- #1

vbrasic

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## Homework Statement

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A block travels in the positive x-direction with some velocity ##v_0##. It is subject to a drag force $$F(v)=-F_0e^{Kv},$$ where ##F_0,\,K## are positive constants. Find the point in time when the velocity of the block momentarily goes to zero. Find how far the block has traveled at this time. Take intial time to be ##t=0## and initial position to be ##x=0##.

## Homework Equations

Just ##F=ma##, and some kinematics equations.

## The Attempt at a Solution

We have that $$F=m\frac{dv}{dt}=-F_0e^{Kv}.$$ Rearranging gives $$\frac{dv}{e^{Kv}}=-\frac{F_0}{m}dt\rightarrow e^{-Kv}dv=-\frac{F_0}{m}dt.$$ We integrate both sides subject to intial conditions as follows: $$\int_{v_0}^{v}e^{-Kv'}dv'=\int_{0}^{t}-\frac{F_0}{m}dt'.$$ This gives, $$-\frac{1}{K}(e^{-Kv}-e^{-Kv_0})=-\frac{F_0}{m}t\rightarrow e^{-Kv}-e^{-Kv_0}=\frac{F_0K}{m}t\rightarrow e^{-Kv}=\frac{F_0K}{m}t+e^{-Kv_0}.$$ Taking the natural logarithm of both sides gives, $$-Kv=\ln{(\frac{F_0K}{m}t+e^{-Kv_0})}\rightarrow v=v_0\ln{(\frac{F_0K}{m}t)}.$$ So evidently the velocity is zero when $$\ln{(\frac{F_0K}{m}t)}\rightarrow t=\frac{m}{F_0K},$$ My problem arises when I try to find the position of the object at this time. We have, $$v=\frac{dx}{dt}=v_0\ln{(\frac{F_0K}{m}t)}.$$ So, $$dx=v_0\ln{(\frac{F_0K}{m}t)}dt.$$ Integrating to initial conditions, we have, $$\int_{0}^{x}dx'=\int_{0}^{t}v_0\ln{(\frac{F_0K}{m}t')}dt'.$$ So, $$x=v_0t(\ln{(\frac{F_0K}{m}t)}-1)\bigg|_{0}^{t}.$$ However, ##\ln{0}## is undefined. So I'm not sure how to proceed. Rather, I'm quite sure I need to use L'Hospital's rule. Just not sure how to get the former term into the form $$\frac{f(x)}{g(x)}.$$

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