# Coefficient of Kinetic Friction on an Incline

• graywolf27

#### graywolf27

Here is the problem: A block is released from rest on an inclined plane and moves 3.5 m during the next 4.8 s at an angle of 31 degrees. The acceleration of gravity is 9.8 m/s^2. What is the magnitude of the acceleration of the block? what is the coefficient of Kinetic Friction Muk for the incline?
- I already identified Xinitial = 0, Vxinitial =0, distance = 3.5 m, time = 4.8 s, theta = 31
- to get the acceleration I used asubx = gsin(31) which gave me 5.05 m/s^2
But then again there is this formula Xfinal = Xinitial + Vxinitial*t +0.5at^2 which at the end becomes a=2distance/t^2 giving me a value for a=0.304 m/s^2 and this is the point were I get confused .

- For the coefficient of Kinetic friction, I used free body diagrams which gave me a formula of Mk=mgsin(theta)/mgcos(theta) then Mk=tan(theta) I substituded Mk=tan(31) and got an answer of Mk=0.601
Am I on the right track? Any help will be appreciated :)