A 70.0-kg person rides in an elevator while standing on a scale. the scale reads 896 n. what is the magnitude and direction acceleration of the elevator?

Respuesta :

If the elevator was stationary, the scale would read exactly the weight of the person:
[tex]W=mg=(70 km)(9.8 m/s^2)=686 N[/tex]
However, the elevator is moving with an acceleration a, so the scale reads a new value of the weight, which we call W'. Newton's second law becomes
[tex]W' = ma'[/tex]
where a' is actually the total acceleration of the person, so it's the sum of the acceleraion of the elevator and of the gravitational acceleration: [tex]a'=a+g[/tex]
Taking the direction of g (downward) as the positive direction, we can find the value of a:
[tex]W' = m(a+g)[/tex]
[tex]a= \frac{W'}{m}-g= \frac{896 N}{70 kg}-9.8 m/s^2=3 m/s^2 [/tex]
So, the acceleration of the elevator is [tex]3 m/s^2[/tex], with same sign of g, so it's directed downward as well.