How to find the coefficient of friction of a ramp

How do I find the coefficient of friction of a wooden ramp (angled at 60 degrees) when a match box car weighing 42 grams accelerates down it at approximately 8.439 m/s/s?

Answers will be greatly appreciated. The problem is that every time I work it out I get a coefficient of friction of about 0.992, which shouldn't be the case considering that it is a very smooth, wooden ramp.

Weight of matchbox = mg = 0.042 x 9.81 = 0.41202N

Component of weight perpendicular to ramp, A = 0.41202cos(60⁰) = 0.20601N
Component of weight parallel to ramp, B = 0.41202sin(60⁰) = 0.35682N

Resultant force on matchbox, F = ma = 0.042 x 8.439 = 0.35444N

The resultant force on matchbox F = B - frictional force
Frictional force = B - F = 0.35682 - 0.35444 = 0.00238N

μk = (frictional force)/(normal force) = 0.00238/0.20601 = 0.01155

This should probably be rounded to 0.012 This seems pretty small, so check my arithmetic. The method is correct.

F=ma=Force down the ramp-Force of friction

F=ma=(0.042kg)(8.349m/s^2)

F down the ramp=m*g*Sin60=[0.042kg(9.8m/s^2)Sin60]

F of Friction=mk*m*g*cos 60=mk*[(40.042kg)*(9.8m/s^2)Cos60]

Masses cancel out

(8.349)=[(9.8)Sin60]-mk*[(9.8)Cos60]

Solving for mk

(8.349)-[(9.8)Sin60]/-[(9.8)Cos60]=mk

mk=0.0282


I haven't done these problems in a while, so I am not sure about the answer.

If it is correct, I am glad that I can assist, but if it is wrong, I warned you. However, I am pretty sure that mk=0.0282 or 0.028 two significant figures.


Best

Net force acting on car = m x a
Net force = weight of car acting down slope (mg.sin60) - friction(F)

ma = mg.sin60 - F
F = mg.sin60 - ma = m (g.sin60 - a) = 0.042kg (9.80.sin60 - 8.44) .. F = 0.002 N

F = μR .. (R = normal reaction (perpendicular) to slope .. R = mg.cos60)

μ = F / R = 0.002N / (0.042 x 9.80 x cos60) .. .. ►μ = .0097 (0.01)