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= 88 x 1/sqrt2 Fn = (14 x9.8) + 62.2 COS TO f = 62.22 N En = 199 N F = (8.4 + 62.22)/cos45

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= 88 x 1/sqrt2 Fn = (14 x9.8) + 62.2 COS TO f = 62.22 N En = 199 N F = (8.4 + 62.22)/cos45 F = 70.62/0.7 F = 99.9 N A 15.0-kg box is released on a 320 incline and accelerates down the incline at 0.30 m/s . Find the friction force impeding its motion. What is the coefficient of kinetic friction? 4 pts

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