Relationship between work and potential energy

Question

Ultimately what I am trying to do here is convince myself that the following relationship holds: $$F = -\frac{dU}{dx}$$ for a force $F$ and a potential function $U$. We have that the work $W$ can be expressed as $$W = \int_C F \cdot dr$$ where $C$ is some sufficiently nice curve. Further, this video from MIT claims that if $F$ is conservative, $$\begin{align}U(x_f) - U(x_i) &= -\int_{x_i}^{x_f}F dx \\ &= \int_{x_i}^{x_f}\frac{dU}{dx}dx \end{align}$$ where the last equality is from the fundamental theorem of calculus. Comparing the last two integrals gives us $$F = -\frac{dU}{dx}$$ but this "derivation" seems to use some relationship between work and potential energy for conservative forces, namely $$W = -\big(U(x_f) - U(x_i)\big) = -\Delta U.$$ I am trying to understand why the last equation is true. Could anyone explain this?

Answer 1

About the concept of potential energy:

Ultimately the basis of the concept of potential energy is the Work-Energy theorem.

The starting point for deriving the Work-Energy theorem is Newton's second law:

$$ F = ma \tag{1} $$

Integrate both sides with respect to the spatial coordinate, integrating from starting point $s_0$ to final point $s$

$$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \tag{2} $$

At this point we can develop the right hand side, by capitalizing on the fact that position and acceleration are not independent of each other.

$$ v = \frac{ds}{dt} \quad \leftrightarrow \quad ds = v \ dt \tag{3} $$

$$ a = \frac{dv}{dt} \quad \leftrightarrow \quad dv = a \ dt \tag{4} $$

I omit the factor $m$ temporarily, it is a multiplicative factor that is just carried over each step

$$ \int_{s_0}^s a \ ds \tag{5} $$

Use (3) to change the differential from $ds$ to $dt$. Since the differential is changed the limits change accordingly.

$$ \int_{t_0}^t a \ v \ dt \tag{6} $$

Change the order:

$$ \int_{t_0}^t v \ a \ dt \tag{7} $$

Change of differential according to (4), with corresponding change of limits.

$$ \int_{v_0}^v v \ dv \tag{8} $$

So we have:

$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \tag{9} $$

We multiply both sides with $m$, and then the right hand side of (9) gives us the right hand side of (2). The result: the Work-Energy theorem:

$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{10} $$

The Work-Energy theorem provides the motivation to define the concept of potential energy.

Define potential energy as the negative of work done.

With that definition in place we have:

$$ - \Delta E_p = \Delta E_k \tag{11} $$

Relation (11) is applicable at any scale, down to infinitisimal scale. It follows that an object moving in a potential gradient will move according to the following differential equation:

$$ \frac{d(-E_p)}{ds} = \frac{dE_k}{ds} \tag{12} $$

Answer 2

For conservative forces , potential energy obeys differential calculus , where as for non conservative forces potential energy does not obey differential calculus .