Spring mass damper system: Distance from equilibrium after applying velocity to mass I have a spring fixed to a wall on one end and a mass object on the other end in its natural resting position. The question is how far does the spring stretch when a velocity $v_0$ is applied to it, assuming there is no friction.
My idea was that the spring will be stretched until the velocity $$v_0=0$$ and the kinetic energy $$k=0$$ resulting in the max $E_{pot}$. 
However I can't figure out a relationship between $E_{kin}$ and $E_{pot}$ other than $E_{kin} + E_{pot} = E_{total}$ and therefore I don't know how to continue from here on.
 A: Initially the mass has kinetic energy $\frac 1 2 mv^2$ and the spring has potential energy $E_{pot}=0$. So $E_{total} = \frac 1 2 mv^2$.
When the spring is fully extended, the velocity of the mass is $0$, so $E_{kin}=0$, and the potential energy is a function $f$ of the spring's maximum  extension $x_{max}$. So $E_{total}=f(x_{max})$.
Since there is no friction, $E_{total}$ is conserved so $\frac 1 2 mv^2 = f(x_{max})$.
Find an expression for the potential energy of the spring $f(x_{max})$ - this will involve the spring constant - and then re-arrange this equation to find $x_{max}$ as a function of $m$, $v$ and the spring constant.
A: With the spring attached to a wall (and not the ceiling) one might assume that the mass is sliding on a friction-less horizontal surface.  Then the starting kinetic energy = the final potential energy: (½) m vo^2 = (½) k x^2  where k is the spring constant and x  is the stretch.  If the spring is hanging vertically, the rest position occurs where its force matches gravity. If it is stretched further down, this matching force is added to, linearly.  The result is that gravity can be ignored (with upward motion treated as compression).  The energy equation remains the same. 
