功与能量的一般关系：公式推导矛盾排查请求

Hey there, let's unpack the contradiction you're running into with gravitational potential energy (GPE) derivations!

First off, your initial derivation using W=Fs is totally valid for near-Earth scenarios where gravity is constant:

• You start with the work formula W=Fs, substitute constant gravitational force $F=mg$, and vertical displacement $s=h$, giving $W=mgh$.
• The key link here is that gravitational potential energy change equals the negative of work done by gravity: $\Delta GPE = -W_{gravity}$. When taking the ground as the 0-potential surface, lifting an object to height $h$ means gravity does negative work, so $GPE = mgh$ checks out mathematically.

Now, if your "more rigorous derivation" is throwing up contradictions, you're almost certainly hitting one of these common pitfalls:

• Mixing up work sign conventions and PE definition
  A super easy mistake: GPE isn't equal to work done by gravity—it's the negative of that work. If you skipped this sign flip in your rigorous work, you'd end up with conflicting results. For example, when an object falls, gravity does positive work, but GPE decreases—this negative relationship is critical to keep straight.

• Applying constant-force formulas to variable-force scenarios
  If your rigorous approach accounts for gravity changing with height (e.g., far from Earth's surface), W=Fs no longer applies—gravity is a variable force here. You need to use integration to calculate work:
  $$ W_{gravity} = \int_{r_0}^{r} -\frac{GMm}{r^2} dr $$
  This leads to the universal GPE formula $GPE = -\frac{GMm}{r}$ (with infinity as the 0-potential surface). The near-Earth mgh is just an approximation of this when $h \ll R_{Earth}$. Using W=Fs here would break the math, creating apparent contradictions.

• Switching potential energy reference surfaces
  You defined the ground as your 0-potential surface initially, but if your rigorous derivation accidentally uses a different reference (like infinity), the formulas will look different. This isn't a contradiction—it's just two expressions tied to different starting points.

• Ignoring vector nature of work
  The scalar W=Fs is a simplification of the dot product $W = \vec{F} \cdot \vec{s}$. If your displacement direction opposes gravity (e.g., lifting an object), the dot product gives a negative value. Forgetting to account for this vector alignment can lead to sign errors that appear as contradictions in your derivation.

If you share the exact steps of your "rigorous derivation," I can pinpoint the exact mistake for you!

内容的提问来源于stack exchange，提问作者Scott Simmons