Is My Derivation of the Power Equation Correct?

In summary, the conversation was about a derivation of the power equation. The equation for mechanical work over a changing path and changing force was taken from Wikipedia and agreed with the speaker's text. To calculate average power, the equation was modified by adding (1/delta t) in front of it. The speaker clarified that this equation calculates the total work done through the motion and divides it by the total time taken. For instantaneous power, the speaker explained that one must limit themselves to an infinitesimal amount of path and the equation involves the dot product of force and velocity. The speaker confirmed that the calculation was correct for average power.
  • #1
Tourniquet
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Hi,

I am working on a derivation of of the power equation. It is very simple, I would like to know if it is correct.

I started with the equation for mechanical work over a changing path, and changing force. This is from Wikipedia, and it agrees with my text.


W = [tex]\int[/tex] F º ds

where: º is the symbol for dot product


F is the force vector; and
s is the position vector.

Then to make it power I just put (1/delta t) in front of it to make:

P = [tex]\frac{1}{\Delta t}[/tex] [tex]\int[/tex] F º ds

Let me know if this checks out.

Thanks!
 
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  • #2
Yes, that is correct, in as much as you want to calculate the average power that the force did during the motion through the specified path. The reason is that you calculated the total work done through the motion, and divided that by the total time it took to complete that motion.

If you want to have the instantaneous power, then you have to limit yourself to an infinitesimal amount of path:

[tex] P_{inst} = \frac{1}{\delta t} \int_{\delta s} F . ds = \frac{F . \delta s}{\delta t} = F . \frac{\delta s}{\delta t} = F . v [/tex]

So the instantaneous power done by a force on a moving point is given by the dot product of the force and the velocity of that point.
 
  • #3
Thanks, I was going for average power.
 

FAQ: Is My Derivation of the Power Equation Correct?

1. What is the power equation derivation?

The power equation derivation is a mathematical process used to derive the equation for power, which is defined as the rate at which work is done or energy is transferred. This equation states that power is equal to the product of force and velocity, or power = force x velocity.

2. How is the power equation derived?

The power equation is derived using the principles of work and energy. Work is defined as the product of force and displacement, and energy is the ability to do work. By combining these definitions and applying the concept of rate, we can derive the power equation as power = work / time = (force x displacement) / time = (force x velocity).

3. Why is the power equation important?

The power equation is important because it allows us to quantify the rate at which work is done or energy is transferred. This is crucial in many scientific and engineering fields, such as mechanics, electricity, and thermodynamics, as power is a fundamental aspect of these systems.

4. What are the units of power?

The units of power are typically joules per second (J/s) or watts (W). This reflects the fact that power is the rate of energy transfer, with one watt being equal to one joule per second.

5. Can the power equation be applied to all systems?

While the power equation can be applied to many systems, it is important to note that it is based on the assumption of constant velocity. Therefore, it may not be applicable to systems with changing velocities or other complex factors. In such cases, alternative equations may need to be used to calculate power.

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