Question

What will a spring scale read for the weight of a 58.0-kg woman in an elevator that moves in free fall?

Answer 1

Answer:

0 N

Explanation:

The elevator is under free fall. So, when a body is under free fall, the acceleration is only due to gravity. So, the acceleration of the elevator or the woman inside it, is acceleration due to gravity in the downward direction.

The spring scale gives the value of the normal force acting on the woman and doesn't give the exact weight of the woman. Under normal conditions, when the spring scale is at rest, then the upward normal force equals the weight and hence weight of a body is equal to the normal force acting on the body.

But, here, the body is not at rest. Weight\tex](mg)[/tex] acts in the downward direction and normal force$(N)$ acts in the upward direction. The woman is moving down with acceleration equal to acceleration due to gravity$(g)$

So, we apply Newton's second law on the woman.

$\textrm{Net force} = \textrm{mass}\times \textrm{acceleration}\\F_{net}=ma$

Net force is equal to the difference of the downward force and upward force.

$F_{net}=mg-N$

Now, $F_{net}=ma$

$mg-N=mg\\N=mg-mg=0$

Therefore, the reading on the spring scale is 0 N.