# Power

Claimed by Aeze3

This topic covers Power.

## The Main Idea

Power is the rate of doing work or the amount of energy consumed over an interval of time.

### A Mathematical Model

When a force is applied over a distance in a unit of time, power is calculated by

- [math]\displaystyle{ power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t} }[/math]

where **F** is force, **Δr** is displacement, **Δt** is the duration of time and **W** is work.

It then follows that instantaneous power is

- [math]\displaystyle{ power = F\cdot v }[/math]

where **v** is velocity.

The SI unit for power is watts (J/s)

### A Computational Model

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

## Examples

Be sure to show all steps in your solution and include diagrams whenever possible

### Simple

A certain motor is capable of doing 3000 J of work in 12 s What is the power output of this motor?

- [math]\displaystyle{ power = \frac{W}{\Delta t} = \frac{3000 J}{12 s} = 250 watts }[/math]

### Middling

Here are questions dealing with human power. **(a)** If you follow a diet of 2000 food calories per day (2000 kC), what is your average rate of energy consumption in watts (power input)? (A food or “large” calorie is a unit of energy equal to 4.2 J; a regular or “small” calorie is equal to 4200 J.) **(b)** How many days of a diet of 2000 large calories are equivalent to the gravitational energy change from sea level to the top of Mount Everest, 8848 m above sea level? Assume your weight is 58 kg. (The body is not anywhere near 100% efficient in converting chemical energy into change in altitude. Also note that this is in addition to your basal metabolism.)

**(a)** [math]\displaystyle{ power = \frac{W}{\Delta t} = \frac{2000 kC}{day} \cdot \frac{4200 J}{1 kC} \cdot \frac{1 day}{24 h} \cdot \frac{1 h}{3600 s} = 97.2 watts }[/math]

**(b)** [math]\displaystyle{ {\frac{97.2 J}{s}} \cdot \frac{3600 s}{1 h} \cdot \frac{24 h}{1 day} = 8398080 J/day }[/math]

- [math]\displaystyle{ \Delta U_g = mg\Delta y = (58 kg)(9.8 m/s^2)(8848 m - 0 m) = 5029203.2 J }[/math]

- [math]\displaystyle{ \Delta t = \frac{W}{power} = \frac{5029203.2 J}{8398080 J/day} = 0.599 days }[/math]

### Difficult

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## History

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## See also

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### Further reading

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### External links

Internet resources on this topic

"General Mechanics/Work and Power." - Wikibooks, Open Books for an Open World. Web. General Mechanics/Work and Power

## References

Chabay, Ruth W.; Sherwood, Bruce A. Matter and Interactions, 4th Edition: 1-2. Wiley.