A 25-kg box slides, from rest, down a 9.0-m-long incline that makes an angle of
15° with the horizontal. The speed of the box when it reaches the bottom of the
incline is 2.4 m/s. (a) What is the coefficient of kinetic friction between the box
and the surface of the incline? (b) How much work is done on the box by the
force of friction? (c) What is the change in the potential energy of the box?
Fx = mg * sin15
Fn = mg * cos15
Ffr = uk * Fn = uk * mg * cos15
Vf2-Vo2 = 2 *a*d
The Attempt at a Solution
Having an issue with kinetic friction calculation here
Taking the force components and applying Newton's 2nd law -> ƩF = ma
Fx - Ffr = max
mg * sin 15 - uk (mg * cos 15) = m * ax
Since I want uk and the final velocity is given, I calculate accel -> Vf2 - Vo2 = 2*a*d so a = 2.4m/s / (2*9.0m) = 0.32 m/s^2
Can I put it into mg * sin 15 - uk (mg * cos 15) = m * ax and end up with uk = 0.2345 ?
I can't get over that ax isn't constant acceleration and I don't think I can use it in this case....