Astudent 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 .

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.

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 :

So, the acceleration of the elevator is . Hence, this is the required solution.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y