Question

A 550 N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is 840 kg. As the elevator starts moving, the scale reads 430 N. Find the acceleration (magnitude and direction) of the elevator, What is the acceleration is the scale reads 670 N?

Answer 1

(a) The elevator has an acceleration of 2.14 m/s² and direction is downward

(b) The acceleration is a=2.14 m/s²

Step-by-step explanation:

Let +y be upward direction

For Part (a)

The mass of student (whose weight is 550N) is:

ms=w/g

ms=550/9.8=56.1 kg

Apply ∑Fy=ma

`n-m_(s)g=m_(s)a\\`

Solve for acceleration a

`a=(n-m_(s)g)/(m_(s))\\ a=(430N-550N)/(56.1kg) \\a=-2.14m/s^(2)`

The elevator has an acceleration of 2.14 m/s² and direction is downward

Part(b)

For the acceleration is the scale reads 670 N

For n=670N

So

`a=(n-m_(s)g)/(m_(s))\\ a=(670N-550N)/(56.1kg)\\ a=2.14m/s^(2)`