Yes! Calculating Friction Coefficient

[extDependency]

Homework Statement

A 30kg object is being pulled across a horizontal surface. The acceleration is 1.2 m/s2. The object is being pulled by a handle that is at 25 degrees. Friction force is 10 N. What forces can we find?

Relevant Equations

Fg = m * g
Fnet = m * a
Fa = Fax / cos(angle)
Fax = Fax leftward - Fax rightward
Fay = Fa * sin(angle)

Fg = m * g:
Fg = 30 * 9.8 = 294 N

Fnet = m * a:
Fnet = 30 * 1.2 = 36 N

Fax = Fnet + Ff:
Fax = 36 + 10 = 46

Fa = Fax / cos(angle):
Fa = 46 / cos(25) = 50.75

Fay = Fa * sin(angle):
Fay = 50.76 * sin(25) = 21.45

Fn = Fg - Fay:
Fn = 294 - 21.45 = 272.55

Ff = mu kinetic * Fn, solving for coefficient:
Coefficient of friction = 10 / 272.55 = .035Does this look correct?

 

[haruspex]

Making assumptions about how you are defining your variables, it all looks ok, except "Fax = Fax leftward - Fax rightward". What did you mean by that?
Your answers show too many significant figures, given the precision of the provided numbers. Round to at most three.