Question

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?

Answer 1

If the elevator was stationary, the scale would read exactly the weight of the person:
$W=mg=(70 km)(9.8 m/s^2)=686 N$
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
$W' = ma'$
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: $a'=a+g$
Taking the direction of g (downward) as the positive direction, we can find the value of a:
$W' = m(a+g)$
$a= \frac{W'}{m}-g= \frac{896 N}{70 kg}-9.8 m/s^2=3 m/s^2$
So, the acceleration of the elevator is $3 m/s^2$, with same sign of g, so it's directed downward as well.