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## Main Question or Discussion Point

Hey, instantaneous power is computed by F dot v, but this equation should only be valid if F is constant correct?

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Hey, instantaneous power is computed by F dot v, but this equation should only be valid if F is constant correct?

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berkeman

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Nope. Why do you say that? F and v can be vector functions of time and space and whatever else.Hey, instantaneous power is computed by F dot v, but this equation should only be valid if F is constant correct?

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P = dW/dt = (dF/dt) dot s + F dot v

The above only simplifies to P = F dot v in the case that dF/dt = 0, or in other words, if F is constant.

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rcgldr

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The equation should end up as: P(t) = F(t) dot V(t). Note that work is defined as the line intergral of F(s) dot ds.The above only simplifies to P = F dot v in the case that dF/dt = 0, or in other words, if F is constant.

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rock.freak667

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W=F.s

When you dot product two vectors you get a scalar.

W=F.s=|F||s|

[tex]P=\frac{W}{t}=\frac{|F||s|}{t}=|F||v|=F.v[/tex]

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but you assumed F was constant in that...

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Stingray

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[tex]

W(x \rightarrow y) = \int_x^y F(z) \cdot \mathrm{d} z ,

[/tex]

where I'm leaving off vector notation. If the path is described parametrically by a function y(t) with y(0) = x, this can be rewritten as

[tex]

W(x \rightarrow y) = \int_0^t F(y(s)) \cdot \frac{\mathrm{d} y(s)}{\mathrm{d} s} \mathrm{d} s .

[/tex]

The path parameter here can be anything, but choose it to be time. Differentiating with respect to t only hits the upper limit of integration. The result is that

[tex]

P = \mathrm{d}W/\mathrm{d}t = F(y(t)) \cdot \frac{\mathrm{d} y(t)}{\mathrm{d} t} = F \cdot v .

[/tex]

The point here is that the definition of work does not explicitly depend on time. The work required to apply a given force along a given path is identical whether that path is traversed in a second or a year. This sets the definitions as I've given them, and does not allow any t-dependence inside the integrand.

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