Summarized Problem: A mass m is constrained to move along the x axis, subject to a force $$F(v) = -F_o e^{v/V}$$ where $$F_o$$ and $$V$$ are constants. (a) Find $$v(t)$$ if the initial velocity is $$v_o > 0$$ at time $$=0$$. (b) At what time does it come instantaneously to rest? (c) Find $$x(t)$$ then determine how far the mass travels before coming instantaneously to rest.

Solution: If $$F(v)$$ is the force, then $$F(v)/m$$ is the acceleration. Therefore, $\frac{F(v)}{m} = \frac{dv}{dt}$ $\frac{mdv}{F(v)} = dt$ $t = m\int_{v_o}^{v} \frac{dv'}{F(v')} = -\frac{m}{F_o} \int_{v_o}^{v} \frac{dv'}{e^{v/V}} = \frac{mV}{F_o} e^{-v/V} - \frac{mV}{F_o} e^{-v_o/V}$ $\implies \frac{mV}{F_o} e^{-v/V} = t + \frac{F_o}{mV} e^{-v_o / V}$ $e^{-v/V} = \frac{F_o t}{mV} + e^{-v_o / V}$ $-\frac{v}{V} = \ln(\frac{F_o t}{mV} + e^{-v_o / V})$ $v = -V \ln(\frac{F_o t}{mV} + e^{-v_o / V})$ The next part of the question (b) is basically asking us to find the time when the mass's velocity is zero. Since $$\ln(1) = 0$$, we really just need to find t such that $$\frac{F_o t}{mV} + e^{-v_o / V} = 1$$. Doing so, we find that the time when the velocity is zero is $t = \frac{mV}{F_o} (1-e^{-v_o / V}).$ For (c), we basically need to find the definite integral of $$v(t)$$ from 0 to $$t$$ to find the distance the mass has travelled at any given $$t$$, then plug in the value for time that we got above. $x = \int_{0}^{t} vdt = -V \int_{0}^{t} \ln(\frac{F_o t}{mV} + e^{-v_o / V}) dt$ Let $$u = \displaystyle \frac{F_o t}{mV} + e^{-v_o / V}$$ and $$du = \displaystyle \frac{F_o}{mV}dt$$. $x = - \frac{mV^2}{F_o} \int_0^t \ln (u) du$ $x = \left. - \frac{mV^2}{F_o} (u \ln (u) - u ) \right|_0^t$ $x = \left. - \frac{mV^2}{F_o} \left[(\frac{F_o t}{mV} + e^{-v_o / V}) \ln (\frac{F_o t}{mV} + e^{-v_o / V}) - \frac{F_o t}{mV} - e^{-v_o / V} \right] \right|_0^t$ $x = \left. - \frac{mV^2}{F_o} \left[(\frac{F_o t}{mV} + e^{-v_o / V}) \ln (\frac{F_o t}{mV} + e^{-v_o / V}) - \frac{F_o t}{mV} - e^{-v_o / V} \right] \right|_0^t$ $x = \left. - \frac{mV^2}{F_o} \left(\frac{F_o t}{mV} + e^{-v_o / V}\right) \left[\ln (\frac{F_o t}{mV} + e^{-v_o / V}) - 1 \right] \right|_0^t$ Now, carry out the evaluation from time 0 to the time we got in (b), and we get $x= \frac{mV^2}{F_o} (1-e^{-v_o/V}) - \frac{mVv_o e^{-v_o/V}}{F_o}$ This is the distance the mass would travel before coming to rest.