Answer to Question #137572 in Physics for Muhammad

Question

A 78.2 Kg person is standing on a scale in an elevator. If the elevator is accelerating downwards at 0.307 m/s^2 what will the reading on the scale be (in kilograms)?

Answer 1

Let’s apply the Newton’s Second Law of Motion:

$$\sum F_y = ma_y,$$ $$N-mg=-ma,$$

here, $N$ is the normal force directed upward, $mg$ is the force of gravity directed downward, $a = 0.307 \ \dfrac{m}{s^2}$ is the acceleration of an elevator (since the acceleration of an elevator directed downward it will be with sign minus) and $g = 9.8 \ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Then, we get:

$N=m(g-a) = 78.2 \ kg \cdot(9.8 \ \dfrac{m}{s^2}-0.307 \ \dfrac{m}{s^2}) = 742.35 \ N.$

Finally, we can find the reading on the scale from the definition of the weight:

$$m = \dfrac{N}{g} = \dfrac{742.35 \ N}{9.8 \ \dfrac{m}{s^2}} = 75.75 \ kg.$$

Answer:

$m = 75.75 \ kg.$