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.