Can someone explain to me work and potential energy and when they are

[jaredvert]

And when they are negative. I have grasped most of my textbook but this seemingly easy concept keeps going over my head. I'm in e and m right now but still can't understand why it is Vba= - integral a to b E times dl.

 

[electricspit]

1. The Work-Energy Theorem states that work and energy are really the same thing!

Work is defined as:

$W = \int \vec{F}\cdot d\vec{x}$

Which we can prove is equivalent to (in a force-field free environment):

$W = \Delta KE$

We know:

$\vec{F} = \frac{d\vec{p}}{dt}$

with $\vec{p}=m\vec{v}$. Then:

$\vec{F} = \frac{d}{dt} (m\vec{v})$

Multiply both sides by $\vec{v}$:

$\vec{F}\cdot\vec{v}=m\vec{v}\frac{d\vec{v}}{dt}$

And we know $\vec{v}=\frac{d\vec{x}}{dt}$:

$\vec{F}\cdot\frac{d\vec{x}}{dt}=m\vec{v}\frac{d\vec{v}}{dt}$

Dropping the differential time:

$\vec{F}\cdot d\vec{x} = mv\cdot dv$

I've also dropped the vectors on the velocity, as $\vec{v}$ and $d\vec{v}$ are in the same direction. If we integrate both sides we find:

$W = \int{\vec{F}\cdot d\vec{x}} = \frac{1}{2} mv^2 = \Delta KE$

2. From above you should notice that a negative change in kinetic energy relates to a negative work. This is all depending on the conventions used. Modern physics uses the convention that when the system loses energy, the work is negative.

For example: pushing a block along a surface. If the block was already moving towards you and you wanted to stop the block it would have to lose energy, and at the same time you would be gaining energy. If the block had large enough kinetic energy, you would move with the block (gaining energy). In fact, the block would be doing work on you, so in other words when the system does work on the force-provider, it is negative work.

3. Using the conservation of energy:

$\Delta KE + \Delta PE = 0$

in a closed system. Therefore:

$\Delta KE = - \Delta PE$

Which leads to the work being defined as:

$W = \Delta KE = -\Delta PE$

So you can see that just by the convention that a system losing kinetic energy is negative work, we arrive to your original problem of why the potential energy is negative the integral of the force. It's there merely by convention and intuition about how things should work. If we look at the electric potential:

$V = k\frac{Q}{r}$

We see:

$\frac{dV}{dr} = -\frac{kQ}{r^2} = -E$

or:

$E(r) = -\frac{dV}{dr}$

If we integrate both sides with respect to $dr$ from point a to point b:

$V(b)-V(a) = -\int\limits_a^b{E\cdot dr}$

4. I'm not sure if you've done this yet in your class, but you will soon. The scalar potential field is conservative, so it doesn't matter which path you take, so you usually see the potential written as a terrifying line-integral:

$V(b)-V(a) = -\int\limits_C{\vec{E}\cdot d\vec{\ell}}$