A block of mass 5 kg rests on a smooth horizontal surface. It is...

The horizontal component of the pulling force is $F_{x} = F \cos 37^{\circ} = 50 \times \dfrac{4}{5} = 40 \text{ N}$.

The vertical component of the pulling force is $F_{y} = F \sin 37^{\circ} = 50 \times \dfrac{3}{5} = 30 \text{ N}$.

The weight of the block is $mg = 5 \times 10 = 50 \text{ N}$. Since $F_{y} < mg$, the block remains on the horizontal surface.

The net force acting on the block in the horizontal direction is $F_{\text{net}} = F_{x} - f_{\text{opposing}}$.

Substituting the given values, $F_{\text{net}} = 40 - 10 = 30 \text{ N}$.

Using Newton's second law, the acceleration of the block is $a = \dfrac{F_{\text{net}}}{m} = \dfrac{30}{5} = 6 \text{ m s}^{-2}$.

Answer: $6 \text{ m s}^{-2}$