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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 .

Respuesta :

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.

[tex]W=mg[/tex]

[tex]m=\dfrac{W}{g}[/tex]

[tex]m=\dfrac{839}{9.8}[/tex]

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 :

[tex]N-mg=ma[/tex]

[tex]N=m(g+a)[/tex]

[tex]924=85.61\times (g+a)[/tex]

[tex]10.79-9.8=a[/tex]

[tex]a=0.99\ m/s^2[/tex]

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

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