Power

Power is how fast energy is transferred or work gets done. The average power over a stretch is the work divided by the time, $P_{avg} = W/t$, and the instantaneous power is the rate right now, $P = \tfrac{dW}{dt}$. The same idea covers any energy transfer: $P_{avg} = \Delta E / \Delta t$. The unit is the watt (W), one joule per second.

Power and work answer different questions: work says how much energy moved, power says how fast. Lifting a crate to the same shelf in $4$ s instead of $8$ s takes the same work at twice the average power. And the energy a steady source delivers is power times time, $E = Pt$, so a $50$ W motor running $60$ s moves $3000$ J while a $200$ W motor running $10$ s moves only $2000$ J: the small motor wins by running longer.

Keep the units apart. A $200$ W motor is not holding $200$ J; it is moving $200$ J every second it runs. Watts are joules per second, a rate, never an amount.

Three habits cost points: treating power and energy as the same number (this section), using the full force no matter its angle to the motion (§2), and averaging the power values off a list or reading constant power as constant speed (§3 and §4). Throughout this lesson, $g = 10$ m/s$^2$ where it is needed.

When a constant force $F$ acts at angle $\theta$ to the velocity, the work over a small step is $F\,d\cos\theta$; divide by the time the step took and the power is $P = Fv\cos\theta$. It is the same cosine filter as the work formula $W = Fd\cos\theta$ from the work topic: only the part of the force along the motion counts.

A force at $90^\circ$ to the velocity delivers zero power, however big it is: the normal force on a sliding block, the tension on a ball whirling in a circle at constant speed, gravity at the very top of a projectile's arc. That matches the energy picture: in each case the speed is not changing at that instant, so no energy is flowing.

For example, a rope with $100$ N of tension at $37^\circ$ above the horizontal pulls a sled moving at $2$ m/s. The power is $P = Fv\cos\theta = (100)(2)(0.8) = 160$ W. Using the full $100$ N would give $200$ W, an overcount; the $\sin\theta$ part of the pull points across the motion and delivers nothing.

Past $90^\circ$ the cosine goes negative and so does the power: a force aimed against the motion, like friction on a sliding box, has $\cos 180^\circ = -1$ and drains energy at the rate $Fv$. Negative power is energy flowing out.

Average power is always the total work over the total time, $P_{avg} = W/t$. It is not the average of the power values on a list, because stretches that last longer count for more.

For example, an engine delivers $60$ W for $10$ s and then $20$ W for $30$ s. The work is $(60)(10) + (20)(30) = 1200$ J over $40$ s, so $P_{avg} = 1200/40 = 30$ W. The midpoint of $60$ and $20$ is $40$ W, and it is wrong here because the engine spends three times longer at $20$ W. The midpoint only works when the times match.

The graph makes this one picture: on a plot of power versus time, the area under the curve is the work, and the average power is the height of the flat line that closes off the same area over the same stretch. A power that climbs steadily from $0$ to $40$ W makes a triangle, so the work is half of peak times time and the average is $20$ W, half the peak.

The check is quick: multiply each power by its own time, add the pieces, divide by the whole time. If you find yourself averaging the powers directly, stop and ask whether the times are equal.

Hold the power steady and the force cannot be: with $P = Fv$ pinned, the force along the motion is $F = P/v$. Every gain in speed leaves less force, and every loss of speed frees up more.

So a cart driven at constant power from a slow start keeps speeding up, but by smaller and smaller amounts: the force, and with it the acceleration, fades as $v$ climbs. The cart does not jump to one cruising speed, and the force does not hold steady. Constant power is neither constant force nor constant speed.

For example, a motor holds $P = 16$ W on a cart. At $v = 1$ m/s the drive force is $F = 16/1 = 16$ N; at $v = 4$ m/s it is down to $4$ N; at $v = 8$ m/s, just $2$ N. The motor feeds the same $16$ J every second; the faster the cart goes, the less force that energy budget can supply.

The energy still adds up: over a time $t$ the motor does $W = Pt$ of work, and with nothing else acting that work is exactly the kinetic energy gained. Steady rate in, steadily rising $K$, falling $F$.

A power question almost always turns on keeping the rate apart from the amount, projecting a force onto the velocity, or averaging over time the right way.

How fast or how much? Power is the rate, $P = W/t$; energy is the amount, $E = Pt$. To compare energy from two sources, multiply each power by its own time first. More watts does not mean more joules until the times are in.

Power at an angle. Project the force onto the velocity before multiplying: $P = Fv\cos\theta$. Along the motion, $\cos\theta = 1$ and $P = Fv$; sideways, $\cos\theta = 0$ and the power is zero; against the motion, the power is negative and energy drains.

The average. Total work over total time, $P_{avg} = W/t$: weight each power by its own time, never average the list. On a $P$ versus $t$ graph the work is the area under the curve, and at constant power the force is whatever $F = P/v$ leaves at the current speed.

Setup. A rope with $100$ N of tension at $60^\circ$ above the horizontal drags a crate along a floor at a steady $2$ m/s. (a) Find the power the rope delivers. (b) Find the work the rope does in $10$ s. (c) On a different stretch the rope delivers $160$ W for $2.5$ s and then $80$ W for $7.5$ s; find the average power over those $10$ s.

(a) Power at an angle. Only the part of the tension along the motion delivers power: $P = Fv\cos\theta = (100)(2)(\cos 60^\circ) = (100)(2)(0.5) = 100$ W. The full tension would give $200$ W, double the truth; the vertical part of the pull lightens the crate but delivers nothing forward.

(b) Work from power. A steady $100$ W means $100$ J every second, so in $10$ s the rope does $W = Pt = (100)(10) = 1000$ J of work. Watts are not joules: the $100$ is a rate, and the time turns it into an amount.

(c) The average. Total work over total time: $W = (160)(2.5) + (80)(7.5) = 400 + 600 = 1000$ J across $10$ s, so $P_{avg} = 1000/10 = 100$ W. The midpoint of $160$ and $80$ is $120$ W, and it is wrong because the low stretch lasts three times longer.

The traps to avoid. Do not read $100$ W as $100$ J (multiply by the time), do not multiply the full force by the speed when it pulls at an angle (project it first with $\cos\theta$), and do not average the power values off the list (weight each one by its own time).