How to find Friction Force and Coefficient of Friction?

AI Thread Summary
To find the friction force and coefficient of friction for a block sliding up a 20-degree incline, a horizontal force of 200 N is applied to a 15 kg block with an acceleration of 25 cm/s². The calculated friction force is 134 N, while the coefficient of friction is determined to be 0.65. The user struggles with the calculations, particularly in determining the normal force and its relationship to the friction force. Confusion arises from the discrepancy between the calculated coefficient of friction (0.97) and the expected value (0.65). Clarification is sought on the correct approach to finding the normal force and resolving the friction calculations.
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Homework Statement



A horizontal force of 200 n is required to cause a 15 kg block to slide up a 20degree incline with acceleration of 25 cm/s^2. Find the friction force on the block and the coefficient of friction.
ANSWERS:
Friction force = 134N
Coefficient of Friction = 0.65

Homework Equations


Ff = u Fn

The Attempt at a Solution


I can't figure out where to begin.
I'm not sure if it's relevant, but I found:
Fg to be 147.15 N
horizontal component to be 50.33 N
Vertical component to be 138.28N
 
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Part one:
Fgcos(20) - Ff = ma
138.28N - Ff = 15 x .25m/s^2
138.28N - Ff = 3.75
-138.28 -138.28
-Ff=-134.53
Ff=134.53N
(Hopefully this is correct?)

Part 2:
Ff = u Fn
u = Ff / Fn
u = 134.53N / 138.28N
u = .97

But the answer should be .65.
What am I doing wrong? The normal force should be 138.28 because 147.15cos(20) is 138.28, right?
Ugh, this is too confusing. Please help!
 

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