Question

A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.79 of her regular weight.
(a) Calculate the magnitude of the acceleration of the elevator.
(b) Find the direction of the acceleration.

Answer 1

Answer:

Part a)

$a = 2.06 m/s^2$

Part b)

Lift is moving downwards as it is coming with negative sign

Explanation:

As we know that the reading of the scale is equal to the normal force on the woman

So we will have

$F_n - mg = ma$

here we know that

$F_n = 0.79 mg$

so we will have

$0.79 mg - mg = ma$

$a = -2.06 m/s^2$

Part a)

magnitude of the acceleration of the lift is given as

$a = 2.06 m/s^2$

Part b)

Lift is moving downwards as it is coming with negative sign