PX155 - C2 - conservative forces, potential energy

conservative forces

work done is independent of the path
- 'conservative' as in energy conserving
conservative if

\int_{A}^{B} \vec{F} d \vec{r} |_{P} = \int_{A}^{B} \vec{F} d \vec{r} |_{P^{'}}

here,

\int_{A}^{B} \vec{F} d \vec{r} |_{P} + \int_{B}^{A} \vec{F} d \vec{r} |_{P^{'}} = 0

this can be written as,

\oint \vec{F} d \vec{r} = 0

- means that the path is a closed loop

work done in reaching any point $P$ from an origin $O$ depends only on the position of $P$
define a function $U = - W (O \to P)$ and set $U (0) = 0$ , where $U$ is the potential energy
- $W (A \to B) = W (A \to 0) + W (0 \to B) = U (A) - U (B)$
  $= - [U (B) - U (A)]$
  $= - Δ U$
- $W (A \to B)$ is also $Δ T$
  $Δ T = - Δ U$
  $Δ T + Δ U = 0$
  $∴ T + U$ , AKA total mechanical energy, is constant/conserved provided that forces act as conservative

deriving a force from a potential

since $d W = F (x) d x = - d U$
- in 1D:

F (x) = - \frac{d U}{d x}

- in 3D:

\vec{F} = - \nabla U = - \frac{δ U}{δ x} \vec{i} - \frac{δ U}{δ y} \vec{j} - \frac{δ U}{δ x} \vec{k}

consider $U = m g h$
- $F (h) = - \frac{d U}{d h} = - m g$
  - acts downwards
  - increases $T$ , decreases $U$
consider $U = \frac{1}{2} k x^{2}$
- $F (x) = - \frac{d U}{d x} = - k x$
  - opposes displacement from $x = 0$ , restoring force

the work that you do

if you apply an equal and opposite force to a conservative force, you are pushing against the conservative force and you increase the potential energy
- eg: raising a mass against gravity
  - you do positive work on the mass
  - gravity does negative work on the mass