Question

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!!

Answer 1

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 \(\mu_k = 0.6325\)

Answer 2

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

Answer 3

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