42.9°
Explanation:
Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:
[tex]x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;[/tex]
[tex]\Rightarrow mg\sin{\theta} = F[/tex]
Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at [tex]\theta.[/tex] Solving for the angle, we get
[tex]\sin{\theta} = \dfrac{F}{mg}[/tex]
or
[tex]\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)[/tex]
[tex]\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right][/tex]
[tex]\;\;\;=42.9°[/tex]
