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)

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.

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

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.

To find the new angular speed, change in kinetic energy, and the work done in pulling the cord, we can use two important concepts: conservation of angular momentum and conservation of mechanical energy.

1. New angular speed (ω2):
We can use the principle of conservation of angular momentum, which states that the angular momentum of a system remains constant unless acted upon by an external torque.
The formula for angular momentum is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
Since the system is frictionless and the block is treated as a particle, the moment of inertia remains constant.
Therefore, we can write: I1ω1 = I2ω2, where I1 and ω1 are the initial moment of inertia and angular speed, and I2 and ω2 are the final moment of inertia and angular speed.
The moment of inertia for a particle rotating about an axis perpendicular to its motion is given by I = mr², where m is the mass and r is the radius.
So, for the initial state, I1 = m1r1², and for the final state, I2 = m1r2², where r1 and r2 are the initial and final radii, respectively.
Plugging in the values, we have (m1r1²)ω1 = (m1r2²)ω2.
Simplifying the equation and solving for ω2 gives us the new angular speed.

2. Change in kinetic energy (ΔKE):
The change in kinetic energy is given by the formula ΔKE = (1/2)I1(ω2² - ω1²), where ΔKE is the change in kinetic energy and I1, ω1, and ω2 are as defined above.
Using the values for I1, ω1, and ω2, we can calculate the change in kinetic energy.

3. Work done in pulling the cord (W):
To find the work done in pulling the cord, we can use the concept of conservation of mechanical energy, which states that the total mechanical energy of a system remains constant in the absence of external forces.
Initially, the block has only kinetic energy (KE1 = (1/2)I1ω1²) and, as there is no work done by external forces, the total mechanical energy is conserved.
After pulling the cord, the block has both kinetic energy (KE2 = (1/2)I2ω2²) and potential energy (PE = mgh, where h is the height difference between the initial and final positions of the block).
Since the surface is frictionless, we can assume that the potential energy is zero.
Therefore, the work done in pulling the cord is given by W = KE2 - KE1.

By applying the above equations and principles, you can calculate the new angular speed, the change in kinetic energy, and the work done in pulling the cord. Make sure to plug in the given values for mass, initial radius, final radius, and initial angular speed to get the specific numerical answers.