I need huge help understanding this type of questions :( Please and thank you The coefficient of static and kinetic frictions between a 15.2-kg box and a desk are 0.55 and 0.36, respectively. What is the net force on the box when a 120.9 N horizontal force is applied to the box while moving?

67.28 N

Explanation:

Given that,

The mass of an box, m = 15.2 kg

The coefficients of static and kinetic frictions for plastic on wood are 0.55 and 0.36, respectively.

The force of static friction,

The force of kinetic friction,

Net force acting on the object is :

F = 120.9 -53.62

= 67.28 N

Hence, this is the required solution.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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c. newton's third law of motion