All questions
Question 1
In two trials, a force does work 2W in time t (Trial 1) and work W in time t/2 (Trial 2). Which trial has greater average power?
- Trial 1, because it has more work
- Trial 2, because it takes less time
- They have equal average power (correct answer)
- 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
A constant horizontal force F accelerates a cart from rest. At one instant the cart's speed is v. What is the instantaneous power delivered by the force then?
- vF
- Fv (correct answer)
- 21mv2
- FΔx
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
A student runs up a flight of stairs, increasing gravitational potential energy by ΔU in time t. A second student increases potential energy by 2ΔU in time t. Whose average power is greater?
- First student, because power depends on time only
- Second student, because more energy is transferred in the same time (correct answer)
- They have the same average power because they both climb stairs
- 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
A constant force F pulls a sled a distance d at constant speed in time t. A second trial uses force 2F and moves the sled distance d at constant speed in time t/2. Which trial has greater average power?
- The first trial, because the distance is the same.
- The second trial, because average power scales as W/t and both W and 1/t increase. (correct answer)
- They have equal average power because both are constant-speed motions.
- 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
A person carries a box horizontally at constant speed across a level floor. The person's force on the box is vertical and equals mg. What is the mechanical power delivered to the box?
- mgv
- mg
- 0 (correct answer)
- mgh
Explanation: This question tests understanding of mechanical power and the direction of force relative to motion. Power is the rate of doing work, calculated as P = F·v·cos(θ), where θ is the angle between force and velocity. The person applies a vertical force (upward) while the box moves horizontally, making θ = 90°. Since cos(90°) = 0, the mechanical power delivered to the box is P = mg·v·0 = 0. Choice A (mgv) incorrectly assumes the force is in the direction of motion. The key insight is that only the component of force in the direction of motion contributes to mechanical power.
Question 6
A winch pulls a sled at constant speed across rough ground with kinetic friction force fk. If the sled moves distance d in time t, what is the winch's average power output?
- tfkd (correct answer)
- fkt
- fkd
- dfk
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.
Question 7
A crane lifts a mass m straight up at constant speed v. Ignoring air resistance, what is the crane's power output?
- mgv (correct answer)
- mg/v
- mg
- mgh
Explanation: This question examines power output in vertical lifting at constant speed. Power is the rate of work done or energy transfer, and for constant velocity, P = F v. The force required to lift mass m at constant speed v is F = m g to counter gravity, so P = m g v. This applies because the crane transfers energy at a rate matching the gain in gravitational potential energy over time. Choice D, m g h, is a distractor representing work, not power, as it omits the time or velocity factor. A transferable strategy is to identify if the motion is at constant speed, then use P = F v with the balancing force.
Question 8
A student runs up a staircase, gaining gravitational potential energy ΔUg in time t. What is the student's average power output?
- ΔUgt
- tΔUg (correct answer)
- ΔUgt
- ΔUg
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 9
An elevator raises a load at constant speed. In trip A it transfers E joules in time t; in trip B it transfers 2E joules in time 2t. How do the average powers compare?
- PB=2PA because more energy is transferred.
- PB=21PA because the time is longer.
- PB=PA because energy and time both double. (correct answer)
- 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 10
A car engine provides a constant driving force F while the car moves at constant speed v for time t. If the time doubles while F and v stay constant, average power is
- doubled
- halved
- unchanged (correct answer)
- 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 11
A constant force does work W on an object. Trial 1 transfers this energy in time t; Trial 2 transfers energy 2W in time 4t. Which trial has greater average power?
- Trial 1, because less energy means greater power
- Trial 2, because more work always means more power
- Trial 1, because tW>4t2W (correct answer)
- 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 12
A constant horizontal force F accelerates a cart from rest on a frictionless track. At time t, the cart's speed is v. At time 2t, its speed is 2v. How does instantaneous power compare at t and 2t?
- Power is the same because the force is constant.
- Power at 2t is twice that at t because P=Fv. (correct answer)
- Power at 2t is four times that at t because kinetic energy quadruples.
- 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 13
Two machines each transfer energy to a spring. Machine 1 increases the spring's elastic potential energy by E in time t; Machine 2 increases it by E in time t/4. Which has greater average power?
- Machine 1, because it uses less power to store the same energy.
- Machine 2, because it transfers the same energy in less time. (correct answer)
- They have equal power because the energy change is the same.
- 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 14
A cyclist rides on level ground at constant speed v against resistive force Fr. Cyclist A rides at speed v; Cyclist B rides at speed 3v, with the same Fr in each case. Which requires greater power?
- Cyclist A, because less speed means more time resisting the force.
- Cyclist B, because P=Frv and tripling v triples power. (correct answer)
- They require equal power because the resistive force is the same.
- 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 15
A cyclist rides at constant speed. On flat ground, the resistive force is Fr and the cyclist's speed is v. On a rougher road, the resistive force becomes 2Fr while the cyclist maintains the same v. How does the cyclist's mechanical power output change?
- It doubles because P=Fv and F doubles while v stays the same. (correct answer)
- It stays the same because the speed is unchanged.
- It halves because more force means less speed.
- 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 16
A winch raises a m-kg bucket vertically at constant speed. Trial 1 lifts it height h in time t; Trial 2 lifts the same bucket height h in time 2t. Compare average power.
- Trial 2 has greater power because it acts longer
- Trial 1 has twice the average power of Trial 2 (correct answer)
- They have equal average power because mgh is the same
- 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 17
A box is pulled at constant speed v by a horizontal force F for time t. What is the average power delivered by F?
- Ft
- Fv (correct answer)
- F/v
- 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 18
A student pushes a crate with horizontal force F at constant speed across a floor. If the student pushes for twice as long at the same speed and force, how does average power change?
- It doubles because the time doubles
- It halves because power is inversely proportional to time
- It stays the same because P=Fv and both are unchanged (correct answer)
- 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 19
A battery transfers energy to a device at a constant rate. In 5 s it transfers E. In 10 s it transfers 2E. What happens to average power?
- It doubles because more energy is transferred
- It halves because the time is longer
- It stays the same because 5E=102E (correct answer)
- 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 20
A motor does work W on a system in time t. A second motor does work 2W in time 4t. Which motor has greater average power?
- The first motor, because it does less work.
- The second motor, because it does more work.
- The first motor, because tW>4t2W. (correct answer)
- 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=tW. For the first motor: P1=tW. For the second motor: P2=4t2W=21tW=2P1. Since P1=2P2, 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=tW explicitly and compare the ratios, not just look at work or time alone.