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. 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 2

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 3

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 4

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 5

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.

Question 6

1. Decrease $v$ while keeping $F$ the same
2. Increase $d$ while keeping $F$ and $v$ the same
3. Increase $v$ while keeping $F$ the same (correct answer)
4. Decrease the time by pushing for a shorter distance

Explanation: This question evaluates the application of power in constant force and velocity pushing. Power is the rate at which work is done, with average P = (F d)/t and since t = d/v, P simplifies to F v. Increasing v while keeping F constant directly increases P, as power scales with velocity. Keeping F and v the same but increasing d increases both work and time proportionally, leaving P unchanged. Choice A is a distractor, as decreasing v would reduce P, mistakenly thinking slower speed means more power due to longer time. Remember, for constant force problems, express power in terms of F and v to see which variables affect it.

Question 7

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 8

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 9

1. $Fv$
2. $Fv\cos\theta$ (correct answer)
3. $Fv\sin\theta$
4. $F\cos\theta$

Explanation: This question evaluates power with angled forces at constant speed. Power is the rate of work done, given by P = F · v = F v cosθ, where θ is the angle between force and velocity. Here, the horizontal velocity v and force at θ above horizontal yield P = F v cosθ, as only the horizontal component contributes to work along the direction of motion. This applies because at constant speed, power matches the rate of energy dissipation, like against friction. Choice A, F v, is a distractor that neglects the angle, assuming full force contributes. A strategy is to use the dot product for power when force and velocity are not parallel.

Question 10

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 11

1. $\Delta U_g t$
2. $\dfrac{\Delta U_g}{t}$ (correct answer)
3. $\dfrac{t}{\Delta U_g}$
4. $\Delta U_g$

Explanation: This question tests understanding of power as the rate of energy transfer. Power is defined as the rate at which energy is transferred or work is done: P = E/t or P = W/t. When the student runs up stairs, they do work against gravity to gain gravitational potential energy ΔU_g. The average power output is therefore P = ΔU_g/t. Choice A (ΔU_g·t) has incorrect units (energy × time) rather than energy/time for power. The strategy is to remember that power always involves dividing energy or work by time.

Question 12

1. $P_A > P_B$ because Sled A has greater speed.
2. $P_B > P_A$ because Sled B has greater force.
3. $P_A = P_B$ because $Fv = (2F)\left(\tfrac{v}{2}\right)$. (correct answer)
4. Not enough information because mass is not given.

Explanation: This question tests understanding of instantaneous power for constant velocity motion. Power is the rate of energy transfer, and for a force applied to an object moving at constant velocity, P = F·v. For Sled A: P_A = F·v. For Sled B: P_B = (2F)·(v/2) = 2F·v/2 = F·v. Therefore, P_A = P_B, showing that doubling force while halving velocity maintains the same power. Choice A incorrectly considers only velocity, ignoring force. The key insight is that power depends on the product of force and velocity, not either factor alone.

Question 13

1. $P_A = P_B$ because the force is the same
2. $P_B = 2P_A$ (correct answer)
3. $P_A = 2P_B$ because car A is slower
4. $P_B > P_A$ only if car B has greater acceleration

Explanation: This question tests power comparison for objects under identical forces at different speeds. Power is the rate of energy transfer, instantaneously given by P = F v when force aligns with velocity. For car A at v, P_A = F v, and for car B at 2v, P_B = F (2v) = 2 F v, so P_B = 2 P_A. This holds regardless of acceleration, as power depends on current velocity. Choice C is a distractor, incorrectly suggesting slower speed means higher power, perhaps confusing with time taken. Always use P = F v for instantaneous power and verify if speeds are instantaneous or average.

Question 14

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 15

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 16

1. Student 1, because the same energy is transferred in less time (correct answer)
2. Student 2, because the same work requires more force over a longer time
3. They have the same average power because the work is the same
4. Student 2, because power depends only on total work done

Explanation: This question tests understanding of power as the rate of energy transfer. Power is defined as P = W/t, where W is work done and t is time taken. Student 1 does work W in time t, giving power P₁ = W/t, while Student 2 does the same work W in time 2t, giving power P₂ = W/(2t) = (1/2)(W/t). Since Student 1 completes the same work in half the time, their power is twice that of Student 2. Choice C incorrectly assumes power depends only on total work, ignoring the crucial time factor. When comparing power outputs, always calculate work/time for each scenario.

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. Power is unchanged because the force is unchanged
2. Power doubles because $P=Fv$ (correct answer)
3. Power halves because the time to go distance $d$ is shorter
4. Power quadruples because kinetic energy increases with $v^2$

Explanation: This question tests understanding of power in terms of force and velocity. Power can be expressed as P = Fv, where F is the applied force and v is the velocity. Initially, the box moves at speed v with force F, giving power P₁ = Fv. When the same force F pulls the box at speed 2v, the new power is P₂ = F(2v) = 2Fv = 2P₁. The power doubles because velocity doubles while force remains constant. Choice A incorrectly ignores the velocity dependence of power. For constant force scenarios, power is directly proportional to velocity.

Question 20

1. $\displaystyle \frac{f_k d}{t}$ (correct answer)
2. $\displaystyle f_k t$
3. $\displaystyle f_k d$
4. $\displaystyle \frac{f_k}{d}$

Explanation: This question tests understanding of power calculation for constant speed motion. At constant speed, the winch force equals the friction force f_k. The work done is W = f_k × d, and power is P = W/t = (f_k × d)/t. This gives the winch's average power output as f_k d/t. Choice C represents work done but not power, missing the crucial time division. For constant speed problems, power equals force times distance divided by time.