ch08-p098 - The power dissipated by the friction must equal...

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98. Since the period T is (2.5 rev/s) 1 = 0.40 s, then Eq. 4-33 leads to v = 3.14 m/s. The frictional force has magnitude (using Eq. 6-2) f = μ k F N = (0.320)(180 N) = 57.6 N.
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Unformatted text preview: The power dissipated by the friction must equal that supplied by the motor, so Eq. 7-48 gives P = (57.6 N)(3.14 m/s) = 181 W....
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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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