Question

A 4.8-kg suitcase is sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the suitcase and the floor is 0.28.

a) If the elevator is moving upward at a constant speed of 1.6 m/s, find the kinetic frictional force acting on the suitcase.


b) If the elevator is accelerating upward at 1.6 m/s?, find the kinetic frictional force acting on the suitcase


c) If the elevator is accelerating downward at 1.6 m/s, find the kinetic frictional force acting on the suitcase.

Answer 1

Below

Step-by-step explanation:

a) At constant speed the only force acting downward is the usual Normal Force due to gravity

Fn = mg = 4.8 * 9.81 =47.1 N

the force of friction is then Ff = Fn * coeff = 47.1 * .28 = 13.2 N

b) If the acceleration is UPWARD this adds to g

Fn then becomes

Fn = 4.8 ( 9.81 + 1.6) = 54.8 N and Ff = 54.8 * .28 = 15.3 N

c) Similar to b) but this time the acceleration of the elevator SUBTRACTS from gravity

Fn = 4.8 ( 9.81 - 1.6) = 39.4 N and Ff = 39.4 * .28 = 11.0 N