# Answer to Question #1067 in Mechanics | Relativity for Micah

Question #1067

A 3kg box slides down a 30.0 degree ramp with an acceleration of 1.4 m/sec^2. Determine the coefficient of kinetic friction between the box and the ramp.

Expert's answer

Let's write the Second Newton's Law for the box:

mg sin(30) - μ N = ma

N = mg cos(30);

mg sin(30) - μ mg cos(30) = ma;

μ = (g sin(30) -a )/(g cos(30)) = 0.41

mg sin(30) - μ N = ma

N = mg cos(30);

mg sin(30) - μ mg cos(30) = ma;

μ = (g sin(30) -a )/(g cos(30)) = 0.41

## Comments

## Leave a comment