5.4.2 Power Delivered by a Force (P=Fv)

Remember that work done W by a force F acting on an object through a distance s is given by

$W=Fs$

If we differentiate both sides of the equation with respect to time, we get

$\begin{aligned}\frac{{dW}}{{dt}}&=\frac{d}{{dt}}(Fs)\\\frac{{dW}}{{dt}}&=F\frac{{ds}}{{dt}}\\P&=Fv\end{aligned}$

This means that if a force F is acting on an object, and that object is moving at speed v, then F is doing work (on the object) at the rate of Fv. In other words, the power P delivered by a force F is given by Fv.

Consider a car fitted with an engine that provides a constant power P_engine. As the car moves, it experiences a drag force D that increases with speed of the car. Let’s assume that D is proportional to v²

$D=k{{v}^{2}}$

This means that the drag force is causing the car to losing energy at the rate of

${{P}_{{loss}}}=Dv=(k{{v}^{2}})(v)=k{{v}^{3}}$

At low speed, the car can gain KE since ${{P}_{{engine}}}>{{P}_{{loss}}}$ . On the other hand, at high speed, the car must lose KE since ${{P}_{{engine}}}<{{P}_{{loss}}}$ . The maximum speed of the car is thus given by

$\begin{aligned}{{P}_{{engine}}}&={{P}_{{loss}}}\\&=k{{v}_{{\max }}}^{3}\\{{v}_{{\max }}}&=\sqrt[3]{{\frac{{{{P}_{{engine}}}}}{k}}}\end{aligned}$

The power-speed graph illustrates the power dynamics pretty neatly.

The force-speed graph is also interesting.

Note that an engine that provides constant power cannot provide the same engine force F_engine at different speeds.

$\begin{aligned}(P&=Fv)\\{{P}_{{engine}}}&={{F}_{{engine}}}v\\{{F}_{{engine}}}&=\frac{{{{P}_{{engine}}}}}{v}\end{aligned}$

In fact, the math shows that F_engine is inversely proportional to v. (That’s why the F-v graph has a $y=\frac{1}{x}$ shape). So as the car accelerates, not only does the retarding drag force increases, the propelling engine force also decreases. Maximum speed occurs at the point where the two graphs meet.