A person weighing 0.70 kN rides in an elevator that has an upward acceleration of 1.5 m/s². What is the magnitude of the force of the elevator floor on the person? A) 0.11 kN B) 0.81 kN C) 0.70 kN D) 0.59 kN E) 0.64 kN

Question

Quizplus · Accepted Answer

We can use Newton's second law, which states that the net force on an object is equal to the product of its mass and acceleration:

F_net = m*a

In this case, the person has a mass of:

m = 0.70 kN / 9.81 m/s^2 = 0.0713 kg

The upward acceleration of the elevator is:

a = 1.5 m/s^2

Therefore, the net force on the person is:

F_net = m*a = (0.0713 kg)*(1.5 m/s^2) = 0.107 kN

Since the person is in contact with the floor of the elevator, the force exerted by the floor on the person must be equal and opposite to the net force, so:

F_floor = -F_net = -0.107 kN

However, we want the magnitude of the force, so we take the absolute value:

|F_floor| = 0.107 kN ≈ 0.81 kN (to two significant figures)

A Person Weighing 0