The material in this section is related to the concept of electrical power.

Power is the rate at which mechanical work is done. The more power we use, the faster we can do a given amount of work.

Power is work divided by time,

$$P = \frac{w}{t}$$

or the amount of change in energy divided by the change in time:

$$P = \frac{\Delta E}{\Delta t}$$

The concept of power allows us to compare the ability of systems to accomplish a certain energy change or a certain amount of physical work.

We can write a useful alternative formula for power by recognizing that average velocity is

$$\bar{v} = \frac{d}{t},$$

where d is distance and t is time. So if

$$P = \frac{F\cdot d}{t}, \; \text{then } \; P = F \bar{v}$$

In this brief section, we'll define power, talk about its units and do a couple of sample calculations.

**Power** is the rate at which mechanical work is done.

Because power is energy divided by time, its SI units are

$$\frac{Kg \, m^2}{s^2 \, s} = \frac{J}{s}$$

This unit is defined as the **Watt**, with symbol **W**.

$$1 \, W = \frac{1 \: J}{s}$$

There are other units of power commonly in use. In the U.S., the unit *horsepower (hp)* is still used to quantify the power of engines and motors, but less so as time goes by (1 hp = 745.7 W).

The Watt is named after James Watt, who invented the steam engine.

Two weight lifters each lift 100 Kg a height of 0.7 m ten times. The first takes 25 s to do it and the second takes 33 s. Calculate the amount of work done by each, and the power expended by each.

**Solution**

$$ \begin{align} w &= F\cdot d \\ &= (100 \; Kg)\left( 9.8 \, \frac{m}{s^2} \right)9.8(7 \; m) \\ &= 6867 \; J = 6.87 \; KJ \end{align}$$

Now the first does that work in 25 s, so the power is

$$P_1 = \frac{\Delta w}{\Delta t} = \frac{6.87 \; KJ}{25 \; s} = 275 \; W$$

The second does the same work in more time. The power is

$$P_2 = \frac{\Delta w}{\Delta t} = \frac{6.87 \; KJ}{33 \; s} = 208 \; W$$

Compare the power output of two cars of identical mass, one of which acclerates from 0 to 60 mi./h in 3.2 seconds, the other of which does it in 4 seconds.

**Solution**

$$ \begin{align} &\frac{60 \;mi.}{1 \; h}\left( \frac{1609 \; m}{1 \; mi} \right)\left( \frac{1 \; h}{3600 \; s} \right) \\ \\ &= 26.82 \, \frac{m}{s} \end{align}$$

Now work is force times distance. The only way to get to the force applied by each car on the road is through the acceleration, so we'll calculate that for each:

$$ \begin{align} a &= \frac{\Delta v}{\Delta t} \\ \\ a_1 &= \frac{26.82 \; m/s}{3.2 \; s} = 8.38 \, \frac{m}{s^2} \\ \\ a_2 &= \frac{26.82 \; m/s}{4 \; s} = 6.7 \, \frac{m}{s^2} \end{align}$$

Now the distance traveled by each car is $d = \frac{1}{2} at^2$:

$$ \begin{align} d_1 &= \frac{1}{2} (8.38 \; m/s^2)(3.2 \;s)^2 = 42.9 \; m\\ \\ d_2 &= \frac{1}{2} (6.7 \; m/s^2)(4 \;s)^2 = 53.6 \; m \end{align}$$

Alternatively, the distance could have been calculated by multiplying the average velocity by the time for each. Now the power: $P = (F\cdot d)/t = [(ma)\cdot d]/t$. We don't have the mass, but we can just leave it as a variable and *compare* the two power values.

$$ \begin{align} P_1 &= \frac{8.38 \cdot 42.9 \cdot m}{3.2} = 112.3 m \; W\\ \\ P_2 &= \frac{6.7 \cdot 53.6 \cdot m}{4} = 89.8 m \; W \end{align}$$

The ratio is $112.3 m / 89.8 m = 1.25$, so car 1 is 25% more powerful than car 2.500 W of power are availaible to do 220 J of work. How long will it take to do the work? How much power would we need in order to cut the time in half?

**Solution**

$$P = \frac{\Delta w}{\Delta t}$$

Rearrange to solve for time:

$$\Delta t = \frac{\Delta w}{P}$$

So the time is

$$t = \frac{220 \, J}{500 \, W} = 0.44 \, s$$

Notice that in our equation for time above, the time depends inversely on power. If power is increased and the amount of work done is constant, then the time decreases proportionally. So in order to cut the time in half, we would have to double the power to 1000 W or 1 KW.

1. |
A power of 230 W is supplied to perform mechanical work for 3.2 seconds. How much total work can be done in that time? ## SolutionIn terms of work, the power is $$P = \frac{\Delta w}{\Delta t}$$ We can rearrange to find the work: $$\Delta w = P \Delta t$$ Now just plug in what we know: $$ \begin{align} \Delta w &= (230 \, W)(3.2 \, s) \\[5pt] &= 736 \, J \end{align}$$ |

2. |
Suppose that 300 J of work is done in 1 minute. What power is required to do this amount of work in this time? ## SolutionThis is just a straightforward use of our definition of power in terms of work: $$ \begin{align} P &= \frac{\Delta w}{\Delta t} \\[5pt] &= \frac{300 \, J}{60 \, s} = 5 \, W \end{align}$$ A lot of work can be done with just a little power if we do it over an extended time. |

3. |
A 25 Kg rocket ascends at a constant speed of 32 m/s. Calculate the power supplied by the rocket engine. ## SolutionIf the rocket is moving at a constant velocity, then the only force it needs to exert is to balance the downward pull of gravity: $$ \begin{align} F = mg &= (25 \, Kg)\left(9.81 \frac{m}{s^2}\right) \\[5pt] &= 245.25 \, N \end{align}$$ If the rocket moves at 32 m/s, then it travels a distance of 32 meters every second. the amount of work done per second is $$ \begin{align} w = Fd &= (245.25 \, N)(32 \, m) \\[5pt] &= 7,358 \, J \end{align}$$ Now the power is just the amount of work done divided by the time, but our time interval is 1 s, so the power is 7,358 W or about |

4. |
A cyclist pedals at 100 rpm (revolutions per minute) with a power output of 320 Watts. If she travels a distance of 1600 m in 4 minutes, how much average force did she exert during the ride? (Recall that work = force × distance.) ## SolutionFirst, the amount of work done in those 4 minutes is $$w = P \Delta t = (320 \, W)(240 \, s) = 76.8 \, KW$$ Now we know that work is force × distance, so $$F = \frac{w}{d} = \frac{76,800 \, J}{1600 \, m} = 48 \, N$$ |

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