# Power

*Power*is the rate at which energy is produced, consumed, or transformed, that is,

….(8) |

A car (1000 kg) traveling 55 mph has a forward cross-sectional area of about 4 m^{2}. What is the power dissipated by air resistance? (Use 1 mph ≈ 0.45 m/s. Recall that the formula for air resistance is *F*_{air }= *C**∉**Av*^{2}, where *∉*, the density of air, is 1.29 kg/m^{3} and *C* ≈ 0.2.)

First, we DRAW A DIAGRAM (Figure 9-10).

**Figure 9-10**

We have only one formula to work with (*P* = Δ*E*/Δ*t*), but we have neither an energy nor a time. But we have several formulas for energy, so let's try to connect it with the force given in the problem. We have

since *cos**φ* = –1. We can substitute for *F*_{air}. And the expression Δ*x/*Δ*t *reminds us of velocity, so we have

*P* = –*F*_{air}*v* = –(*C**ρ**Av*^{2})*v*

= –1.6 x 10^{4} W.

The minus sign indicates that the energy is dissipated by the force.

The same car is traveling 65 mph. What is the power dissipated by air resistance?

P = –2.6 x 104 W.

Why is there such a large difference?