Saturday
April 19, 2014

Homework Help: AP Physics

Posted by Hannah on Friday, March 22, 2013 at 11:44pm.

A small block with mass 0.035 kg slides in a vertical circle of radius 0.425 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 4.05 N. In this same revolution, when the block reaches the top of its path, point B, the magnitude of the normal force exerted on the block has magnitude 0.660 N.

How much work was done on the block by friction during the motion of the block from point A to point B?

Okay, so I started off by doing N+mg=m(v^2)/r to get the velocity at point A. I plugged in my variables like this: 0.0350(v^2/.425)+9.8*.035=.660+.343
.0350(v^2/.425)=.660
v=2.831 m/s

Am I on the right track? I tried doing a similar thing for point B and got v=6.709 m/s. Once I got those two I plugged them into K1+U1+W=K2+mgh but I have no idea if I'm doing all of this right. Any help/confirmation would be much appreciated!

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