Mass attached to a spring rotating in a circle

[jono240]

Homework Statement

A mass of 2kg rotates at 1m/s in a horizontal circle on a table at the end of a spring with an elastic constant of 50N/m. If the original length of the spring is 2m, find the extension of the spring.

Given - M=2kg, V=1m/s, k=50N/m, original length of x=2m
find x'

Homework Equations

Ek + Ee = Ek' + Ee'

Fc = mv² / r

The Attempt at a Solution

Im starting the question by finding the extension without taking in account for the attraction to the centre by the centripetal force. I know I will need to do this later.
This is where I am stuck

Ek + Ee = Ek' + Ee'
Since there is no energy when the thing is at rest I use
0 = Ek' + Ee'
0 = .5mv'² + .5kx²
-.5mv'² = .5kx²
SQRT [ (-.5mv'²) / (.5k) ] = x

Problem is you can't sqrt a negative number.. I am stumped. I don't know where else to start
please help guys

 

[256bits]

I am not sure energy balance equation works here as you found out.

Try finding the radial acceleration, the radial force, and then the spring extension.