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.

Power can also be defined as the rate of change of energy transfer:

[math]\displaystyle{ P = \frac{\Delta E}{\Delta t} }[/math]

When there is no heat transfer (Q = 0), the change in energy is equal to the work done. Therefore,

[math]\displaystyle{ P = \frac{\Delta E}{\Delta t} = \frac{W}{\Delta t} }[/math]

Since work is the dot product of force and displacement, we can further expand the equation:

[math]\displaystyle{ P = \frac{\vec{F}\cdot\Delta\vec{r}}{\Delta t} }[/math]

Rearranging gives:

[math]\displaystyle{ P = \vec{F}\cdot\frac{\Delta\vec{r}}{\Delta t} }[/math]

It then follows that instantaneous power is

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

where v is velocity.

The SI unit for power is watts (J/s). The units can also be expressed in terms of fundamental SI units:

[math]\displaystyle{ Watt = \frac{J}{s} = \frac{N\cdot m}{s} = \frac{kg\cdot m^2}{s^3} }[/math]

Note that W can represent either watts or work, depending on the context.

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

A 1200 kg elevator is lifted vertically from rest through a height of 45 m in 15 seconds. At the end of the lift, the elevator is moving upward at a speed of 6 m/s. Assuming the motor operates at a constant power output, calculate the average power output required to lift the elevator. Ignore friction and air resistance.

‘’‘Step 1:’’’ Identify the work done by the motor.

The motor must provide enough energy to increase both the gravitational potential energy and the kinetic energy of the elevator.

W = \Delta U_g + \Delta K

‘’‘Step 2:’’’ Calculate the change in gravitational potential energy.

\Delta U_g = mg\Delta y

= (1200 kg)(9.8 m/s^2)(45 m)

= 529200 J

‘’‘Step 3:’’’ Calculate the change in kinetic energy.

Since the elevator starts from rest, its initial velocity is zero.

\Delta K = \frac{1}{2}m(v_f^2-v_i^2)

= \frac{1}{2}(1200 kg)(6^2-0^2)

= 21600 J

‘’‘Step 4:’’’ Calculate the total work performed by the motor.

W = \Delta U_g + \Delta K

= 529200 J + 21600 J

= 550800 J

‘’‘Step 5:’’’ Calculate the average power output.

Using the relationship between work, time, and power:

power = \frac{W}{\Delta t}

= \frac{550800 J}{15 s}

= 36720 watts

‘’‘Final Answer:’’’

\boxed{power = 36,720 \text{ W} = 36.72 \text{ kW}}

Connectedness

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History

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See also

• Work
• Gravitational Potential Energy

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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.