A mass is initially moving at a velocity v x (0) =v o . The frictional force is

Question

A mass is initially moving at a velocity vx(0) =vo. The frictional force is-Foekv. Find the velocity andposition as a function of time and analyze the results.

Explanation / Answer

m*a = Force m*dv/dt = -F0*exp(kv) dv/dt = (-F0/m)*exp(kv) let u = exp(kv) du/dt = k*exp(kv)*dv/dt = k*u*dv/dt => dv/dt =(1/[k*u])*du/dt Sub into dv/dt = (-F0/m)*exp(kv) (1/(k*u))*du/dt = (-F0/m)*u du/dt = (-F0*k/m)*u2 du/u2 = (-F0*k/m)*dt -1/u = (-F0*k/m)*t + C1 1/(exp(kv)) = (F0*k/m)*t + C2                              1                         1 v = ---- * ln ---------------------------        k                  C     +   (F0*k/m)*t At t = 0, v = v0 v0 = (1/k)*ln[1 / (C + 0)] k*v0 = ln(1/C) C = exp(-k*v0) For position, x = integral [dv/dt* dt] = (1/k)*integral [ ln(1/ (C + C2*t))]dt                where C2 = F0*k/m x = -(1/k)* integral [ln(C + C2*t)]dt let r = C + C2*t         dr= C2*dt =>   dt = dr/C2 x = -(1/[C2*k]) *integral (ln (r)) dr let u =ln(r)             dv = dr    du =dr/r               v = r x = -(1/[C2*k]) *[r*ln(r) - integral [r*dr/r] x = -(1/[C2*k]) *[r*ln(r) - r + C3] x = -(1/[C2*k]) *[(C+C2*t)*ln(C+C2*t) - (C+C2*t) + C3]               (exp(-kv0) + F0*k*t/m) {ln[exp(-kv0) + F0*k*t/m] - 1} + C3 x =    -------------------------------------------------------------------------                                              k*F0*k/m C3 is an unknown constant, because you are not given the initialposition.

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