Answer: Option A
:
A
Let the contact force on the block by the surface be F which makes an angle $\theta$with the vertical

The component of F perpendicular to the contact surface is the normal reaction N and the component of F parallel to the surface is the well-known frictional force f. As the surface is horizontal, N is vertically upward. For vertical equilibrium,
N = mg = (0.400 kg) $\frac{10m}{s^2}$ = 4.0 N.
The frictional force is f = 3.0 N.
(a)
Or, θ = ${tan}^{-1}$( $\frac{3}{4}$) = ${37}^{\circ }$.
(b)The magnitude of the contact force is
F =$\sqrt{N^2+f^2}$
=$\sqrt{{\left(4.0N\right)}^2+{\left(3.0N\right)}^2}$ = 5.0 N.