Answer:
The net force acting on the box is 550 N (going leftwards) and the box's acceleration is 22 meters per second squared.
Explanation:
A)
The net force acting on the box is the sum of the forces. Since Yumyum pulls with a force of 300 N and Mamu pulls with a force of 250 N, both to the left, the net force (in the x-direction) will be:
[tex]\displaystyle \sum F _x = (300 \text{ N}) + (250 \text{ N}) = 550 \text{ N}[/tex]
B)
The net force is equal to the object's mass multiplied by its acceleration. That is:
[tex]\displaystyle \sum F_x = ma[/tex]
Substitute known values:
[tex]\displaystyle (550 \text{ N}) = (25 \text{ kg})a[/tex]
Find a:
[tex]\displaystyle a = \frac{550 \text{ N}}{25 \text{ kg}} = 22\text{ m/s$^2$}[/tex]
In conclusion, the net force acting on the box is 550 N (going leftwards) and the box's acceleration is 22 meters per second squared.
