Instantaneous Power problem

It can still be valid because you use instantaneous velocity and instantaneous force, not constant velocity or constant force. It gives the power output for one instant in time regardless of the operations of the system over time.

 

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.

 

Well here is what I think it is.

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]

 

The work to move something from point x to point y is given by the line integral
[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.

 

during time interval dt, force F can be regarded as a constant, the work done in dt is dW = F dot ds, power is P = dW/dt = F dot ds/dt = F dot v. so even though force is not constant, P = F dot v is still valid.