URGENT QUESTION PLEASE SHOW STEPS!
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URGENT QUESTION PLEASE SHOW STEPS!

[From: ] [author: ] [Date: 13-11-05] [Hit: ]
(120-150*uk)/15.3kg = 3m/s² => 120N - 150*uk = 3m/s²*15.3kg = 45.(120N-45.9N)/150N = uk = 0.4938 -coefficient of kinetic friction is 0.......
A steel box is being pushed along a wooden table top by an applied force of 120N forwards. If the normal force between the box and table is 150N, and it accelerates at 3.0 m/s^2 [F], what is the coefficient of kinetic friction?

Show your work please :)

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We know that Fnet/mass = 3m/s²
We know 150N = mass*g => mass = 150/g = 15.3kg
(120-150*uk)/15.3kg = 3m/s² => 120N - 150*uk = 3m/s²*15.3kg = 45.9N
(120N-45.9N)/150N = uk = 0.4938 <----------------

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coefficient of kinetic friction is 0.5
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