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Finding the coefficient of kinetic friction 
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#1
Jan2812, 08:00 PM

P: 2

1. The problem statement, all variables and given/known data
A construction worker passes a brick to another worker by pushing it across a wooded floor. the brick has a mas of 2.7kg. The force required making the brick move with acceleration of 0.42m/s^{2} is 28 N. What is the coefficient of kinetic friction acting on the brick? 3. The attempt at a solution FN=Fg = mass X acceleration =2.7kg X 0.42m/s^{2} =1.134 N F_{kf}= μ_{k} X FN μ_{k} = F_{kf} [itex]/[/itex] FN = 28N [itex]/[/itex] 1.134N =24.69 I'm pretty sure I got the acceleration part wrong. What am I supposed to do to find the normal force? 


#2
Jan2812, 08:21 PM

HW Helper
P: 6,202

0.42 is the resultant acceleration so multiplying that by 2.7 kg will give you the resultant force.
28 N is the applied force, you are correct in that F_{kf} = μ_{k }X F_{N}. So what equation can you make to relate the 28 and F_{kf} ? (Hint: A free body diagram will help you if you can't visualize it in your head) 


#3
Jan2812, 08:27 PM

Mentor
P: 11,676

The normal force for a mass M on a horizontal surface has magnitude M*g. Where you started to calculate FN above you used the horizontal acceleration of the brick rather than g = 9.81 m/s^{2}. So what you got instead of the normal force was the net horizontal force acting on the brick (the net horizontal force that results in the given acceleration of the brick of mass M). The net horizontal force should also be equal to the difference between the applied force and the force due to friction: [itex] F_{net} = f_{applied}  \mu_k F_N [/itex] 


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