Physics, 30.07.2019 02:30 aylengarcia090

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?

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Answer from: babymom74

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, $W = 0.70 \;\rm kN = 0.70 \times 1000=700 \;\rm N$ .

The magnitude of upward acceleration is, $a = 1.5 \;\rm m/s^{2}$ .

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,

$W =mg\\\\700 = m \times 9.8\\\\m=71.42 \;\rm kg$

Substitute the values in equation (1) as,

$N-W =ma \\N -700 = 71.42 \times 1.5\\N = 807.13 \;\rm N$

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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Answer from: ijohnh14

First of all, we can find the mass of the person, since we know his weight W:
W=mg=0.70 kN=700 N

And so
$m= \frac{W}{g}= \frac{700 N}{9.81 m/s^2}=71.4 kg$

We know for Newton's second law that 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:
$\sum F = ma$
There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as
R_v -W = ma

We know the acceleration of the system, a=1.5 m/s^2

(upward, so with same sign of Rv), so we can solve to find the value of Rv, the normal force exerted by the elevator's floor on the person:
R_v = ma+W=(71.4 kg)(1.5 m/s^2)+700 N =807N

R_v = ma+W=(71.4 kg)(1.5 m/s^2)+700 N =807N

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Aperson weighing 0.70 kn rides in an elevator that has an upward acceleration of 1.5 m/s2. what is t...