Find the friction force impeding its motion.

AI Thread Summary
To find the friction force on a 15.0 kg box on a 32-degree incline accelerating at 0.30 m/s², the normal force is calculated as 127 N. The equation ma = mgsin(theta) - umgcos(theta) can be used to determine the friction force and the coefficient of kinetic friction. By rearranging this equation, one can solve for the coefficient of kinetic friction (u) and the friction force. The discussion highlights the confusion among students due to inadequate teaching methods. Understanding the relationship between forces on an incline is essential for solving such problems.
nickrace09
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Ok, our teacher is pretty much insane and doesn't teach anything at all. I'm sure this problem is incredibly simple, but I'm just not sure how to do it.

Original Problem: A 15.0 kg box is released on a 32 degree incline and accelerates down the incline at .30m/s/s. Find the friction force impeding its motion. What is the coefficient of kinetic friction?

What I Did: I found Normal Force to be 127 N with Wcos(theta) but I don't know where to go from there. Any help would be much appreciated.
 
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nickrace09 said:
Ok, our teacher is pretty much insane and doesn't teach anything at all. I'm sure this problem is incredibly simple, but I'm just not sure how to do it.

Original Problem: A 15.0 kg box is released on a 32 degree incline and accelerates down the incline at .30m/s/s. Find the friction force impeding its motion. What is the coefficient of kinetic friction?

What I Did: I found Normal Force to be 127 N with Wcos(theta) but I don't know where to go from there. Any help would be much appreciated.

my teacher doesn't teach us anything either...except this ma=mgsin(theta)-umgcos(theta) then you can use that to solve for coefficient of kinetic friction and friction force. I think.
 
thanks dude that works
 
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