Question

A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836 N.
a) As the elevator moves up, the scale reading increases to 935 N, then decreases back to 836 N. Find the acceleration of the elevator.
b) As the elevator approaches the 74th floor, the scale reading drops as low as 782 N. What is the acceleration of the elevator?

Answer 1

The answers to the problem are as follows:

  a) use f=ma. Mass of student is 836/9.8=85.3kg, force is 935-836=99N. Put it into the formula you will get a=99/85.3 

b)Use same formula. Force is 782-836=-54 so put it into formula a=-54/85.3 

c)Stopping would take longer as the acceleration is smaller 


I hope my answer has come to your help. God bless and have a nice day ahead!

Answer 2

Answer:

Part a)

$a = 1.16 m/s^2$

Part b)

$a = 0.63 m/s^2$

Explanation:

Part a)

When elevator is moving upwards then reading of the scale is measurement of normal force

it is given as

$F_n - mg = ma$

$935 - 836 = \frac{836}{9.81} a$

$a = 1.16 m/s^2$

Part b)

When elevator decelerate then the scale reading is less than the actual weight as its acceleration will have reverse direction

So we will have

$F_n - mg = -ma$

$782 - 836 = -\frac{836}{9.81} a$

$a = 0.63 m/s^2$