Question

A 9.10-kg box is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the box and the floor is 0.375. 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 1.69 m/s2, and (c) accelerating downward with an acceleration whose magnitude is 1.69 m/s2

Answer 1

Answer

given,

mass of the box = 9.1 kg

kinetic friction coefficient = 0.375

acceleration = 1.69 m/s²

kinetic frictional force on the box = ?

If the elevator is stationary, the kinetic frictional force is

F = µ m g

F = 0.375 × 9.1 × 9.81

F = 33.48 N.

If the elevator is accelerating upward at 1.88 m/s^2, we add the acceleration to the acceleration due to gravity. The force is

F = µ m (g+a)

F = 0.375 x  9.1 x (9.81 + 1.69)

F = 0.375 x  9.1 x 11.5

  F= 42.31 N

If the elevator is accelerating downward, we subtract.

F = µ m (g-a)

F = 0.375 x  9.1 x (9.81 - 1.69)

F = 0.375 x  9.1 x 8.12

  F= 27.71 N