Force , power and Work ?

[sAXIn]

I have a question where I have a force F(t) acting on particle ,
I intgrate 2 times and find X(t) , i used innitial conditions given x=0 , t=0 , v=0 .

1) now they ask me to write the work done by the power between t=0 , t=<epsilon> I know that dw=F dot DX but here the f and x are both functions of t so .... ???

2) after that i need to write the power of the force on time t>0 i think interval t=0 , t=t ... ???

thank's in advance

[ramollari]

1)
You have dx/dt = ... -> dx = ...dt
W = \int_{x_0}^{x}Fdx = \int_{t_0}^{t}F(t)...dt

You know F in terms of t to get the solution.

2)
P = Fv = F(t)v(t), which you already have.

[sAXIn]

yes but you don't put in consideration that x is function of t it\s integral of F(t)d(x(t)) or something

2) I know that p=FV when the force is constant here it isn't ?!

[marlon]

sAXIn,

There are two options actually.

Suppose (i take x and F both along the x-axis, but you know that the integrandum of the work is a SCALAR product of two vectors)
F = 2
x = t²

You just need to substitute everything towards t...
Then W = \int Fdx = \int 2d(t^2) = \int 4tdt

Make sure that you adapt the boundaries.

or, since you know the velocity , you can calculate the kinetic energy and you know that work is equal to the difference in kinetic energy between the end and beginning of the motion : W = E_{k}^{FINAL} - E_k^{BEGINNING}

marlon

[ramollari]

1) Of course it is used:

W = \int_{t_0}^{t}F(t)v(t)dt

where dx = v(t)dt. So what's the problem.

2)
You need instantaneous power, which is the product of instantaneous force at time t and instantaneous velocity at time t. If you had used

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

then it would have been average power that doesn't apply in this problem.

If you're skeptic:

P = \frac{d}{dt}[W(t)] = \frac{d}{dt}(\int_{}^{}F(t)vdt) = Fv


where you well know that dx = vdt.

[sAXIn]

I got it's the same : I can integrate f(t)v(t)and then def. by dt so its = power or just skip it and f(t)v(t) is my power !!! thanks a lot ...