Mathematical misunderstanding of Work-Potential Energy Theorem?

Question

This is a relatively basic question, but I don't understand why it is the case. This is from my dynamics book and is mainly a mathematical misunderstanding.

$$ \ dU = F\cos\theta ds $$

Which means the integral should be:

$$ \int_{u_1}^{u_2}dU = \int_{s_1}^{s_2}F\cos\theta ds $$

Therefore:

$$ U_2 - U_1 = \int_{s_1}^{s_2}F\cos\theta ds $$

However, the book says it is:

$$ U_{1-2} = \int_{s_1}^{s_2}F\cos\theta ds $$

I understand why it should be $ U_1 - U_2$ for conservation of energy reasons, but I don't see it in the math... This way we can say the sum of initial energy (PE,KE,etc.) is equal to the sum of final energy (PE,KE,etc.).

Am I missing something somewhere!? The book does not give any hints to this (as far as i can tell). I've posted the most basic portion, but everything else is derived from it. I understand the equations, but I just don't understand why it's $U_1 - U_2$.

Answer 1

Your first line is actually wrong. The corrected one should be

$$ \ dU = - F\cos\theta ds $$ .

Since you're not adding/removing any energy, the force (whatever your potential energy is coming from) should increase the kinetic energy and decrease the potential energy.

Answer 2

Take the example of gravity. Gravitational potential energy is defined as $$U_g=-\frac{GM}{r}$$ This means that $$U_g=-\int_{r_1}^{r_2}F_g \cos \theta dr$$ Because $F_g=mg$, $$U_g=-\int_{r_1}^{r_2}mg \cos \theta dr$$ Integrate and you find $$U=-mg\cos\theta \left (\left.r\right|_{r_1}^{r_2}\right)$$ and $$U=-mg\cos\theta (r_2-r_1)$$ and if the object is going straight up and down, $$U=-mg \Delta r$$