5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
Students X and Y each lift an identical box from the floor to the same shelf. Each box starts and ends at rest. X takes 2 s and Y takes 4 s. Which statement is correct?
Answer and reasoning
AX transfers more energy to the box–Earth system, because X lifts it faster. A student who thinks doing a task faster takes more energy picks this. Both boxes rise the same height and end at rest, so both systems gain mgΔy; X transfers it at a greater rate, not in a greater amount.
BX transfers the same energy as Y at twice Y’s rate, because X takes half the time.Correct Each box–Earth system gains the same energy, mgΔy (the box starts and ends at rest, so ΔK = 0). X transfers it in half the time, so X's average power, ΔE/Δt, is twice Y's.
CY transfers more energy to the box–Earth system, because Y pushes for longer. A student who thinks the energy a force transfers depends on how long it acts picks this. Work depends on force and displacement; both systems gain the same mgΔy.
DX and Y deliver the same power, because they do the same amount of work. A student who treats power as the same thing as work picks this. Power is work divided by time: with the same work, X, taking half the time, delivers twice the average power.
A 60 kg sprinter starts from rest and reaches a speed of 8.0 m/s in 2.5 s along a level track. What is the average rate at which the sprinter's kinetic energy increases during the 2.5 s?
Answer and reasoning
A9.6 × 10¹ W A student who treats kinetic energy as proportional to speed picks this: (1/2)(60)(8.0) = 240, divided by 2.5 s. K = (1/2)mv² uses the square of the speed.
B1.5 × 10³ W A student who writes the kinetic energy as mv² picks this: 60 × 64 = 3840 J, divided by 2.5 s. With K = (1/2)mv², ΔK = 1920 J.
C7.7 × 10² WCorrect ΔK = (1/2)(60)(8.0)² = 1920 J, gained in 2.5 s, so the average power is ΔK/Δt = 1920/2.5 = 768 W ≈ 7.7 × 10² W.
D4.8 × 10³ W A student who multiplies the energy by the time picks this: 1920 J × 2.5 s. Power is energy per unit time, so the energy is divided by the time.
Working ΔK = (1/2)(60 kg)(8.0 m/s)² − 0 = 1920 J. Pavg = ΔK/Δt = 1920 J/2.5 s = 768 W ≈ 7.7 × 10² W.
A winch pulls a crate along a level floor with a horizontal force F in the direction of the crate's motion. The graph shows F as a function of the crate's position x. The crate moves from x = 0 to x = 12 m in 5.0 s. What is the average power delivered to the crate by the winch's force during this time?
Answer and reasoning
A9.6 × 10² W A student who takes the work to be the largest force times the distance picks this: 400 N × 12 m = 4800 J, divided by 5.0 s. The force is less than 400 N over the first 8 m, so the work is the area, 4000 J.
B2.0 × 10⁴ W A student who multiplies the work by the time picks this: 4000 J × 5.0 s. Power is work per unit time, so the work is divided by the time.
C4.0 × 10³ W A student who treats power as the same thing as work picks this, giving the 4000 J of work as the power. Average power is that work divided by the 5.0 s it took.
D8.0 × 10² WCorrect The work is the area under the graph: (1/2)(200 + 400)(8) = 2400 J from 0 to 8 m, plus 400 × 4 = 1600 J from 8 to 12 m, so W = 4000 J. The average power is W/Δt = 4000 J/5.0 s = 8.0 × 10² W.
Working W = area under F(x): trapezoid from 0 to 8 m, (1/2)(200 N + 400 N)(8 m) = 2400 J, plus rectangle from 8 to 12 m, (400 N)(4 m) = 1600 J; W = 4000 J. Pavg = W/Δt = 4000 J/5.0 s = 8.0 × 10² W.
A car of mass m starts from rest on a level road. Its engine delivers a constant power P to the car, and friction and air resistance are negligible. What is the car's speed at time t?
Answer and reasoning
A√(2Pt/m)Correct The engine's work is the only work done on the car, so dK/dt = P. With P constant and K = 0 at t = 0, K = Pt. Then (1/2)mv² = Pt gives v = √(2Pt/m).
B√(Pt/m) A student who assumes a constant force and acceleration picks this: F = ma = mv/t, and P = Fv = mv²/t gives v = √(Pt/m). With constant power the force P/v falls as the car speeds up, so the acceleration is not constant; integrate dK/dt = P instead.
C2Pt/m A student who treats kinetic energy as proportional to speed picks this: (1/2)mv = Pt. This does not even have units of speed; K = (1/2)mv² needs a square root.
DPt/m A student who treats the power as if it were a constant force picks this: a = P/m and v = at = Pt/m. Power is not a force, and the expression does not have units of speed; the driving force is P/v.
Working P = dW/dt = dK/dt, constant, so K = Pt (K = 0 at t = 0). (1/2)mv² = Pt, so v = √(2Pt/m). The force P/v decreases as v increases, so the acceleration is not constant.
A motor raises a load vertically at constant speed, delivering power P to it. Air resistance is negligible. The motor then raises a load of twice the mass at a constant speed 1.5 times as great. What power does the motor now deliver to the load?
Answer and reasoning
A2.0P A student who thinks the power depends only on the force picks this: the force doubles. P = Fv, and the speed is also 1.5 times as great.
B3.0PCorrect At constant velocity the motor's upward force equals the load's weight, and it is parallel to the velocity, so P = Fv = mgv. Doubling m and multiplying v by 1.5 multiplies P by 2 × 1.5 = 3.0.
C4.5P A student who thinks the power needed scales with the load's kinetic energy picks this: 2 × 1.5² = 4.5. At constant velocity the motor's force is mg and P = mgv, proportional to v, not v².
D1.5P A student who thinks the mass cancels picks this, scaling the power with the speed alone. The force needed is mg, so a load of twice the mass needs twice the force and, at the same speed, twice the power.
Working At constant velocity the motor's force equals the load's weight, mg, and is parallel to the velocity, so P = Fv = mgv. New power: (2m)g(1.5v) = 3.0mgv = 3.0P.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
3.5.A.1 Power (P) Fix
Power (P)
The rate at which energy changes with respect to time, either by transfer into or out of a system or by conversion from one type to another within it. A scalar. SI unit: watt (W), 1 W = 1 J/s.
Students often think Doing a task faster takes more energy, so the person who lifts a load more quickly transfers more energy to it. In fact No, if the box starts and ends at rest. The energy transferred is mgΔy in both cases; the faster person transfers it in less time, that is, at a greater rate (greater power).
Students often think Power describes only energy crossing a system's boundary; a conversion inside a system, with no energy entering or leaving, involves no power. In fact No. Power is the rate at which energy changes, either by transfer into or out of a system or by conversion from one type to another within it. For example, the rate at which Ug becomes K inside a falling ball–Earth system is a power.
3.5.A.2 Average power (Pavg) Fix
Average power (Pavg)
The energy transferred or converted divided by the time the transfer or conversion took: Pavg = ΔE/Δt. On a graph of energy against time, it is the slope of the line joining the start and end of the interval. SI unit: W.
Students often think Power is energy multiplied by time (and so energy is power divided by time). In fact No. Power is energy divided by time, Pavg = ΔE/Δt (units J/s = W). Energy is power multiplied by time, ΔE = Pavg Δt, which is why a kilowatt-hour is a unit of energy.
Students often think The rate at which energy changes is greatest where the energy is greatest, so the height of an energy–time graph shows the power, and a rate falls as the amount falls. In fact No. Power is the rate of change of energy, the slope of an energy–time graph, not its height. A system with a large energy can be changing slowly or not at all, and one with little energy can be changing quickly.
3.5.A.3 Average power from work Fix
Average power from work
Because work is the energy transferred to a system by a force, the average power of a force is the total work it does divided by the time taken: Pavg = W/Δt. SI unit: W.
Students often think Power is the same thing as work or energy, so equal work means equal power and more energy transferred means more power, however long the transfer takes. In fact No. Power is the rate of doing work, Pavg = W/Δt. The same work done in half the time means twice the average power.
Students often think The work done (the energy transferred) depends on how long a force acts, so pushing or lifting for a longer time transfers more energy. In fact Not in itself. The energy a force transfers is the work it does, W = ∫F⃗ · dr⃗, which depends on the force and the displacement, not on the time. Time enters the power, Pavg = W/Δt.
3.5.A.4 Instantaneous power (Pinst) Fix
Instantaneous power (Pinst)
The power delivered by a force at one instant: Pinst = dW/dt, the slope of the tangent to a graph of work done against time. It equals the average power at every instant of an interval only when the power is constant. SI unit: W.
Energy from a power–time graph
Since P = dW/dt, the energy transferred between two times is W = ∫P dt, the area under the graph of power against time between those times. SI unit: J.
Students often think Instantaneous and average power are the same thing, so the power at one instant (for example at the end of an interval) can be used as the average over the interval, and the average can be used as the value at an instan… In fact No. The instantaneous power is Pinst = dW/dt at one instant; the average power is the total work divided by the time, W/Δt. The instantaneous power equals the average power throughout the interval only when the power is constant.
Students often think Constant power means constant force and constant acceleration, so the constant-acceleration equations apply, with the force found as F = mv/t. In fact No. With constant power P = Fv, the driving force P/v falls as the speed rises, so the acceleration decreases. From rest, with no other work done, P = dK/dt gives K = Pt and v = √(2Pt/m).
3.5.A.5 Power delivered by a constant force Fix
Power delivered by a constant force
Pinst = F∥v = Fv cos θ, where θ is the angle between the force and the object's velocity. The power is positive when θ < 90°, zero when the force is perpendicular to the velocity, and negative when θ > 90°.
Students often think The power a force delivers depends only on the size of the force, so a constant force delivers constant power, doubling a force doubles the power whatever the motion, and power can be used as if it were a force. In fact No. P = Fv cos θ: the power depends on the force, the object's speed and the angle between them. A constant force delivers more power as the object speeds up, and a force of fixed size can deliver more power than a larger one if the object moves faster.
Students often think Mass always cancels in energy and power problems, as it does in free fall, so the power needed to move a load depends only on how fast it moves. In fact No. To lift a load at constant velocity, the force needed is mg, so P = Fv = mgv, which is proportional to the mass. Mass cancels only in special cases, such as an object–Earth system converting Ug into K.
7 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 7
A ball is released from rest and falls. Air resistance is negligible. Consider the ball–Earth system while the ball falls. Which statement about power for this system is correct?
Answer and reasoning
AUg is converted into K at a rate that increases steadily as the ball speeds up.Correct Power includes conversion within a system. The rate at which Ug becomes K is dK/dt = Fg v = mgv, and v increases steadily during free fall, so the rate increases.
BNo power is involved, since no energy enters or leaves the ball–Earth system. A student who thinks power describes only transfers across a system's boundary picks this. Power is also the rate of conversion within a system; here Ug turns into K at the rate mgv.
CUg is converted into K at a constant rate, since the gravitational force is constant. A student who thinks power depends only on the force picks this. The rate is Fg v = mgv; the force is constant but the speed increases, so the rate increases.
DUg is converted into K at a rate that decreases as the value of Ug becomes smaller. A student who links a rate to the amount remaining picks this. The rate of change of Ug is −mgv, whose size grows as the ball speeds up, whatever the value of Ug.
The graph shows the kinetic energy K of a cart as a function of time t, with three time intervals, I, II and III, marked. Which ranking of the average power delivered to the cart by the net force during the three intervals, PI, PII and PIII, is correct?
Answer and reasoning
APII = PIII > PI A student who reads the power from the height of the graph picks this: K reaches 80 J at the end of II and III but only 40 J at the end of I. Power is the rate of change of energy, the slope, not the value.
BPI = PII > PIII A student who ranks by the energy transferred, ignoring the time, picks this: ΔK is 40 J in both I and II. Interval II lasts twice as long, so its average power is half that of I.
CPII > PI > PIII A student who multiplies the energy by the time instead of dividing picks this: 40 J × 2 s = 80 J·s for I, 40 J × 4 s = 160 J·s for II and 0 for III. Power is ΔK/Δt, the energy divided by the time.
DPI > PII > PIIICorrect The net force's work equals ΔK, so the average power is the slope of the K(t) graph over each interval: 40 J/2 s = 20 W in I, 40 J/4 s = 10 W in II, and 0 in III.
Working By the work–energy theorem the net force's work equals ΔK. I: 40 J/2 s = 20 W. II: (80 − 40) J/4 s = 10 W. III: 0 J/2 s = 0. So PI > PII > PIII. m04 (ΔK × Δt): 80, 160, 0 → PII > PI > PIII.
A pump takes water at rest from the surface of a lake, raises it through a height of 6.0 m, and ejects it from the end of a pipe with a speed of 8.0 m/s. Water passes through the pump at a steady rate of 15 kg/s. Friction and air resistance are negligible. Use g = 10 m/s². What is the average power the pump delivers to the water?
Answer and reasoning
A9.0 × 10² W A student who counts only the gain in gravitational potential energy picks this: 15 kg/s × 10 m/s² × 6.0 m = 900 W. The water also leaves the pipe moving, so the pump supplies (1/2)v² = 32 J of kinetic energy per kilogram as well.
B1.9 × 10³ W A student who writes the kinetic energy as mv² picks this, giving each kilogram v² = 64 J instead of (1/2)v² = 32 J of kinetic energy: 15 × (60 + 64) = 1860 W ≈ 1.9 × 10³ W.
C1.4 × 10³ WCorrect Each second, 15 kg of water is raised through 6.0 m and given a speed of 8.0 m/s from rest, so it gains 15 × (10 × 6.0 + 8.0²/2) = 15 × 92 = 1380 J. The average power is therefore 1380 W ≈ 1.4 × 10³ W.
D9.6 × 10² W A student who treats kinetic energy as proportional to speed picks this, using (1/2)v = 4.0 m/s in place of the kinetic energy per kilogram: 15 × (60 + 4.0) = 960 W. A speed cannot be added to gh, an energy per kilogram; the kinetic energy per kilogram is (1/2)v² = 32 J/kg.
Working Each second 15 kg gains gh + v²/2 = 60 + 32 = 92 J/kg, so Pavg = 15 × 92 = 1380 W ≈ 1.4 × 10³ W.
A person slowly compresses an ideal spring from its equilibrium length by a distance x in a time t, doing work at an average rate P. The person then compresses the same spring from its equilibrium length by a distance 2x in a time 2t. What is the person's average rate of doing work in the second case?
Answer and reasoning
A4.0P A student who treats power as the same thing as work picks this, scaling the power with the 4 times larger work and ignoring the doubled time.
B2.0PCorrect The work done equals the energy stored, (1/2)k(Δx)², so compressing to 2x takes 4 times the work. It takes twice the time, so the average power, W/Δt, is 4/2 = 2.0 times as large.
C1.0P A student who thinks the energy stored in a spring is proportional to its compression picks this: twice the compression would mean twice the work, in twice the time. Us depends on (Δx)², so the work is 4 times as large.
D8.0P A student who multiplies the energy by the time picks this: 4 times the work × 2 times the time. Average power is work divided by time.
Working Compressed slowly, the work equals the stored energy, (1/2)k(Δx)². Doubling Δx multiplies the work by 4; the time doubles. Pavg = W/Δt: 4/2 = 2, so the new average power is 2.0P.
A 6.0 kg cart starts from rest on a level, frictionless track. The graph shows the power delivered to the cart by the net force as a function of time. What is the cart's speed at t = 4.0 s?
Answer and reasoning
A4.0 m/s A student who uses the power at t = 4.0 s as the average over the whole 4.0 s picks this: 12 W × 4.0 s = 48 J. The power rises from zero, so its average is 6 W and the energy is 24 J.
B2.0 m/s A student who reads the power at t = 4.0 s, 12 W, as the energy delivered, 12 J, picks this. Power is a rate; the energy is the area under the graph, 24 J.
C2.8 m/sCorrect Since P = dW/dt, the work done by the net force is the area under the P(t) graph: (1/2)(4.0 s)(12 W) = 24 J. That equals the cart's gain in kinetic energy, so (1/2)(6.0)v² = 24 and v = √8.0 ≈ 2.8 m/s.
D8.0 m/s A student who treats kinetic energy as proportional to speed picks this: (1/2)(6.0)v = 24 J gives 8.0 m/s. With K = (1/2)mv², v = √8.0 ≈ 2.8 m/s.
Working Energy delivered = ∫P dt = area under P(t) = (1/2)(4.0 s)(12 W) = 24 J = ΔK. (1/2)(6.0 kg)v² = 24 J, so v = √8.0 ≈ 2.8 m/s.
The diagram shows the path of a ball thrown from level ground; T is the highest point of the path. Air resistance is negligible. Let v⃗ be the ball's velocity. A student claims that the gravitational force F⃗g on the ball delivers zero power to it at T. Which statement correctly supports the claim?
Answer and reasoning
AAt T, the ball is momentarily at rest, so its speed v is zero there. A student who thinks a projectile stops at the top of its path picks this. The ball lands some distance from its launch point, so it has a horizontal velocity that stays constant; at T only the vertical component of v⃗ is zero.
BAt T, F⃗g has done no work on the ball since the launch, so its power is zero. A student who treats power as the same thing as work picks this, and the premise is also false: the ball has risen to T, so gravity has done negative work on it since launch. Power is the rate of doing work at an instant, P = Fv cos θ, which is zero at T because θ = 90°.
CAt T, the upward force from the throw just balances F⃗g, so their powers cancel. A student who thinks the ball carries a force from the throw picks this. After release no upward force acts on the ball; gravity is the only force, and its power is zero at T because it is perpendicular to the velocity.
DAt T, the force F⃗g is perpendicular to v⃗, so it has no component along v⃗.Correct The power delivered by a constant force is P = Fv cos θ. The path shows the ball still moving sideways at T, so v⃗ is horizontal there, while F⃗g points straight down: θ = 90° and cos θ = 0. Gravity delivers zero power at that instant, although Fg and v are both nonzero.
A ball of mass m is released from rest. Air resistance is negligible, and g is the magnitude of the gravitational field. At what rate is the gravitational force doing work on the ball at the instant the ball has fallen a distance h?
Answer and reasoning
Amg√(2gh)Correct After falling h from rest, (1/2)mv² = mgh gives v = √(2gh). The gravitational force mg is constant and points along the velocity, so its instantaneous power is P = Fv cos 0° = mg√(2gh).
Bmg√(gh) A student who writes the kinetic energy as mv² picks this: mv² = mgh gives v = √(gh). With K = (1/2)mv², v = √(2gh).
Cmg√(gh/2) A student who gives the average power over the fall picks this: mgh divided by the fall time √(2h/g). The question asks for the rate at one instant, P = Fg v, which is twice this average.
D2mg²h A student who treats kinetic energy as proportional to speed picks this: (1/2)mv = mgh gives v = 2gh, and P = mg(2gh). That does not even have units of power; v = √(2gh).
Working Energy (ball–Earth system): (1/2)mv² = mgh, so v = √(2gh). The gravitational force mg is constant and parallel to the velocity, so P = Fg v cos 0° = mg√(2gh).
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
