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Correlation between conservative forces, non-conservative forces and potential energy
So I recently learned the definition of conservative forces, and how the work done by such forces depends only on the initial and final position of the particle but then we learnt about definition of potential energy as the capacity of a system to increase the K.E of the system on the expense of potential energy via internal conservative forces.
The thing that I am not understanding is what is the correlation between the former definition of conservative force and that used to define potential energy?
Is the text trying to say only when we do work against internal conservative forces only then the system can store potential energy and if that is so, why can't non-conservative forces do so?
And just as a side question, if gravity and electromagnetic forces are conservative that is if the fundamental forces itself are conservative,how can non-conservative forces even arise?
newtonian-mechanics work potential-energy definition conservative-field
edited Sep 8 '21 at 2:52
Qmechanic♦
Destiny-wolfDestiny-wolf
It isn't the fundamental force that is conservative or dissipative, it's the nature of the interaction.
In short: a conservative interaction is one in which a small number of fields act on an object such that the object is accelerated smoothly in the direction of the field lines.
A dissipative interaction is one in which many different fields act on an object in random directions such that the object or parts of the object must take a longer path to get there and impart energy to other objects along the way. We approximate large numbers of random interactions as dissipative forces.
Toy Example: The Peg Force
Consider a ball rolled down a ramp. Suppose the coefficient of rolling friction is zero and there is no air resistance. The ball must have kinetic energy (rotational plus translational) equal to its original potential energy at the top of the ramp.
Now consider a ball rolled down a ramp that has a long wooden peg drilled into it every few inches. The ball bounces and jiggles all over down the ramp and eventually trickles out with barely any kinetic energy. The extra energy has been dissipated as heat in the ball and the pegs from all the collisions.
In both cases, gravity is doing the work and electricity (electrons in the ball repelling electrons in the wood) are setting the path. In the toy example, we can see each bounce of the ball, so we would tend to treat it as many collisions, not some smoothly valued Peg Force, but if we had a mile long, ten inch wide track, we would ignore the width of the track and invent a Peg Force to predict how the ball would accelerate.
Real examples:
Electric Resistance works almost exactly like our Peg Force. Instead of cylindrical pegs and a spherical ball, we have fuzzy spherical atoms and countless point-like electrons, and instead of gravity as the motive force pushing the ball, we have the electric force pushing electrons. The electrons bounce off of the atoms, the atoms bounce and jiggle, they make their neighbors bounce and jiggle, dissipating energy as heat.
Friction is like trying to drag a saw across another saw, edge to edge. The jagged structure of matter at very small scales makes atoms at the point of contact take a much longer path, bouncing and jiggling along with countless collisions. These in turn make their neighbors bounce and jiggle, dissipating energy as heat.
g sg s
...potential energy as the capacity of a system to increase the K.E of the system on the expense of potential energy via internal conservative forces.
The second Newton's Law is always valid, but for non conservative forces as friction, $F_{net}$ is known by measuring $ma$. On the other hand, for forces that depends on the position, both sides of the equation can be known independly.
$$F = -\frac{\partial V}{\partial x} = m \frac{dv}{dt}$$
If we multiply both sides by an infinitesimal displacement $dx$:
$$Fdx = -\frac{\partial V}{\partial x}dx = -dV = m \frac{dv}{dt}dx = m dv\frac{dx}{dt} = mvdv$$
Integrating both sides:
$$-\Delta V = \Delta \frac{1}{2}mv^2 = \Delta KE$$
as you wrote.
Claudio SaspinskiClaudio Saspinski
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