How do Electrons perform work in circuits?

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Electrons in a circuit do perform work, but their kinetic energy and mass are not the primary sources of energy transfer; rather, it is the electric field that drives the movement of electrons. The equation for kinetic energy, E = 1/2 mv^2, is derived from calculus and indicates that the mass of electrons contributes minimally to the overall energy in a circuit. The work done by electrons is primarily related to changes in potential energy, with resistance in the circuit converting some energy into heat. The drift velocity of electrons is very slow, and their kinetic energy is negligible compared to the total electrical energy delivered. Ultimately, the electric charge and the electric field are more significant in performing work within circuits than the mass of the electrons themselves.
  • #31
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.
 
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  • #32
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.
 
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  • #33
marmoset said:
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
 
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