Estimate for energy dissipated by a damper/dashpot I have a system with a mass $m$ attached to the end of a cable. The cable mass is assumed negligible.  The cable is attached to the ground at the one end while the other, with the attached mass $m$, is moving vertically with some known velocity $v$.  The cable is modelled as a 2nd order system with known values for the damping $b$ and spring $k$ coefficients.  I'm trying to use the conservation of energy principle to determine the impulse force resulting in the cable (at the moving end specifically) as the cable becomes taut, but I'm having trouble accounting for the energy dissipated due to the cable damping.  The cable is well damped and this significantly reduces the impulse force generated.  I've setup up the following system of equations, does anyone have any advice on how I can go about solving them? I'm guessing that I need to follow some sort of iterative process? Or is there an alternative method that I could follow that would be simpler?
$$ E_1 = E_2 $$
$$E_1 = \frac{1}{2}mv_1^2$$
$$E_2 = \frac{1}{2}kx^2 +mgx + \int_0^t bv(t)^2 dt$$
$$v(t) = v_0 + \int_0^ta(t) dt$$
$$a(t) = F(t)/m$$
$$F(t) = kx(t) + bv(t)+mg$$
$E_1$ is the kinetic energy the instant before the cable begins to stretch, while $E_2$ is the energy of the system when the mass has come to rest with the cable having stretched vertically some distance $x$.  The energy in the system at $E_2$ is equal to to the stored energy in the spring, the increase in potential energy due to the cable stretching a distance $x$ and the energy dissipated by the damper/dashpot. I'm trying to solve Equation 3 to calculate the change in cable length $x$, which I'll then use to calculate the impulse force from,
$$F_{impulse} = kx$$
Does the answer perhaps lie with using these equations with the conservation of momentum principle?
$$mv_1 = F_{impulse}*t$$
The mass is also assumed to not rotate on impact and I neglect the energy dissipation due to the propagation of lateral or longitudinal waves in the cable.

 A: If initially the mass is at $x=0$ and the initial velocity is $V$ then the (underdamped) position response is:
$$ x(t) = X \exp(-\beta t)\sin(\omega t) = \frac{V}{\omega} \mathrm{e}^{-\zeta \omega_n t} \sin(\omega t) $$
where $$\begin{aligned} \omega_n & = \sqrt{\frac{k}{m}} \\ \zeta & = \frac{d}{2 m \omega_n} = \frac{d}{2 \sqrt{k m}} \\ \omega &= \omega_n \sqrt{1-\zeta^2} = \sqrt{\frac{k}{m} - \frac{d^2}{4 m^2}} \end{aligned} $$
The force on the rope is $F=k x + d \dot{x}$ and the impulse is $J=\int F\,\mathrm{d}t$ defined over half a cycle of oscillation. Plugging the position response yields
$$ J = \int \limits_0 ^ {\frac{\pi}{\omega}} k x + d \dot{x} \,\mathrm{d} t = \\
 = V m \left(1 + \mathrm{e}^{-\pi \frac{\zeta}{\sqrt{1-\zeta^2}}} \right)$$
So without damping $\zeta=0$ and $J=2 V m$ for a perfect "bounce" and with critical damping $\zeta=1$ and $J=V m$ with a "plastic" response. The above can be redefined as a coefficient of restitution $\epsilon$ with
$$ \epsilon = \mathrm{e}^{-\pi \frac{\zeta}{\sqrt{1-\zeta^2}}} $$
The kinetic energy is $E=\frac{1}{2} m \dot{x}^2$ and its value at the $n$-th half cycle of oscillation is $$E_n = \frac{1}{2} m V^2 \epsilon ^ {2 n} $$ and since the coefficient of restitution is $0 \leq \epsilon \leq 1$ then $E_n \leq E_0=\frac{1}{2}m V^2$
A: It seems to me you are making this more complicated than it needs to be. When the cable first becomes taut, the spring force is not yet in play and the only force will be $v\cdot k$ - by the definition of the drag in the dash pot.  You can compute the subsequent motion by solving the damped harmonic oscillator. 
Let me know if this is enough?
