Verified answer

The force of gravity on the block straight down is:

Fg = m×g

The component of Fg that is parallel down the slope is:

Fp = m×g×sin(θ)

The component of Fg that is perpendicular to the slope is:

Fp = m×g×cos(θ)

The force of friction between the slope and the block is:

Ff = μ×Fp

Ff = μ×m×g×cos(θ)

The net force on the block, parallel down the slope is:

Fnet = Fp - Ff

Fnet = m×g×sin(θ) - μ×m×g×cos(θ)

Fnet = m×g×(sin(θ) - μ×cos(θ))

Also:

Fnet = m×a

so

a = g×(sin(θ) - μ×cos(θ))

a/g = sin(θ) - μ×cos(θ)

a/g - sin(θ) = -μ×cos(θ)

sin(θ) - a/g = μ×cos(θ)

(sin(θ) - a/g) / cos(θ) = μ

μ = (sin(θ) - a/g) / cos(θ)

Now... let's fill in the numbers...

μ = (sin(27.4) - 1.56/9.8) / cos(24.4)

μ = 0.33 < - - - - - - ANSWER

In dynamics ∑F=ma

Use a coordinate sysem such that the x axis is paralell with the incline

First sum the forces in the x direction to get Ff (the force exerted by friction)

∑Fx=ma

Ff-mgSin27.4=ma

therefore you can find Ff=mgSin27.4+ma

Then sum the forces in the y direction to find Fn (the normal force exerted perpendicularly by the block)

∑Fy=0

Fn-mgCos27.4=0

thus Fn=mgCos27.4

Now that you know Ff and Fn you can find u from this equation-- Ff=uFn, where u is the friction coeficient