Question

A 3.9kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.438. 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 2.92 m/s^2, and (c) accelerating downward with an acceleration whose magnitude is 2.92 m/s^2.

Answer 1

Step-by-step explanation:

There are three forces acting on the box:

Weight force mg pulling down,

Normal force N pushing up,

and friction force F = Nμ pushing horizontally.

Sum the forces in the +y direction.

∑F = ma

N − mg = ma

N = mg + ma

N = m (g + a)

Friction force is:

F = Nμ

Substituting:

F = m (g + a) μ

(a) Vertical acceleration is 0 m/s².

F = (3.9 kg) (9.8 m/s² + 0 m/s²) (0.438)

F = 16.7 N

(b) Vertical acceleration is +2.92 m/s².

F = (3.9 kg) (9.8 m/s² + 2.92 m/s²) (0.438)

F = 21.7 N

(c) Vertical acceleration is -2.92 m/s².

F = (3.9 kg) (9.8 m/s² − 2.92 m/s²) (0.438)

F = 11.8 N