Question Number 200428 by Spillover last updated on 18/Nov/23
Answered by ajfour last updated on 18/Nov/23
![F=mg−((kv^2 )/2) ((vdv)/dx)=g−((kv^2 )/(2m))=−(k/(2m))(v^2 −((2mg)/k)) (1/2)∫_0 ^( v) ((2vdv)/(v^2 −((2mg)/k)))=−(k/(2m))∫_0 ^( x) dx ln [((((2mg)/k)−v^2 )/((((2mg)/k))))]=−((kx)/m) 1−((kv^2 )/(2mg))=e^(−((kx)/m)) v^2 =((2mg)/k)(1−e^(−((kx)/m)) ) (v^2 )∣_(max) =((2mg)/k) for x→∞](https://www.tinkutara.com/question/Q200436.png)
$${F}={mg}−\frac{{kv}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{{vdv}}{{dx}}={g}−\frac{{kv}^{\mathrm{2}} }{\mathrm{2}{m}}=−\frac{{k}}{\mathrm{2}{m}}\left({v}^{\mathrm{2}} −\frac{\mathrm{2}{mg}}{{k}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:{v}} \frac{\mathrm{2}{vdv}}{{v}^{\mathrm{2}} −\frac{\mathrm{2}{mg}}{{k}}}=−\frac{{k}}{\mathrm{2}{m}}\int_{\mathrm{0}} ^{\:{x}} {dx} \\ $$$$\mathrm{ln}\:\left[\frac{\frac{\mathrm{2}{mg}}{{k}}−{v}^{\mathrm{2}} }{\left(\frac{\mathrm{2}{mg}}{{k}}\right)}\right]=−\frac{{kx}}{{m}} \\ $$$$\mathrm{1}−\frac{{kv}^{\mathrm{2}} }{\mathrm{2}{mg}}={e}^{−\frac{{kx}}{{m}}} \\ $$$${v}^{\mathrm{2}} =\frac{\mathrm{2}{mg}}{{k}}\left(\mathrm{1}−{e}^{−\frac{{kx}}{{m}}} \right) \\ $$$$\left({v}^{\mathrm{2}} \right)\mid_{{max}} =\frac{\mathrm{2}{mg}}{{k}}\:\:\:\:\:{for}\:\:{x}\rightarrow\infty \\ $$$$ \\ $$