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

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