Find Position of 2nd Node from Velocity of 1st: System of 3 Dampers

[c8lin13]

Homework Statement

there is a system of 3 dampers, attached to a fixed surface, with 2 dampers in parallel, and one that's in series to the other 2 which r attached to the fixed surface.
The equation asks if the velocity of the first node(node attached only to single damper) is a m/s, then what is the position of the second node(node between single dampr, and parallel dampers)

Homework Equations

Ca=500Ns/m
Cb=100Ns/m
Cc=150Ns/m
(Cb and Cc are in parallel))

The Attempt at a Solution

Ive tried applying Newtons second, but I am pretty sure that isn't right, or I am doing it wrong. The question I'm wondering about is (b), i also don't know how to start c, and d...so help with those would be great too

 

[Valkarie]

. a) Find the equivalent spring constant for the systemKeq = (Ca*Cb*Cc) / (Cb + Cc)Keq = (500*100*150) / (100 + 150)Keq = 75000 N/mb) Find the position of the second node given that the velocity at the first node is 0.1 m/sx2 = (1/Keq)*v1x2 = (1/75000)*0.1x2 = 1.33 x 10-5 m