help with a physics question please?

Question

A person stands on a scale in an elevator. The maximum and minimum scale readings are 771.7 N and 318.8 N, respectively.
The acceleration of gravity is 9.8 m/s^2.
Assume the magnitude of the acceleration is the same during starting and stopping, and determine the acceleration of the elevator. Answer in units of m/s^2.

Answer 1

The first step is to set up two simultaneous equations for the acceleration upward and the acceleration downward

For upward:
mg+ma=771.7
for downward
mg-ma=318.8

add the two equations together:
2mg=771.7+318.8

m=(771.7+318.8)/(2*9.8)
=55.64

Then solve for a in either equation and check with the other

a=771.7/55.64 - 9.8
=4.1

check
a=9.8-318.8/55.64
a=4.1

j

Answer 2

Hi RedCap!
As a matter of fact: When no acceleration (a=0) the scale reads p0=m*g, m is your mass, g=9.8m/s^2; Now
When up (a>0) the scale reads p1=m*(g+a)=771.7,
when down the scale reads p2=m*(g-a)=318.8.
Thus p1-p2=m*(g+a)-m*(g-a)=2*m*a, and p1+p2=m*(g+a)+m*(g-a)=2*m*g
So (p1-p2)/(p1+p2)=a/g or a=g*(p1-p2)/(p1+p2).