Question

When my book says that when a mass is under work when the mass is accelerated the kinetic energy therefore increase. When we study a ideal system, there is no losses and all the work are stored as kinetic energy according to the first law of thermodynamics dE/dt = F*v.

What I don't understand is how the change in energy over time is equal to the force applied times the velocity of the mass can be true.

Answer 1

Should I move the dt to the other side of the equal sign so a integration over energy is equal to integration over time on the F*v?

Answer 2

Ok so you get the following:\[\frac{ dE }{ dt }=Fv\]\[dE=Fvdt\]\[E=\int\limits_{t _{i}}^{t _{f}}Fvdt=\int\limits_{t _{i}}^{t _{f}}F \left( \frac{ dx }{ dt } \right)dt\]Do you see now?

Answer 3

Not really, sorry
So I get that \[\int\limits_{t0}^{t1} F \frac{ dx }{ dt }dt = \int\limits_{x1}^{x0}Fdx\]
which should equal to
\[\int\limits_{x0}^{x1} madx = \int\limits_{x0}^{x1} m \frac{ d ^{2} x }{ dt ^{2} } dx\]

Answer 4

But there it gets out of hand

Answer 5

Ah, but the force is constant over time. If the force were varying in time as was the velocity, you need to solve a differential equation. Simple integration wouldn't get you the answer.

Answer 6

Oh I get it.
The acceleration is equal to a = dv/dt and the dt cancel eachother out when we integrat over dt we are left with int(v(t))_(v(0)) mvdv and due to m being a constant we ar eleft it vdv inside the integration. So m/2(v(t)^2 - v(0)^2) which is the classic formula for kinetik energy.

But the equation I had in my quaestion is a differential function, right?

Answer 7

Yes, that equation is a differential equation, but it yields to simple integration when F is constant. When F isn't constant you end up with a second order differential equation which will require different methods for solving.