How do Electrons perform work in circuits?

[marmoset]

About the integral explanation of the 1/2 in the expression for kinetic energy, I understand the integration, but why does integrating mass x velocity with respect to velocity give kinetic energy?

I had always explained the half by saying that to stop an object mass m traveling at a velocity u, you apply a force which gives the object an acceleration of -F/m. Then I use v2 = u2 + 2as to work out the distance (s) this object travels before it stops. So 0 = u2 - 2Fs/m, and so Fs = mu2/2. Fs expresses the work done to stop the object, and so from conservation of energy the object's initial KE must have beeen mu2/2.

 

[nuby]

When one object(a) is pushing on another object(b) (accelerating it). Isn't 1/2 of the force pushing object (a) from (b), and the other 1/2 (of the total force in system) pushing (b) from (a)? This is how I see it. Similar to centrifugal and centripetal force.

 

[Snazzy]

About the integral explanation of the 1/2 in the expression for kinetic energy, I understand the integration, but why does integrating mass x velocity with respect to velocity give kinetic energy?

I had always explained the half by saying that to stop an object mass m traveling at a velocity u, you apply a force which gives the object an acceleration of -F/m. Then I use v2 = u2 + 2as to work out the distance (s) this object travels before it stops. So 0 = u2 - 2Fs/m, and so Fs = mu2/2. Fs expresses the work done to stop the object, and so from conservation of energy the object's initial KE must have beeen mu2/2.

$$F \cdot dx = F \cdot vdt=\frac{d(mv)}{dt}vdt=v\cdot d(mv)=mv\cdot dv$$

$$KE=\int F\cdot dx = \int mv\cdot dv = \frac{1}{2}mv^2 + C$$

$$KE(0) = 0$$

$$\frac{1}{2} m0^2 + C = 0$$

$$C = 0$$

$$KE = \frac{1}{2}mv^2$$