physics problem dealing with elevators?

Question

A student stands on a bathroom scale in an
elevator at rest on the 64th °oor of a building.
The scale reads 849 N. As the elevator moves
up, the scale reading increases to 942 N, then
decreases back to 849 N.
The acceleration of gravity is 9:8 m=s2 :
Find the acceleration of the elevator. An-
swer in units of m=s2.

As the elevator approaches 74th °oor the scale
reading drops as low as 774 N.
What is the acceleration of the elevator?
Answer in units of m=s2.

Answer 1

As the elevator moves up with an increase in the weight of the student from 849N to 942N, there was a net force upward acting on the student. This net force equals 942-849=93N. If the scale read 849 originally when the elevator was at rest then the mass of the student must have been 849/9.8=86.6kg. Now solve for the acceleration using the formula F=ma, where F is the net force, m the mass of the student, and a the acceleration.

F=ma
93=86.6a
a=93/86.6
=1.07m/s^2

To bring the reading back to 849, will need a net downward force of 942-849=93N. m remains the same. With the same net force as in the above, but in the downward direction, the acceleration must be the same in magnitude, i.e, 1.07m/s^2. But this time it is deceleration because the elevator would be slowing down, the net force being in the downward direction.

The elevator decelerates as it approaches the 74th floor because of the lower reading of 774N. The net force is 849-774=75N downward.

F=ma
75=86.6a
a=75/86.6
=0.86m/s^2.

Answer 2

F1=m*(g+a1), F1=942 N, m*g=849 N. thus a1=(F1-m*g)/m, where m=849/g. pulling up.
F2=m*(g+a2), F2=774 N, a2=(774/849-1)*g=-0.87 m/s^2 - falling down with brake.