# A 64kg person is standing on a scale in an elevator which moves down at a constant velocity. It eventually slows and comes to a stop. ?

What is the reading on the scale when while the elevator is accelerating?

The magnitude of the elevator's acceleration is 0.73m/s^2.

### 2 Answers

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- ♥Astrid♥Lv 71 month agoFavorite Answer
Newton's Second Law:

F = ma

Applied to the person's weight:

W = mg

W = (64 kg)(9.81 m/s²)

W = 628 N = 141 lbs

Now add the acceleration of the elevator:

W = (64)(9.81 + 0.73)

W = 675 N = 152 lbs

- billrussell42Lv 71 month ago
from your first sentence, the elevator is de-accelerating

scale reads weight plus force needed to de-accelerate

weight is 64x9.8 newtons

force = ma = 64x0.73 newtons

total is the sum of those two.

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