Question

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?

Answer 1

(a) 1.1
`m/sec^2`

(b) 0.611
`m/sec^2`

Step-by-step explanation:

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

So W = 834 N

Acceleration due to gravity
`g=9.8m/sec^2`

We know that W = mg

So
`m=(W)/(g)=(834)/(9.8)=85.1020kg`

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

Scale reading W = 928 N

Mass = 85.1020 kg

So
`g+a=(928)/(85.1020)=10.9045m/sec^2`

So a = 10.9045 -9.8 = 1.1
`m/sec^2`

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

Scale reading = 782 N

So
`g-a=(782)/(85.1020)=8.1819m/sec^2`

So a =9.8 -9.1819 = 0.611
`m/sec^2`