1 problem found
A particle of unit mass is projected vertically upwards with speed \(u\). At 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 this height. Show that \[ F= {\textstyle \frac12} \beta v^2+ \alpha - {\textstyle \frac12} \beta u^2 \;, \] where \(v\) is the speed of the particle, and explain why \( \alpha = \frac12 \beta u^2 +g\), 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.
Solution: Considering the energy of the particle, we have initial kinetic energy of \(\frac12 u^2\) and final energy is \(\frac12 v^2\), the change in energy is the work done by the force, \begin{align*} &&\text{Work done against resistance} &= \text{loss in kinetic energy} \\ &&\int F \, \d x &= \int \alpha e^{-\beta x} \, \d x \\ &&&= \frac{\alpha}{\beta} \l 1 - e^{-\beta x} \r \\ &&&= \frac{1}{\beta} \l \alpha - F\r \\ &&&= \frac12 u^2 - \frac12 v^2 \\ \Rightarrow && F &= \frac12 \beta v^2 + \alpha - \frac12 \beta u^2 \end{align*} When \(v = 0\) there is no air resistance, ie \(F = g\), but \(g = 0 + \alpha - \frac12 \beta u^2 \Rightarrow \alpha = g + \frac12 \beta u^2\) \(F = \frac12 \beta v^2 + g\), ie air resistance is \(\frac12 \beta v^2\) Looking at forces acting on the particle when it's descending, \begin{align*} && v \frac{dv}{dx} &= g - \frac12 \beta v^2 \\ \Rightarrow && \frac{v}{g - \frac12 \beta v^2} \frac{dv}{dx} &= 1 \\ \Rightarrow && \int \frac{v}{g - \frac12 \beta v^2} \, dv &= \int dx \\ \Rightarrow && \frac1{\beta}\l\ln(g - \frac12\beta v^2) - \ln(g)\r &= y\\ \Rightarrow && \ln \l 1 - \frac12 \frac{\beta}{g}v^2 \r &= \beta y \\ \Rightarrow && \frac{g}{\beta} \l 1-e^{-\beta y} \r = \frac12 v^2 \end{align*} Since force is the rate of change of work, we can say that the force is \(ge^{-\beta y}\) and the air resistance is \(g \l 1-e^{-\beta y} \r\)