# 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 Pengine. 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 Fengine 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 Fengine 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.

Video Explanation

Power Delivered by a Force

Concept Test

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