Answer:
Approximately [tex]2.00\; {\rm m\cdot s^{-2}}[/tex], assuming that the friction between the box and the floor is negligible.
Explanation:
The external force on the box in this question is much smaller than the weight of the box. Hence, the box will stay on the ground, and the box will accelerate in the horizontal direction but not in the vertical direction.
To find the acceleration of the box in the horizontal component, divide the horizontal component net force on the box by the mass of the box. Since the only force on this box in the horizontal direction is the external force, the net force on the box in the horizontal direction would be equal to the horizontal component of that force. At [tex]37.0^{\circ}[/tex] above the horizontal, the horizontal component of this [tex]100\; {\rm N}[/tex] force would be:
[tex](100\; {\rm N})\, \cos(37^{\circ}) \approx 79.86355\; {\rm N}[/tex].
In other words, the net force on this box in the horizontal component would be approximately [tex]79.86355\; {\rm N}[/tex].
Divide the horizontal component of the net force by mass to find the horizontal component of acceleration:
[tex]\displaystyle \frac{79.86355\; {\rm N}}{40.0\; {\rm kg}} \approx 2.00\; {\rm m\cdot s^{-2}}[/tex].
Since the vertical component of acceleration is zero, the overall acceleration of the box would be equal to the acceleration in the horizontal direction: approximately [tex]2.00\; {\rm m\cdot s^{-2}}[/tex].
