Awood block, after being given a starting push, slides down a ramp at a constant speed. what is the angle of the ramp above horizontal? suppose that μk = 0.35.

The solution for the problem is:

Constant speed means Fnet = 0. 
Let m = mass of wood block and Θ = angle of ramp; then if µk = 0.35 …

The computation would be:

Fnet = 0 = mg (sin Θ) - (µk) (mg) (cos Θ) 
mg (sin Θ) = µk (mg) (cos Θ) 
µk = tan Θ 
Θ = arctan(µk)

= arctan (0.35)

≈ 19.3°

i think the answer would be b

unbalanced force

explanation: