Question

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?

Answer 1

The magnitude of the force of the elevator floor on the person is; 807.13 N

What is the Magnitude of force?

We are given;

Weight; W = 0.7 KN

Acceleration; a = 1.5 m/s²

There are only two forces acting on the person: which are his weight W (downward) and the normal force N of the floor against the body (upward). Thus the equilibrium equation here is;

N - W = F

N - W = ma

Here, m is the mass of body, and its value is calculated as;

W = mg

0.7 = 9.81m

m = 0.7/9.81

m = 71.42 kg

Thus;

N - 0.7 = 71.42 * 1.5

N = 0.7 + (71.42 * 1.5)

N = 807.13 N

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