You are riding in an elevator on the way to the eighteenth floor of your dormitory. the elevator's upward acceleration is a = 1.94 m/s2 . beside you is the box containing your new computer; box and contents have a total mass of 28.4 kg . while the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. if the coefficient of kinetic friction between the box and elevator floor is μk = 0.35, what magnitude of force must you apply?

116.795N

Explanation:

Hello! to solve this problem the following steps are necessary

1. Find the total force to the box to the floor of the elevator, it is found remembering the second law of Newton that states that the sum of forces is a body is equal to the product of mass by acceleration.

Therefore, the downward force of the box would be equal to the sum of the weight and the effect of the product of the mass due to acceleration, the following equation is inferred.

F=ma+mg

F=total force

m=mass of box=28.4kg

g=gravity=9.81m/S^2

a=1.94m/S^2

F=(28.4kg)(1.94)+(28.4)(9.81)=333.7N

2.

Now we remember that the minimum force to move the box horizontally is the product of the friction coficient (μk = 0.35)by the total downward force found in the previous step

Fh=F(μk)

Fh=333.7N(0.35)=116.795N

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.