Question

Heuristic Energy Explanation -- This is not a question, so please refrain from treating it as such.

Answer 1

@Hope99

Answer 2

We begin from Newton's 2nd Law of Motion: F = ma. Because I know you have at least a cursory understanding of calculus, I will rewrite this as
\[F = m \frac{dv}{dt} \]
where v is the particle's velocity. Since m is in general a constant, we can draw it inside the derivative:
\[F = \frac{d(mv)}{dt} \]
And last but not least, I will define the momentum of the object by
\[p = mv\]
leaving me with
\[F = \frac{dp}{dt} \]
This is a better way to write Newton's 2nd Law, which I will now interpret.

In nature, there are influences called Forces. These forces are the rate of change of a particle's momentum, where I define momentum as the product between the object's mass and velocity. If I have several forces acting on an object, I first add them up. Then, the result is the rate of change of the particle's momentum.

In a special case, if the sum of all the forces on an object is zero, the rate of change of momentum is zero, which means that the momentum is constant.

Answer 3

As it turns out, if a particle is accelerating with acceleration "a", the following equation is true (the equation can be derived with some calculus):

\[ \Delta(v^2) = v_f^2-v_i^2 = 2a \Delta x \]
Using some algebra, I can rewrite this as
\[ \Delta( \frac{1}{2}m v^2) = ma \Delta x \]
Therefore, identifying ma with some force acting on an object, we find that
\[\Delta(\frac{1}{2} m v^2) = F\Delta x\]
I call the quantity
\[\frac{1}{2} mv^2\]
the "Kinetic Energy" of the particle.

Therefore, if some force F acts on a particle over a distance "delta x", it produces a change in the kinetic energy of the particle.

I should note here that this implies that the force is constant. If the force varies with position, the real expression is
\[\Delta(\frac{1}{2} mv^2) = \int_{x_i}^{x_f} F\cdot dx \]

Answer 4

work as an integral :)

Answer 5

Therefore, if there are no net forces acting on the system, then the kinetic energy of the particle is constant. This is all well and good, but that's a pretty restrictive condition. We shall seek to make it a little more useful.

Imagine that I have a gravitational force acting on my particle. The force is
\[F = -mg \]
Plugging that into my equation, and denoting 1/2 mv^2 by T for simplicity,
\[\Delta(T) = -mg\Delta x = \Delta(-mgx) \]
where x is now the height above the ground, since that's the direction that gravity acts. If I move everything to the left side,
\[ \Delta( T + mgx) = 0 \]
So if gravity is the only thing acting on the particle, this strange quantity
\[T + mgx = \frac{1}{2} mv^2 + mgx\]
is constant. We will call this quantity the total energy, and denote it by E.
\[E = \frac{1}{2} mv^2 + mgx \]

Answer 6

and that constant strange quantity is the potential energy!

Answer 7

What is this quantity mgx? Well, because it represents total energy that has not yet been converted to kinetic energy, we call it potential energy. More specifically, because that is the potential that is associated with the gravitational force, it is the gravitational potential energy, denoted by U_g

\[U_g = mgx\]

A force due to a spring is more complicated because unlike gravity, it is not constant -- it varies with how far the spring has been stretched or compressed. If I denote how far it's stretched or compressed by l, and its stiffness by k, it turns out that the corresponding potential energy is
\[U_s = \frac{1}{2} k l^2 \]
I can add this term to my total energy:

\[E = T + U_g + U_s = \frac{1}{2} mv^2 + mgx + \frac{1}{2}k l^2 \]

Now, my total energy is conserved as long as the only forces acting on my system are gravity or springs.

We can continue in this way for other forces. However, NOT ALL FORCES can be written this way, so potential energy only makes sense for certain forces. We typically call such forces conservative (they conserve energy). As I said before, pushing a box obviously cannot be derived from a potential so it's a non-conservative force.

Hopefully you have a better understanding now of why I continue to say that saying a system has more force than another is a nonsensical question.

Answer 8

Since the magnetic force is NOT conservative this is idea is not applied it it? And is also of the reason you said its more complicated huh?

Answer 9

this idea*, applied to it* ,and aslo* said is more*

Answer 10

Generally speaking magnetic forces are not conservative... however, through some vector calculus sorcery, it turns out that the magnetic force between two current loops can be treated as conservative. The magnetic field produce by a loop of electrical current looks like this: