Work Done By Conservative Forces

[Heisenberg7]

I did classical mechanics a while ago and I was going over some stuff that I wasn't sure if I understood correctly and now I've come over this one. It says that work done by conservative forces is equal to the negative difference in potential energy. Or, $W_c = - \Delta U$. And I've really been trying to make sense of this. I know that when energy is conserved we have $\Delta E = 0 \implies \Delta K + \Delta U = 0 \implies \Delta K = - \Delta U$. Does that have something to do with this equation? Would that also mean that work done by conservative forces is equal to the change in kinetic energy? What I am thinking: It kind of does make sense. Let's say we have a conservative force. Then work done by it is $W_c = \int_{a}^{b} \vec{F} \cdot \vec{dl}$. If $W_c > 0$ then it helps object's motion so work done by it would actually increase object's velocity. So if there are no other forces, all the work done would go into change of kinetic energy. But how does that relate to the change in potential energy? I would just like to hear a kind of intuitive argument to why this connects to the previous equation ($\Delta K = - \Delta U$) if I'm right. Not just mathematically, but an example.

Thanks in advance.

 

[kuruman]

One of the ways a conservative force $\mathbf F$ is defined is that it can be derived by a scalar potential energy $U$. In three dimensions, one would write $$\mathbf F=-\mathbf{\nabla}U.$$ So it's a matter of definition. Note that you can the potential energy if you know the force using a line integral, $$U(\mathbf r)=-\int_{\mathbf{r}_{\text{ref}}}^{\mathbf r}\mathbf F\cdot d\mathbf r$$where ${\mathbf{r}_{\text{ref}}}$ is the point where the potential energy is assumed to be zero.

Consider this example. A mass fall from rest and hits the floor at distance $h$ below. Find the landing speed of the mass.

You can use the work-energy theorem and say that the change in kinetic energy of the mass is equal to the net work done on it. Here gravity is the only force that does work. So you write $$\begin{align}
& \Delta K=W_{\text{grav}} \nonumber \\
& \left(\frac{1}{2}mv^2-0\right)=mgh.
\end{align}$$If you want to use energy considerations, you would write
$$\begin{align}
& \Delta K+\Delta U=0 \nonumber \\
& \left(\frac{1}{2}mv^2-0\right)+\left(0-mgh\right) =0.
\end{align}$$ Equations (1) and (2) are algebraically identical. However, the work done by gravity on the right side of equation (1) becomes the negative of the change in potential energy on the left side of equation (2).