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Coefficient of kinetic friction 
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#1
Oct505, 04:16 PM

P: 16

A 12kg block is released from rest on an inclined plane with angle 35. Acceleration of the block is 1.23457 The acceleration of gravity is 9.8m/2^2.
What is the coefficient of kinetic friction for the incline? I found the normal force (12*9.8*cos35)=96.33228 Fnet=ma=12*1.23457=14.81484 Ff=NFnet=81.51744 So, I thought mu=Fnet/N=0.8462, but that answer is apparently wrong. Any ideas? 


#2
Oct505, 04:25 PM

P: 666

You know that the mass is 12kg, the and the acceleration 1.23457 m/s^{2}. You know that the product of those two is equal to the frictional force + the downward force. The downward force is the sum of the gravitational force and the normal force. Make a few substitutions and you should have your answer. Alex 


#3
Oct505, 04:38 PM

P: 16

Ok, so gravitational force is 9.8*12, right? That would be 117.6N. So the downward force is 117.6+96.33228=213.93228. But that means ma does not equal the product of the two. Where am I messing up?



#4
Oct505, 04:43 PM

P: 666

Coefficient of kinetic friction
1. Draw a freebody diagram. 2. Write Newton's Second Law in vector form (important!). 3. Write an expression for each of the forces. 4. Write a scalar equation for each. 5. Express any constraints on your equations. 6. Solve the final system(s) of equations. 


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