4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
A ball falls freely toward the ground, and air resistance is negligible. A student claims: “For the ball–Earth system, the power is zero, because no energy is transferred into or out of the system.” Which statement correctly evaluates the student’s claim?
Answer and reasoning
AIt is correct: power describes energy supplied by devices such as motors. A student who thinks power applies to machines and appliances picks this. Power is the rate of any energy change, including a conversion inside a system; the falling ball converts Ug to K at a nonzero rate with no device involved.
BIt is incorrect: Ug is converted to K within the system at a nonzero rate.Correct Power is the rate at which energy changes, by transfer across the system boundary or by conversion within the system. As the ball falls, the system's gravitational potential energy decreases and its kinetic energy increases at the same rate, so the rate of conversion, a power, is not zero even though the total energy is constant.
CIt is incorrect: the power equals the gravitational force on the ball. A student who treats power as the size of a force picks this. The claim is wrong, but not for this reason: a force in newtons cannot equal a power in watts. The rate of conversion is the gravitational force multiplied by the ball's speed, mgv.
DIt is incorrect: the power equals the ball's kinetic energy at each instant. A student who thinks power and energy are the same thing picks this. The claim is wrong, but not for this reason: kinetic energy, in joules, is an amount, while power, in watts, is the rate at which that amount changes.
Working The gravitational force is internal to the ball–Earth system, so no energy crosses the system boundary, and the total energy is constant. But Ug decreases and K increases: energy is converted from one type to another within the system at a nonzero rate (mgv at speed v). Power includes the rate of such a conversion, so the claim is incorrect.
The graph shows the power P delivered by a motor as a function of time t. The power is constant during each of the two intervals shown. How much energy does the motor transfer from t = 0 to t = 6.0 s?
Answer and reasoning
A3.6 × 10³ J A student who multiplies the largest power by the whole time calculates (600 W)(6.0 s) = 3.6 × 10³ J. The motor delivers 600 W for only the first 2.0 s; for the next 4.0 s it delivers 300 W, so each interval must be treated separately.
B2.7 × 10³ J A student who averages the two power values, (600 W + 300 W)/2 = 450 W, and multiplies by 6.0 s gets 2.7 × 10³ J. The 300 W interval lasts twice as long as the 600 W interval, so the simple average of the two values overstates the energy.
C3.8 × 10² J A student who divides power by time instead of multiplying calculates 600 W ÷ 2.0 s + 300 W ÷ 4.0 s ≈ 3.8 × 10² J. Since P = ΔE/Δt, the energy is ΔE = PΔt: power multiplied by time, which is the area under the graph.
D2.4 × 10³ JCorrect The energy is the area under the power–time graph. From 0 to 2.0 s: (600 W)(2.0 s) = 1200 J. From 2.0 s to 6.0 s: (300 W)(4.0 s) = 1200 J. The total is 2400 J = 2.4 × 10³ J.
Working For each interval of constant power, ΔE = PΔt (the area of a rectangle under the graph). 0 to 2.0 s: (600 W)(2.0 s) = 1200 J. 2.0 s to 6.0 s: (300 W)(4.0 s) = 1200 J. Total ΔE = 2400 J = 2.4 × 10³ J.
A cart of mass m is initially at rest on a level track with negligible friction. A motor pulls the cart with a light horizontal string that exerts a constant force, and after a time interval Δt the cart's speed is v. Which expression gives the average power delivered to the cart by the string during Δt?
Answer and reasoning
Amv²/Δt A student who takes the average power to be the power at the end of the interval uses P = Fv with F = mv/Δt and the final speed v, getting mv²/Δt. The cart's speed rises from zero, so the power delivered rises too, and the average over Δt is half the final value.
Bmv/(2Δt) A student who treats kinetic energy as proportional to speed, as if K = (1/2)mv, takes the work done to be (1/2)mv and divides by Δt. Kinetic energy depends on v², so the work done on the cart is (1/2)mv² and the average power is mv²/(2Δt).
Cmv²/(2Δt)Correct The work done by the string equals the cart's change in kinetic energy, W = (1/2)mv², because it is the only horizontal force. Average power is the work divided by the time: Pavg = W/Δt = mv²/(2Δt).
Dmv²Δt/2 A student who multiplies the work by the time instead of dividing picks this. Power is the rate of doing work, Pavg = W/Δt, so the time belongs in the denominator: mv²/(2Δt).
Working The string's force is the only horizontal force, so by the work-energy theorem the work it does equals the cart's change in kinetic energy: W = (1/2)mv² − 0. Pavg = W/Δt = mv²/(2Δt). (Check: with a constant force, the average speed is v/2 and F = mv/Δt, so Pavg = F(v/2) = mv²/(2Δt).)
A puck on a horizontal table with negligible friction is tied to a string whose other end is fixed to a peg in the table. The puck moves in a circle around the peg at constant speed v, and the tension in the string is FT. What is the power delivered to the puck by the tension?
Answer and reasoning
AZero, since the tension is perpendicular to the velocity.Correct The string pulls the puck toward the center of the circle, and the puck's velocity is tangent to the circle, so θ = 90° and P = FT v cos 90° = 0. This agrees with the puck's constant speed: its kinetic energy does not change.
BFT v, the product of the tension and the speed of the puck. A student who multiplies the full force by the speed without considering the angle picks this. Only the component of the tension along the velocity delivers power, and here that component is zero because the tension is perpendicular to the velocity.
CPositive, since the tension is exerted on the moving puck. A student who thinks any force on a moving object delivers power to it picks this. A force delivers power only if it has a component along the velocity; the tension is perpendicular to the velocity, so it changes the puck's direction but not its speed.
DNegative, since the tension opposes the puck's outward motion. A student who pictures the puck being thrown outward by a centrifugal force picks this. The puck does not move outward: its velocity is tangent to the circle, and the tension, toward the center, is perpendicular to it, so the power is zero.
Working The tension points along the string, toward the peg at the center of the circle. The velocity is tangent to the circle, so the angle between them is 90°: P = FT v cos 90° = 0. (Consistent with the constant speed: the puck's kinetic energy does not change.)
In preparation: 0 of 4 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, unit: W) Fix
Power (P, unit: W)
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 of energy to another within a system. Power is a scalar.
Watt (W)
The SI unit of power: 1 W = 1 J/s = 1 kg·m²/s³. A process with a power of 60 W transfers or converts 60 J of energy each second.
Energy conversion within a system
A change of energy from one type to another inside a system, such as gravitational potential energy becoming kinetic energy as an object falls in an object–Earth system. The total energy of the system need not change, but the rate of conversion is still a power.
Students often think Power and energy are the same thing: a process with more power involves more energy, and processes that involve equal amounts of energy have equal power. In fact No. Energy, in joules, is an amount; power, in watts, is the rate at which energy is transferred or converted, the energy per unit time. Two processes that involve the same energy have different powers if they take different times.
Students often think Power is a measure of force: a stronger push or pull means more power, and the power delivered by a force equals the force, or its component along the motion, with no factor for the object's speed. In fact No. Force, in newtons, describes an interaction; power, in watts, is the rate at which energy is transferred or converted. A force delivers power only when the object it is exerted on moves with a velocity component along the force: P = F∥v.
3.5.A.2 Average power (Pavg, unit: W) Fix
Average power (Pavg, unit: W)
The energy transferred or converted divided by the time the transfer or conversion takes: Pavg = ΔE/Δt. The same energy transferred in less time means a greater average power.
Energy from a power–time graph
For an interval of constant power, the energy transferred is ΔE = PΔt, the area of a rectangle under the power–time graph. When the power changes from one interval to the next, the energy for each interval is found separately and the results are added.
Students often think Power is energy multiplied by time (so energy is power divided by time), and a process that lasts longer has more power. In fact No. Average power is energy divided by time, Pavg = ΔE/Δt, and energy is power multiplied by time. For the same energy, a longer time means less power.
Students often think Doing the same task faster requires more work or energy, so a student who runs up a flight of stairs increases Ug more than a student of equal mass who walks up it. In fact No. At constant speed, the increase in gravitational potential energy is mgΔy whatever the time taken. Climbing faster transfers the same energy in less time, which is a greater power.
3.5.A.3 Average power from work Fix
Average power from work
Because work is the change in energy of an object or system due to a force, the average power delivered by a force is the work it does divided by the time taken: Pavg = W/Δt.
Students often think The average power over an interval equals the instantaneous power at the end of the interval, found from P = Fv with the final speed. In fact Not when the speed changes. For a constant force on an object that speeds up from rest, the instantaneous power F∥v grows from zero, and the average power over the interval is half the instantaneous power at the end.
Students often think Kinetic energy, and so the work needed to speed an object up, is proportional to speed, as if K = (1/2)mv. In fact No. K = (1/2)mv² depends on the square of the speed, so the work needed to bring an object from rest to speed v is (1/2)mv²; bringing it to 2v needs four times as much work, not twice as much.
3.5.A.4 Instantaneous power (Pinst, unit: W) Fix
Instantaneous power (Pinst, unit: W)
The power at a single instant. For a constant force exerted on an object, Pinst = F∥v = Fv cos θ, where v is the object's speed at that instant and θ is the angle between the force and the velocity. If the speed changes, the instantaneous power changes even when the force is constant.
Angle between force and velocity (θ)
The angle between the direction of a force and the direction of the object's velocity. If θ = 90°, the force delivers no power; if θ is less than 90°, it delivers energy to the object; if θ is greater than 90°, it takes energy from the object.
Students often think The sine and cosine in P = Fv cos θ are interchangeable, or the angle can be measured from any convenient line, so F sin θ can be used as the component of the force along the velocity. In fact No. θ is the angle between the force and the velocity, and the component of the force along the velocity is F cos θ. For a rope at 37° above a horizontal velocity, the component along the velocity is F cos 37°.
Students often think The power delivered by a force is Fv, using the full magnitude of the force, whatever its direction relative to the velocity. In fact No. Only the component of the force parallel to the velocity delivers power: P = F∥v = Fv cos θ. A component perpendicular to the velocity delivers no power.
4 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 4
Students A and B have equal masses. Each climbs the same flight of stairs at a constant speed: A runs up in 5.0 s, and B walks up in 10 s. Consider the increase ΔUg in the gravitational potential energy of each student–Earth system and the average rate at which Ug increases (the average power). Which statement correctly compares the two climbs?
Answer and reasoning
AA's average power is twice B's, as the same ΔUg occurs in half the time.Correct Both students have the same mass and climb the same height, so ΔUg = mgΔy is the same for both. A takes 5.0 s and B takes 10 s, so Pavg = ΔUg/Δt is twice as large for A.
BA's ΔUg is greater than B's, since A climbs the stairs much faster than B. A student who thinks doing a task faster takes more energy picks this. ΔUg = mgΔy depends only on the mass and the height climbed, which are the same for both; running changes the rate of the energy transfer, not its amount.
CTheir average powers are equal, since the two values of ΔUg are equal. A student who treats power and energy as the same thing picks this. The energies are equal, but power is energy per unit time: A transfers the same energy in half the time, so A's average power is twice B's.
DB's average power is greater, since B's climb lasts for a longer time interval. A student who multiplies energy by time instead of dividing picks this. Average power is ΔUg/Δt, so for the same energy the longer climb has the smaller power: B's average power is half of A's.
Working ΔUg = mgΔy is the same for both students (equal masses, same height). Pavg = ΔUg/Δt: A's time is half of B's, so A's average power is twice B's.
A crate of mass m is at rest on the ground. A vertical rope then lifts the crate so that it moves upward with a constant acceleration of magnitude g/2. Air resistance is negligible. Which expression gives the power delivered to the crate by the rope at time t after the crate starts moving?
Answer and reasoning
A0.25 mg²t A student who uses the net force on the crate, m(g/2), in P = Fv picks this: (mg/2)(gt/2) = 0.25 mg²t. That is the rate at which the crate's kinetic energy increases, not the power delivered by the rope. The rope's force is the tension, 1.5mg, which also supplies the increase in Ug.
B0.50 mg²t A student who takes the rope's tension to equal the crate's weight, mg, picks this: (mg)(gt/2) = 0.50 mg²t. The crate speeds up as it rises, so the rope must pull harder than the crate's weight: FT = mg + m(g/2) = 1.5mg.
C0.38 mg²t A student who takes the average power over the first time t as the power at time t picks this. The rope does work FT d = (1.5mg)(gt²/4) = 0.375 mg²t² in that time, so the average power is 0.375 mg²t, about 0.38 mg²t. The crate's speed, and with it the power, increases steadily, so the power at time t is twice the average: 0.75 mg²t.
D0.75 mg²tCorrect The rope's tension satisfies FT − mg = m(g/2), so FT = 1.5mg. At time t the crate's speed is (g/2)t, and the tension is along the velocity, so P = FT v = (1.5mg)(gt/2) = 0.75 mg²t.
Working Newton's second law for the crate, upward positive: FT − mg = m(g/2), so the rope's tension is FT = (3/2)mg. Starting from rest with constant acceleration g/2, the crate's speed at time t is v = (g/2)t. The tension is parallel to the velocity (θ = 0), so Pinst = FT v = (3/2)mg × (g/2)t = 0.75 mg²t. (Energy check: the net force, mg/2, makes K increase at a rate (mg/2)(gt/2) = 0.25 mg²t, and Ug increases at a rate mg(gt/2) = 0.50 mg²t; the rope supplies both, 0.75 mg²t.)
A student pulls a box across a level floor at constant speed with a rope. The diagram shows the tension in the rope, the angle of the rope above the horizontal, and the box's velocity. Use cos 37° = 0.80 and sin 37° = 0.60. What is the power delivered to the box by the tension in the rope?
Answer and reasoning
A30 W A student who uses sin 37° instead of cos 37° calculates (25 N)(2.0 m/s)(0.60) = 30 W. The angle is measured between the rope and the horizontal velocity, so the component along the velocity is FT cos 37°, not FT sin 37°.
B50 W A student who multiplies the full tension by the speed calculates (25 N)(2.0 m/s) = 50 W. Part of the tension is vertical, perpendicular to the velocity, and delivers no power; only FT cos 37° = 20 N does.
C40 WCorrect Only the component of the tension along the velocity delivers power. The velocity is horizontal and the rope is 37° above the horizontal, so P = FT v cos θ = (25 N)(2.0 m/s)(0.80) = 40 W.
D20 W A student who treats power as a force picks the horizontal component of the tension, (25 N)(0.80) = 20 N, and labels it in watts. Power is that force component multiplied by the box's speed: (20 N)(2.0 m/s) = 40 W.
Working The velocity is horizontal, so the angle between the tension and the velocity is 37°. P = FT v cos θ = (25 N)(2.0 m/s)(0.80) = 40 W.
A cart starts from rest on a level track with negligible friction. A constant horizontal force is exerted on the cart, and no other horizontal force acts on it. At time t after the cart starts, the power delivered to the cart by the force is P. The power delivered by the force at time 2t is how many times P?
Answer and reasoning
A1 A student who thinks a constant force delivers a constant power picks this. P = Fv, and although F does not change, the cart's speed doubles between t and 2t, so the power doubles too.
B2Correct With a constant force and no other horizontal force, the acceleration is constant, so the speed grows in proportion to time: the speed at 2t is twice the speed at t. P = Fv with F constant, so the power doubles.
C4 A student who treats power as the same thing as energy follows the kinetic energy, which is proportional to v² and so becomes four times as large. The power is the rate at which energy is delivered, Fv, which is proportional to v and only doubles.
D8 A student who multiplies work by time instead of dividing reasons that the work done by 2t is four times the work done by t and multiplies by twice the time, giving 8. Power is a rate, not work × time; at each instant it is Fv, which doubles.
Working The net force is constant, so the acceleration is constant and v = at: the speed at 2t is twice the speed at t. Pinst = Fv with F constant, so the power at 2t is 2P.
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
