A particle of unit mass is projected vertically upwards with speed u. At any height x, while the particle is moving upwards, it is found to experience a total force F, due to gravity and air resistance, given by $F = \alpha e^{-\beta x}$, where $\alpha$ and $\beta$ are positive constants. Calculate the energy expended in reaching a given height z. Show that
$F=\frac 1 2 \beta v^2 + \alpha - \frac 1 2 \beta u^2$
where $v$ is the speed of the particle, and explain why $\alpha = \frac 1 2 \beta u^2 + g$ where $g$ is the acceleration due to gravity.
Determine an expression, in terms of $y$, $g$ and $\beta$, for the air resistance experienced by the particle on its downward journey when it is at a distance $y$ below its highest point.
Fairly long question but I can solve it for the most part.
The problem I have is the last part.
The model solution requires I use $v\frac {dv} {dy}= g - \frac 1 2 \beta v^2$ (since the air res. is going the other way) and it leads to $\frac 1 2 \beta v^2 = g(1-e^{-\beta y})$.
When I was solving it I tried to use conservation of energy, $mgy = \frac 1 2 v^2$, where $m$ = 1. Which lead to $\frac 1 2 \beta v^2 = \beta g y$. Why is this wrong?
Many thanks in advance,
Chris
Air resistance (also any sort of friction) renders the use of conservation of energy useless for the problem, because these constitute the "loss" of energy to some ignored sink. (For example, air resistance heats up the air, but you never carry that into your calculations, it would be impossible.)
This is why they call these "nonconservative" forces.
Of course, energy is conserved in the real world, but in just the problem you are considering, energy is magically "lost" because of the nonconservative force of friction or resistance.