Aperson weighing 0.70 kn rides in an elevator that has an upward acceleration of 1.5 m/s2. what is the magnitude of the normal force of the elevator's floor on the person?

The magnitude of the normal force of the elevator's floor on the person is of 807.13 N.

Given data:

The weight of person is, .

The magnitude of upward acceleration is, .

The given problem is based on the Newton's Second law and normal force. As per the Newton's second law,  the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:

Then,

F = ma

Since, there are only two forces acting on the person: his weight W (downward) and the normal force N of the floor against the body (upward). So we can rewrite the previous equation as,

N - W = F

N - W = ma ......................................................(1)

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

Substitute the values in equation (1) as,

Thus,  the magnitude of the normal force of the elevator's floor on the person is of 807.13 N.

Learn more about the normal force here:

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it breaks and dirupture other materials and other sencews from the bottom

if they have a finite resultant