Answer:
3.339 N
Explanation:
In this problem, the box is moving at constant velocity, so the box is in motion and therefore we can consider only the kinetic friction.
The resultant of the forces acting on the box must be equal to the product of mass and acceleration:
[tex]F_{net}=ma[/tex]
However, the box is moving at constant velocity, so the acceleration is zero: a=0. Therefore,
[tex]F_{net}=0[/tex] (1)
There are two forces acting on the box:
- The push: F
- The kinetic friction on the box, equal to
[tex]-\mu_k mg[/tex]
where [tex]\mu_k = 0.1113[/tex] is the coefficient of kinetic friction and (mg)=30 N is the weigth of the box, and where the negative sign means it is in the opposite direction of the motion.
Therefore, eq.(1) becomes
[tex]F-\mu mg =0\\F= \mu mg=(0.1113)(30 N)=3.339 N[/tex]
