ch06-p047 - valley, we reverse both the normal force...

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47. At the top of the hill, the situation is similar to that of Sample Problem 6-7 but with the normal force direction reversed. Adapting Eq. 6-19, we find F N = m ( g – v 2 /R ). Since F N = 0 there (as stated in the problem) then v 2 = gR . Later, at the bottom of the
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Unformatted text preview: valley, we reverse both the normal force direction and the acceleration direction (from what is shown in Sample Problem 6-7) and adapt Eq. 6-19 accordingly. Thus we obtain F N = m ( g + v 2 /R ) = 2 mg = 1372 N 1.37 10 3 N....
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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