy of the surfaces and the air, and some is carried away as sound.
Students often think Force and energy are the same kind of thing: energy can turn into a force, such as the friction that stops an object, and a force can be stored in an object. In fact No. A force is exerted by one object on another only while they interact; it is not stored and is not a type of energy. A force can transfer energy by doing work, but it is the energy, not the force, that is stored or converted.
3.4.C.2 Internal and external forces Fix
Internal and external forces
Internal forces are exerted on each other by objects inside the system; external forces are exerted on the system's objects by objects outside it. Only work done by external forces transfers energy into or out of the system.
Conservation of mechanical energy
If no work is done on a system by external forces and there are no nonconservative interactions within it, Ki + Ui = Kf + Uf: kinetic and potential energy change only into each other.
Nonconservative interaction
An interaction, such as kinetic friction or air resistance, whose work depends on the path taken and that has no potential energy associated with it. It reduces the mechanical energy of the objects involved.
Students often think Work done by forces between objects inside the system transfers energy into or out of the system, so any change in kinetic energy shows that energy came in or went out, even when it is matched by a change in potential e… In fact No. With Earth in the system, gravity is an internal force; a gain in kinetic energy can come entirely from a decrease in gravitational potential energy, with no energy crossing the boundary. Only work done by external forces transfers energy into or out of the system.
Students often think Mechanical energy is always conserved, so the kinetic plus potential energy of a system stays the same even when friction or another external force acts. In fact No. A system's mechanical energy is constant only if no work is done on it by external forces and there are no nonconservative interactions, such as friction, within it. Otherwise its mechanical energy changes, although energy is still conserved.
3.4.C.3 Dissipated energy Fix
Dissipated energy
Mechanical energy that nonconservative forces turn into thermal energy (of the surfaces and the air) and sound. The energy still exists, but it is no longer kinetic or potential energy.
Students often think A supporting force, such as the normal force or a string's tension, always equals the object's weight mg. In fact No. The normal force takes whatever value the situation needs. On an incline at angle θ, with no other force perpendicular to the surface and no acceleration in that direction, FN = mg cos θ, which is less than mg.
Students often think Zero potential energy means the system has no energy left, or has lost all its energy. In fact No. The zero of potential energy is chosen by whoever analyzes the situation; Ug = 0 at some position says nothing about the total energy there. The system can still have kinetic energy, and Ug can even be negative.
11 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 11
A block of mass m is held against a compressed spring on a frictionless horizontal surface and released. The graph shows the magnitude F of the spring's force on the block as a function of the block's displacement x from its release point, up to the point where the block leaves the spring. Taking the block alone as the system, what is the block's speed when its displacement is D/2?
Answer and reasoning
A1.00√(F₀D/m) A student who keeps the starting acceleration, F₀/m, for the whole displacement picks this: v² = 2(F₀/m)(D/2) = F₀D/m. The spring's force falls as the block moves, so the acceleration falls too; the work is the area under the graph, 3F₀D/8, not F₀(D/2).
B0.71√(F₀D/m) A student who takes the work done by a varying force to be its final value times the distance picks this: W = (F₀/2)(D/2) = F₀D/4, so (1/2)mv² = F₀D/4 and v = √(F₀D/(2m)). The force falls from F₀ to F₀/2 over the interval, so the work is the trapezoidal area, 3F₀D/8.
C0.61√(F₀D/m) A student who writes the kinetic energy as mv² picks this: mv² = 3F₀D/8, so v = √(3F₀D/(8m)). With K = (1/2)mv², the speed is larger by a factor of √2: √(3F₀D/(4m)) = 0.87√(F₀D/m).
D0.87√(F₀D/m)Correct With the block alone as the system, the spring is outside it, and the spring's work is the energy transferred into the system. That work is the area under the graph from 0 to D/2, a trapezoid with parallel sides F₀ and F₀/2: (1/2)(F₀ + F₀/2)(D/2) = 3F₀D/8. All of it becomes kinetic energy: (1/2)mv² = 3F₀D/8, so v = √(3F₀D/(4m)) = 0.87√(F₀D/m).
Working System: block alone, so its only energy is K, and the spring's force is external. From the graph, F = F₀(1 − x/D). Work done by the spring from 0 to D/2 = area under the F(x) graph = ∫₀D/2 F₀(1 − x/D) dx = F₀(D/2) − F₀(D/2)²/(2D) = 3F₀D/8 (a trapezoid with parallel sides F₀ and F₀/2, width D/2). ΔK = W: (1/2)mv² = 3F₀D/8, so v = √(3F₀D/(4m)) = 0.87√(F₀D/m). Distractors: starting acceleration F₀/m kept over D/2 gives v² = 2(F₀/m)(D/2) = F₀D/m, 1.00√(F₀D/m); final force F₀/2 times D/2 gives (1/2)mv² = F₀D/4, 0.71√(F₀D/m); K = mv² gives mv² = 3F₀D/8, 0.61√(F₀D/m).
A block of mass m is released from rest a height h above the top of an ideal vertical spring. The block lands on the spring and compresses it by a maximum distance d before momentarily stopping. Air resistance is negligible, and g is the magnitude of the gravitational field. What is the spring constant of the spring?
Answer and reasoning
A2mgh/d² A student who counts the drop only as far as the top of the spring picks this: mgh = (1/2)kd². The block keeps falling while it compresses the spring, so the system loses mg(h + d) of gravitational potential energy.
Bmg(h+d)/d² A student who takes the energy stored in the spring to be the largest spring force times the compression, kd × d, picks this: mg(h + d) = kd². The spring force rises from zero, so the stored energy is the area (1/2)kd².
C2mg(h+d)/d²Correct Take the block, spring and Earth as the system; its mechanical energy is constant. The block is at rest at release and at the lowest point, and it drops h + d in total, so mg(h + d) = (1/2)kd², giving k = 2mg(h + d)/d².
Dmg/d A student who thinks the block stops where the spring force balances its weight picks this: kd = mg. At that point the block is moving at its greatest speed; it stops only farther down, where the spring force exceeds mg.
Working System: block, spring and Earth; no external work and no nonconservative forces, so its mechanical energy is constant. From release to the lowest point, K = 0 at both, the block drops h + d, so ΔUg = −mg(h + d) and ΔUs = (1/2)kd². 0 = −mg(h + d) + (1/2)kd², so k = 2mg(h + d)/d².
A small bob of mass m on a string is pulled aside to the position shown and released from rest. Air resistance is negligible, and g is the magnitude of the gravitational field. What is the tension in the string as the bob passes through its lowest point?
Answer and reasoning
Amg(3 − 2cos θ)Correct The bob drops L(1 − cos θ), so v² = 2gL(1 − cos θ). At the lowest point the net force points toward the pivot: T − mg = mv²/L = 2mg(1 − cos θ), so T = mg(3 − 2cos θ), greater than mg.
Bmg(2cos θ − 1) A student who thinks the net force on an object in circular motion points outward picks this: mg − T = mv²/L gives T = mg − 2mg(1 − cos θ). The bob's acceleration at the bottom points up, toward the pivot, so T is greater than mg.
C2mg(1 − cos θ) A student who thinks the string alone supplies the centripetal force picks this: T = mv²/L. The net force is mv²/L, and it is made up of the tension upward and the weight downward, so T = mg + mv²/L.
Dmg(3 − 2sin θ) A student who takes the bob's height above the lowest point as L(1 − sin θ) picks this. θ is measured from the vertical, so the vertical side next to it is L cos θ and the drop is L(1 − cos θ).
Working Bob–Earth system, mechanical energy constant (tension ⟂ motion, no work). Drop from release to lowest point: L − L cos θ. (1/2)mv² = mgL(1 − cos θ), so v² = 2gL(1 − cos θ). At the lowest point the net force points up, toward the pivot: T − mg = mv²/L = 2mg(1 − cos θ). T = mg(3 − 2cos θ).
A block on a frictionless horizontal surface is pushed against an ideal spring, compressing it by Δx, and released; it leaves the spring with speed v. A second block, with twice the mass, is pushed against the same spring, compressing it by 3Δx, and released. Its speed on leaving the spring is closest to which of the following?
Answer and reasoning
A1.2v A student who thinks a spring's energy is proportional to its compression picks this: the energy triples, so v² × 3/2 and the speed is √1.5 ≈ 1.2 times v. Us depends on (Δx)², so the energy is 9 times as large.
B2.1vCorrect Us = (1/2)k(Δx)² becomes K = (1/2)mv², so v² ∝ (Δx)²/m. Tripling Δx multiplies the energy by 9 and doubling m halves v² for a given energy: v² × 9/2, so the speed is 3/√2 ≈ 2.1 times v.
C4.5v A student who treats kinetic energy as proportional to speed picks this: the energy is 9 times as large and the mass twice as large, so v × 9/2. With K = (1/2)mv², the speed is the square root of that factor: √4.5 ≈ 2.1.
D3.0v A student who thinks mass always cancels in energy problems picks this: v ∝ Δx gives 3v. The spring supplies a fixed energy, and a heavier block gets a smaller speed from it: v ∝ Δx/√m.
Working (1/2)k(Δx)² = (1/2)mv², so v ∝ Δx/√m. New speed = v × 3/√2 = 2.12v ≈ 2.1v.
A 1.0 kg block on a frictionless horizontal surface is attached to one end of a horizontal spring of spring constant 200 N/m; the other end of the spring is fixed to a wall. The block is at rest with the spring at its equilibrium length when a person starts to pull the block away from the wall. The graph shows the person's force F on the block as a function of the block's displacement x. Taking the block and spring as the system, what is the block's speed at x = 0.20 m?
Answer and reasoning
A2.0 m/sCorrect The person transfers 6.0 J into the system (the area under the graph: 2.0 J for the sloping part and 4.0 J for the flat part). The spring then stores (1/2)(200)(0.20)² = 4.0 J, so K = 2.0 J and v = √(2 × 2.0/1.0) = 2.0 m/s.
B3.5 m/s A student who thinks all the work done on a system becomes kinetic energy picks this: K = 6.0 J gives v = 3.5 m/s. The spring is part of the system, and 4.0 J of the energy transferred is stored as elastic potential energy.
C2.8 m/s A student who takes the person's work to be the largest force times the distance, 40 N × 0.20 m = 8.0 J, picks this: K = 8.0 − 4.0 = 4.0 J. The force rises from zero over the first 0.10 m, so the work is the area under the graph, 6.0 J.
D1.4 m/s A student who writes the kinetic energy as mv² picks this: (1.0)v² = 2.0 J gives 1.4 m/s. With K = (1/2)mv², 2.0 J gives v = 2.0 m/s.
Working Energy transferred into the block–spring system by the person = area under F(x): (1/2)(40 N)(0.10 m) + (40 N)(0.10 m) = 2.0 J + 4.0 J = 6.0 J. The wall does no work (its point of contact does not move). ΔUs = (1/2)(200 N/m)(0.20 m)² = 4.0 J. ΔK = 6.0 J − 4.0 J = 2.0 J = (1/2)(1.0 kg)v², so v = 2.0 m/s.
A probe is launched straight up from the surface of a planet of mass M and radius R. The planet has no atmosphere, and its rotation and the effects of other bodies are negligible. What launch speed must the probe have so that it just reaches a distance 2R from the planet's center?
Answer and reasoning
A√(2GM/R) A student who uses ΔUg = mgΔy with the surface value g = GM/R² for a rise of R picks this: (1/2)mv₀² = m(GM/R²)R. The field weakens as the probe rises, so less energy is needed; Ug = −GMm/r gives √(GM/R).
B√(GM/(2R)) A student who writes the kinetic energy as mv² picks this: mv₀² = GMm/(2R). With K = (1/2)mv₀², the speed is √2 times larger, √(GM/R).
C√(3GM/(2R²)) A student who uses the force law's 1/r² for the potential energy picks this: (1/2)mv₀² = GMm(1/R² − 1/(4R²)). Ug = −GMm/r goes as 1/r; this expression does not even have units of speed.
D√(GM/R)Correct With Ug = −GMm/r, the system's mechanical energy is constant: (1/2)mv₀² − GMm/R = −GMm/(2R). So (1/2)v₀² = GM/(2R) and v₀ = √(GM/R).
Working Probe–planet system: no external work, no nonconservative forces, so K + Ug is constant, with Ug = −GMm/r. (1/2)mv₀² − GMm/R = 0 − GMm/(2R). (1/2)v₀² = GM/R − GM/(2R) = GM/(2R). v₀ = √(GM/R).
A sled slides down a snowy hill. The graph shows the kinetic energy K and the gravitational potential energy Ug of the sled–Earth system as functions of the sled's horizontal position x. A student claims that energy was transferred between the sled–Earth system and its surroundings as the sled moved from x = 0 to x = 40 m. Which statement from the graph correctly supports the claim?
Answer and reasoning
AK rises by 4 kJ between x = 0 and 40 m, so energy was transferred into the system. A student who thinks a gain in kinetic energy shows energy coming in from outside, even though Earth is in the system, picks this. The rise in K comes from the fall in Ug inside the system, and the net transfer is out of the system, not into it: K + Ug falls by 2 kJ.
BUg falls by 6 kJ between x = 0 and 40 m, so the mechanical energy fell by 6 kJ. A student who equates mechanical energy with potential energy picks this. Mechanical energy also includes K, which rose by 4 kJ, so the fall is only 2 kJ; a fall in Ug alone could all have become kinetic energy.
CUg falls by 6 kJ, but K rises by only 4 kJ, so K + Ug is 2 kJ lower at 40 m.Correct If no energy crossed the boundary of the sled–Earth system and nothing inside it were nonconservative, K + Ug would be constant. It falls from 6 kJ to 4 kJ, so 2 kJ left the system: the snow and the air did negative work on the sled, and the energy became thermal energy and sound.
DUg is 0 kJ at x = 40 m, which shows that the sled–Earth system has no energy left there. A student who reads zero potential energy as 'no energy' picks this. The zero of Ug is a chosen reference level, and the system still has 4 kJ of kinetic energy at x = 40 m.
Working From the graph: at x = 0, K + Ug = 0 + 6 kJ = 6 kJ; at x = 40 m, K + Ug = 4 kJ + 0 = 4 kJ. Mechanical energy fell by 2 kJ, so with no nonconservative interactions inside the sled–Earth system, 2 kJ was transferred out by the negative work of the snow and air.
Three identical balls are thrown with the same speed from the same point at the top of a cliff: ball U straight up, ball H horizontally, and ball D straight down. Air resistance is negligible. Which ranking of the speeds vU, vH and vD with which the balls reach the level ground below is correct?
Answer and reasoning
AvD > vH > vU A student who thinks a downward throw adds to the effect of gravity picks this. The direction of the throw changes the path and the time of flight, not the energy: all three balls start with the same K and drop the same height.
BvD = vH = vUCorrect Each ball–Earth system has constant mechanical energy. All three start with the same kinetic energy (the same speed; energy is a scalar, so direction does not matter) and the same Ug, and end at the same height, so they hit the ground with the same speed.
CvU > vH > vD A student who thinks the speed gained depends on the length of the path picks this, since ball U travels farthest. With no nonconservative forces, the speed depends only on the net change in height, which is the same for all three.
DvD = vU > vH A student who takes the impact speed to be the vertical speed alone picks this: balls U and D reach the ground with a larger vertical velocity than H. Speed includes the horizontal component; H keeps its horizontal velocity, and its total speed equals the others'.
Working For each ball–Earth system, (1/2)mvf² = (1/2)mv₀² + mg|Δy| with the same v₀ and the same |Δy|, so the three final speeds are equal.
A 2.0 kg block is released from rest and slides 5.0 m down a rough incline that makes an angle of 37° with the horizontal. The coefficient of kinetic friction between the block and the incline is 0.25. Air resistance is negligible. Use g = 10 m/s², sin 37° = 0.60 and cos 37° = 0.80. What is the block's speed after it has slid 5.0 m?
Answer and reasoning
A7.7 m/s A student who thinks mechanical energy is always conserved picks this, converting all 60 J of Ug into K. Friction does −20 J of work on the block, so only 40 J becomes kinetic energy.
B5.9 m/s A student who takes the normal force to be mg = 20 N picks this: friction 5.0 N does −25 J, leaving K = 35 J. On the incline FN = mg cos 37° = 16 N, so friction is 4.0 N.
C6.3 m/sCorrect The block–Earth system loses mg(5.0 × 0.60) = 60 J of Ug. Kinetic friction, 0.25 × mg cos 37° = 4.0 N, does −20 J of work on it over 5.0 m, transferring that energy out (it is dissipated as thermal energy and sound). So K = 40 J and v = √(2 × 40/2.0) ≈ 6.3 m/s.
D8.9 m/s A student who uses the 5.0 m slid along the incline as the drop in height picks this: ΔUg = −100 J and K = 80 J. The drop is 5.0 m × sin 37° = 3.0 m.
Working System: block and Earth; the incline is outside, and its friction does negative work. Drop: (5.0 m)(0.60) = 3.0 m, so ΔUg = −(2.0)(10)(3.0) = −60 J. FN = mg cos 37° = 16 N; Ff = 0.25 × 16 N = 4.0 N; Wf = −(4.0 N)(5.0 m) = −20 J. ΔK = 60 J − 20 J = 40 J = (1/2)(2.0 kg)v², v = √40 ≈ 6.3 m/s.
A small block of mass m is released from rest at a height h above the bottom of a vertical circular loop of radius R and slides without friction along the track and around the inside of the loop. Air resistance is negligible, g is the magnitude of the gravitational field, and h is large enough that the block stays in contact with the track. What is the magnitude of the normal force exerted on the block by the track when the block is at the top of the loop?
Answer and reasoning
Amg(2h/R − 5)Correct The block–Earth system’s mechanical energy is constant, and the top is 2R above the bottom, so v² = 2g(h − 2R). At the top the weight and the normal force both point toward the center: N + mg = mv²/R, so N = 2mg(h − 2R)/R − mg = mg(2h/R − 5).
Bmg(2h/R − 4) A student who thinks the track’s force alone supplies the centripetal force, so the weight plays no part, picks this: N = mv²/R = 2mg(h − 2R)/R. At the top the weight also points toward the center and supplies part of mv²/R, so N is smaller by mg.
Cmg(2h/R − 3) A student who uses the bottom-of-the-loop result N = mg + mv²/R at the top picks this: N = 2mg(h − 2R)/R + mg = mg(2h/R − 3). At the top the track pushes down, the same way as the weight, so N + mg = mv²/R.
Dmg(2h/R − 1) A student who thinks all of the starting potential energy has become kinetic energy at the top picks this: v² = 2gh, so N = 2mgh/R − mg. The block is still 2R above the bottom at the top, so only mg(h − 2R) has become kinetic energy.
Working System: block and Earth. The normal force is perpendicular to the motion, so it does no work, and there is no friction: mechanical energy is constant. The top of the loop is 2R above the bottom: (1/2)mv² = mg(h − 2R), so v² = 2g(h − 2R). At the top both the weight and the normal force point down, toward the center: N + mg = mv²/R = 2mg(h − 2R)/R. N = mg(2h/R − 4) − mg = mg(2h/R − 5). Distractors: N = mv²/R gives mg(2h/R − 4); the bottom-of-the-loop result N = mg + mv²/R gives mg(2h/R − 3); v² = 2gh (no Ug at the top) gives mg(2h/R − 1). Checked with sympy.
A uniform flexible rope of mass M and length L lies straight on a frictionless horizontal table, perpendicular to the table’s edge, with one-quarter of its length hanging vertically over the smooth edge. The rope is released from rest and slides off the table. Air resistance is negligible, and g is the magnitude of the gravitational field. What is the speed of the rope at the instant its upper end leaves the table?
Answer and reasoning
A0.68√(gL) A student who writes the rope’s kinetic energy as Mv² picks this: Mv² = 15MgL/32, so v² = 15gL/32. Kinetic energy is (1/2)Mv², so the speed is larger by a factor of √2.
B0.97√(gL)Correct Using centers of mass, the rope–Earth system’s Ug falls from −(M/4)g(L/8) = −MgL/32 to −MgL/2, a decrease of 15MgL/32. This equals (1/2)Mv², so v² = 15gL/16 and v = 0.97√(gL).
C1.22√(gL) A student who finds the work done by a varying force as its final value times the distance picks this: Mg(3L/4) = (1/2)Mv², so v² = 3gL/2. The pull of gravity on the hanging part grows from Mg/4 to Mg as the rope slides, so the work done is less than Mg(3L/4).
D0.61√(gL) A student who keeps the starting acceleration g/4 for the whole motion picks this: v² = 2(g/4)(3L/4) = 3gL/8. The hanging part, and so the net force and the acceleration, grow as the rope slides, so the rope ends up faster than this.
Working System: rope and Earth; the table’s normal force does no work and there is no friction, so K + Ug is constant. Take Ug = 0 at the table top. Initially the hanging quarter (mass M/4) has its center of mass L/8 below the table: Ui = −(M/4)g(L/8) = −MgL/32. Finally the whole rope hangs with its center of mass L/2 below the table: Uf = −MgL/2. (1/2)Mv² = Ui − Uf = 15MgL/32, so v² = 15gL/16 and v = √(15/16)√(gL) = 0.97√(gL). Check by integration: W = ∫03L/4 (Mg(L/4 + s)/L) ds = 15MgL/32, the same. Distractors: K = Mv² gives v² = 15gL/32, v = 0.68√(gL); the final pull Mg times the distance 3L/4 gives v² = 3gL/2, v = 1.22√(gL); the starting acceleration g/4 kept over 3L/4 gives v² = 3gL/8, v = 0.61√(gL). Checked with sympy.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
