# Kinetic Energy of Springs with Mass

1. Oct 14, 2005

### fusebox13

$$\frac{1}{6} m_s v^2$$
Incase my latex is broken, its (1/6)(m_s)(v^2). Where m_s is the mass of the spring, and V is the velocity of the spring at it's tip.
I'm looking for a derivation for this equation, because it was just handed to me and I don't like using equations without knowing where they came from. Can anyone point me in the right direction?

2. Oct 15, 2005

### Staff: Mentor

Think of the spring (length L) in segments of mass dm. One end of the spring is fixed, the other moves at speed $v_s$. So a mass element at position x from the fixed end has speed $(x/L) v_s$, and thus a KE of $1/2 (dm) (x/L)^2 v_s^2$. Express dm in terms of dx: $dm = (m_s/L) dx$ and integrate the previous expression (from 0 to L) to find the total KE.

3. Oct 15, 2005

### fusebox13

Thanks

So, essentially we are just summing up all the tiny dKE's. That makes a whole lot of sense now. Thanks Doc.

:!!)