Student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. the scale reads 834 n. (a) as the elevator moves up, the scale reading increases to 928 n. find the acceleration of the elevator. 1.11 correct: your answer is correct. m/s2 (b) as the elevator approaches the 74th floor, the scale reading drops to 782 n. what is the acceleration of the elevator?

(a) 1.1

(b)  0.611

Explanation:

We have given that when the elevator is at rest the scales reads 834 N

So W = 834 N

Acceleration due to gravity

We know that W = mg

So

(a)Now as the elevator moves upward so effective acceleration = g+a

Scale reading W = 928 N

Mass = 85.1020 kg

So

So  a = 10.9045 -9.8 = 1.1

(b) As the scale reading decreases effective acceleration = g-a

Scale reading = 782 N

So

So a =9.8 -9.1819 = 0.611