A 6.0-kg box slides down an inclined plane that slopes 39° to the horizontal. The box accelerates at a rate of 3.1 m/s2. What is the coefficient of kinetic friction between the box and the surface of the inclined plane?

Question

anonymous · Answer

Use F-R=ma, where F is the driving force, in this case the one due to weight, which is mgsin@ or 6*9.81*sin39=37N and R is resistive force, that is due to friction. Deeply talking about the R, we first have to find the Normal Contact force or N=mgcos@ or 6*9.81*cos39=45.7N and Force due to friction or the R is then $N, where $ is the coefficient of friction and N is contact force. Therefore R=45.7$ Newtons.
Now for the grand finale... F-R=ma or 37-45.7$=6*3.1, then$=0.402