Power in physics is defined as the rate at which work is done. Mathematically this can simply be stated as the derivative of work, or[br][br][center][math]P=\frac{dW}{dt}=\frac{d}{dt}\left(\vec{F}\cdot\vec{ds}\right).[/math][/center][br] [br]If power is large, it means work is done rapidly, or over a short duration of time. There are useful simplifications of the definition above. For the average power, all we need is the total work done over some longer time period [math]\Delta t[/math] and we can write [math]P_{avg} = \frac{W}{\Delta t}.[/math] This is mathematically true because if we add lots of dW terms in the numerator by integrating dW, we get W. If we add lots of time intervals dt by integrating, we get [math]t_f-t_i=\Delta t.[/math] [br][br]For instantaneous power we must assume that the force and angle between the vectors are unchanging over an infinitesimal duration, and therefore, starting with our work definition we may write:[br][br][center][math] [br]W=\int\vec{F}\cdot \vec{ds} \\[br]dW=\vec{F}\cdot \vec{ds} \\[br]dW=\vec{F}\cdot \vec{v}dt \\[br]P=\frac{dW}{dt}=\vec{F}\cdot\vec{v} \\[br][/math][/center]

[i] [color=#1e84cc]EXAMPLE: A car starts from rest and gets up to [math]60 mi/hr = 26.8 m/s[/math] in a time of 4.0 s. If the car's mass is 1300 kg, how much average power did this require?[/color][br][br][color=#1e84cc]SOLUTION: We need to use the work-kinetic energy principle to find W first. Then we can use the definition of average power to get the answer. [math]W=\Delta K = \frac{1}{2} 1300kg ((26.8m/s)^2 - 0^2) = 4.66\times 10^5 J.[/math] Then we get [math]P_{avg} = 4.66\times 10^5J/4.0s = 1.17\times 10^5 W.[/math][/color][/i][br][br]This value of 117kW can be converted to horsepower (hp) by using 746W=1hp. This tells us that this nearly 3000lb car manages to go from zero to sixty in 4.0 seconds with only 156 horsepower. If you know cars this rings untrue. So what's going on?[br][br]The first problem is that advertisers will tell you the power the engine makes outside of the vehicle. When placed in a vehicle and tasked with turning bearings, transmission with fluid and differentials, typically 20-25% of that power is lost before the rubber hits the road.[br][br]The second issue is that engines only make their rated power at a singe, rather high engine speed. You will see power of an engine reported as, for example, 240hp at 5800rpm. What this means is that this engine makes 240 horsepower when the crankshaft is rotating at a rate of 5800 revolutions per minute. Refer back to the chapter on propulsion to see such a curve of power versus engine speed. [br][br]AN ASIDE: Note that while engine speed is measured in revolutions per minute, it shouldn't be, given the definition of revolution as compared to rotation. [br][br]In any case, at all other engine speeds between idle and red line the engine makes less power than this. Right at idle it might be just a few tens of horsepower. As a car accelerates up to speed, the engine spends most of its time making considerably less power than the peak rating.[br][br]So our car needs 156hp at the wheels, and if we assume 20% of that is lost to friction before reaching the wheels, so that only 80% makes it to the wheels, this implies [math]156hp = 80\%P_{engine}[/math] where [math]P_{engine}[/math] is the horsepower the engine would have if removed from the car and not tasked with turning all the gears, etc. This amounts to [math]195hp.[/math] Since this is the average power at all engine speeds sampled while accelerating, it might correspond to a car having considerably more peak power than this... maybe 50% more. This puts the number around 300hp. Assuming this car has no further issues preventing rapid acceleration such as traction limitations, it is feasible for a car of this weight and power to accelerate to 60 mph in 4.0s. It would certainly need all wheel drive to avoid traction limitations. [br][br]One last comment is that we are neglecting air drag and rolling resistance, neither of which is large at these speeds, but both of which will further inflate the required power number by perhaps 15-20 hp.[br]

[i][color=#1e84cc]EXAMPLE: Another application of the power equation might come from an example like the following: A cyclist in the Tour de France is riding up a climb of 15% slope (grade). How fast can they ride up this mountain given a combined rider plus bike mass of 77 kg and a sustained power output of 420 W?[/color][br][br][color=#1e84cc]SOLUTION: In this case, we need to use the instantaneous power equation, [math]P=\vec{F}\cdot\vec{v}.[/math] The force that the rider exerts to climb the hill is exactly equal to and opposite to the force of gravity trying to draw them down the hill if they are traveling at constant speed. Recall that the force of gravity down the slope is [math]F=mg\sin\theta.[/math] So [math]P=mg\sin\theta v,[/math] and we don't have to worry about the dot product since the two vectors are parallel. We are neglecting air drag here since the speed is low. Otherwise on a more gradual climb, this might not be a good thing to do. Our result here is [math]v=3.8 \frac{m}{s}.[/math][/color][/i]

Our world is full of vehicles. We discussed in a previous chapter that such vehicles - whether trains, airplanes, cars or bikes - are subject to air drag as the major resistive force while they are traveling at typical cruising speeds. Recall that the force of air drag is proportional to speed squared, or [math]F_{drag}=\tfrac{1}{2}\rho c_dAv^2.[/math] Since instantaneous power is given by [math]P=\vec{F}\cdot\vec{v}[/math] the power to overcome air drag is given by [math]P_{drag}=\tfrac{1}{2}\rho c_dAv^3.[/math] It is assumed that the velocity is parallel to the tractive force that propels the vehicle along, and that at cruise this tractive force equals the resistive force of drag, since we are neglecting other less significant resistive forces. This cubic dependence means that it takes a tremendous increase in power to double your speed on a bicycle or to increase the top speed of a car. Let's take a look at this.

[color=#1e84cc][br]EXAMPLE: Suppose a certain car has a drag-limited top speed of 100 mph (miles per hour) and can produce 100 hp (horsepower). I mention drag-limited because many cars these days have electronically-limited top speeds. How much power does this same car need in order to reach a drag-limited top speed of 200 mph like a typical supercar?[br][br]SOLUTION: To do this, we don't need to convert to SI units. We can just use ratios. It can be set up like this: [br][br][center][math]\frac{P_2}{P_1}=\frac{\tfrac{1}{2}\rho c_d A v_2^3}{\tfrac{1}{2}\rho c_d A v_1^3} \\[br]\frac{P_2}{P_1}=\frac{v_2^3}{v_1^3} \\[br]\frac{P_2}{100 hp}=\frac{(200 mph)^3}{(100 mph)^3} \\[br]P_2 = 2^3(100 hp) = 800 hp. [br][/math][/center][br]It takes eight times the power to go twice the speed! The same holds true for cyclists, which is why any person in decent condition can ride a bike close to 20mph and yet elite-level cyclists struggle to exceed 30mph for extended periods. In this case going 1.5 times as fast requires [math]1.5^3=3.375[/math] times more power. That's a tall order for even a world-class cyclist.[/color]