Topic 3.5 Power: define power as the rate of energy transfer, distinguish average from instantaneous power, and compute it from $P = dW/dt$ and $P = \vec{F}\cdot\vec{v}$.

What this topic is asking

The College Board (Topic 3.5) wants you to define power as the rate of energy transfer, to distinguish average from instantaneous power, and to compute it from $P = \dfrac{dW}{dt}$ and from $P = \vec{F}\cdot\vec{v}$. Power is how the energy concepts of Unit 3 connect to time: two processes can transfer the same energy but at very different rates, and the exam often probes the constant-power case where the available force changes with speed.

Defining power

Power answers "how fast?" rather than "how much?". A weightlifter and a forklift can each raise a load the same height, doing the same work, but the forklift may do it far faster and so deliver far more power. The watt is small enough that engines are often rated in kilowatts or horsepower ($1$ hp $\approx 746$ W). Because energy transferred equals power times time ($E = P\,\Delta t$ for constant power), the kilowatt-hour is an energy unit, not a power unit, a distinction the exam likes to test.

Average versus instantaneous power

Average power uses the total energy transferred over a whole interval, $P_{avg} = \Delta E/\Delta t$. Instantaneous power is the value at a single moment, the time derivative $P = dW/dt$. They differ whenever the rate of energy transfer changes during the interval, just as average and instantaneous velocity differ when the speed varies. A car accelerating at constant engine power delivers the same average and instantaneous power (both constant), but a car whose throttle varies has an instantaneous power that swings while the average smooths it out.

Power from force and velocity

A particularly useful form expresses instantaneous power through the force doing the work and the velocity of the object:

Only the component of the force along the velocity delivers power; a force perpendicular to the motion (like the centripetal force) delivers zero power. This form is the key to constant-power problems. If an engine delivers a fixed power $P$, then $F = P/v$: at low speed the force is large (strong acceleration off the line), and as the speed climbs the available force shrinks, so the acceleration tapers off and the car approaches a top speed where the driving force just balances the resistance.

Try this

Q1. A motor does $6000$ J of work in $4.0$ s. Calculate its average power. [2 points]

• Cue. $P_{avg} = W/\Delta t = 6000/4.0 = 1500$ W.

Q2. A car's engine delivers $30$ kW. Calculate the driving force when the car moves at $25$ m/s at full power. [2 points]

• Cue. $F = P/v = 30\,000/25 = 1200$ N.