A person weighing 0.6 kN rides in an elevator that has a downward acceleration of 1.3 m/s 2 . The acceleration of gravity is 9.8 m/s 2 . What is the magnitude of the force of the elevator floor on the person? Answer in units of kN.

520.41 N.

Explanation:

From newton's second law of motion,

The force of the elevator floor on a person is given as,

R = m(g-a)

R = W-ma Equation 1

Where R = Force of the elevator floor, m = mass of the person, a = acceleration of the elevator, W = weight of the person.

But,

W = mg,

where g = acceleration due to gravity.

m = W/g Equation 2

Given: W = 0.6 kN, = 600 N, g = 9.8 m/s²

m = 600/9.8

m = 61.22 kg.

Given: W = 0.6 kN = 600 N, m = 61.22 kg, a = 1.3 m/s²

Substitute into equation 1

R = 600-61.22(1.3)

R = 600-79.586

R = 520.41 N.

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