Help with a physics homework invloving friction and motion...

The problem states: A 75 kg box slides down a 25.0 degree ramp with an acceleration of 3.60 m/s^2. Find the coefficient of kinetic friction between the box and the ramp.

I have tried everything and cannot get the right answer of .061. This isn't graded its just for review before a test which is why I have the answer. Please help!!!

Using Newton’s 2nd law (and taking down the incline as positive) where the forces acting on the box are the parallel component of the box’s weight down the incline (mgsinθ) and the friction force up the incline (-µmgcosθ):

ΣF(x) = ma = mgsinθ - µmgcosθ ----------------->mass divides out
µ = (gsinθ - a) / gcosθ
= (9.80m/s²sin25.0° - 3.60m/s²) / 9.80m/s²cos25.0°
= 0.0610

Note that the friction force is the product of the friction coefficient and the normal force, and since the only two forces acting perpendicular to the incline are the normal force and the perpendicular component of the box’s weight, the normal force must equal this perpendicular component of weight: mgcosθ.

Hope this helps.

friction f=(µk)*N & N=mg
=> ma=mg*sin(θ) - f
=> ma=mg*sin(θ) - (µk)mg
=> a=g*sin(θ) -µk*g
=> µk=sin(θ) -(a/g)
=> µk=sin (25 degree) - (3.60/9.8)
=> µk=0.055
It doesn't look like 0.061 is the right answer for given data.