Posted by Christina on Wednesday, November 7, 2007 at 4:54pm.
A small block on a frictionless horizontal surface has a mass of 2.50×10−2 Kg. It is attached to a massless cord passing through a hole in the surface. The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 1.75 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. You may treat the block as a particle.
What is the new angular speed?
Find the change in kinetic energy of the block.
How much work was done in pulling the cord?
(I'm having problems setting up/figuring out which equations to use)

Physics  drwls, Wednesday, November 7, 2007 at 5:09pm
Don't think about equations. Think about conservation laws.
They expect you to treat this as a conservation of angular mometum problem. The block will speed up so that the product M R^2 w is conserved (constant). Calculate the new w (angular speed).
Once you have the new w, calculate the new kinetic energy (1/2)M(rw)^2
r decreases by a factor of 2
w increases by a factor of 4
KE increases by a factor of ?
The change in KE equals the work done. 
Physics  Christina, Wednesday, November 7, 2007 at 5:25pm
I'm on the change in KE part and this is what I get:
KEf=(1/2)(2.5*10^2)(.150*7)^2
=.01378125
KEi=(1/2)(2.5*10^2)(.300*1.75)^2
=.0034453125
I'm not quite sure what to do with these two numbers...
I subtracted KEi from KEf, but was told I had "rounding error" with my value of .01 
Physics  drwls, Thursday, November 8, 2007 at 3:02am
Yes, the KE increases by a factor of 4. I have not checked your numbers, but the difference between the KE values will be the work done. You should only carry three significant figures.
The work done should be three times the initial kinetic energy.