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**A 1.2·kg block is tied to a string that is wrapped around the rim of a pulley of radius 7·cm. The block is released from rest.**

Assuming the pulley is a uniform disk with a mass of 0.37·kg, find the speed of the block after it has fallen through a height of 0.43·m.

Assuming the pulley is a uniform disk with a mass of 0.37·kg, find the speed of the block after it has fallen through a height of 0.43·m.

**2. F=mrω^2 and possibly F=mg and v=r*omega**

**3. I wasn't sure how to do this problem at all. I looked around online for help and got a little bit but I'm still unsure. I used F=mg to determine the F then used that to determine ω^2. Is that correct? Obviously there would be more to the problem but this is as far as I got and I'm not even sure if it's correct.**

So I got F=11.76 by multiplying 1.2kg * 9.8m/s^2 which I then put in the equation for 11.76=(.37kg)(.07m)w^2 and then got 21.3 for omega. This seems high and is probably due to the conversion from CM to M which I'm unsure is correct. Assuming it is I moved on and put that omega value 21.3 into v=wr for v=21.3*.07 and got 1.49 which is incorrect.

So I got F=11.76 by multiplying 1.2kg * 9.8m/s^2 which I then put in the equation for 11.76=(.37kg)(.07m)w^2 and then got 21.3 for omega. This seems high and is probably due to the conversion from CM to M which I'm unsure is correct. Assuming it is I moved on and put that omega value 21.3 into v=wr for v=21.3*.07 and got 1.49 which is incorrect.

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