A box slides down a 37.9-degree ramp with an acceleration of 1.13 m/s/s. Determine the coefficient of kinetic friction between the box and the ramp. Enter your answer accurate to the fourth decimal place please!!

Question

compphysgeek · Answer

|dw:1384125977417:dw|
First of all draw an image of the situation with all the forces. You can see then that the force accelerating the box is $$ma = mg \sin(37.9) - mg \cos(37.9) \mu_k \ \Rightarrow \mu_k = \frac{g\sin(37.9) - a}{g\cos(37.9)}$$ $a = 1.13 m/s^2$ and g are given.

I'm getting $\mu_k = 0.6325$

anonymous · Answer

oh my gosh, thank you so much!!!! i really appreciate it :)

compphysgeek · Answer

just try to understand what I did and see if you get the same answer ;)