Question

A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 839 N. As the elevator moves up, the scale reading increases to 924 N, then decreases back to 839 N. The acceleration of gravity is 9.8 m/s 2 . Find the acceleration of the elevator. Answer in units of m/s 2 .

Answer 1

Step-by-step explanation:

It is given that,

As the elevator moves the scale reading increases to 924 N, then decreases back to 839 N.

When the elevator is at rest, the scale reads 839 N.

`W=mg`

`m=(W)/(g)`

`m=(839)/(9.8)`

m = 85.61 kg

We need to find the acceleration of the elevator. When the elevator moves up, the force acting on it can be written as :

`N-mg=ma`

`N=m(g+a)`

`924=85.61* (g+a)`

`10.79-9.8=a`

`a=0.99\ m/s^2`

So, the acceleration of the elevator is
`0.99\ m/s^2`. Hence, this is the required solution.