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A 7.74-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.392. Determine the kinetic frictional force that acts on the box when the elevator is:
(A) : Stationary,
(B) : Accelerating upward with an acceleration whose magnitude is 2.71 m/s2,
(C) : Accelerating downward with an acceleration whose magnitude is 2.71 m/s2

[terryg752]

Here is how to do it:

Let mass be m = 7.74

1. Normal reaction = mg

Hence Friction = .392 mg

2. Net force on box = 2.71 m (upward)

Hence Normal Reaction = (mg + 2.71 m)

Friction = .392 (mg + 2.71 m)


3. Net force on box = 2.71 m (downward)

Hence Normal Reaction = (mg - 2.71 m)

Friction = .392 (mg - 2.71 m)