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Coefficient Of Kinetic Friction Question 
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#1
Nov1310, 04:23 PM

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1. The problem statement, all variables and given/known data
A 25 kg box slides down a 26 degree ramp with an acceleration of 1.32 m/s^{2}. The acceleration of gravity is 9.81 m/s^{2} Find the coefficient of kinetic friction between the box and the ramp. 2. Relevant equations mass = m = 25 kg angle = (theta) = 26 degrees acceleration = a = 1.32 m/s^{2} gravity = F_{g} = 9.81 m/s^{2} coefficient of kinetic friction = M_{k} = F_{k}/F_{n} F_{k} = resistance, friction? Force opposite Mass*Acceleration. F_{n} = perpendicular to the object's motion, upward 3. The attempt at a solution F_{µ} = F_{n}  COS26(I don't know...) Attached is a pic of my teacher's process for solving the same problem, only with different values. I'm clueless. Teacher said this is the toughest of the 14 problems that are due in 6 hours, so I thought I might seek assistance in learning how to do it. Please and thanks. :) 


#2
Nov1310, 05:26 PM

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The force of kinetic friction is proportional to the normal force Fn, and its acts along the slope in opposite direction as the body moves. Fn, the normal force is the component of mg which is perpendicular to the slope,
Fn=mgcos(theta). The force of kinetic friction is F_{k}=μ mg cos(theta). The component of mg which is parallel with the slope is Fp=mg sin(theta), and it points down the slope. The force of kinetic friction points in the opposite direction. The resultant of these forces will accelerate the body down the slope ma=mg sin(theta)μ mg cos(theta). Given m, a, theta, g, find μ. ehild 


#3
Nov1310, 09:00 PM

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#4
Nov1310, 10:47 PM

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P: 10,757

Coefficient Of Kinetic Friction Question
It is the coefficient of friction. You used the notation M_{k}.
ehild 


#5
Nov1310, 11:33 PM

P: 3




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