Two cars have identical engines producing the same power output. Car A climbs a steep hill slowly, exerting a large force. Car B drives on a flat road at high speed, exerting a small forward force. Which statement is correct?
ACar A transfers more energy per second because the hill requires a larger force
BCar B transfers more energy per second because its speed is higher
CBoth cars transfer energy at the same rate, since they have the same power output
DCar A has greater power because it exerts more force against gravity
Power is the rate of energy transfer: P = F · v. Two machines with identical power output transfer energy at the same rate, even if they operate at very different forces and speeds. Car A uses high force × low velocity; Car B uses low force × high velocity — the product F · v is the same for both. Options A and D confuse force magnitude with power; a larger force does not imply a higher power output if the speed is proportionally lower.
Question 2 Multiple Choice
A centripetal force keeps a satellite moving in a circular orbit at constant speed. How much power does the centripetal force deliver to the satellite?
AP = mv²/r, since that is the centripetal force magnitude times the speed
BA large positive power, because the satellite is moving at high speed
CZero, because the centripetal force is always perpendicular to the velocity
DNegative power, because the centripetal force acts inward while the satellite moves tangentially outward
Power is P = F · v = |F||v|cos(θ), where θ is the angle between force and velocity. For circular motion, the centripetal force always points toward the center while the velocity is always tangential — these are perpendicular, so θ = 90° and cos(90°) = 0. Therefore P = 0. This is consistent with the work-energy theorem: the satellite's speed (and kinetic energy) doesn't change during uniform circular motion, confirming that no net energy is transferred. Option A multiplies the force magnitude by speed without accounting for the perpendicularity.
Question 3 True / False
A machine that does twice as much total work in the same amount of time has twice the power.
TTrue
FFalse
Answer: True
Average power is defined as P̄ = ΔW/Δt — total work done divided by the time interval. If two machines operate over the same time interval and one does twice the work, it is delivering energy at twice the rate, so it has twice the average power. This follows directly from the definition. Note the distinction from instantaneous power P = dW/dt = F · v, which can vary moment to moment; the statement here concerns average power over a fixed interval.
Question 4 True / False
A high-power engine generally does more total work than a lower-power engine.
TTrue
FFalse
Answer: False
Power is the *rate* of doing work, not the total amount. A low-power motor running for a long time can do far more total work than a high-power motor that runs briefly. For example, a 1 W motor running for a year does about 31.5 MJ of work; a 1 MW motor running for 1 second does only 1 MJ. Total work equals power multiplied by time (W = P · t for constant power), so time horizon matters as much as power level. The two quantities answer different questions.
Question 5 Short Answer
Why is instantaneous power given by the dot product F · v rather than simply the product of magnitudes |F| × |v|?
Think about your answer, then reveal below.
Model answer: Only the component of force parallel to the velocity does work and transfers energy. A force perpendicular to motion — like the normal force on a surface, or centripetal force in circular motion — does no work and delivers no power, even though it may be large. The dot product F · v = |F||v|cos(θ) automatically accounts for this by including only the parallel component (|F|cos(θ)). Using the product of magnitudes alone would incorrectly count perpendicular forces as contributing to energy transfer.
This is a direct consequence of the definition of work: dW = F · ds. Dividing both sides by dt gives P = F · (ds/dt) = F · v. A force perpendicular to ds contributes zero to dW regardless of its magnitude, so it contributes zero to power. The dot product is not a mathematical convenience — it encodes the physical fact that energy transfer depends on alignment between force and motion.