# 5b - 60° 6 kg 100 N Solution Free body diagram f F v F h...

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PHY2053 – Section 6168 Quiz #5 Friday, February 13, 2004 Name______________ KEY __________ UFID___________________ Please read the problem carefully and be sure to show all of your work. Partial credit will be given. A 6.0 kg block is pushed along the ceiling with a constant applied force of 100.0 N that acts at an angle of 60° with the horizontal, as in the figure. The block accelerates to the right at 7.0 m/s 2 . Determine the coefficient of kinetic friction between the block and the ceiling.

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Unformatted text preview: 60° 6 kg 100 N Solution: Free body diagram: f F v F h mg N a F v = 100 sin 60° = 100 sin 60° F h = 100 cos 60° Vertical: F v = mg + N Solving for N gives N = F v- mg Horizontal: F h – f = ma but f = µ k N = k (F v – mg) Substituting expressions for F v into equation for f , followed by substitution of f and F v into equation for horizontal motion gives: 100 cos 60° - k ( 100 sin 60° – mg) = ma Solving for k gives: mg ma k − − = o o 60 sin 100 60 cos 100 = 0.28...
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## This note was uploaded on 09/25/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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5b - 60° 6 kg 100 N Solution Free body diagram f F v F h...

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