A 8.kg box is sliding across the horizontal floor of an elevator the coefficient of kinetic friction between the box and the floor is 0.310. Determine the kinetic frictional force that acts on the box when the elevator is accelerating downward with an acceleration whose magnitude is 1.5m/s
= (0.310)(8)(9.8*1.5)
To determine the kinetic frictional force that acts on the box when the elevator is accelerating downward, we need to use Newton's second law of motion, which states that F = ma, where F is the net force acting on an object, m is its mass, and a is its acceleration.
In this case, the net force acting on the box is the difference between the force of gravity and the force of kinetic friction. The force of gravity is given by Fg = mg, where m is the mass of the box and g is the acceleration due to gravity (9.8 m/s²).
Fg = (8 kg)(9.8 m/s²) = 78.4 N
The force of kinetic friction is given by Ff = μkN, where μk is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the force of gravity acting on the box in this case since the elevator is not moving vertically.
N = Fg = 78.4 N
Therefore, the force of kinetic friction can be calculated as:
Fk = μkN = (0.310)(78.4 N) = 24.264 N
So, the kinetic frictional force that acts on the box when the elevator is accelerating downward with an acceleration of 1.5 m/s² is approximately 24.264 N.