AP Physics 1 Quiz: Power

What this quiz covers

This quiz focuses on Power, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Trial 1, because it has more work
2. Trial 2, because it takes less time
3. They have equal average power (correct answer)
4. Cannot be determined without the force

Explanation: This question tests understanding of average power as work divided by time. Power is defined as P = W/t, where W is work done and t is time taken. In Trial 1, power is P₁ = 2W/t. In Trial 2, power is P₂ = W/(t/2) = 2W/t. Both trials have the same average power because doubling the work while doubling the time (Trial 1) gives the same result as halving the work while halving the time (Trial 2). Choice A incorrectly focuses only on work without considering time. To compare powers, always calculate the ratio of work to time for each situation.

Question 2

1. $\dfrac{F}{v}$
2. $Fv$ (correct answer)
3. $\dfrac{1}{2}mv^2$
4. $F\Delta x$

Explanation: This question tests understanding of instantaneous power. Power is the rate of energy transfer or work done, and instantaneous power is P = F·v when force F acts on an object moving with velocity v. At the instant when the cart has speed v, the instantaneous power delivered by the constant force F is P = Fv. Choice C (½mv²) represents kinetic energy, not power, confusing energy with the rate of energy transfer. The key insight is that instantaneous power depends on the current velocity, even though the force has been constant throughout the acceleration.

Question 3

1. First student, because power depends on time only
2. Second student, because more energy is transferred in the same time (correct answer)
3. They have the same average power because they both climb stairs
4. Cannot be determined without knowing their masses

Explanation: This question tests understanding of power as the rate of energy transfer. Power is defined as P = ΔE/t, where ΔE is energy transferred and t is time. The first student increases potential energy by ΔU in time t, giving power P₁ = ΔU/t. The second student increases potential energy by 2ΔU in the same time t, giving power P₂ = 2ΔU/t = 2P₁. The second student has twice the power because they transfer twice the energy in the same time. Choice D incorrectly suggests mass information is needed when the energy changes are already given. To compare power, divide energy transferred by time taken.

Question 4

1. The first trial, because the distance is the same.
2. The second trial, because average power scales as $W/t$ and both $W$ and $1/t$ increase. (correct answer)
3. They have equal average power because both are constant-speed motions.
4. They have equal average power because power depends only on force.

Explanation: This question assesses the concept of power in AP Physics 1, which is the rate at which work is done or energy is transferred. Average power is P_avg = W/Δt, with work W = F*d for constant force over distance d. In the first trial, W = F d and Δt = t, so P = F d / t; in the second, W = 2F d and Δt = t/2, so P = (2F d)/(t/2) = 4 (F d / t), which is greater. The increased force and halved time both contribute to higher power. Choice A distracts by focusing only on equal distance, ignoring changes in force and time. Always compute work and divide by time separately for each scenario to compare average powers accurately.

Question 5

1. $P_B = 2P_A$ because more energy is transferred.
2. $P_B = \dfrac{1}{2}P_A$ because the time is longer.
3. $P_B = P_A$ because energy and time both double. (correct answer)
4. Cannot be determined without the elevator's speed.

Explanation: This question tests understanding of power when both energy and time scale proportionally. Power is defined as P = E/t, where E is energy transferred and t is time. In trip A, PA = E/t, while in trip B, PB = 2E/(2t) = E/t = PA. Since both energy and time double by the same factor, their ratio (power) remains constant. Choice A incorrectly focuses only on energy doubling without considering time also doubles. When both work and time change by the same factor, power remains unchanged.

Question 6

1. doubled
2. halved
3. unchanged (correct answer)
4. quadrupled

Explanation: This question tests understanding of average power versus instantaneous power. Power is the rate of energy transfer, calculated as P = Fv when force and velocity are constant and aligned. Since both F and v remain constant, the instantaneous power P = Fv is constant throughout the motion. Average power equals total work divided by total time, but when instantaneous power is constant, average power equals instantaneous power regardless of duration. Choice A incorrectly assumes power depends on time duration. When force and velocity are both constant, power remains constant regardless of how long the process continues.

Question 7

1. Trial 1, because less energy means greater power
2. Trial 2, because more work always means more power
3. Trial 1, because $\frac{W}{t} > \frac{2W}{4t}$ (correct answer)
4. They have the same average power because both involve constant force

Explanation: This question tests understanding of average power calculation. Power is work divided by time: P = W/t. Trial 1 has power P₁ = W/t. Trial 2 has power P₂ = 2W/(4t) = (1/2)(W/t) = P₁/2. Trial 1 has greater average power because W/t > 2W/(4t), as correctly stated in choice C. Choice B incorrectly assumes more work always means more power without considering time. When comparing powers, always calculate the work-to-time ratio for each case.

Question 8

1. Power is the same because the force is constant.
2. Power at $2t$ is twice that at $t$ because $P=Fv$. (correct answer)
3. Power at $2t$ is four times that at $t$ because kinetic energy quadruples.
4. Power at $2t$ is half that at $t$ because the acceleration decreases.

Explanation: This question assesses the concept of power in AP Physics 1, which is the rate at which work is done or energy is transferred. Instantaneous power for constant force is P = F v, where v changes due to acceleration. With constant acceleration a = F/m, speed at time t is v = a t, and at 2t is 2v, so power at t is F v and at 2t is F (2v) = 2 F v, twice as much. This reflects the increasing rate of kinetic energy transfer as speed grows. Choice C is a distractor, confusing power with kinetic energy, which quadruples but isn't directly the ratio for power. For accelerating objects, calculate instantaneous power using current velocity in P = F v as a reliable method.

Question 9

1. Machine 1, because it uses less power to store the same energy.
2. Machine 2, because it transfers the same energy in less time. (correct answer)
3. They have equal power because the energy change is the same.
4. Machine 1, because power depends on energy only, not time.

Explanation: This question assesses the concept of power in AP Physics 1, which is the rate at which work is done or energy is transferred. Average power is P_avg = ΔE/Δt, with both machines increasing elastic potential energy by E. Machine 1 takes time t, so P = E/t; Machine 2 takes t/4, so P = E/(t/4) = 4(E/t), greater power. The shorter time amplifies the rate of energy transfer. Choice C distracts by equating equal energy to equal power, but time differentiates them. For energy transfer comparisons, always divide the energy change by the specific time interval to determine power.

Question 10

1. Cyclist A, because less speed means more time resisting the force.
2. Cyclist B, because $P=F_r v$ and tripling $v$ triples power. (correct answer)
3. They require equal power because the resistive force is the same.
4. They require equal power because both move at constant speed.

Explanation: This question assesses the concept of power in AP Physics 1, which is the rate at which work is done or energy is transferred. At constant speed, power to overcome resistance is P = F_r v, where F_r is the resistive force. Both cyclists face the same F_r, but Cyclist B at 3v requires P = F_r (3v) = 3 F_r v, triple that of Cyclist A's F_r v. This assumes F_r is independent of speed, as stated. Choice C misleads by claiming equal power from equal force, ignoring velocity's multiplication in the formula. To handle resistance problems, apply P = F v and note how changes in velocity affect power requirements.

Question 11

1. It doubles because $P=Fv$ and $F$ doubles while $v$ stays the same. (correct answer)
2. It stays the same because the speed is unchanged.
3. It halves because more force means less speed.
4. It becomes zero because constant speed implies zero power.

Explanation: This question tests understanding of power in steady-state motion against resistance. At constant speed, the driving force must equal the resistive force, so the cyclist's force doubles from Fr to 2Fr. Power equals force times velocity: P = Fv. Initially P₁ = Frv, and on the rougher road P₂ = 2Frv = 2P₁, so power doubles. Choice B incorrectly assumes constant speed means constant power, missing that force must increase to maintain speed against greater resistance. For constant velocity against resistance, power equals resistive force times speed.

Question 12

1. Trial 2 has greater power because it acts longer
2. Trial 1 has twice the average power of Trial 2 (correct answer)
3. They have equal average power because $mgh$ is the same
4. Trial 2 has twice the average power of Trial 1

Explanation: This question tests understanding of power as the rate of doing work. Power equals work divided by time: P = W/t. In both trials, the winch lifts the same mass m through the same height h, so the work done against gravity is W = mgh in both cases. Trial 1 completes this work in time t, giving power P₁ = mgh/t. Trial 2 takes twice as long (2t), giving power P₂ = mgh/(2t) = (1/2)(mgh/t) = P₁/2. Therefore, Trial 1 has twice the average power of Trial 2. Choice C incorrectly assumes equal work means equal power, ignoring time. To find power, always divide work by the time taken to do that work.

Question 13

1. $Ft$
2. $Fv$ (correct answer)
3. $F/v$
4. $F$

Explanation: This question tests the concept of average power in the context of constant velocity motion. Power is defined as the rate at which work is done, given by P = W/t, where W is work and t is time. In this scenario, the work done by the force F over distance d = v t is W = F d = F v t, so average power P = (F v t)/t = F v. This shows that power depends on force and velocity, not directly on time for the average over that interval. A common distractor like choice A, Ft, might confuse power with impulse, which is force times time, but power involves energy transfer rate, not momentum change. To approach similar problems, always derive power from work divided by time or use P = F v for constant speed cases.

Question 14

1. It doubles because the time doubles
2. It halves because power is inversely proportional to time
3. It stays the same because $P=Fv$ and both are unchanged (correct answer)
4. It quadruples because work doubles and time doubles

Explanation: This question tests understanding of power at constant velocity. Power equals force times velocity: P = Fv. Since both the force F and speed v remain constant throughout, the power P = Fv stays the same regardless of how long the student pushes. Doubling the time doubles the work done but doesn't change the rate at which work is done. Choice B incorrectly assumes power depends on total time rather than instantaneous conditions. For constant force and velocity, power remains constant regardless of duration.

Question 15

1. It doubles because more energy is transferred
2. It halves because the time is longer
3. It stays the same because $\frac{E}{5} = \frac{2E}{10}$ (correct answer)
4. It cannot be determined without the force

Explanation: This question tests understanding of average power with constant energy transfer rate. Power is the rate of energy transfer, so average P = ΔE / Δt. In 5 s, P = E/5, and in 10 s, P = 2E/10 = E/5, remaining the same as expected for constant rate. This shows average power is consistent when energy scales linearly with time. Choice A is a distractor, mistakenly thinking more total energy means higher power without considering extended time. For such problems, check if rates are constant by seeing if energy over time is proportional.

Question 16

1. The first motor, because it does less work.
2. The second motor, because it does more work.
3. The first motor, because $\frac{W}{t} > \frac{2W}{4t}$. (correct answer)
4. They have equal average power because both involve work and time.

Explanation: This question tests understanding of average power calculations. Power is the rate of doing work, defined as $P = \frac{W}{t}$. For the first motor: $P_1 = \frac{W}{t}$. For the second motor: $P_2 = \frac{2W}{4t} = \frac{1}{2} \frac{W}{t} = \frac{P_1}{2}$. Since $P_1 = 2 P_2$, the first motor has greater average power. Choice B incorrectly focuses on total work rather than the rate of work. The strategy is to always calculate $P = \frac{W}{t}$ explicitly and compare the ratios, not just look at work or time alone.

Question 17

1. $mgv$ (correct answer)
2. $mg$
3. $mgh$
4. $\dfrac{mgh}{v}$

Explanation: This question tests understanding of power in lifting at constant velocity. Power is the rate of doing work or transferring energy. When lifting at constant velocity, the net force is zero, so the upward tension equals the downward weight mg. The crane does work at rate P = F·v, where F is the tension force and v is the upward velocity. Since tension F = mg, the power output is P = mgv. Choice C (mgh) represents potential energy, not power, missing that power requires dividing by time. The strategy is to identify the force doing work and multiply by velocity for constant-speed motion.

Question 18

1. Pump A, because it transfers more total energy
2. Pump B, because $\frac{E/2}{t/4} = \frac{2E}{t}$ (correct answer)
3. They have the same average power because both lift water
4. Cannot be determined without the water's mass

Explanation: This question tests understanding of power as energy transfer rate. Power equals energy divided by time: P = E/t. Pump A has power P_A = E/t. Pump B has power P_B = (E/2)/(t/4) = (E/2) × (4/t) = 2E/t = 2P_A. Pump B has twice the power of Pump A because it transfers half the energy in one-quarter the time. Choice A incorrectly focuses on total energy rather than the rate of energy transfer. To find power, always divide energy by time, not just compare total energies.

Question 19

1. $Fv$ (correct answer)
2. $Ft$
3. $\dfrac{F}{t}$
4. $F$

Explanation: This question tests understanding of power for constant velocity motion. Power is the rate of energy transfer, calculated as P = W/t where W is work done. For constant velocity horizontal motion, the applied force F equals the friction force, and work done is W = F·d where d is distance. Since the cart moves at constant speed v for time t, the distance is d = vt, making work W = F(vt). Therefore, average power is P = W/t = F(vt)/t = Fv. Choice B (Ft) incorrectly represents impulse, not power. The key insight is that for constant velocity, instantaneous power P = Fv equals average power.

Question 20

1. Both have the same power because the speed is the same.
2. Student 2 has greater power because $P=Fv$ and $2F$ at the same $v$ doubles $P$. (correct answer)
3. Student 1 has greater power because using less force is more efficient.
4. Student 2 has the same power because power depends only on work, not force.

Explanation: This question assesses the concept of power in AP Physics 1, which is the rate at which work is done or energy is transferred. Power is defined as P = W/Δt, but for constant force and velocity, it simplifies to P = Fv, where F is the force and v is the velocity. In this scenario, both students pull carts at the same constant speed v, but Student 2 applies twice the force, 2F, to overcome presumably greater friction or other resistance. Therefore, Student 2's power is 2Fv, which is double that of Student 1's Fv. A common distractor is choice A, which incorrectly assumes power depends only on speed, ignoring the role of force in the P = Fv formula. To approach similar problems, always identify whether power is calculated using energy transfer over time or force times velocity, depending on the given constants.