Question

A person who normally weighs 250 pounds is standing on a scale inside an elevator. The elevator is moving upward with a speed of 7 m/s and then begins to decelerate at a rate of 5 m/s²(0.5 g). What is the reading on the scale before the elevator begins to decelerate? What is the reading on the scale during the deceleration?

Answer 1

Initially , the elevator is moving at a constant velocity 7 m/s (no acceleration) ..So , there is no force on the person due to acceleration .

So , the reading would be same as the normal weight.

2 ).

Now, the elevator is decelerating at the rate of 5 m/s² .i.e it would have a acceleration of 5 m/s² in downward direction.

The man is at rest with respect to the elevator which is accelerated downwards at the rate of 5 m/s² with respect to the ground .

The forces on the man are

(1) mg downward (by the earth) and (m = 250 lb)

(2) N upward (by the floor ).

The equation of motion of the block is , therefore

250 lb * 9.8 m/s² - N = 250 lb * 5 m/s²

N = 250 lb * (9.8 - 5 ) m/s²

N = 250* 4.8 lb m /s²

As we know that the scale gives reading of mass here the N gives effective weight.

So, the reading of the scale will give 250* 4.8/9.8 lb = 122.45 lb