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 accounts for the decrease in the kinetic energy of this system while the ball rises?
Answer and reasoning
AThe ball's kinetic energy is converted into gravitational potential energy stored in the ball. A student who thinks a single object can store gravitational potential energy picks this. Ug belongs to the ball–Earth system. With the ball alone as the system there is no potential energy, so the decrease in K must be energy transferred out by an external force.
BEarth's gravitational force does negative work, moving energy out of the system.Correct The system is the ball alone, so it can have only kinetic energy, and Earth's gravitational force on the ball is external. While the ball rises, that force points opposite to the ball's displacement, so it does negative work: energy is transferred out of the system, and K decreases by exactly that amount.
CThe upward force that the throw gave the ball is gradually used up as the ball rises. A student who thinks a thrown ball carries the force of the throw with it picks this. The hand stops exerting a force when the ball leaves it. The only force on the rising ball is Earth's gravitational force, which does negative work on it.
DThe ball's kinetic energy is converted into the downward force of gravity on the ball. A student who treats force and energy as the same kind of thing picks this. Gravity is a force (in N), not an energy (in J), and it acts on the ball whether the ball rises or falls; K decreases because this force does negative work on the ball.
A cart moves along a level track with negligible friction and hits a barrier fixed at the end of the track. In trial 1 the barrier is an ideal spring; in trial 2 it is a block of soft clay, which stays squashed. The system is the cart and the barrier. Which statement correctly compares the system in the two trials at the instant the cart first comes to rest?
Answer and reasoning
AOnly in trial 1 does the system have potential energy, as only the spring can spring back.Correct An ideal spring changes shape reversibly, so in trial 1 the cart's kinetic energy is stored as Us, which the spring returns as it pushes the cart back. The clay stays squashed: in trial 2 the cart's kinetic energy has been dissipated as thermal energy of the clay and cart, and some sound, not stored as potential energy.
BThe system has potential energy in both trials, as each barrier has been squashed by the cart. A student who thinks every squashed object stores elastic potential energy picks this. Potential energy is stored only in a reversible change of shape. The clay does not push the cart back, because its energy has become thermal energy.
CThe system has no energy in either trial, as the cart is at rest at that instant. A student who links energy only to motion picks this. At that instant the cart's K is zero, but in trial 1 the spring stores Us, and in trial 2 the clay and cart have gained thermal energy.
DOnly in trial 1 does the system have any energy; in trial 2 the clay has used it up. A student who thinks energy can be used up picks this. Energy is conserved in all interactions: in trial 2 the cart's kinetic energy has become thermal energy of the clay and cart, and a little sound; it has not disappeared.
Working Trial 1: the ideal spring deforms reversibly, so at the instant the cart stops, the cart's initial kinetic energy is stored as Us in the cart–spring system. Trial 2: the clay deforms irreversibly and does not push back; the cart's kinetic energy has become thermal energy of the clay and cart (with a little sound), and the system has no potential energy.
A 2.0 kg ball is moving at 4.0 m/s at a point 3.0 m above the floor. A student defines Ug = 0 at the height of a tabletop 1.0 m above the floor. Use g = 10 m/s². What is the mechanical energy of the ball–Earth system when the ball is at that point?
Answer and reasoning
A76 J A student who measures Ug from the floor, ignoring the stated zero, uses 3.0 m, finds Ug = 60 J and gets a total of 76 J. The zero is at the tabletop, so the ball is 2.0 m above it.
B72 J A student who leaves out the ½ in K = (1/2)mv² finds K = 32 J and a total of 72 J. With the ½, K = 16 J.
C20 J A student who uses the mass in place of the weight finds Ug = (2.0)(2.0) = 4.0 and a total of 20 J. Ug = mgy; leaving out g = 10 m/s² makes it 10 times too small.
D56 JCorrect Mechanical energy is K + Ug. K = (1/2)(2.0 kg)(4.0 m/s)² = 16 J. Measured from the chosen zero, the ball is 2.0 m up, so Ug = (2.0 kg)(10 m/s²)(2.0 m) = 40 J. The sum is 56 J.
Working K = (1/2)mv² = (1/2)(2.0 kg)(4.0 m/s)² = 16 J. The ball is 3.0 m − 1.0 m = 2.0 m above the chosen zero, so Ug = mgy = (2.0 kg)(10 m/s²)(2.0 m) = 40 J. Mechanical energy = K + Ug = 16 J + 40 J = 56 J.
A pendulum bob is released from rest at point X, as shown in the diagram. Point X is 0.80 m above the lowest point Y of the bob's path, and point Z is 0.35 m above Y. Air resistance is negligible. Use g = 10 m/s². What is the speed of the bob as it passes point Z?
Answer and reasoning
A3.0 m/sCorrect The string pulls perpendicular to the bob's motion, so it does no work, and the bob–Earth system's mechanical energy is constant. From X to Z the bob drops 0.80 m − 0.35 m = 0.45 m, so (1/2)mv² = mg(0.45 m) and v = √(2 × 10 × 0.45) m/s = 3.0 m/s.
B2.1 m/s A student who leaves out the ½ in the kinetic energy writes mv² = mg(0.45 m) and gets v = √4.5 m/s = 2.1 m/s. With K = (1/2)mv², the speed is √2 times larger.
C9.0 m/s A student who treats kinetic energy as proportional to speed writes (1/2)mv = mg(0.45 m) and gets 2 × 10 × 0.45 = 9.0. The speed comes from a square root, v = √(2g|Δy|).
D4.0 m/s A student who uses the whole height of X above Y finds the speed at Y: √(2 × 10 × 0.80) m/s = 4.0 m/s. At Z the bob has dropped only 0.45 m, the change in height between X and Z.
Working System: bob and Earth. The string's force is perpendicular to the bob's motion, so it does no work, and air resistance is negligible, so mechanical energy is constant. From X to Z the bob drops 0.80 m − 0.35 m = 0.45 m: (1/2)mv² = mg(0.45 m), v = √(2 × 10 m/s² × 0.45 m) = √(9.0 m²/s²) = 3.0 m/s.
A cart attached to a horizontal ideal spring oscillates back and forth along a level track with negligible friction. The other end of the spring is fixed to a wall. Which choice of system has a total energy that stays constant during the motion?
Answer and reasoning
AThe cart alone, as the spring's force on it is a conservative force. A student who thinks a conservative force keeps the energy of the object it acts on constant picks this. For the cart alone, the spring's force is external and does work on it, so the cart's energy rises and falls as it speeds up and slows down.
BAny choice of system, as energy is conserved in every interaction. A student who reads conservation of energy as 'every system's energy is constant' picks this. The cart alone gains and loses kinetic energy as the spring does work on it; only a system on which no net work is done has constant energy.
CNo choice of system, as the energy is zero whenever the cart stops. A student who links energy only to motion picks this. When the cart stops at each end of its motion, K is zero, but the spring is stretched or compressed, so the cart–spring system still has all its energy, as Us.
DThe cart and the spring, as the spring's force is then internal.Correct With the cart and the spring as the system, the spring's force on the cart is internal. The wall's force on the spring does no work because that end of the spring does not move, and gravity and the normal force are perpendicular to the motion. No work is done on the system, so its energy is constant, moving between K and Us.
A student lifts a box of mass m vertically, starting from rest. When the box is a height h above its starting point, it is moving upward with speed v. Air resistance is negligible. Which expression gives the energy transferred to the box–Earth system by the student's lifting force up to that instant?
Answer and reasoning
Amgh + ½mv²Correct The box–Earth system gains kinetic energy (1/2)mv² and gravitational potential energy mgh. Gravity is internal to this system, so the only external force doing work is the student's: the energy it transfers equals the total gain, mgh + (1/2)mv².
Bmgh + mv² A student who drops the ½ from the kinetic energy picks this. The box's kinetic energy is (1/2)mv², so the energy transferred is mgh + (1/2)mv².
Cgh + ½v² A student who cancels the mass because it cancels for a falling object picks this. Here nothing on the other side of the equation contains m to cancel it; gh + (1/2)v² has units of J/kg, not J.
D2mgh + ½mv² A student who counts gravity twice picks this: the increase in Ug, and also gravity's negative work, −mgh, treated as energy removed that the student must replace. Earth is in the system, so gravity's effect is counted once, as ΔUg = mgh.
Working System: box and Earth. Gravity is internal; the only external force that does work is the student's lifting force. Energy transferred = change in the system's energy = ΔK + ΔUg = (1/2)mv² + mgh.
A student releases a pendulum and uses sensor data to find the mechanical energy of the pendulum–Earth system at several times. The graph shows the results. The student claims: “The data show that energy is not conserved.” Which response to the claim is correct?
Answer and reasoning
AIt is right: air resistance and friction have used up some of the pendulum's energy. A student who thinks energy can be used up picks this. The pendulum–Earth system does lose mechanical energy, but that energy still exists as thermal energy of the air and pivot, and as sound; the total energy of the pendulum and its surroundings is unchanged.
BIt is wrong: mechanical energy stays constant, so the data must contain errors. A student who treats 'mechanical energy is conserved' as a law for every situation picks this. Mechanical energy is constant only if there are no nonconservative interactions; air resistance and pivot friction are present, so the steady fall in the data is real.
CIt is wrong: the lost mechanical energy was dissipated as thermal energy and sound.Correct The graph shows the mechanical energy falling steadily, from 0.40 J to 0.28 J in 50 s. Energy is conserved in all interactions: air resistance and friction at the pivot are nonconservative, and the mechanical energy they remove becomes thermal energy of the air and pivot, and sound.
DIt is wrong: the missing mechanical energy is stored in the string as its tension force. A student who thinks a force can store energy picks this. Tension is a force, not an energy, and a string that does not stretch stores no energy. The mechanical energy lost has been dissipated as thermal energy and sound.
A block of mass m is released from rest and slides a distance L down a ramp inclined at angle θ above the horizontal. Friction is negligible. Which expression gives the speed of the block after it has slid the distance L?
Answer and reasoning
A√(2gL cos θ) A student who finds the drop in height with cos θ picks this. L is the hypotenuse of the ramp's triangle, and the vertical side is opposite θ, so the drop is L sin θ; L cos θ is the horizontal distance covered.
B√(4gL sin θ) A student who counts gravity twice, adding gravity's work mgL sin θ to the decrease in Ug, writes (1/2)mv² = 2mgL sin θ and picks this. For the block–Earth system gravity is internal, so its effect is counted once, as the decrease in Ug.
C(2gL sin θ)² A student who squares instead of taking the square root turns v² = 2gL sin θ into this. The energy equation gives v², so v is its square root; this expression does not even have units of speed.
D√(2gL sin θ)Correct The block drops a vertical height L sin θ. With no friction, and the normal force perpendicular to the motion, the block–Earth system's mechanical energy is constant: (1/2)mv² = mgL sin θ, so v = √(2gL sin θ).
Working Vertical drop: L sin θ. For the block–Earth system no external force does work (the normal force is perpendicular to the motion) and there is no friction, so (1/2)mv² = mgL sin θ and v = √(2gL sin θ).
A 4.0 kg block slides across a rough, level floor, and its speed decreases from 5.0 m/s to 3.0 m/s. The system is the block alone. What is the change in the total energy of the system?
Answer and reasoning
A+32 J A student who ignores the sign of an energy change picks this. The block ends with less energy than it started with, so the change is negative: 32 J has been transferred out of the system.
B−32 JCorrect The system is a single object, so its only energy is kinetic: ΔE = (1/2)(4.0 kg)(3.0 m/s)² − (1/2)(4.0 kg)(5.0 m/s)² = 18 J − 50 J = −32 J. The negative sign shows that friction, an external force, has done negative work and transferred 32 J out of the system.
C−64 J A student who leaves out the ½ in K = (1/2)mv² finds kinetic energies of 100 J and 36 J and a change of −64 J. With the ½ they are 50 J and 18 J, and the change is −32 J.
D+18 J A student who gives the block's kinetic energy at the end, (1/2)(4.0 kg)(3.0 m/s)² = 18 J, as the change picks this. A change is the final energy minus the initial energy: 18 J − 50 J = −32 J.
Working A single-object system has only kinetic energy. ΔE = ΔK = (1/2)(4.0 kg)(3.0 m/s)² − (1/2)(4.0 kg)(5.0 m/s)² = 18 J − 50 J = −32 J. The floor's friction force is external and does −32 J of work on the block, so 32 J is transferred out of the system, where it appears as thermal energy (and a little sound).
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 containing one object modeled as having 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.
Kinetic energy (K)
Energy a system has because its objects are moving: 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 converts 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 can have only kinetic energy; Earth's gravitational force on the ball is then an external force that does work on it.
Students often think A thrown object carries the force of the throw with it, and this force is gradually used up as the object moves, which is why it slows down. In fact No. The hand exerts a force on the ball only while they are in contact. After release, the only force on the ball (air resistance negligible) is Earth's gravitational force, which slows the ball as it rises.
3.4.A.2 Potential energy (U) Fix
Potential energy (U)
Energy associated with the configuration (relative positions or shape) of objects in a system that interact through conservative forces or that change shape reversibly. It belongs to the system, not to any one object. SI unit: J.
Gravitational potential energy (Ug)
The potential energy of a system of an object and Earth. Near Earth's surface, ΔUg = mgΔy; the zero level is chosen by whoever analyzes the situation. SI unit: J.
Elastic potential energy (Us)
The 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 The elastic potential energy of a spring is proportional to its compression or stretch, so doubling Δx doubles the energy stored. In fact No. Us = (1/2)k(Δx)², so the energy depends on the square of the compression or stretch: doubling Δx multiplies Us by 4.
Students often think Energy is associated only with motion or activity, so a system whose objects are momentarily at rest has no energy at that instant. In fact No. At that instant the kinetic energy is zero, but the system can have potential energy, for example in a compressed spring, or thermal energy, so its total energy need not be zero.
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 (or the drop in height) 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 energy by 4 only doubles the speed.
Students often think The numerical factor ½ in the kinetic or elastic potential energy can be dropped, for example by writing K = mv². In fact No. The ½ is part of each expression. Leaving it out of one term changes the result: dropping it from K alone makes a calculated speed too small by a factor of √2.
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. Conversions alone leave the system's total energy unchanged.
Energy bar chart
A representation of a system's energies at two instants, with a bar for each type (K, Ug, Us) and a bar W for the work done on the system by external forces in between. The bars balance: initial energies + W = final energies, plus any energy dissipated.
Students often think Every projectile is momentarily at rest at the highest point of its path, so its kinetic energy there is zero. In fact Only if it was launched straight up. A projectile launched at an angle above the horizontal still moves horizontally at the top of its path, so its speed and its kinetic energy there are not zero.
Students often think The mass of an object never affects the result of an energy calculation, because it always cancels, as it does for a falling object. In fact No. Mass cancels only when every term in the energy equation contains the same mass, as for a freely falling object. When a spring's energy lifts a ball, the spring's energy does not depend on the ball's mass, so a heavier ball rises less high.
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, and so 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 energy is transferred across its boundary, for example by including both interacting objects so that the forces between them are internal. Energy can still change type within it.
Students often think Conservation of energy means that the total energy of any system is 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 by work. Only some choices of system have a constant total energy.
Students often think A conservative force conserves the energy of any object it acts on, so a system consisting of that object alone has constant energy. In fact No. A spring force does work on a cart and changes its kinetic energy. The energy is constant only for a system containing both interacting objects, such as the cart and the spring, so that the spring's force is internal and the energy moves between K and Us.
3.4.B.4 Energy transfer by work Fix
Energy transfer by work
Energy moved into or out of a system by an external force doing work on it. Positive work transfers energy in, negative work transfers it out, and the change in the system's total energy equals the energy transferred. SI unit: J.
Students often think A height or energy value at a single point, such as the height of the start above the lowest point or the final value of Ug, can be used in place of the change between the two points being compared. In fact No. Energy equations use changes: ΔUg = mgΔy depends on the change in height between the two points being compared, and the energy transferred equals the change in the system's energy, not its value at the end.
Students often think The energy transferred to a system equals the change in its kinetic energy, so if the speed of its objects is constant, no energy is transferred and the system's energy does not change. In fact No. A system's energy can change with no change in speed, for example when Ug increases as a crate is lifted at constant speed, or when thermal energy increases as a box is pushed across a rough floor. The energy transferred equals the change in the system's total energy, not just in its 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 as friction and air resistance act: the energy an object loses no longer exists anywhere. In fact No. Every interaction conserves energy. The mechanical energy that seems to disappear becomes thermal energy of the surfaces or the air, and some is carried away as sound.
Students often think Force and energy are the same kind of thing: a force can be stored in an object or turn into energy, and energy can turn into a force such as gravity or tension. 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.
Nonconservative interaction
An interaction, such as kinetic friction or air resistance, whose work depends on the path and which converts mechanical energy into thermal energy and sound. No potential energy is associated with it.
Students often think Work done by forces between objects inside the system transfers energy into or out of the system, so it is counted in addition to the energy changes it produces, for example counting both the work done by gravity and th… In fact No. Forces that objects in the system exert on each other are internal; their effect is already accounted for as changes in the system's potential or thermal energy. Only work done by external forces transfers energy across the system boundary.
Students often think Mechanical energy is always conserved, so the kinetic plus potential energy of a system stays the same even when friction, air resistance or an external push 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 or air resistance, within it. Otherwise its mechanical energy changes, even though energy is still conserved.
3.4.C.3 Surroundings (environment) Fix
Surroundings (environment)
Everything outside the chosen system. Work done on the system by objects in the surroundings transfers energy between the two.
Thermal energy
Energy associated with the random motion of the particles that make up objects. Mechanical energy dissipated by friction or air resistance appears as thermal energy of the surfaces or the air (and as sound). SI unit: J.
Students often think Work and changes in energy are simply amounts, always positive, so the sign, and with it the direction of the transfer, need not be tracked. In fact Yes. A positive change means energy has been transferred into the system; a negative change means energy has been transferred out. A block that slows to rest on a rough floor has a negative change in energy, because energy leaves it.
8 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 8
A ball is released from rest and falls toward the ground. Air resistance is negligible. A student chooses the ball alone as the system and draws the energy bar chart shown, for the instant of release and the instant just before the ball reaches the ground. Which statement about the student's chart is correct?
Answer and reasoning
AThe chart is correct, as the ball's own Ug has all been converted into K by the end. A student who thinks a single object can have gravitational potential energy picks this. The chart would be right for the ball–Earth system, but the student chose the ball alone; for that system the energy comes in as work done by gravity.
BA W bar for the work done by gravity should be added, and the Ug bar kept too. A student who counts gravity twice picks this. Gravity is shown either as work (Earth outside the system) or as Ug (Earth inside), never both. With both, the chart could not balance: 3 units of Ug plus 3 units of work cannot equal 3 units of K.
CThe final K bar should be drawn below zero, as the ball is moving downward. A student who gives kinetic energy the sign of the velocity picks this. K = (1/2)mv² depends on the square of the speed, so it is never negative, whichever way the ball moves.
DThe Ug bar should be replaced by a positive W bar, for the work of gravity.Correct With the ball alone as the system, Earth is outside it, so the system has no Ug. Earth's gravitational force is external and does positive work on the falling ball. The chart should show no initial energy, a W bar of 3 units, and a final K bar of 3 units.
A block of mass m slides with speed v along a horizontal surface with negligible friction and runs into an ideal spring. The block is momentarily at rest when the spring is compressed by a distance Δx. The system is the block and the spring. Which expression gives the spring constant k of the spring?
Answer and reasoning
Amv²/Δx A student who takes the spring's energy to be proportional to Δx writes (1/2)mv² = (1/2)kΔx and gets this. The energy stored in a spring depends on the square of the compression, Us = (1/2)k(Δx)².
B2mv²/Δx² A student who drops the ½ from the kinetic energy writes mv² = (1/2)k(Δx)² and gets this. Both energies carry a factor ½, and they cancel.
Cmv²/Δx²Correct The block's kinetic energy (1/2)mv² is stored as the spring's elastic potential energy (1/2)k(Δx)² when the block is momentarily at rest. The factors of ½ cancel, so k = mv²/Δx².
Dmv²/(2Δx) A student who equates the spring's force with the block's energy writes kΔx = (1/2)mv² and gets this. kΔx is a force, in N; the energy stored in the spring is (1/2)k(Δx)², in J.
Working No external force does work on the block–spring system: the wall holding the spring does not move, and gravity and the normal force are perpendicular to the motion. With no friction, mechanical energy is constant: (1/2)mv² = (1/2)k(Δx)², so k = mv²/Δx².
A ball is launched from level ground at an angle above the horizontal. Air resistance is negligible. A student draws the energy bar chart shown for the ball–Earth system, from the instant of launch to the instant the ball is at the highest point of its path. Which statement about the chart is correct?
Answer and reasoning
AThe chart is correct, as the ball is momentarily at rest at the top of its path. A student who thinks every projectile stops at the top picks this. That is true only for a ball thrown straight up. This ball keeps moving horizontally, so its kinetic energy at the top is not zero.
BA W bar for the negative work done by gravity should be added to the chart. A student who treats gravity as an external force here picks this. Earth is in the system, so gravity is internal, and its effect is already shown by the increase in Ug; a work bar would count it twice.
CThe final K bar is wrong: the ball still moves horizontally at the top.Correct At the highest point only the vertical component of the velocity is zero; the horizontal component keeps its launch value. The ball still has kinetic energy there, so the final K bar should be above zero and the Ug bar shorter than 4 units, with the total still 4 units.
DThe final bars should add to less than the initial bars, as energy is used up. A student who thinks energy is used up as the ball rises picks this. With air resistance negligible, no work is done on the ball–Earth system from outside, so its total energy at the top equals its total at launch.
A spring launcher points straight up. When its ideal spring is compressed by Δx, it launches a ball of mass m that rises to a maximum height H above its release point. The spring is then compressed by 2Δx and used to launch a ball of mass 2m. Air resistance is negligible. How high above its release point does the second ball rise?
Answer and reasoning
AH, as doubling Δx doubles the stored energy, which the doubled mass cancels A student who takes the spring's energy to be proportional to Δx thinks doubling the compression doubles the stored energy, which the doubled mass then cancels. Us depends on (Δx)², so the stored energy is four times as large.
B2H, as the spring stores four times the energy but the mass is doubledCorrect For the ball–spring–Earth system, (1/2)k(Δx)² = mgH, so H = k(Δx)²/(2mg). Doubling Δx multiplies the stored energy by 4, and doubling the mass halves the height a given energy can lift: 4 ÷ 2 = 2, so the ball rises 2H.
C4H, as the stored energy quadruples and the ball's mass cancels out A student who thinks mass always cancels in energy problems gets 4H. Mass cancels for a falling object because it appears in every term; here the spring's energy does not depend on the ball's mass, so a heavier ball rises less high.
D8H, as the quadrupled energy is doubled again by the doubled mass A student who multiplies by the mass factor instead of dividing gets 4 × 2 = 8H. A heavier ball needs more energy for each meter it rises, so doubling the mass halves the height: 4 ÷ 2 = 2.
Working System: ball, spring and Earth; no external work and no friction, so (1/2)k(Δx)² = mgH and H = k(Δx)²/(2mg). With 2Δx and 2m: H' = k(2Δx)²/(2(2m)g) = (4/2)H = 2H.
A rope lifts a crate vertically at constant speed. The energy bar chart shown gives the kinetic energy and the gravitational potential energy of the crate–Earth system at two instants during the lift. The bar for the work W done on the system by the rope is left blank. What value of W completes the chart?
Answer and reasoning
A0 J, as the crate's kinetic energy does not change at all A student who thinks no energy is transferred when the speed is constant picks this. The kinetic energy is unchanged, but Ug rises from 20 J to 80 J, and that energy must come from the rope's work.
B+60 J, the increase in the total energy of the crate–Earth systemCorrect The chart must balance: initial energies + W = final energies, so 10 J + 20 J + W = 10 J + 80 J and W = +60 J. The rope is the only external force that does work on the crate–Earth system, and the energy it transfers appears as the 60 J increase in Ug.
C+120 J, the rise in Ug plus the 60 J that gravity takes out A student who counts gravity twice, once as the 60 J rise in Ug and again as −60 J of work done by gravity, writes 10 J + 20 J + W − 60 J = 10 J + 80 J and gets W = +120 J. Earth is inside the crate–Earth system, so gravity's effect appears only as the rise in Ug; the rope transfers +60 J.
D+80 J, the gravitational potential energy at the end of the lift A student who reads the final Ug bar instead of the change picks this. The system already had 20 J of Ug at the start, so the rope transfers only the increase, 80 J − 20 J = 60 J.
Working The chart balances: Ki + Ugi + W = Kf + Ugf, so 10 J + 20 J + W = 10 J + 80 J and W = +60 J.
Blocks 1 and 2 are released from rest at the same height on the two tracks shown in the diagram, and each slides down to the finish level, a height h below the start. Block 1 has twice the mass of block 2. Friction and air resistance are negligible. How do the speeds of the blocks at the finish level compare?
Answer and reasoning
AThe two speeds are equal, as blocks 1 and 2 drop the same height h.Correct For each block–Earth system, only gravity does work (the normal force is perpendicular to the track) and there is no friction, so (1/2)mv² = mgh. The mass cancels and the shape of the track does not enter: both blocks reach v = √(2gh).
BBlock 1 is faster, as it has twice the mass of block 2. A student who thinks heavier objects reach greater speeds picks this. Block 1 starts with twice the Ug and gains twice the K, so its speed is the same as block 2's.
CBlock 2 is faster, as it slides farther along its track. A student who thinks speed builds up with the distance traveled picks this. Block 2 speeds up in the dip but slows again as it climbs back out; with no friction, only the net drop h counts.
DBlock 1 is faster, as the normal force in the dip slows block 2. A student who thinks the normal force does work picks this. The track's normal force is perpendicular to the block's motion everywhere, so it transfers no energy. Block 2 slows while climbing out of the dip because of gravity, and it ends with the same speed as block 1.
Working For each block–Earth system, the normal force is perpendicular to the track and does no work, and there is no friction, so mechanical energy is constant: (1/2)mv² = mgh and v = √(2gh). The mass cancels and the path does not enter, so blocks 1 and 2 reach the same speed.
A child slides from rest down a straight slide with negligible friction and reaches a speed v at the bottom. A second straight slide has the same steepness, but its top is four times as high above its bottom. What speed does the child reach at the bottom of the second slide?
Answer and reasoning
A4v, as the speed is proportional to the height dropped A student who treats kinetic energy as proportional to speed takes the speed to be proportional to the height and picks 4v. The kinetic energy, not the speed, is four times as large; the speed is √4 = 2 times as large.
Bv, as the speed is set by the steepness of the slide A student who confuses speed with acceleration picks this. The same steepness gives the same acceleration, but the child accelerates along a slide four times as long, and the final speed depends on the drop: v = √(2gh).
C2v, as the speed depends on the square root of the heightCorrect For the child–Earth system with negligible friction, (1/2)mv² = mgh, so v = √(2gh): the speed is proportional to the square root of the drop. Four times the height gives √4 = 2 times the speed.
D16v, as the speed depends on the square of the height A student who applies a square instead of a square root picks 16v. The speed goes as √h, so four times the height gives 2v; 16v would need 256 times the height.
Working (1/2)mv² = mgh gives v = √(2gh), so v is proportional to √h. With four times the height, v' = √4 × v = 2v. The steepness sets the acceleration, not the final speed.
A person pushes a box across a rough, level floor at constant speed. The system is the box and the floor. Which statement correctly describes the change in the total energy of the system?
Answer and reasoning
AIt does not change, as the box keeps moving at a constant speed. A student who thinks the energy transferred equals the change in kinetic energy picks this. The box's K is constant, but the thermal energy of the box and floor rises, by exactly the energy the push transfers in.
BIt does not change, as friction removes the energy that the push adds. A student who treats friction between two objects inside the system as a transfer out picks this. The box and the floor are both in the system, so friction is internal; it does not remove energy from the system but turns it into thermal energy of the box and floor.
CIt decreases, as friction destroys energy while the box slides. A student who thinks friction destroys energy picks this. Energy is conserved in all interactions: friction turns mechanical energy into thermal energy, and here the system also gains energy from the push.
DIt increases, as the person's push does positive work on the system.Correct The person is outside the system, and the push acts in the direction of the box's displacement, so it does positive work and transfers energy in. Friction between the box and floor is internal: it transfers nothing out, but turns the incoming energy into thermal energy of the box and floor, so the speed stays constant while the system's energy rises.
Compiled from the AP Physics 1 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
