Power: Difference between revisions
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When a force is applied over a distance in a unit of time, power is calculated by | When a force is applied over a distance in a unit of time, power is calculated by | ||
:<math>power = \frac{F \Delta r}{\Delta t} | [[File:power(1).jpg |border|right]] | ||
:<math>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 | It then follows that instantaneous power is | ||
:<math>power = F\cdot v</math> where '''v''' is velocity. | :<math>power = F\cdot v</math> | ||
where '''v''' is velocity. | |||
The SI unit for power is watts (J/s) | |||
===A Computational Model=== | ===A Computational Model=== | ||
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===Simple=== | ===Simple=== | ||
A certain motor is capable of doing 3000 J of work in 12 s | A certain motor is capable of doing 3000 J of work in 12 s | ||
What is the power output of this motor? | What is the power output of this motor? | ||
:<math>power = \frac{W}{\Delta t} | :<math>power = \frac{W}{\Delta t} = \frac{3000 J}{12 s} = 250 watts </math> | ||
===Middling=== | ===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>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> {\frac{97.2 J}{s}} \cdot \frac{3600 s}{1 h} \cdot \frac{24 h}{1 day} = 8398080 J/day </math> | |||
:<math> \Delta U_g = mg\Delta y = (58 kg)(9.8 m/s^2)(8848 m - 0 m) = 5029203.2 J </math> | |||
:<math> \Delta t = \frac{W}{power} = \frac{5029203.2 J}{8398080 J/day} = 0.599 days </math> | |||
===Difficult=== | ===Difficult=== | ||
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== See also == | == See also == | ||
*[[Work]] | |||
*[[Gravitational Potential Energy]] | |||
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context? | Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context? | ||
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Internet resources on this topic | Internet resources on this topic | ||
"General Mechanics/Work and Power." - Wikibooks, Open Books for an Open World. Web. [https://en.wikibooks.org/wiki/General_Mechanics/Work_and_Power General Mechanics/Work and Power] | |||
==References== | ==References== | ||
Latest revision as of 09:52, 2 December 2015
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
.jpg)
- [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
Connectedness
- How is this topic connected to something that you are interested in?
- How is it connected to your major?
- Is there an interesting industrial application?
History
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.
See also
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?
Further reading
Books, Articles or other print media on this topic
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.