Finding coefficient of friction for block pulled up an incline at constant speed

[seizureboi]

Homework Statement

A 3kg block is pulled by a rope up a 37 degree incline at constant speed. If the tension in the rope is 45N and the rope is parallel to the incline, calculate the coefficient of kinetic friction between the block and the incline.

Homework Equations

a = acceleration (0 m/s₂)
Ө = theta (37 degrees)
g = 9.8 m/s²
m = mass
T = Tension
μ = coefficient of friction
N_F = Normal Force
friction = μN_F
N_F = mgcosӨ

The Attempt at a Solution

My attempt was setting up the equation: TsinӨ - mgsinӨ - friction = ma
OR 45sin(37) - (3)(9.8)sin(37) - μ(3)(9.8)cos(37) = 0
OR 27.1 - 17.7 = μ23.5
OR 9.4 = μ23.5
Finally, μ = .4

Just wondering if it's correct. =)

 

[jegues]

TsinӨ - mgsinӨ - friction = ma

You don't need a sinӨ multiplying the tension, the force of the tension is given parallel to the plane so you wouldn't need to factor an angle in, make sense?

 

[seizureboi]

Cool, because I did try it that way and i got μ = 1.16 and I thought I was wrong so I concluded with adding a random sinӨ lol.