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

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.

$W=mg$

$m=\dfrac{W}{g}$

$m=\dfrac{839}{9.8}$

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 :

$N-mg=ma$

$N=m(g+a)$

$924=85.61\times (g+a)$

$10.79-9.8=a$

$a=0.99\ m/s^2$

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